Network Theory Pune University MCQs

Network Theory Pune University MCQs

This set of Network Theory Multiple Choice Questions & Answers focuses on “Basic Network Concepts”.

1. Energy per unit charge is ____________

a) Power

b) Voltage

c) Current

d) Capacitance

Answer: b

Explanation: The work or energy per unit charge utilised in the process of separation of charges is known as Voltage or Potential difference. The phenomenon of transfer of charge from one point to another is termed Current. The rate at which the work is done is called Power. Charge per unit voltage is Capacitance.

2. A conductor is said to have resistance of one ohm if a potential difference of one volt across its terminals causes a current of X ampere to flow through it. What will be the value of X?

a) 4

b) 2

c) 3

d) 1

Answer: d

Explanation: Ohm’s law states that the potential difference across a conductor is proportional to the current through it. The constant of proportionality is called the Resistance.

According to Ohm’s law, V = IR .

–> I = V/R = 1V/1Ω = 1A.

3. Resistance depends on the temperature of the conductor.

a) True

b) False

Answer: a

Explanation: Resistance is directly proportional to its length, inversely proportional to the area of cross section of the conductor, depends on the nature of the material and on the temperature of the conductor.

4. A 25 Ω resistor has a voltage of 150 sin377 t. Find the corresponding power.

a) 900 sin 2 337 t

b) 90 sin 2 337 t

c) 900 sin 2 377 t

d) 9 sin 2 337 t

Answer: c

Explanation: Given R = 25 Ω and v = 150 sin 377 t

i = \

= 900 sin 2 377 t.

5. Unit of inductance is ________

a) Weber

b) Henry

c) Farad

d) Tesla

Answer: b

Explanation: The unit of inductance is Henry. Weber is the unit of magnetic flux. Tesla is the unit of flux density. Farad is the unit of capacitance.

6. Inductance of an inductor is inversely proportional to its ___________

a) Number of turns

b) Area of cross section

c) Absolute permeability

d) Length

Answer: d

Explanation: Inductance of an inductor, L = µN 2 A/l

From the above equation, Inductance of an inductor is inversely proportional to its length.

7. Energy stored in an inductor is ________

a) LI

b) LI 2

c) LI/2

d) LI 2 /2

Answer: d

Explanation: V = L \(\frac{di}{dt}\)

dE = Vidt = L \(\frac{di}{dt} idt\) = Lidt

E = \(\int_0^I dE = \int_0^I Lidt = \frac{1}{2} LI^2\).

8. An inductor of 3mH has a current i = 5(1 – e -5000t ). Find the corresponding maximum energy stored.

a) 37.5 mJ

b) 375 J

c) 37.5 kJ

d) 3.75 mJ

Answer: a

Explanation: Given L = 3 mH, i = 5(1 – e -5000t )

V = L \

] = 75 e^{-5000t}\)

I = i = 5(1 – e -∞ ) = 5 A

E = \(\frac{1}{2}\) LI 2 = 0.5 × 3 × 10 -3 × 5 2 = 37.5 mJ.

9. The capacitance of a capacitor does not depend on the absolute permittivity of the medium between the plates.

a) True

b) False

Answer: b

Explanation: C = Ɛ \(\frac{A}{d}\)

Where d is the distance between the plates, A is the cross-sectional area of the plates and Ɛ is absolute permittivity of the medium between the plates.

Hence, the capacitance of a capacitor depends on the absolute permittivity of the medium between the plates.

10. Which of the following is not the energy stored in a capacitor?

a) \

\(\frac{QC}{2}\)

Answer: d

Explanation: Energy stored in a capacitor, E = \(\frac{CV^2}{2}\)

Since C = Q/V

E = \(\frac{CV^2}{2} = \frac{QV}{2} = \frac{QC}{2}\).

11. A voltage is defined by \=

\) for \

µ i = 0 for t < 0

b) i = 20µA for 0 < t < 2s

c) i = 40e t-2 µA for t > 2s

d) i = -40e t-2 µA for t > 2s

Answer: c

Explanation: Using i = C \(\frac{dv}{dt}\)

For t < 0, i = 0

For 0 < t < 2s, v = 2t; i = 10 × 10 -6 \

= 20 µA

For t > 2s, v = 4e -

i = 10 × 10 -6 \(\frac{d}{dt}\)[4e - ] = 10 × 10 -6 [-4e - ] = -40e - µA.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Sources and Some Definitions”.

1. Source is a basic network element which supplies power to the networks.

a) True

b) False

Answer: b

Explanation: The basic network element which supplies energy to the networks is Source. Hence, it is true.

2. The dependent sources are of _____________ kinds.

a) 5

b) 2

c) 3

d) 4

Answer: d

Explanation: The dependent sources are of four kinds, depending on whether the control variable is voltage or current and the controlled source is a voltage source or current source. They are VCVS, VCCS, CCVS and CCCS.

3. The constant g m has dimension of ___________

a) Ampere per volt

b) Ampere

c) Volt

d) Volt per ampere

Answer: a

Explanation: Transconductance is the ratio of current to voltage. Hence, the constant g m has a dimension of ampere per volt or siemens .

4. In CCVS, voltage depends on the control current and the constant called __________

a) Transconductance

b) Transresistance

c) Current Gain

d) Voltage Gain

Answer: b

Explanation: In CCVS, voltage is directly proportional to the control current. The constant of proportionality is called Transresistance.

V = ri.

5. Every circuit is a network, but all networks are not circuits.

a) True

b) False

Answer: a

Explanation: The interconnection of two or more circuit elements is called a Network. If the network contains at least one closed path, it is called a Circuit.

6. Which of the following is not an example of a linear element?

a) Resistor

b) Thermistor

c) Inductor

d) Capacitor

Answer: b

Explanation: If the resistance, inductance or capacitance offered by an element does not change linearly with the change in applied voltage or circuit current, the element is termed as a linear element. Resistor, Inductor and Capacitor are examples of linear elements. Thermistor is an example of Non-Linear element.

7. Find the odd one out.

a) Resistor

b) Voltage-dependent resistor

c) Temperature-dependent resistor

d) Light-dependent resistor

Answer: a

Explanation: A non-linear circuit element is one in which the current does not change linearly with the change in applied voltage. Examples of non-linear elements are VDR, thermistor and LDR. Hence resistor is the odd one.

8. Which of the following is an Active element?

a) Resistor

b) Inductor

c) Capacitor

d) OP-AMP

Answer: d

Explanation: OP-AMP is an active element because it can be used for the amplification or generation of signals. All the other circuit elements are passive elements.

9. A semiconductor diode is an ____________ element.

a) Bilateral

b) Unilateral

c) Active

d) Passive

Answer: b

Explanation: In semiconductor diode, current flows through the diode only in one direction. Hence, it is a unilateral element.

10. Example of distributed element is ___________

a) Resistor

b) Thermistor

c) Semiconductor diode

d) Transmission lines

Answer: d

Explanation: Distributed elements are those which are not separable for analysis purposes. Examples of distributed elements are transmission lines in which the resistance, inductance and capacitance are distributed along its length.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Circuit Elements and Kirchhoff’s Laws”.

1. Potential difference in electrical terminology is known as?

a) Voltage

b) Current

c) Resistance

d) Conductance

Answer: a

Explanation: Potential difference in electrical terminology is known as Voltage and is denoted either by V or v. It is expressed in terms of energy per unit charge.

2. The circuit in which current has a complete path to flow is called ______ circuit.

a) short

b) open

c) closed

d) open loop

Answer: c

Explanation: The circuit in which current has a complete path to flow is called a closed circuit. When the current path is broken so that current cannot flow, the circuit is called an open circuit.

3. If the voltage-current characteristics is a straight line through the origin, then the element is said to be?

a) Linear element

b) Non-linear element

c) Unilateral element

d) Bilateral element

Answer: a

Explanation: If the voltage-current characteristic is a straight line through the origin, then the element is said to be Linear element. The difference in potential energy of charges is called Potential difference.

4. The voltage across R 1 resistor in the circuit shown below is?

network-theory-questions-answers-circuit-kirchhoff-laws-q4

a) 10

b) 5

c) 2.5

d) 1.25

Answer: b

Explanation: According to voltage divider rule, 10v is divide equally across resistors R 1 and R 2 . So the voltage across R 1 will be 5v.

5. The energy stored in the inductor is?

a) Li²/4

b) Li²/2

c) Li²

d) Li²/8

Answer: b

Explanation: The energy stored in the inductor the area under the power of the inductor and is given by W = ʃpdt = ʃLidi = Li²/2.

6. How many types of dependent or controlled sources are there?

a) 1

b) 2

c) 3

d) 4

Answer: d

Explanation: There are 4 dependent or controlled sources. They are VCVS, VCCS.

7. Find the voltage V x in the given circuit.

network-theory-questions-answers-circuit-kirchhoff-laws-q7

a) 10

b) 20

c) 30

d) 40

Answer: a

Explanation: From the circuit applying Kirchhoff’s voltage law, we can write 50 = 15 + 10 + 15 + V x => V x = 10V.

8. If the resistances 1Ω, 2Ω, 3Ω, 4Ω are parallel, then the equivalent resistance is?

a) 0.46Ω

b) 0.48Ω

c) 0.5Ω

d) 0.52Ω

Answer: b

Explanation: The equivalent resistance 1/R t = (1/R 1 )+(1/R 2 )+(1/R 3 )+(1/R 4 ). And R 1 , R 2 , R 3 , R 4 are 1Ω, 2Ω, 3Ω, 4Ω respectively. => R t = 0.48Ω.

9. Find total current in the circuit.

network-theory-questions-answers-circuit-kirchhoff-laws-q9

a) 1

b) 2

c) 3

d) 4

Answer: a

Explanation: R 2 is parallel to R 3 . So equivalent resistance of R 2 and R 3 is 1K. The total resistance in the circuit is K = 3K.Current in the circuit is 3V/3KΩ = 1mA.

10. If the resistances 3Ω, 5Ω, 7Ω, 9Ω are in series, then their equivalent resistance is?

a) 9

b) 20

c) 24

d) 32

Answer: c

Explanation: If the resistances are in series, then equivalent resistance is the sum of all the resistances that are in series. Equivalent resistance is Ω = 24Ω.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Voltage and Current Sources”.

1. Pick the incorrect statement among the following.

a) Inductor is a passive element

b) Current source is an active element

c) Resistor is a passive element

d) Voltage source is a passive element

Answer: b

Explanation: Energy sources are active elements, capable of delivering power to some external device.

2. For a voltage source to be neglected, the terminals across the source should be ___________

a) replaced by inductor

b) short circuited

c) replaced by some resistance

d) open circuited

Answer: b

Explanation: If the voltage source is to be neglected, it can be replaced simply by means of a wire i.e, it should be short circuited.

3. Voltage source and terminal voltage can be related as ___________

a) terminal voltage is higher than the source emf

b) terminal voltage is equal to the source emf

c) terminal voltage is always lower than source emf

d) terminal voltage cannot exceed source emf

Answer: c

Explanation: A practical voltage source can be represented with a resistance in series with the source. Hence, there would be some voltage drop at the resistor and the terminal voltage is always lower than the source emf.

4. In case of ideal current sources, they have ___________

a) zero internal resistance

b) low value of voltage

c) large value of currrent

d) infinite internal resistance

Answer: d

Explanation: For the ideal current sources, the current is completely independent of voltage and it has infinte internal resistance.

5. In a network consisting of linear resistors and ideal voltage source, if the value of resistors are doubled, then voltage across each resistor ___________

a) increases four times

b) remains unchanged

c) doubled

d) halved

Answer: b

Explanation: Even on changing the values of linear resistors, the voltage remains constant in case of ideal voltage source.

6. A practical current source can also be represented as ___________

a) a resistance in parallel with an ideal voltage source

b) a resistance in parallel with an ideal current source

c) a resistance in series with an ideal current source

d) none of the mentioned

Answer: b

Explanation: A practical current source could be represented with a resistor in parallel with an ideal current source.

7. A practical voltage source can also be represented as ___________

a) a resistance in series with an ideal current source

b) a resistance in series with an ideal voltage source

c) a resistance in parallel with an ideal voltage source

d) none of the mentioned

Answer: b

Explanation: A practical voltage source could be represented with a resistor in series with an ideal voltage source.

8. Constant voltage source is ___________

a) active and bilateral

b) passive and bilateral

c) active and unilateral

d) passive and unilateral

Answer: c

Explanation: Voltage source is an active element and is unilateral.

9. Which of the following is true about an ideal voltage source?

a) zero resistance

b) small emf

c) large emf

d) infinite resistance

Answer: a

Explanation: An ideal voltage source with zero internal resistance.

10. A dependent source ___________

a) may be a current source or a voltage source

b) is always a voltage source

c) is always a current source

d) none of the mentioned

Answer: a

Explanation: Dependent sources can either be current sources or voltage sources.

11. With some initial change at t = 0+, a capacitor will act as ___________

a) open circuit

b) short circuit

c) a current source

d) a voltage source

Answer: d

Explanation: At t=0+, the capacitor starts charging to a particular voltage and acts as a voltage source.

12. If a current source is to be neglected, the terminals across the source are ___________

a) replaced by a source resistance

b) open circuited

c) replaced by a capacitor

d) short circuited

Answer: b

Explanation: As the ideal current source has infinite resistance, it can be neglected by open circuiting the terminals.

13. A constant current source supplies a electric current of 200 mA to a load of 2kΩ. When the load changed to 100Ω, the load current will be ___________

a) 9mA

b) 4A

c) 700mA

d) 12A

Answer: b

Explanation: From Ohm’s law, resistance is inversely proportional to the current.

14. A voltage source having an open circuit voltage of 200 V and internal resistance of 50Ω is equivalent to a current source of ___________

a) 4A with 50Ω in parallel

b) 4A with 50Ω in series

c) 0.5A with 50Ω in parallel

d) none of the mentioned

Answer: a

Explanation: A voltage source with resistance in series can be replaced with a current source with the resistance in parallel.

15. A voltage source of 300 V has internal resistance of 4Ω and supplies a load having the same resistance. The power absorbed by the load is?

a) 1150 W

b) 1250 W

c) 5625 W

d) 5000 W

Answer: c

Explanation: Power absorbed = I 2 R.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Ohm’s Law”.

1. Resistance of a wire is yΩ. The wire is stretched to triple its length, then the resistance becomes ___________

a) y/3

b) 3y

c) 6y

d) y/6

Answer: b

Explanation: Resistance of a conductor is directly proportional to its length. That is, when the length of conductor is tripled, its resistance also gets tripled.

2. An electric current of 10 A is the same as ___________

a) 10 J/C

b) 10 V/C

c) 10C/sec

d) 10 W/sec

Answer: c

Explanation: Mathematically, electric current can be defined as the ratio of the charge to the time in which charge flows.

3. Consider a circuit with two unequal resistances in parallel, then ___________

a) large current flows in large resistor

b) current is same in both

c) potential difference across each is same

d) smaller resistance has smaller conductance

Answer: c

Explanation: In parallel combination of resistors, the potential difference across each resistors is the same.

4. In which of the following cases is Ohm’s law not applicable?

a) Electrolytes

b) Arc lamps

c) Insulators

d) Vacuum ratio values

Answer: c

Explanation: According to Ohm’s law, it is applicable only to conductors. Hence, Ohm’s law is not applicable in case of insulators.

5. A copper wire of length l and diameter d has potential difference V applied at its two ends. The drift velocity is V. If the diameter of wire is made d/4, then drift velocity becomes ___________

a) V/16

b) 16V

c) V

d) V/4

Answer: b

Explanation: Drift velocity is inversely propotional to area of material i.e, V=I/nAq.

6. Which of the following bulbs will have high resistance?

a) 220V, 60W

b) 220V,100W

c) 115V,60W

d) 115V,100 W

Answer: a

Explanation: Resistance is directly proportional to square of voltage and inversely proportional to the power.

7. Ohm’s law is not applicable to ___________

a) dc circuits

b) high currents

c) small resistors

d) semi-conductors

Answer: d

Explanation: Ohm’s law is not applicable to semi-conductors and insulators.

8. Conductance is expressed in terms of ___________

a) mho

b) mho/m

c) ohm/m

d) m/ohm

Answer: a

Explanation: Conductance is the reciprocal of resistance and is expressed in terms of mho.

9. Resistivity of a wire depends on ___________

a) length of wire

b) cross section area

c) material

d) all of the mentioned

Answer: c

Explanation: Resistivity of a wire is a constant and it depends on the type of material used.

10. In a current-voltage relationship graph of a linear resistor, the slope of the graph will indicate ___________

a) conductance

b) resistance

c) resistivity

d) a constant

Answer: a

Explanation: The slope of the graph is the ratio of current to a voltage which indicates conductance.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Kirchhoff’s Current Law”.

1. Kirchhoff’s Current law is based on the law of conservation of ___________

a) energy

b) momentum

c) mass

d) charge

Answer: d

Explanation: Kirchhoff’s current law is based on the law of conservation of charge i.e, charge that flows in = charge that flows out.

2. The current law represents a mathematical statement of fact that ___________

a) voltage cannot accumulate at node

b) charge cannot accumulate at node

c) charge at the node is infinite

d) none of the mentioned

Answer: b

Explanation: Charge cannot accumulate at the node, it can only flow in and out of the node.

3. Kirchhoff’s current law is applied at ___________

a) loops

b) nodes

c) both loop and node

d) none of the mentioned

Answer: b

Explanation: Kirchhoff’s current law can be applicable to nodes only.

4. Determine the current in all resistors in the circuit shown below.

network-theory-questions-answers-kirchhoffs-current-law-q4

a) 2A, 4A, 11A

b) 5A, 4.8A, 9.6A

c) 9.3A, 20.22A, 11A

d) 10.56A, 24.65A, 14.79A

Answer: d

Explanation: All the resistors are in parallel, so the voltage across each resistor is the same V.

i 1 =V/7, i 2 = V/3, i 3 =V/5. By current law, 50A = V/7 + V/3 + V/5. On solving, we obtain V and then values of i 1 , i 2 , i 3 .

5. For the circuit below , find the voltage across 5Ω resistor and the current through it.

network-theory-questions-answers-kirchhoffs-current-law-q5

a) 1.93 V

b) 2.83 V

c) 3.5 V

d) 5.7 V

Answer: b

Explanation: Here all the resistors are connected in parallel and let the voltage be V. Hence, i 15 =V/15, i 5 =V/5, i 2 =V/2, i 1 =V/1. By kirchhoff’s current law, V/15 + V/5 + V/2 V/1 +5 = 10. On solving equation, we obtain the value of V. As all resistors are in parallel, voltage across each is same as V.

6. Determine the current through the resistor R3 shown in the figure using KCL.

network-theory-questions-answers-kirchhoffs-current-law-q6

a) 25mA

b) 10mA

c) 20mA

d) 35mA

Answer: a

Explanation: Using KCL, 60mA = 10mA + 25mA + i 3 .

7. Find the current i3 in the circuit shown below.

network-theory-questions-answers-kirchhoffs-current-law-q7

a) 2A

b) 1A

c) 3A

d) 0.5A

Answer: c

Explanation: By applying the KCL at the node in the circuit , 5A = 2A +i 3 .

8. Kirchhof’s current law can be mathematically stated as ___________

a) ∑ k=0 n I = 0

b) i 2 ∑ k=0 n I = 0

c) i∑ k=0 n I = 0

d) none of the mentioned

Answer: a

Explanation: KCL states that the sum of currents entering and leaving a node is equal to zero.

9. Determine the current if a 20 coulomb charge passes a point in 0.25 seconds.

a) 10 A

b) 20 A

c) 2 A

d) 80 A

Answer: d

Explanation: By the definition of electric current, I=q/t.

10. Find the current through the branch containing resistance R3.

network-theory-questions-answers-kirchhoffs-current-law-q10

a) 2A

b) 3.25A

c) 2A

d) 2.75A

Answer: d

Explanation: By KCL, 5A = 0.25A + 2A + i 3 .

This set of Network Theory Multiple Choice Questions & Answers focuses on “Kirchhoff’s Voltage Law”.

1. Kirchhoff’s voltage law is based on principle of conservation of ___________

a) energy

b) momentum

c) mass

d) charge

Answer: a

Explanation: KVL is based on the law of conservation of energy.

2. In a circuit with more number of loops, which law can be best suited for the analysis?

a) KCL

b) Ohm’s law

c) KVL

d) None of the mentioned

Answer: c

Explanation: KVL can be best suited for circuits with more number of loops.

3. Mathematically, Kirchhoff’s Voltage law can be _____________

a) ∑_ n = 0

b) V2∑_ n = 0

c) V∑_ n = 0

d) None of the mentioned

Answer: a

Explanation: According to KVL, the sum of all voltages of branches in a closed loop is zero.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Tree and Co-Tree”.

1. A graph is said to be a directed graph if ________ of the graph has direction.

a) 1 branch

b) 2 branches

c) 3 branches

d) every branch

Answer: d

Explanation: If every branch of the graph has direction, then the graph is said to be a directed graph. If the graph does not have any direction then that graph is called undirected graph.

2. The number of branches incident at the node of a graph is called?

a) degree of the node

b) order of the node

c) status of the node

d) number of the node

Answer: a

Explanation: Nodes can be incident to one or more elements. The number of branches incident at the node of a graph is called degree of the node.

3. If no two branches of the graph cross each other, then the graph is called?

a) directed graph

b) undirected graph

c) planar graph

d) non-planar graph

Answer: c

Explanation: If a graph can be drawn on a plane surface such that no two branches of the graph cross each other, then the graph is called planar graph.

4. Consider the graph given below. Which of the following is a not a tree to the graph?

network-theory-questions-answers-tree-co-tree-q4

Answer: d

Explanation: Tree is sub graph which consists of all node of original graph but no closed paths. So, ‘d’ is not a tree to the graph.

5. Number of twigs in a tree are?

a) n

b) n+1

c) n-1

d) n-2

Answer: c

Explanation: Twig is a branch in a tree. Number of twigs in a tree are n-1. If there are 4 nodes in a tree then number of possible twigs are 3.

6. Loops which contain only one link are independent are called?

a) open loops

b) closed loops

c) basic loops

d) none of the mentioned

Answer: c

Explanation: The addition of subsequent link forms one or more additional loops. Loops that contain only one link are independent are called basic loops.

7. If the incident matrix of a graph is given below. The corresponding graph is?

a b c d e f

1 +1 0 +1 0 0 +1

2 -1 -1 0 +1 0 0

3 0 +1 0 0 +1 -1

4 0 0 -1 -1 -1 0

Answer: b

Explanation: For the given incidence matrix,

a b c d e f

1 +1 0 +1 0 0 +1

2 -1 -1 0 +1 0 0

3 0 +1 0 0 +1 -1

4 0 0 -1 -1 -1 0

the corresponding graph is

considering the directions specified in the graph.

8. If A represents incidence matrix, I represents branch current vectors, then?

a) AI = 1

b) AI = 0

c) AI = 2

d) AI= 3

Answer: b

Explanation: If A represents incidence matrix, I represents branch current vectors, then the relation is AI= 0 that is its characteristic equation must be equated to zero

9. If a graph consists of 5 nodes, then the number of twigs in the tree is?

a) 1

b) 2

c) 3

d) 4

Answer: d

Explanation: Number of twigs = n-1. As given number of nodes are 5 then n = 5. On substituting in the equation, number of twigs = 5 -1 = 4.

10. If there are 4 branches, 3 nodes then number of links in a co-tree are?

a) 2

b) 4

c) 6

d) 8

Answer: a

Explanation: Number of links = b-n+1. Given number of branches = 4 and number of nodes = 3. On substituting in the equation, number of links in a co-tree = 4 – 3 + 1 = 2.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Link Currents: Tie-Set Matrix”.

1. The current in a closed path in a loop is called?

a) loop current

b) branch current

c) link current

d) twig current

Answer: c

Explanation: In a loop there exists a closed path and a circulating current which is called link current. The current in any branch can be found by using link currents.

2. Tie-set is also called?

a) f loop

b) g loop

c) d loop

d) e loop

Answer: a

Explanation: The fundamental loop formed by one link has a unique path in the tree joining the two nodes of the link. This loop is also called f-loop.

3. Consider the graph shown below. If a tree of the graph has branches 4, 5, 6, then one of the twigs will be?

network-theory-questions-answers-link-tie-set-q3

a) 1

b) 2

c) 3

d) 4

Answer: d

Explanation: Branches of the tree are called twigs. So 4, 5, 6 are the twigs of the tree. The current in any branch of a graph can be found by using link currents.

4. Consider the graph shown below. If a tree of the graph has branches 4, 5, 6, then one of the links will be?

network-theory-questions-answers-link-tie-set-q3

a) 3

b) 4

c) 5

d) 6

Answer: a

Explanation: The branches of the co-tree are called links. So the links will be 1, 2, 3. For a given tree of a graph addition of each link between any two nodes form a loop called fundamental loop.

5. The loop current direction of the basic loop formed from the tree of the graph is?

a) same as the direction of the branch current

b) opposite to the direction of the link current

c) same as the direction of the link current

d) opposite to the direction of the branch current

Answer: c

Explanation: The loop current direction of the basic loop formed from the tree of the graph is same as the direction of the link current.

6. Consider the graph shown below. The direction of the loop currents will be? .

network-theory-questions-answers-link-tie-set-q6

a) I 1 ACW

b) I 2 ACW

c) I 3 CW

d) I 4 ACW

Answer: a

Explanation: The direction of the loop current will be along the direction of the link current in a basic loop. So I 1 – ACW, I 2 – CW, I 3 – ACW, I 4 – CW.

7. For Tie-set matrix, if the direction of current is same as loop current, then we place ______ in the matrix.

a) +1

b) -1

c) 0

d) +1 or -1

Answer: a

Explanation: For Tie-set matrix, if the direction of current is same as loop current, then we place +1 in the matrix.

8. If a row of the tie set matrix is as given below, then its corresponding equation will be?

1 2 3 4 5 6 7 8

I1 -1 +1 0 0 +1 0 0 0

a) -V 1 +V 2 +V 3 =0

b) -I 1 +I 2 +I 3 =0

c) -V 1 +V 2 -V 3 =0

d) -I 1 +I 2 -I 3 =0

Answer: a

Explanation: KVL equations are derived from tie set matrix and these include voltages not currents. So, -V 1 +V 2 +V 3 =0.

9. The matrix formed by link branches of a tie set matrix is?

a) Row matrix

b) Column matrix

c) Diagonal matrix

d) Identity matrix

Answer: d

Explanation: As the direction of the basic loops of the tree are taken along the direction of the link currents, then the matrix formed by the link currents will be an identity matrix.

10. The number of tie set matrices formed from a graph is?

a) N N-1

b) N N

c) N N-2

d) N N+1

Answer: c

Explanation: For every tree, there will be a unique tie set matrix. So there will be N N-2 tie set matrices.

This set of Network Theory Interview Questions and Answers focuses on “Cut-Set and Tree Branch Voltages”.

1. What is the direction of the cut-set?

a) same as the direction of the branch current

b) opposite to the direction of the link current

c) same as the direction of the link current

d) opposite to the direction of the branch current

Answer: a

Explanation: A cut-set is a minimal set of branches of a connected graph such that the removal of these branches causes the graph to be cut into exactly two parts. The direction of the cut-set is same as the direction of the branch current.

2. Consider the graph shown below. The direction of the cut-set of node ‘a’ is?

network-theory-interview-questions-answers-q2

a) right

b) left

c) upwards

d) downwards

Answer: c

Explanation: The direction of the cut set at node ‘a’ will be the direction of the branch current at node ‘a’. So the direction of the current will be upwards.

3. Consider the graph shown below. The direction of the cut-set at node ‘b’ will be?

network-theory-interview-questions-answers-q2

a) upwards

b) right

c) downwards

d) left

Answer: b

Explanation: The direction of the current will be towards right. The direction of the cut set at node ‘b’ will be the direction of the branch current at node ‘b’. So the direction of the current will be towards right.

4. In the graph shown below, the direction of the cut-set at node ‘c’ is?

network-theory-interview-questions-answers-q2

a) downwards

b) upwards

c) left

d) right

Answer: b

Explanation: The direction of the cut set at node ‘c’ will be the direction of the branch current at node ‘c’. So the direction of the current will be upwards.

5. In the graph shown below, the direction of the cut-set at node ‘d’ will be?

network-theory-interview-questions-answers-q2

a) left

b) downwards

c) right

d) upwards

Answer: c

Explanation: The direction of the cut set at node ‘d’ will be the direction of the branch current at node ‘d’. So the direction of the current will be upwards.

6. The row formed at node ‘a’ in the cut set matrix in the figure shown below is?

network-theory-interview-questions-answers-q2

a) +1 +1 +1 +1 0 0 0 0

b) +1 0 0 0 +1 0 0 +1

c) -1 0 0 0 -1 0 0 -1

d) -1 -1 0 0 -1 -1 0 0

Answer: b

Explanation: The direction of the cut set at node ‘a’ is towards node ‘a’. So the current direction of I 1 is same as cut set direction. So it is +1. Similarly for all other currents.

7. The row formed at node ‘c’ in the cut set matrix in the following figure?

network-theory-interview-questions-answers-q2

a) -1 -1 0 0 +1 -1 0 0

b) 0 0 +1 0 0 -1 -1 0

c) +1 0 0 0 +1 0 0 +1

d) -1 0 0 0 -1 0 0 -1

Answer: b

Explanation: The direction of the cut set at node ‘c’ is away from node ‘c’. So the current direction of I 3 is same as cut set direction. So it is +1. Similarly for all other currents.

8. The number of cut set matrices formed from a graph is?

a) N N-1

b) N N

c) N N-2

d) N N+1

Answer: c

Explanation: For every tree, there will be a unique cut set matrix. So there will be N N-2 cut set matrices.

9. For every tree there will be _____ number of cut set matrices.

a) 1

b) 2

c) 3

d) 4

Answer: a

Explanation: For every tree, there will a unique cut set matrix. So, number of cut-set matrices for every tree = 1.

10. If a row of the cut set matrix formed by the branch currents of the graph is shown below. Then which of the following is true?

I1 I2 I3 I4 I5 I6 I7 I8

-1 -1 0 0 +1 -1 0 0

a) -V 1 -V 2 +V 5 -V 6 =0

b) -I 1 -I 2 +I 5 -I 6 =0

c) -V 1 +V 2 +V 5 -V 6 =0

d) -I 1 +I 2 +I 5 -I 6 =0

Answer: b

Explanation: KCL equations are derived from cut set matrix and these include currents not voltages. So, -I 1 -I 2 +I 5 -I 6 =0.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Mesh Analysis”.

1. If there are M branch currents, then we can write ___________ number of independent equations.

a) M-2

b) M-1

c) M

d) M+1

Answer: c

Explanation: If there are M branch currents, then we can write M number of independent equations. Number of independent equations = M.

2. If there are M meshes, B branches and N nodes including reference node, the number of mesh currents is given as M=?

a) B +

b) B +

c) B-

d) B-

Answer: d

Explanation: If there are M meshes, B branches and N nodes including reference node, the number of mesh currents is given as M = B-.

3. Determine the current I 1 in the circuit shown below using mesh analysis.

network-theory-questions-answers-mesh-analysis-ac-q3

a) 0.955∠-69.5⁰

b) 0.855∠-69.5⁰

c) 0.755∠-69.5⁰

d) 0.655∠-69.5⁰

Answer: b

Explanation: The equation for loop 1 is I 1 + 6(I 1 -I 2 ) = 5∠0⁰. The equation for loop 2 is 6(I 1 -I 2 ) + I 2 + I 2 = 0. Solving the above equations, I 1 = 0.855∠-69.5⁰.

4. In the circuit shown below. Find the current I 2 .

network-theory-questions-answers-mesh-analysis-ac-q3

a) 0.5∠-90⁰

b) 0.6∠-90⁰

c) 0.7∠-90⁰

d) 0.8∠-90⁰

Answer: b

Explanation: The equation for loop 1 is I 1 + 6(I 1 -I 2 ) = 5∠0⁰. The equation for loop 2 is 6(I 1 -I 2 ) + I 2 + I 2 = 0. Solving the above equations, I 2 = 0.6∠-90⁰.

5. Find Z 11 , Z 12 , Z 13 obtained from the mesh equations in the circuit shown below.

network-theory-questions-answers-mesh-analysis-ac-q5

a) Ω, 5 Ω, 0Ω

b) Ω, 5 Ω, 0Ω

c) Ω, – 5 Ω, 0Ω

d) Ω, -5 Ω, 0Ω

Answer: d

Explanation: Z 11 = self impedance of loop 1 = Ω. Z 12 = Impedance common to both loop 1 and loop2 = -5Ω. Z 13 = No common impedance between loop1 and loop 3 = 0Ω.

6. Determine Z 21 , Z 22 , Z 23 in the circuit shown below.

network-theory-questions-answers-mesh-analysis-ac-q5

a) 5Ω, Ω, j6 Ω

b) -5Ω, Ω, j6 Ω

c) -5Ω, Ω, j6 Ω

d) -5Ω, Ω, – j6 Ω

Answer: b

Explanation: Z 21 = common impedance to loop 1 and loop 2 = -5 Ω. Z 22 = self impedance of loop2 = Ω. Z 23 = common impedance between loop2 and loop 3 = – Ω.

7. Find Z 31 , Z 32 , Z 33 in the circuit shown below.

network-theory-questions-answers-mesh-analysis-ac-q5

a) 0Ω, j6Ω, Ω

b) 0Ω, -j6Ω, Ω

c) 0Ω, -j6Ω, Ω

d) 0Ω, j6Ω, Ω

Answer: a

Explanation: Z 31 = common impedance to loop 3 and loop 1 = 0 Ω. Z 32 = common impedance between loop3 and loop 2 = – Ω. Z 33 = self impedance of loop 3 = Ω.

8. Find the values of Z 11 , Z 22 , Z 33 in the circuit shown below.

network-theory-questions-answers-mesh-analysis-ac-q8

a) Ω, Ω, Ω

b) Ω, Ω, Ω

c) Ω, Ω, Ω

d) Ω, Ω, Ω

Answer: b

Explanation: Z 11 = self impedance of loop 1 = Ω. Z 22 = self impedance of loop2 = Ω. Z 33 = self impedance of loop 3 = Ω.

9. Find the common impedances Z 12 , Z 13 , Z 21 , Z 23 , Z 31 , Z 32 respectively in the circuit shown below.

network-theory-questions-answers-mesh-analysis-ac-q8

a) -j3Ω, 0Ω, -j3Ω, j5Ω, 0Ω, j5Ω

b) j3Ω, 0Ω, -j3Ω, j5Ω, 0Ω, j5Ω

c) j3Ω, 0Ω, -j3Ω, j5Ω, 0Ω,- j5Ω

d) j3Ω, 0Ω, -j3Ω, -j5Ω, 0Ω, j5Ω

Answer: a

Explanation: The common impedances Z 12 and Z 21 are Z 12 = Z 21 = -j3Ω. The common impedances Z 13 and Z 31 are Z 13 = Z 31 = 0Ω. The common impedances Z 23 and Z 32 are Z 23 = Z 32 = j5Ω.

10. Find the value V 2 in the circuit shown below if the current through Ω is zero.

network-theory-questions-answers-mesh-analysis-ac-q8

a) 16∠-262⁰

b) 17∠-262⁰

c) 18∠-262⁰

d) 19∠-262⁰

Answer: b

Explanation: The three loop equations are I 1 – I 2 = 20∠0⁰. I 1 + I 2 + I 3 = 0. I 2 + I 3 = -V 2 . The current through Ω is zero, I 2 = ∆ 2 /∆ = 0

4+j3 20∠0⁰ 0

∆2 = | -j3 0 j5 |

0 -V2 5-j5

On solving, V 2 = 17∠-262⁰.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Supermesh Analysis”.

1. Consider the circuit shown below. Find the current I 1 .

network-theory-questions-answers-supermesh-analysis-q1

a) 1

b) 1.33

c) 1.66

d) 2

Answer: b

Explanation: Applying Super mesh analysis, the equations will be I 2 -I 1 =2 -10+2I 1 +I 2 +4=0. On solving the above equations, I 1 =1.33A.

2. Consider the circuit shown below. Find the current I 2 .

network-theory-questions-answers-supermesh-analysis-q1

a) 1.33

b) 2.33

c) 3.33

d) 4.33

Answer: c

Explanation: Applying Super mesh analysis, the equations will be I 2 -I 1 =2

-10+2I 1 +I 2 +4=0. On solving the above equations, I 2 =3.33A.

3.Consider the circuit shown below. Find the current I 1 .

network-theory-questions-answers-supermesh-analysis-q3

a) -1

b) -2

c) -3

d) -4

Answer: c

Explanation: Applying Super mesh analysis, the equations will be I 1 +I 1 +10+I 2 +I 2 =0. I 1 +I 2 =-5. I 2 -I 1 =1. On solving, I 1 =-3A.

4. Consider the circuit shown below. Find the current I 2 .

network-theory-questions-answers-supermesh-analysis-q3

a) -2

b) -1

c) 2

d) 1

Answer: a

Explanation: Applying Super mesh analysis, the equations will be I 1 +I 1 +10+I 2 +I 2 =0. I 1 +I 2 =-5. I 2 -I 1 =1. On solving, I 2 =-2A.

5. Find the power supplied by the voltage source in the following figure.

network-theory-questions-answers-supermesh-analysis-q3

a) 0

b) 1

c) 2

d) 3

Answer: a

Explanation: I 3 -I 2 =2. As I 2 =-2A, I 3 =0A. Th term power is the product of voltage and current. So, power supplied by source = 10×0=0W.

6. Find the current i 1 in the circuit shown below.

network-theory-questions-answers-supermesh-analysis-q6

a) 8

b) 9

c) 10

d) 11

Answer: c

Explanation: The current in the first loop is equal to 10A. So the current i 1 in the circuit is i 1 = 10A.

7. Find the current i 2 in the circuit shown below.

network-theory-questions-answers-supermesh-analysis-q6

a) 6.27

b) 7.27

c) 8.27

d) 9.27

Answer: b

Explanation: For 2nd loop, 10 + 2(i 2 -i 3 ) + 3(i 2 -i 1 ) = 0. For 3rd loop, i 3 + 2(i 3 -i 2 )=10. As i 1 =10A, On solving above equations, we get i 2 =7.27A.

8. Find the current i 3 in the circuit shown below.

network-theory-questions-answers-supermesh-analysis-q6

a) 8.18

b) 9.18

c) 10.18

d) 8.8

Answer: a

Explanation: For 2nd loop, 10 + 2(i 2 -i 3 ) + 3(i 2 -i 1 ) = 0. For 3rd loop, i 3 + 2(i 3 -i 2 )=10. As i 1 =10A, On solving above equations, we get i 3 =8.18A.

9. Find the current I 1 in the circuit shown below.

network-theory-questions-answers-supermesh-analysis-q9

a) 8

b) -8

c) 9

d) -9

Answer: b

Explanation: Applying Super Mesh analysis, I 1 – 10(I 2 ) – 5(I 3 ) = 50. 2(I 2 ) + I 3 + 5(I 3 -I 1 ) + 10(I 2 -I 1 ) = 0. I 2 – I 3 = 2. On solving above equations, we get I 1 =-8A.

10. Find the current I 2 in the circuit shown below.

network-theory-questions-answers-supermesh-analysis-q9

a) 5.3

b) -5.3

c) 7.3

d) -7.3

Answer: d

Explanation: Applying Super Mesh analysis, I 1 -10(I 2 )-5(I 3 ) = 50. 2(I 2 ) + I 3 + 5(I 3 -I 1 ) + 10(I 2 -I 1 ) = 0. I 2 – I 3 = 2. On solving above equations, we get I 2 =-7.3A.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Nodal Analysis”.

1. If there are N nodes in a circuit, then the number of nodal equations that can be formed are?

a) N+1

b) N

c) N-1

d) N-2

Answer: c

Explanation: If there are N nodes in a circuit, then the number of nodal equations that can be formed are N-1. Number of nodal equations = N-1.

2. In the network shown below, find the voltage at node ‘a’.

network-theory-questions-answers-nodal-analysis-ac-q2

a) 5.22∠104.5⁰

b) 5.22∠-104.5⁰

c) 6.22∠104.5⁰

d) 6.22∠-104.5⁰

Answer: b

Explanation: Applying nodal analysis at node ‘a’, (V a -10∠0 o )/j6+V a /+(V a -Vb)/3=0. Applying nodal analysis at node ‘b’, (V b -V a )/3+V b /j4+V b /j1=0. Solving the above equations we get, V a = 5.22∠-104.5⁰V.

3. Determine the voltage at node ‘b’ in the circuit shown below.

network-theory-questions-answers-nodal-analysis-ac-q2

a) -1.34∠-180⁰

b) 1.34∠-180⁰

c) -0.34∠-180⁰

d) 0.34∠-180⁰

Answer: a

Explanation: Applying nodal analysis at node ‘a’, (V a -10∠0 o )/j6+V a /+(V a -V b )/3=0. Applying nodal analysis at node ‘b’, (V b -V a )/3+V b /j4+V b /j1=0. Solving the above equations we get, V b = -1.34∠-180⁰V.

4. In the circuit shown below we get a nodal equation as V a —V b =x. Find the value of ‘x ‘ ‘.

network-theory-questions-answers-nodal-analysis-ac-q4

a) (5∠0 o )/3

b) – (5∠0 o )/3

c) (10∠0 o )/3

d) – (10∠0 o )/3

Answer: c

Explanation: The general equations are Y aa V a +Y ab V b = I 1 , Y ba V a +Y bb V b = I 2 . We get Y aa =1/3+1/j4+1/ and the self admittance at node a is the sum of admittances connected to node a. Y ab =-). I 1 = (10∠0 o )/3=x.

5. Find the value of ‘y’ in the equation –V a +V b =y obtained from the following circuit.

network-theory-questions-answers-nodal-analysis-ac-q4

a) (10∠30 o )/5

b) -(10∠30 o )/5

c) (5∠30 o )/5

d) (-5∠30 o )/5

Answer: b

Explanation: We got Y bb =1/5+1/j5-1/j6 and Y ab =–. The mutual admittance between node b and a is the sum of the admittances between nodes b and a. I 2 =-(10∠30 o )/5=y.

6. In the circuit shown below find the power dissipated by 2Ω resistor.

network-theory-questions-answers-nodal-analysis-ac-q6

a) 16.24

b) 17.24

c) 18.24

d) 19.24

Answer: c

Explanation: Applying nodal analysis at node ‘a’, (Va-20∠30 o )/3+Va/+Va/=0. On solving, Va = 16.27∠18.91⁰. Current in 2Ω resistor I 2 = Va/ = (16.27∠18.91 o )/(5.38∠68.19 o )=3.02∠-49.28 o . Power dissipated in 2Ω resistor P=I 2 2 R=3.02 2 ×2 = 18.24W.

7. In the circuit shown below. Determine the power dissipated in 3Ω resistor.

network-theory-questions-answers-nodal-analysis-ac-q6

a) 7.77

b) 8.77

c) 9.77

d) 10.77

Answer: b

Explanation: Current in 3Ω resistor I 3 = (-20∠30 o +16.27∠18.91 o )/3=1.71∠-112 o . Power dissipated in 3Ω resistor P=I 3 2 R=1.71 2 ×3=8.77W.

8. In the following circuit. Find the power output of the source.

network-theory-questions-answers-nodal-analysis-ac-q6

a) 27

b) 28

c) 29

d) 30

Answer: a

Explanation: Total power delivered by the source is the product of voltage and current and is given by power output of the source VIcosφ = 20 x 1.71cos142⁰ = 26.95W.

9. For the circuit shown below, find the voltage across the resistance R L if R L is infinite.

network-theory-questions-answers-nodal-analysis-ac-q9

a) 3

b) 2

c) 1

d) 0

Answer: d

Explanation: If R L is infinite, the voltage across it will be 0. So the voltage across the resistance R L if R L is infinite is zero.

10. Find the voltage V ab in the circuit shown below.

network-theory-questions-answers-nodal-analysis-ac-q9

a) 21.66∠-45.02⁰

b) 20.66∠-45.02⁰

c) 21.66∠45.02⁰

d) 20.66∠45.02⁰

Answer: c

Explanation: Applying nodal analysis at node ‘a’, (Va-20∠0 o )/+(Va-20∠90 o )/=0. On solving, we get Va = 21.66∠45.02⁰V.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Supernode Analysis”.

1. Consider the figure shown below. Find the voltage at node 1.

network-theory-questions-answers-supernode-analysis-q1

a) 13

b) 14

c) 15

d) 16

Answer: b

Explanation: Applying Super Node Analysis, the combined equation of node 1 and node 2 is (V 1 -V 3 )/3+3+(V 2 -V 3 )/1-6+V 2 /5=0. At node 3, (V 3 -V 1 )/3+(V 3 -V 2 )/1+V 3 /2=0. Also V 1 -V 2 =10. On solving above equations, we get V 1 = 13.72V ≈ 14V.

2. Consider the figure shown below. Find the voltage at node 2.

network-theory-questions-answers-supernode-analysis-q1

a) 3

b) 4

c) 5

d) 6

Answer: b

3. Consider the figure shown below. Find the voltage at node 3.

network-theory-questions-answers-supernode-analysis-q1

a) 4.5

b) 5.5

c) 6.5

d) 7.5

Answer: a

4. Consider the figure shown below. Find the power delivered by the source 6A.

network-theory-questions-answers-supernode-analysis-q1

a) 20.3

b) 21.3

c) 22.3

d) 24.3

Answer: c

Explanation: The term power is defined as the product of voltage and current and the power delivered by the source = V 2 x6 = 3.72×6 = 22.32W.

5. Find the voltage at node 1 in the circuit shown below.

network-theory-questions-answers-supernode-analysis-q5

a) 18

b) 19

c) 20

d) 21

Answer: b

Explanation: The equation at node 1 is 10 = V 1 /3+(V 1 -V 2 )/2. According to super Node analysis, (V 1 -V 2 )/2=V 2 /1+(V 3 -10)/5+V 3 /2V 2 -V 3 =20. On solving, we get, V 1 =19V.

6. Consider the figure shown below. Find the voltage at node 2.

network-theory-questions-answers-supernode-analysis-q5

a) 11.5

b) 12

c) 12.5

d) 13

Answer: a

Explanation: The equation at node 1 is 10 = V 1 /3+(V 1 -V 2 )/2

According to super Node analysis, (V 1 -V 2 )/2=V 2 /1+(V 3 -10)/5+V 3 /2V 2 -V 3 =20. On solving, we get, V 2 =11.5V.

7. Find the voltage at node 3 in the figure shown below.

network-theory-questions-answers-supernode-analysis-q7

a) 18

b) 20

c) 22

d) 24

Answer: a

Explanation: At node 1, (V 1 -40-V 3 )/4+(V 1 -V 2 )/6-3-5=0. Applying Super Node Analysis at node 2 and 3, (V 2 -V 1 )/6+5+V 2 /3+V 3 /5+(V 3 +40-V 1 )/4=0. Also, V 3 -V 2 =20. On solving above equations, V 3 = 18.11V ≈ 18V.

8. Find the power absorbed by 5Ω resistor in the following figure.

network-theory-questions-answers-supernode-analysis-q7

a) 60

b) 65.5

c) 70.6

d) 75

Answer: b

Explanation: The current through 5Ω resistor = V 3 /5=18.11/5=3.62A. The power absorbed by 5Ω resistor = 2 )×5=65.52W.

9. Find the value of the voltage in the equivalent voltage source of the current source shown below.

network-theory-questions-answers-supernode-analysis-q9

a) 20

b) 25

c) 30

d) 35

Answer: c

Explanation: The value of the voltage in the equivalent voltage source of the current source the voltage across the terminals A and B is = 30V.

10. Find the value of the current in the equivalent current source of the voltage source shown below.

network-theory-questions-answers-supernode-analysis-q10

a) 1

b) 2

c) 3

d) 4

Answer: b

Explanation: The value of the current in the equivalent current source of the voltage source the short circuit current at the terminals A and B is I=60/30=2A.

This set of Network Theory Questions and Answers for Freshers focuses on “Star-Delta Transformation”.

1. If a resistor R x is connected between nodes X and Y, R y between X and Y, R z between Y and Z to form a delta connection, then after transformation to star, the resistor at node X is?

a) R x R y /(R x +R y +R z )

b) R x R z /(R x +R y +R z )

c) R z R y /(R x +R y +R z )

d) (R x +R y )/(R x +R y +R z )

Answer: a

Explanation: After transformation to star, the resistor at node X is R x R y /(R x +R y +R z ) and this resistance lies between R x , R y in star connection.

2. If a resistor R x is connected between nodes X and Y, R y between X and Y, R z between Y and Z to form a delta connection, then after transformation to star, the resistance at node Y is?

a) R z R y /(R x +R y +R z )

b) R z R x /(R x +R y +R z )

c) R x R y /(R x +R y +R z )

d) (R z +R y )/(R x +R y +R z )

Answer: b

Explanation: After transformation to star, the resistor at node Y is R z R x /(R x +R y +R z ) and this resistance lies between R x , R z in star connection.

3. If a resistor R x is connected between nodes X and Y, R y between X and Y, R z between Y and Z to form a delta connection, then after transformation to star, the resistance at node Z is?

a) R y R x /(R x +R y +R z )

b) R y R x /(R x +R y +R z )

c) R z R y /(R x +R y +R z )

d) (R z +R x )/(R x +R y +R z )

Answer: c

Explanation: After transformation to star, the resistor at node Y is R z R y /(R x +R y +R z ) and this resistance lies between R z , R y in star connection.

4. If the resistors of star connected system are R 1 , R 2 , R 3 then the resistance between 1 and 2 in delta connected system will be?

a) (R 1 R 2 + R 2 R 3 + R 3 R 1 )/R 3

b) (R 1 R 2 + R 2 R 3 + R 3 R 1 )/R 1

c) (R 1 R 2 + R 2 R 3 + R 3 R 1 )/R 2

d) (R 1 R 2 + R 2 R 3 + R 3 R 1 )/(R 1 +R 2 )

Answer: a

Explanation: After transformation to delta, the resistance between 1 and 2 in delta connected system will be (R 1 R 2 + R 2 R 3 + R 3 R 1 )/R 3 and this resistance lies between R 1 , R 2 in delta connection.

5. If the resistors of star connected system are R 1 , R 2 , R 3 then the resistance between 2 and 3 in delta connected system will be?

a) (R 1 R 2 + R 2 R 3 + R 3 R 1 )/R 3

b) (R 1 R 2 + R 2 R 3 + R 3 R 1 )/R 2

c) (R 1 R 2 + R 2 R 3 + R 3 R 1 )/R 1

d) (R 1 R 2 + R 2 R 3 + R 3 R 1 )/(R 3 +R 2 )

Answer: c

Explanation: After transformation to delta, the resistance between 2 and 3 in delta connected system will be (R 1 R 2 + R 2 R 3 + R 3 R 1 )/R 1 and this resistance lies between R 3 , R 2 in delta connection.

6. If the resistors of star connected system are R 1 , R 2 , R 3 then the resistance between 3 and 1 in delta connected system will be?

a) (R 1 R 2 + R 2 R 3 + R 3 R 1 )/R 1

b) (R 1 R 2 + R 2 R 3 + R 3 R 1 )/R 3

c) (R 1 R 2 + R 2 R 3 + R 3 R 1 )/R 2

d) (R 1 R 2 + R 2 R 3 + R 3 R 1 )/(R 3 +R 1 )

Answer: c

Explanation: After transformation to delta, the resistance between 2 and 3 in delta connected system will be (R 1 R 2 + R 2 R 3 + R 3 R 1 )/R 2 and this resistance lies between R 1 , R 3 in delta connection.

7. Find the equivalent resistance at node A in the delta connected circuit shown in the figure below.

network-theory-questions-answers-freshers-q7

a) 1

b) 2

c) 3

d) 4

Answer: d

Explanation: Performing delta to star transformation we obtain the equivalent resistance at node A is R =/=4Ω.

8. Find the equivalent resistance at node C in the delta connected circuit shown below.

network-theory-questions-answers-freshers-q7

a) 3.66

b) 4.66

c) 5.66

d) 6.66

Answer: b

Explanation: Performing delta to star transformation we obtain the equivalent resistance at node A is R =/=4.66Ω.

9. Find the equivalent resistance between node 1 and node 3 in the star connected circuit shown below.

network-theory-questions-answers-freshers-q9

a) 30

b) 31

c) 32

d) 33

Answer: c

Explanation: The equivalent resistance between node 1 and node 3 in the star connected circuit is R =/10=32Ω.

10. Find the equivalent resistance between node 1 and node 2 in the star connected circuit shown below.

network-theory-questions-answers-freshers-q9

a) 2

b) 29

c) 30

d) 31

Answer: b

Explanation: The equivalent resistance between node 1 and node 3 in the star connected circuit is R =/11=29Ω.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Superposition Theorem”.

1. Superposition theorem states that the response in any element is the ____________ of the responses that can be expected to flow if each source acts independently of other sources.

a) algebraic sum

b) vector sum

c) multiplication

d) subtraction

Answer: b

Explanation: The superposition theorem is used to analyze ac circuits containing more than one source. Superposition theorem states that the response in any element is the vector sum of the responses that can be expected to flow if each source acts independently of other sources.

2. Superposition theorem is valid for only linear systems.

a) true

b) false

Answer: a

Explanation: Superposition theorem is valid for only linear systems. Superposition theorem is not valid for non-linear systems. In a network containing complex impedance, all quantities must be treated as complex numbers.

3. Determine the current through Ω impedance considering 50∠0⁰ voltage source.

network-theory-questions-answers-superposition-theorem-ac-q3

a) 6.42∠77.47⁰

b) 6.42∠-77.47⁰

c) 5.42∠77.47⁰

d) 5.42∠-77.47⁰

Answer: d

Explanation: According to the superposition theorem the current due to the 50∠0 o source is I 1 with the current source 20∠30⁰ A short-circuited. I 1 = (50∠0 o )/ = 5.42∠-77.47 o A.

4. Find the voltage across Ω impedance considering 50∠0⁰ voltage source.

network-theory-questions-answers-superposition-theorem-ac-q3

a) 30.16∠-9.28⁰

b) 30.16∠9.28⁰

c) 29.16∠-9.28⁰

d) 29.16∠9.28⁰

Answer: c

Explanation: Voltage across Ω impedance considering 50∠0⁰ voltage source is V 1 = 5.42∠-77.47 o = 29.16∠-9.28 o V.

5. Find the current through Ω impedance considering 20∠30⁰ voltage source.

network-theory-questions-answers-superposition-theorem-ac-q3

a) 8.68∠-42.53⁰

b) 8.68∠42.53⁰

c) 7.68∠42.53⁰

d) 7.68∠-42.53⁰

Answer: b

Explanation: The current through Ω impedance considering 20∠30⁰ voltage source is I 2 = 20∠30 o ×j4/ = 8.68∠42.53 o A.

6. Determine the voltage across Ω impedance considering 20∠30⁰ voltage source.

network-theory-questions-answers-superposition-theorem-ac-q3

a) 45.69∠-110.72⁰

b) 45.69∠110.72⁰

c) 46.69∠-110.72⁰

d) 46.69∠110.72⁰

Answer: d

Explanation: The voltage across Ω impedance considering 20∠30⁰ voltage source is V 2 = 8.68∠42.53 o = 46.69∠110.72⁰V.

7. Find the voltage across Ω impedance using Superposition theorem.

network-theory-questions-answers-superposition-theorem-ac-q3

a) 40.85∠72.53⁰

b) 40.85∠-72.53⁰

c) 41.85∠72.53⁰

d) 41.85∠-72.53⁰

Answer: a

Explanation: The voltage across Ω impedance using Superposition theorem is the sum of the voltages V 1 and V 2 . V = V 1 +V 2 = 29.16∠-9.28 o +46.69∠110.72 o =40.85∠72.53 o V.

8. Determine the voltage V ab considering the source 50∠0⁰V.

network-theory-questions-answers-superposition-theorem-ac-q3

a) 50∠0⁰

b) 4∠0⁰

c) 54∠0⁰

d) 46∠0⁰

Answer: a

Explanation: Let source 50∠0⁰ act on the circuit and set the source 4∠0⁰A equal to zero. If the current source is zero, it becomes open-circuited. Then the voltage across ‘ab’ is 50∠0⁰.

9. Determine the voltage V ab considering the source 4∠0⁰A in the circuit shown above.

network-theory-questions-answers-superposition-theorem-ac-q3

a) 46∠0⁰

b) 4∠0⁰

c) 54∠0⁰

d) 50∠0⁰

Answer: b

Explanation: Set the voltage source 50∠0⁰V to zero, and is short-circuited. So, the voltage drop across ‘ab’ is zero.

10. Find the voltage V ab in the circuit shown above using Superposition theorem.

network-theory-questions-answers-superposition-theorem-ac-q3

a) 4∠0⁰

b) 50∠0⁰

c) 54∠0⁰

d) 46∠0⁰

Answer: b

Explanation: The total voltage is the sum of the two voltages V 1 and V 2 => V ab = V 1 +V 2 = 50∠0⁰V.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Thevenin’s Theorem”.

1. Thevenin’s voltage is equal to the _____________ voltage across the _______________ terminals.

a) short circuit, input

b) short circuit, output

c) open circuit, output

d) open circuit, input

Answer: c

Explanation: Thevenin’s voltage is equal to open circuit voltage across output terminals not the short circuit voltage across output terminals.

2. Consider the circuit shown below. The expression of Thevenin’s voltage (V Th ) is?

network-theory-questions-answers-thevenins-theorem-ac-q2

a) V(Z 1 /(Z 1 +Z 2 ))

b) V(Z 2 /(Z 1 +Z 2 ))

c) V(Z 1 )

d) V(Z 2 )

Answer: b

Explanation: Thevenin’s theorem gives us a method for simplifying a given circuit. The thevenin’s voltage is V Th = V(Z 2 /(Z 1 +Z 2 )).

3. The value of Z Th in the circuit shown below is?

network-theory-questions-answers-thevenins-theorem-ac-q2

a) Z 3 +(Z 1 Z 2 /(Z 1 +Z 2 ))

b) Z 1 +(Z 3 Z 2 /(Z 3 +Z 2 ))

c) Z 2 +(Z 1 Z 3 /(Z 1 +Z 3 ))

d) (Z 1 Z 2 /(Z 1 +Z 2 ))

Answer: a

Explanation: The thevenin’s equivalent form of any complex impedance consists of an equivalent voltage source and an equivalent impedance. The thevenin’s impedance is Z Th = Z 3 +(Z 1 Z 2 /(Z 1 +Z 2 )).

4. In the circuit shown below, find the thevenin’s voltage across ‘ab’ terminals.

network-theory-questions-answers-thevenins-theorem-ac-q4

a) 48.5∠40.35⁰

b) 48.5∠-40.35⁰

c) 49.5∠-40.35⁰

d) 49.5∠40.35⁰

Answer: d

Explanation: Though the thevenin’s equivalent circuit is not same as its original circuit, the output current and voltage are the same in both cases. The thevenin’s voltage is equal to the voltage across Ω impedance. V Th =50∠0 o ×/+)=49.5∠40.35⁰V.

5. Find the thevenin’s impedance in the circuit shown below.

network-theory-questions-answers-thevenins-theorem-ac-q4

a) 4.83∠-1.13⁰

b) 5.83∠1.13⁰

c) 4.83∠1.13⁰

d) 5.83∠-1.13⁰

Answer: a

Explanation: The impedance is equal to the impedance seen into the network across the output terminals. Z Th = +/ = 4.83∠-1.13⁰Ω.

6. Determine the Thevenin’s voltage across ‘ab’ terminals in the circuit shown below.

network-theory-questions-answers-thevenins-theorem-ac-q6

a) 41.86∠0⁰

b) 42.86∠0⁰

c) 43.86∠0⁰

d) 44.86∠0⁰

Answer: b

Explanation: The voltage across the points a and b is called thevenin’s equivalent voltage. Thevenin’s equivalent voltage V ab =100∠0 o ×j3/ = 42.86∠0 o V.

7. Find the Thevenin’s impedance across ‘ab’ terminals in the circuit shown below.

network-theory-questions-answers-thevenins-theorem-ac-q6

a) j4.71

b) j5.71

c) j6.71

d) j7.71

Answer: c

Explanation: The impedance is equal to the impedance seen into the network across the output terminals. Z ab =j5 + /j7 = j6.71Ω.

8. Determine the load current across j5Ω in the circuit shown below.

network-theory-questions-answers-thevenins-theorem-ac-q2

a) 3.66∠90⁰

b) 3.66∠-90⁰

c) 4.66∠90⁰

d) 4.66∠-90⁰

Answer: b

Explanation: The load current is the ratio of thevenin’s equivalent voltage and thevenin’s equivalent impedance. The load current I L = (42.86∠0 o )/ = 3.66∠-90 o A.

9. Determine the thevenin’s voltage in the circuit shown below.

network-theory-questions-answers-thevenins-theorem-ac-q9

a) 18∠146.31⁰

b) 18∠-146.31⁰

c) 19∠146.31⁰

d) 19∠-146.31⁰

Answer: a

Explanation: The voltage across Ω resistor is = (5∠90 o )/) = 7.07∠-45 o V. The voltage across ‘ab’ = -10∠0 o +5∠90 o -7.07∠-45 o =18∠146.31 o V.

10. Find the Thevenin’s impedance in the following circuit.

network-theory-questions-answers-thevenins-theorem-ac-q9

a) 11.3∠45⁰

b) 12.3∠45⁰

c) 11.3∠-45⁰

d) 12.3∠-45⁰

Answer: c

Explanation: The impedance is equal to the impedance seen into the network across the output terminals. Z ab = 4 + / = 11.3∠-45 o Ω.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Norton’s Theorem”.

1. Norton’s current is equal to the current passing through the ___________ circuited ___________ terminals.

a) short, input

b) short, output

c) open, output

d) open, input

Answer: b

Explanation: Norton’s current is equal to the current passing through short circuited output terminals not the current through open circuited output terminals.

2. The expression of Norton’s current (I N ) in the circuit shown below is?

network-theory-questions-answers-nortons-theorem-ac-q2

a) V/Z 1

b) V/Z 2

c) V(Z 2 /(Z 1 +Z 2 ))

d) VZ 1 /(Z 1 +Z 2 )

Answer: a

Explanation: The Norton’s equivalent form of any complex impedance circuit consists of an equivalent current source and an equivalent impedance. The expression of Norton’s current is I N = V/Z 1 .

3. The expression of equivalent impedance (Z N ) in the circuit shown below is?

network-theory-questions-answers-nortons-theorem-ac-q2

a) (Z 1 +Z 2 )/Z 1

b) (Z 1 +Z 2 )/Z 2

c) Z 1 Z 2 /(Z 1 +Z 2 )

d) Z 1 +Z 2

Answer: c

Explanation: The impedance between the points a and b with the source replaced by a short circuit is Norton’s equivalent impedance. The Norton’s equivalent impedance is Z N = Z 1 Z 2 /(Z 1 +Z 2 ).

4. Determine Norton’s equivalent current in the circuit shown below.

network-theory-questions-answers-nortons-theorem-ac-q4

a) 5∠53.13⁰

b) 4∠53.13⁰

c) 4∠53.13⁰

d) 5∠-53.13⁰

Answer: d

Explanation: The current through the terminals a and b is the Norton’s equivalent current. The Norton’s equivalent current is I = (25∠0 o )/ = 5∠-53.13⁰A.

5. The Norton’s equivalent impedance in the circuit shown below.

network-theory-questions-answers-nortons-theorem-ac-q4

a) 4.53∠9.92⁰

b) 4.53∠-9.92⁰

c) 5.53∠9.92⁰

d) 5.53∠-9.92⁰

Answer: a

Explanation: The Norton’s equivalent impedance is Z = /+)=4.53∠9.92 o Ω. The impedance between the points a and b with the source replaced by a short circuit is Norton’s equivalent impedance.

6. Determine the Norton’s impedance seen from terminals ‘ab’.

network-theory-questions-answers-nortons-theorem-ac-q6

a) 6∠90⁰

b) 7∠90⁰

c) 6∠-90⁰

d) 7∠-90⁰

Answer: c

Explanation: The impedance between the points a and b with the source replaced by a short circuit is Norton’s equivalent impedance. The Norton’s impedance is Z ab =/-)=6∠-90 o Ω.

7. Find the Norton’s current passing through ‘ab’ in the circuit shown below.

network-theory-questions-answers-nortons-theorem-ac-q6

a) 4.16∠126.8⁰

b) 5.16∠126.8⁰

c) 5.16∠-126.8⁰

d) 4.16∠-126.8⁰

Answer: d

Explanation: The Norton’s current is equal to the current passing through the short circuit between the points a and b. The Norton’s current is I=(10∠0 o )/(3∠90 o )+(5∠90 o )/(2∠-90 o ) =4.16∠-126.8 o A.

8. Find the load current in the circuit shown below.

network-theory-questions-answers-nortons-theorem-ac-q6

a) 3.19∠166.61⁰

b) 3.19∠-166.61⁰

c) 4.19∠166.61⁰

d) 4.19∠-166.61⁰

Answer: b

Explanation: The load current in the circuit is given by I L = I×(6∠-90 o )/(5+6∠-90 o ) = 3.19∠-166.61 o A.

9. Determine Norton’s equivalent impedance in the circuit shown below.

network-theory-questions-answers-nortons-theorem-ac-q9

a) Ω

b) Ω

c) Ω

d) Ω

Answer: a

Explanation: The impedance seen from the terminals when the source is reduced to zero is Z = Ω.

10. Find the Norton’s current in the circuit shown below.

network-theory-questions-answers-nortons-theorem-ac-q9

a) 40∠30⁰

b) 40∠-30⁰

c) 30∠30⁰

d) 30∠-30⁰

Answer: c

Explanation: The Norton’s current is equal to the current passing through the short circuit between the points a and b. So the current passing through the short circuited terminals ‘a’ and ‘b’ is I = 30∠30⁰A.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Reciprocity Theorem”.

1. To check for the Reciprocity Theorem we consider ______ of response to excitation.

a) ratio

b) addition

c) product

d) subtraction

Answer: a

Explanation: For the Reciprocity Theorem to satisfy the ratio of response to the excitation of the circuit should be equal to the ratio of response to excitation after the source is replaced.

2. For the Reciprocity Theorem to satisfy the ratio of response to excitation before and after the source is replaced should be?

a) different

b) same

c) before source is replaced is greater than after the source is replaced

d) before source is replaced is less than after the source is replaced

Answer: b

Explanation: For the Reciprocity Theorem to satisfy the ratio of response to excitation before and after the source is replaced should be same and if that condition satisfies the reciprocity theorem is valid for the given circuit.

3. The circuit which satisfies Reciprocity Theorem is called?

a) Short circuit

b) Open circuit

c) Linear circuit

d) Non-linear circuit

Answer: c

Explanation: The circuit which satisfies Reciprocity Theorem is called linear circuit. A linear circuit is an electronic circuit in which, for a sinusoidal input voltage of frequency f, any steady-state output of the circuit is also sinusoidal with frequency f.

4. Find the current through the 2Ω resistor in the circuit shown below.

network-theory-questions-answers-reciprocity-theorem-q4

a) 0.143

b) 1.43

c) 14.3

d) 143

Answer: b

Explanation: Total resistance in the circuit = 2+[3||│├]=3.5Ω. The current drawn by the circuit (I t )=20/3.5=5.71Ω. The current drawn by 2Ω resistor = 1.43A.

5. In the following circuit, the current drawn by 2Ω resistor after the source is replaced is?

network-theory-questions-answers-reciprocity-theorem-q4

a) 143

b) 14.3

c) 1.43

d) 0.143

Answer: c

Explanation: The circuit after the source is replaced is

network-theory-questions-answers-reciprocity-theorem-q5

Total resistance = 3.23Ω. The current drawn by the circuit =20/3.23=6.19A. The current in branch a-b is 1.43A.

6. The following circuit satisfies Reciprocity Theorem.

network-theory-questions-answers-reciprocity-theorem-q4

a) True

b) False

Answer: a

Explanation: The ratio of response to excitation before the source is replaced is equal to 0.0715. And the ratio of response to excitation before the source is replaced is equal to 0.0715. So, the circuit satisfies the Reciprocity theorem.

7. Find the current through 3Ω resistor in the circuit shown below.

network-theory-questions-answers-reciprocity-theorem-q7

a) 1

b) 2

c) 3

d) 4

Answer: b

Explanation: The 6Ω resistor is parallel to 3Ω resistor and the resultant is in series with 2Ω resistor. Total current from source = 12/│)=3A. Current through 3Ω resistor = 3 × 6/=2A.

8. Find the current through 2Ω resistor after source is replaced in the below circuit.

network-theory-questions-answers-reciprocity-theorem-q7

a) 4

b) 3

c) 2

d) 1

Answer: c

Explanation: The circuit after source is replaced is

network-theory-questions-answers-reciprocity-theorem-q8

Total current from the source =12/│)=2.67A. Current through 2Ω resistor=2.67× 6/=2A.

9. The following circuit satisfies the reciprocity theorem.

network-theory-questions-answers-reciprocity-theorem-q7

a) False

b) True

Answer: b

Explanation: The ratio of response to excitation before the source is replaced is equal to 0.167. And the ratio of response to excitation before the source is replaced is equal to 0.167. So, the circuit satisfies the Reciprocity theorem.

10. While considering Reciprocity theorem, we consider ratio of response to excitation as ratio of?

a) voltage to voltage

b) current to current

c) voltage to current

d) none of the mentioned

Answer: c

Explanation: While considering Reciprocity theorem, we consider ratio of response to excitation as ratio of voltage to current or current to voltage.

This set of Network Theory Interview Questions and Answers for freshers focuses on “Compensation Theorem”.

1. Reciprocity Theorem is applied for _____ networks.

a) Linear

b) Bilateral

c) Linear bilateral

d) Lumped

Answer: c

Explanation: Reciprocity Theorem is applied for linear bilateral networks, not for linear or for linear bilateral or for lumped networks.

2. Reciprocity Theorem is used to find the change in _______ when the resistance is changed in the circuit.

a) Voltage

b) Voltage or current

c) Current

d) Power

Answer: b

Explanation: Reciprocity Theorem is used to find the change in voltage or current when the resistance is changed in the circuit. If reciprocity theorem is satisfied the ratio of response to excitation is same for the two conditions.

3. Find the current through 3Ω resistor in the circuit shown below.

network-theory-interview-questions-answers-freshers-q3

a) 1

b) 2

c) 3

d) 4

Answer: a

Explanation: Total resistance in the circuit = 2+[3||] = 3.5Ω. The total current drawn by the circuit =10/ = 1.67A. Current through 3Ω resistor = 1.11A ≅1A.

4. Determine the current flowing in the ammeter having 1Ω internal resistance connected in series with the 3Ω resistor as shown in the below circuit.

network-theory-interview-questions-answers-freshers-q3

a) 0.91

b) 0.92

c) 0.93

d) 0.94

Answer: c

Explanation: Current through 3Ω resistor = 1.11A. So voltage drop across 1Ω resistor = 1.11×1 = 1.11V. Now the circuit can be modified as

network-theory-interview-questions-answers-freshers-q4

Now current through 3Ω resistor = 0.17A. This current is opposite to the current calculated before. So ammeter reading = = 0.94A.

5. Find the current through 6Ω resistor in the circuit shown below.

network-theory-interview-questions-answers-freshers-q5

a) 0.33

b) 0.44

c) 0.55

d) 0.66

Answer: c

Explanation: Total resistance in the circuit = 4+6||3Ω. The total current drawn by the circuit = 10/=1.67A. Current through 6Ω resistor = 0.55A.

6. Determine the current flowing in the ammeter having 1Ω internal resistance connected in series with the 6Ω resistor as shown in the below circuit.

network-theory-interview-questions-answers-freshers-q5

a) 0.4

b) 0.45

c) 0.9

d) 0.95

Answer: b

Explanation: Current through 3Ω resistor = 0.55A. So voltage drop across 1Ω resistor = 0.55×1 = 0.55V.

Now the circuit can be modified as

network-theory-interview-questions-answers-freshers-q6

Now current through 6Ω resistor = 0.094A. This current is opposite to the current calculated before. So ammeter reading = = 0.45A.

7. Find the current through 6Ω resistor in the circuit shown below.

network-theory-interview-questions-answers-freshers-q7

a) 0.11

b) 0.22

c) 0.33

d) 0.44

Answer: c

Explanation: Total current in the circuit = 10/=2A. Current through 6Ω resistor = 2×/=0.33A.

8. Consider the following circuit. Determine the current flowing in the ammeter having 1Ω internal resistance in series with the 6Ω resistor.

network-theory-interview-questions-answers-freshers-q7

a) 0.1

b) 0.2

c) 0.3

d) 0.4

Answer: c

Explanation:

network-theory-interview-questions-answers-freshers-q8

New total current = 0.33/=0.04A. Now reading of ammeter = 0.33-0.04=0.29A ≅ 0.3A.

9. Find the current through 3Ω resistor in the circuit shown below.

network-theory-interview-questions-answers-freshers-q9

a) 0.45

b) 0.56

c) 0.67

d) 0.78

Answer: c

Explanation: Total current = 10 / = 2A. Current through 3Ω resistor= 2 x /) = 0.67A.

10. Consider the following circuit. Determine the current flowing in the ammeter having 1Ω internal resistance in series with the 3Ω resistor.

network-theory-interview-questions-answers-freshers-q9

a) 0.6

b) 0.7

c) 0.8

d) 0.9

Answer: a

Explanation: The current flowing in the ammeter having 1Ω internal resistance in series with the 3Ω resistor shown in the circuit is 0.6 A.

network-theory-interview-questions-answers-freshers-q10

Current through 3Ω resistor = 0.67/) = 0.08A. Ammeter reading = 0.67 – 0.08 = 0.59 ≅ 0.6A.

This set of Network Theory test focuses on “Maximum Power Transfer Theorem”.

1. The condition for maximum voltage to be transferred to the load is?

a) Source resistance greater than load resistance

b) Source resistance less than load resistance

c) Source resistance equal to load resistance

d) Source resistance greater than or equal to load resistance

Answer: b

Explanation: Our aim is to find the necessary conditions so that the power delivered by the source to the load is maximum. The condition for maximum voltage to be transferred to the load is source resistance less than load resistance.

2. The condition for maximum current to be transferred to the load is?

a) Source resistance greater than or equal to load resistance

b) Source resistance equal to load resistance

c) Source resistance less than load resistance

d) Source resistance greater than load resistance

Answer: d

Explanation: The condition for maximum current to be transferred to the load is source resistance greater than load resistance. For many applications an important consideration is the maximum power transfer to the load.

3. The condition for maximum power to be transferred to the load is?

a) Source resistance equal to load resistance

b) Source resistance greater than load resistance

c) Source resistance greater than or equal to load resistance

d) Source resistance less than load resistance

Answer: a

Explanation: The condition for maximum power to be transferred to the load is source resistance equal to load resistance. Maximum power transfer is desirable from the output amplifier to the speaker of an audio sound system.

4. In the circuit shown determine the value of load resistance when the load resistance draws maximum power?

network-theory-questions-answers-test-q4

a) 50

b) 25

c) 75

d) 100

Answer: b

Explanation: The source delivers maximum power when load resistance is equal to source resistance. So, load resistance = 25Ω.

5. Find the value of the maximum power in the circuit shown below.

network-theory-questions-answers-test-q4

a) 25

b) 50

c) 75

d) 100

Answer: a

Explanation: Current = 50/ = 1A. Maximum power delivered to load = 2 × R L . On substituting the values obtained and given we get maximum power in the circuit is 2 × 25 = 25W.

6. If the source Z S is complex, then the condition for the maximum power to be transferred is?

a) Z L =Z S

b) Z L =Z S *

c) Z L =-Z S

d) Z L =-Z S *

Answer: b

Explanation: If the source Zs is complex, then the condition for the maximum power to be transferred is Z L =Z S * that is load impedance is complex conjugate of source impedance.

7. If Z S =R S +jX S , Z L =R L , then condition for maximum power to be transferred is?

a) R L =|Z S |

b) R L =Z S

c) R L =-|Z S |

d) R L =-Z S

Answer: a

Explanation: If Z S =R S +jX S , Z L =R L , then condition for maximum power to be transferred is R L =|Z S | that is maximum power is transferred when the load resistance is equal to the magnitude of the source impedance.

8. Consider the following circuit. Find the load resistance so that the load draws maximum power.

a) 6

b) 7

c) 8

d) 9

Answer: d

Explanation: The condition for the load to draw maximum power is by open circuiting terminals a and b and the load resistance is R=/+/=8.96Ω≅9Ω.

9. Find the maximum power that is delivered by the source in the circuit shown below.

a) 50

b) 51

c) 52

d) 53

Answer: b

Explanation: V a =50×6/=18.75V, V b =50×8/=17.39V. V ab = V a – V b = 18.75 – 17.39 = 1.36V. I=1.36/=75mA. P=I 2 R= 2 ×9=0.051W ≅ 51mW.

10. If Z S = R S +jX S , Z L =R L +jX L , then if R L is fixed, the condition for maximum power to be transferred is?

a) X S =X L

b) X S =-X L

c) X S +X L =0

d) None of the mentioned

Answer: b

Explanation: If Z S = R S +jX S , Z L =R L +jX L , then if R L is fixed, the condition for maximum power to be transferred is X S =-X L .

This set of Network Theory Multiple Choice Questions & Answers focuses on “Tellegen’s theorem”.

1. The dual pair of current is?

a) voltage

b) current source

c) capacitance

d) conductance

Answer: a

Explanation: In an electrical circuit itself there are pairs of terms that can be interchanged to get new circuits. The dual pair of current is voltage. And the dual pair of voltage is current.

2. The dual pair of capacitance is?

a) capacitance

b) resistance

c) current source

d) inductance

Answer: d

Explanation: The dual pair of inductance is capacitance. And the dual pair of capacitance is inductance. In an electrical circuit itself, there are pairs of terms which can be interchanged to get new circuits.

3. The dual pair of resistance is?

a) inductance

b) capacitance

c) conductance

d) current

Answer: c

Explanation: The dual pair of resistance is conductance. And the dual pair of conductance is resistance.

4. The dual pair of voltage source is?

a) voltage

b) current source

c) current

d) resistance

Answer: b

Explanation: The dual pair of voltage source is current source. And the dual pair of current source is voltage source.

5. The dual pair of KCL is?

a) KVL

b) current

c) voltage

d) current source

Answer: a

Explanation: In an electrical circuit itself there are pairs of terms which can be interchanged to get new circuits. The dual pair of KCL is KVL. And the dual pair of KVL is KCL.

6. Tellegen’s Theorem is valid for _____ network.

a) linear or non-linear

b) passive or active

c) time variant or time invariant

d) all of the mentioned

Answer: d

Explanation: Tellegen’s Theorem is valid for any lumped network. So, Tellegan’s theorem is valid for linear or non-linear networks, passive or active networks and time variant or time invariant networks.

7. For Tellegan’s Theorem to satisfy, the algebraic sum of the power delivered by the source is _____ than power absorbed by all elements.

a) greater

b) less

c) equal

d) greater than or equal

Answer: c

Explanation: For Tellegan’s Theorem to satisfy, algebraic sum of the power delivered by the source equal to power absorbed by all elements. All branch currents and voltages in that network must satisfy Kirchhoff’s laws.

8. Consider the circuit shown below. Find whether the circuit satisfies Tellegan’s theorem.

network-theory-questions-answers-tellegans-theorem-q8

a) satisfies

b) does not satisfy

c) satisfies partially

d) satisfies only for some elements

Answer: a

Explanation: i 1 =i 2 =2A, i 3 =2A. V 1 =-2V, V 2 =-8V, V 3 =10V. Algebraic sum = network-theory-questions-answers-tellegans-theorem-q8a

9. The circuit shown below satisfies Tellegen’s theorem.

network-theory-questions-answers-tellegans-theorem-q9

a) True

b) False

Answer: a

Explanation: i 1 =i 2 =4A, i 3 =4A. V 1 =-20V, V 2 =0V, V 3 =20V. Algebraic sum = network-theory-questions-answers-tellegans-theorem-q9a

10. If two networks have same graph with different type of elements between corresponding nodes, then?

a) \

\(\Sigma_{k=1}^{b}\)V1ki2k ≠ 0, \(\Sigma^{b}_{k=1}\)V2ki1k ≠ 0

Answer: a

Explanation: If two networks have same graph with different type of elements between corresponding nodes, then \(\Sigma^{b}_{k=1}\)V1ki2k = 0, \(\Sigma^{b}_{k=1}\)V2ki1k = 0.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Millman’s Theorem”.

1. According to Millman’s Theorem, if there are n voltage sources with n internal resistances respectively, are in parallel, then these sources are replaced by?

a) single current source I’ in series with R’

b) single voltage source V’ in series with R’

c) single current source I’ in parallel to R’

d) single voltage source V’ in parallel to R’

Answer: b

Explanation: Millman’s Theorem states that if there are voltage sources V1, V 2 ,…… Vn with internal resistances R 1 , R 2 ,…..R n , respectively, are in parallel, then these sources are replaced by single voltage source V’ in series with R’.

2. According to Millman’s Theorem, if there are n voltage sources with n internal resistances respectively, are in parallel, then the value of equivalent voltage source is?

a) V ‘ =(V 1 G 1 +V 2 G 2 +⋯.+V n G n )

b) V ‘ =((V 1 G 1 +V 2 G 2 +⋯.+V n G n ))/((1/G 1 +1/G 2 +⋯1/G n ))

c) V ‘ =((V 1 G 1 +V 2 G 2 +⋯.+V n G n ))/(G 1 +G 2 +⋯G n )

d) V ‘ =((V 1 /G 1 +V 2 /G 2 +⋯.+V n /G n ))/( G 1 +G 2 +⋯G n )

Answer: c

Explanation: The value of equivalent voltage source is V ‘ = ((V 1 G 1 +V 2 G 2 +⋯.+V n G n ))/(G 1 +G 2 +⋯G n ).

3. According to Millman’s Theorem, if there are n voltage sources with n internal resistances respectively, are in parallel, then the value of equivalent resistance is?

a) R’=G 1 +G 2 +⋯G n

b) R’=1/G 1 +1/G 2 +⋯1/G n

c) R’=1/((G 1 +G 2 +⋯G n ))

d) R’=1/(1/G 1 +1/G 2 +⋯1/G n )

Answer: c

Explanation: Let the equivalent resistance is R’. The value of equivalent resistance is R’=1/((G 1 +G 2 +⋯G n )).

4. According to Millman’s Theorem, if there are n current sources with n internal conductances respectively, are in series, then these sources are replaced by?

a) single voltage source V’ in parallel with G’

b) single current source I’ in series with G’

c) single current source I’ in parallel with G’

d) single voltage source V’ in series with G’

Answer: c

Explanation: Millman’s Theorem states that if there are current sources I 1 ,I 2 ,…… In with internal conductances G 1 ,G 2 ,…..G n , respectively, are in series, then these sources are replaced by single current source I’ in parallel with G’.

5. According to Millman’s Theorem, if there are n current sources with n internal conductances respectively, are in series, then the value of equivalent current source is?

a) I ‘ =((I 1 R 1 +I 2 R 2 +⋯.+I n R n ))/(R 1 +R 2 +⋯R n )

b) I’=I 1 R 1 +I 2 R 2 +⋯.+I n R n

c) I’=((I 1 /R 1 +I 2 /R 2 +⋯.+I n /R n ))/(R 1 +R 2 +⋯R n )

d) I’=I 1 /R 1 +I 2 /R 2 +⋯.+I n /R n

Answer: a

Explanation: The value of equivalent current source is I ‘ =((I 1 R 1 +I 2 R 2 +⋯.+I n R n ))/(R 1 +R 2 +⋯R n ).

6. According to Millman’s Theorem, if there are n current sources with n internal conductances respectively, are in series, then the value of equivalent conductance is?

a) G’=R 1 +R 2 +⋯R n

b) G’=1/(1/R 1 +1/R 2 +⋯1/R n )

c) G’=1/((R 1 +R 2 +⋯R n ))

d) G’=1/R 1 +1/R 2 +⋯1/R n

Answer: c

Explanation: Let the equivalent conductance is G’. The value of equivalent conductance is G’=1/((R 1 +R 2 +⋯R n )).

7. Calculate the current through 3Ω resistor in the circuit shown below.

network-theory-questions-answers-millmans-theorem-q7

a) 1

b) 2

c) 3

d) 4

Answer: c

Explanation: Applying Nodal analysis the voltage V is given by /2+/5=V/3. V=8.7V. Now the current through 3Ω resistor in the circuit is I = V/3 = 8.7/3 = 2.9A ≅ 3A.

8. Find the current through 3Ω resistor in the circuit shown below using Millman’s Theorem.

network-theory-questions-answers-millmans-theorem-q7

a) 4

b) 3

c) 2

d) 1

Answer: b

Explanation: V ‘ =((V 1 G 1 +V 2 G 2 ))/(G 1 +G 2 )=+20)/=12.86V. R’=1/((G 1 +G 2 ))=1/=1.43Ω. Current through 3Ω resistor=I=12.86/=2.9A≅3A.

9. Consider the circuit shown below. Find the current through 4Ω resistor.

network-theory-questions-answers-millmans-theorem-q9

a) 2

b) 1.5

c) 1

d) 0.5

Answer: b

Explanation: Applying Nodal analysis the voltage V is given by /1+/3=V/4. V=6V. The current through 4Ω resistor I = V/4 = 6/4 = 1.5A.

10. In the following circuit. Find the current through 4Ω resistor using Millman’s Theorem.

network-theory-questions-answers-millmans-theorem-q9

a) 0.5

b) 1

c) 1.5

d) 2

Answer: c

Explanation: V ‘ =((V 1 G 1 +V 2 G 2 ))/(G 1 +G 2 )=+10)/=6.25V. R’=1/((G 1 +G 2 ))=1/=0.75Ω. I=6.25/=1.5A.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Thevenin Theorem Involving Dependent and Independent Sources”.

1. A circuit is given in the figure below. The Thevenin equivalent as viewed from terminals x and x’ is ___________

network-theory-questions-answers-thevenin-theorem-involving-dependent-independent-sources-q1

a) 8 V and 6 Ω

b) 5 V and 6 Ω

c) 5 V and 32 Ω

d) 8 V and 32 Ω

Answer: b

Explanation: We, Thevenized the left side of xx’ and source transformed right side of yy’.

network-theory-questions-answers-thevenin-theorem-involving-dependent-independent-sources-q1a

V xx’ = V th = \(\displaystyle\frac{\frac{4}{8} + \frac{8}{24}}{\frac{1}{8} + \frac{1}{24}}\) = 5V

∴ R th = 8 ||

= \(\frac{8×24}{8+24}\) = 6 Ω.

2. For the circuit given in figure below, the Thevenin equivalent as viewed from terminals y and y’ is _________

network-theory-questions-answers-thevenin-theorem-involving-dependent-independent-sources-q1

a) 8 V and 32 Ω

b) 4 V and 32 Ω

c) 5 V and 6 Ω

d) 7 V and 6 Ω

Answer: d

Explanation: We, Thevenized the left side of xx’ and source transformed right side of yy’.

network-theory-questions-answers-thevenin-theorem-involving-dependent-independent-sources-q1a

Thevenin equivalent as seen from terminal yy’ is

V xx’ = V th = \(\displaystyle\frac{\frac{4}{24} + \frac{8}{8}}{\frac{1}{24} + \frac{1}{8}}\) = 5V

= \(\frac{0.167+1}{0.04167+0.125}\) = 7 V

∴ R th = || 8

= \(\frac{24×8}{24+8}\) = 6 Ω.

3. In the following circuit, when R = 0 Ω, the current IR equals to 10 A. The value of R, for which maximum power is absorbed by it is ___________

network-theory-questions-answers-thevenin-theorem-involving-dependent-independent-sources-q3

a) 4 Ω

b) 3 Ω

c) 2 Ω

d) 1 Ω

Answer: c

Explanation: The Thevenin equivalent of the circuit is as shown below.

network-theory-questions-answers-thevenin-theorem-involving-dependent-independent-sources-q3a

Therefore from the figure we can infer that R th = 2 Ω

4. In the following circuit, when R = 0 Ω, the current I R equals to 10 A. The maximum power will be?

network-theory-questions-answers-thevenin-theorem-involving-dependent-independent-sources-q3

a) 50 W

b) 100 W

c) 200 W

d) 400 W

Answer: a

Explanation: The Thevenin equivalent of the circuit is as shown below.

I = 10 A, R th = 2 Ω

∴ P max =

2 × 2

= 5×5×2 = 50 W.

5. For the circuit given below, the Thevenin resistance across the terminals A and B is _____________

network-theory-questions-answers-thevenin-theorem-involving-dependent-independent-sources-q5

a) 5 Ω

b) 7 kΩ

c) 1.5 kΩ

d) 1.1 kΩ

Answer: b

Explanation: Let V AB = 1 V

5 V AB = 5

Or, 1 = 1 × I 1 or, I 1 = 1

Also, 1 = -5 + 1(I – I 1 )

∴ I = 7

Hence, R = 0.2 kΩ.

6. For the circuit given below, the Thevenin voltage across the terminals A and B is ____________

network-theory-questions-answers-thevenin-theorem-involving-dependent-independent-sources-q5

a) 1.25 V

b) 0.25 V

c) 1 V

d) 0.5 V

Answer: d

Explanation: Current through 1 Ω = \(\frac{5}{2}\) – I 1

Using source transformation to 5 V sources, V OC = 1 × I 1

V OC = -5 V OC + (\(\frac{5}{2}\) – I 1 ) × 1

Eliminating I1, we get, V OC = 0.5 V.

7. In the following circuit, the value of open circuit voltage and the Thevenin resistance between terminals a and b are ___________

network-theory-questions-answers-thevenin-theorem-involving-dependent-independent-sources-q7

a) V OC = 100 V, R TH = 1800 Ω

b) V OC = 0 V, R TH = 270 Ω

c) V OC = 100 V, R TH = 90 Ω

d) V OC = 0 V, R TH = 90 Ω

Answer: d

Explanation: By writing loop equations for the circuit, we get,

V S = V X , I S = I X

V S = 600(I 1 – I 2 ) + 300(I 1 – I 2 ) + 900 I 1

= I 1 – 600I 2 – 300I 3

= 1800I 1 – 600I 2 – 300I 3

I 1 = I S , I 2 = 0.3 V S

I 3 = 3I S + 0.2V S

V S = 1800IS – 600(0.01V S ) – 300(3I S + 0.01V S )

= 1800I S – 6V S – 900I S – 3V S

10V S = 900I S

For Thevenin equivalent, V S = R TH I S + V OC

So, Thevenin voltage V OC = 0

Resistance R TH = 90Ω.

8. In the circuit given below, it is given that V AB = 4 V for R L = 10 kΩ and V AB = 1 V for R L = 2kΩ. The values of the Thevenin resistance and voltage for the network N are ____________

network-theory-questions-answers-thevenin-theorem-involving-dependent-independent-sources-q8

a) 16 kΩ and 30 V

b) 30 kΩ and 16 V

c) 3 kΩ and 6 V

d) 50 kΩ and 30 V

Answer: b

Explanation: When R L = 10 kΩ and V AB = 4 V

Current in the circuit I = \(\frac{V_{AB}}{R_L} = \frac{4}{10}\) = 0.4 mA

Thevenin voltage is given by V TH = I (R TH + R L )

= 0.4(R TH + 10)

= 0.4R TH + 4

Similarly, for R L = 2 kΩ and V AB = 1 V

So, I = \(\frac{1}{2}\) = 0.5 mA

V TH = 0.5(R TH + 2)

= 0.5 R TH + 1

∴ 0.1R TH = 3

Or, R TH = 30 kΩ

And V TH = 12 + 4 = 16 V.

9. For the circuit shown in figure below, the value of the Thevenin resistance is _________

network-theory-questions-answers-thevenin-theorem-involving-dependent-independent-sources-q9

a) 100 Ω

b) 136.4 Ω

c) 200 Ω

d) 272.8 Ω

Answer: a

Explanation: I X = 1 A, V X = V test

V test = 100(1-2I X ) + 300(1-2I X – 0.01V S ) + 800

Or, V test = 1200 – 800I X – 3V test

Or, 4V test = 1200 – 800 = 400

Or, V test = 100V

∴ R TH = \(\frac{V_{test}}{1}\) = 100 Ω.

10. For the circuit shown in the figure below, the Thevenin voltage and resistance looking into X-Y are __________

network-theory-questions-answers-thevenin-theorem-involving-dependent-independent-sources-q10

a) \

4V and \

4 V and 2 Ω

Answer: d

Explanation: \(R_{TH} = \frac{V_{OC}}{I_{SC}}\)

V TH = V OC

Applying KCL at node A, \(\frac{2I-V_{TH}}{1} + 2 = I + \frac{V_{TH}}{2}\)

Or, I = \(\frac{V_{TH}}{1}\)

Putting, 2V TH – V TH + 2 = V TH + \(\frac{V_{TH}}{2}\)

Or, V TH = 4 V.

∴ R TH = 4/2 = 2Ω.

11. In the figure given below, the value of the source voltage is ___________

network-theory-questions-answers-thevenin-theorem-involving-dependent-independent-sources-q11

a) 12 V

b) 24 V

c) 30 V

d) 44 V

Answer: c

Explanation: By applying KCL, \(\frac{V_P-E}{6} + \frac{V_P}{6}\) – 1 = 0

Or, 2 V P – E = 6

Where,

= 2

∴ E – V P = 12

Or, V P = 18 V

∴ E = 30V.

12. In the figure given below, the value of Resistance R by Thevenin Theorem is ___________

network-theory-questions-answers-thevenin-theorem-involving-dependent-independent-sources-q12

a) 10

b) 20

c) 30

d) 40

Answer: b

Explanation: \(\frac{V_P-100}{10} + \frac{V_P}{10}\) + 2 = 0

Or, 2V P – 100 + 20 = 0

∴ V P = 80/2 = 40V

∴ R = 20Ω.

13. In the figure given below, the Thevenin’s equivalent pair, as seen at the terminals P-Q, is given by __________

network-theory-questions-answers-thevenin-theorem-involving-dependent-independent-sources-q13

a) 2 V and 5 Ω

b) 2 V and 7.5 Ω

c) 4 V and 5 Ω

d) 4 V and 7.5 Ω

Answer: a

Explanation: For finding V TH ,

network-theory-questions-answers-thevenin-theorem-involving-dependent-independent-sources-q13a

V TH = \(\frac{4 ×10}{10+10}\) = 2V

For finding R TH ,

network-theory-questions-answers-thevenin-theorem-involving-dependent-independent-sources-q13b

R TH = 10 || 10

= \(\frac{10×10}{10+10}\) = 5 Ω.

14. The Thevenin equivalent impedance Z between the nodes P and Q in the following circuit is __________

network-theory-questions-answers-thevenin-theorem-involving-dependent-independent-sources-q14

a) 1

b) 1 + s + \

2 + s + \

3 + s + \(\frac{1}{s}\)

Answer: a

Explanation: To calculate the Thevenin resistance, all the current sources get open-circuited and voltage source short-circuited.

∴ R TH =

= \(\frac{\leftMissing or unrecognized delimiter for \right×}{\leftMissing or unrecognized delimiter for \right+}\)

= \(\frac{\frac{1}{s}+1+1+s}{\frac{1}{s}+1+1+s}\) = 1

So, R TH = 1.

15. While computing the Thevenin equivalent resistance and the Thevenin equivalent voltage, which of the following steps are undertaken?

a) Both the dependent and independent voltage sources are short-circuited and both the dependent and independent current sources are open-circuited

b) Both the dependent and independent voltage sources are open-circuited and both the dependent and independent current sources are short-circuited

c) The dependent voltage source is short-circuited keeping the independent voltage source untouched and the dependent current source is open-circuited keeping the independent current source untouched

d) The dependent voltage source is open-circuited keeping the independent voltage source untouched and the dependent current source is short-circuited keeping the independent current source untouched

Answer: c

Explanation: While computing the Thevenin equivalent voltage consisting of both dependent and independent sources, we first find the equivalent voltage called the Thevenin voltage by opening the two terminals. Then while computing the Thevenin equivalent resistance, we short-circuit the dependent voltage sources keeping the independent voltage sources untouched and open-circuiting the dependent current sources keeping the independent current sources untouched.

This set of Network Theory Questions & Answers for Exams focuses on “Norton’s Theorem Involving Dependent and Independent Sources”.

1. The circuit shown in figure has a load equivalent to _________

network-theory-exam-questions-answers-q1

a) \

4 Ω

d) 2 Ω

Answer: b

Explanation: Applying KCL in the given circuit, we get, \(\frac{V}{4} + \frac{V-2I}{2}\) = I

Or, \(\frac{3V-4I}{4}\) = I

Or, 3V = 8I

∴ \(\frac{V}{I} = \frac{8}{3}\) Ω.

2. In the following circuit, the value of Norton’s resistance between terminals a and b are ___________

network-theory-exam-questions-answers-q2

a) R N = 1800 Ω

b) R N = 270 Ω

c) R N = 90 Ω

d) R N = 90 Ω

Answer: d

Explanation: By writing loop equations for the circuit, we get,

V S = V X , I S = I X

V S = 600(I 1 – I 2 ) + 300(I 1 – I 2 ) + 900 I 1

= I 1 – 600I 2 – 300I 3

= 1800I 1 – 600I 2 – 300I 3

I 1 = I S , I 2 = 0.3 V S

I 3 = 3I S + 0.2V S

V S = 1800IS – 600(0.01V S ) – 300(3I S + 0.01V S )

= 1800I S – 6V S – 900I S – 3V S

10V S = 900I S

For Voltage, V S = R N I S + V OC

Here V OC = 0

So, Resistance R N = 90Ω.

3. For the circuit shown in figure below, the value of Norton’s resistance is _________

network-theory-exam-questions-answers-q3

a) 100 Ω

b) 136.4 Ω

c) 200 Ω

d) 272.8 Ω

Answer: a

Explanation: I X = 1 A, V X = V test

V test = 100(1-2I X ) + 300(1-2I X – 0.01V S ) + 800

Or, V test = 1200 – 800I X – 3V test

Or, 4V test = 1200 – 800 = 400

Or, V test = 100V

∴ R N = \(\frac{V_{test}}{1}\) = 100 Ω.

4. For the circuit shown in the figure below, the Norton Resistance looking into X-Y is __________

network-theory-exam-questions-answers-q4

a) 2 Ω

b) \

2 Ω

Answer: d

Explanation: \(R_N = \frac{V_{OC}}{I_{SC}}\)

V N = V OC

Applying KCL at node A, \(\frac{2I-V_N}{1} + 2 = I + \frac{V_N}{2}\)

Or, I = \(\frac{V_N}{1}\)

Putting, 2V N – V N + 2 = V N + \(\frac{V_N}{2}\)

Or, V N = 4 V.

∴ R N = 4/2 = 2Ω.

5. In the figure given below, the value of Resistance R by Norton’s Theorem is ___________

network-theory-exam-questions-answers-q5

a) 40

b) 20

c) 50

d) 80

Answer: b

Explanation: \(\frac{V_P-100}{10} + \frac{V_P}{10}\) + 2 = 0

Or, 2V P – 100 + 20 = 0

∴ V P = 80/2 = 40V

∴ R = 20Ω

6. In the figure given below, the Norton Resistance, as seen at the terminals P-Q, is given by __________

network-theory-exam-questions-answers-q6

a) 5 Ω

b) 7.5 Ω

c) 5 Ω

d) 7.5 Ω

Answer: a

Explanation: For finding V N ,

network-theory-exam-questions-answers-q6a

V N = \(\frac{4 × 10}{10+10}\) = 2V

For finding R N ,

network-theory-exam-questions-answers-q6b

R N = 10 || 10

= \(\frac{10×10}{10+10}\) = 5 Ω.

7. The Norton equivalent impedance Z between the nodes P and Q in the following circuit is __________

network-theory-exam-questions-answers-q7

a) 1

b) 1 + s + \

2 + s + \

3 + s + \(\frac{1}{s}\)

Answer: a

Explanation: To calculate the Norton resistance, all the current sources get open-circuited and voltage sources get short-circuited.

∴ R N =

= \(\frac{\leftMissing or unrecognized delimiter for \right×}{\leftMissing or unrecognized delimiter for \right+}\)

= \(\frac{\frac{1}{s}+1+1+s}{\frac{1}{s}+1+1+s}\) = 1

So, R N = 1.

8. In the circuit given below, it is given that V AB = 4 V for R L = 10 kΩ and V AB = 1 V for R L = 2kΩ. The value of Norton resistance for the network N is ____________

network-theory-exam-questions-answers-q8

a) 16 kΩ

b) 30 kΩ

c) 3 kΩ

d) 50 kΩ

Answer: b

Explanation: When R L = 10 kΩ and V AB = 4 V

Current in the circuit \(\frac{V_{AB}}{R_L} = \frac{4}{10}\) = 0.4 mA

Norton voltage is given by V N = I (R N + R L )

= 0.4(R N + 10)

= 0.4R N + 4

Similarly, for R L = 2 kΩ and V AB = 1 V

So, I = \(\frac{1}{2}\) = 0.5 mA

V N = 0.5(R N + 2)

= 0.5 R N + 1

∴ 0.1R N = 3

Or, R N = 30 kΩ.

9. For the circuit given below, the Norton’s resistance across the terminals A and B is _____________

network-theory-exam-questions-answers-q9

a) 5 Ω

b) 7 kΩ

c) 1.5 kΩ

d) 1.1 kΩ

Answer: b

Explanation: Let V AB = 1 V

5 V AB = 5

Or, 1 = 1 × I 1 or, I 1 = 1

Also, 1 = -5 + 1(I – I 1 )

∴ I = 7

Hence, R = 7 kΩ.

10. A circuit is given in the figure below. The Norton equivalent as viewed from terminals x and x’ is ___________

network-theory-exam-questions-answers-q10

a) 6 Ω and 1.333 A

b) 6 Ω and 0.833 A

c) 32 Ω and 0.156 A

d) 32 Ω and 0.25 A

Answer: b

Explanation: We, draw the Norton equivalent of the left side of xx’ and source transformed right side of yy’.

network-theory-exam-questions-answers-q10a

V xx’ = V N = \(\displaystyle\frac{\frac{4}{8} + \frac{8}{24}}{\frac{1}{8} + \frac{1}{24}}\) = 5V

∴ R N = 8 ||

= \(\frac{8×24}{8+24}\) = 6 Ω

∴ \(I_N = \frac{V_N}{R_N} = \frac{5}{6}\) = 0.833 A.

11. For the circuit given in figure below, the Norton equivalent as viewed from terminals y and y’ is _________

network-theory-exam-questions-answers-q10

a) 32 Ω and 0.25 A

b) 32 Ω and 0.125 A

c) 6 Ω and 0.833 A

d) 6 Ω and 1.167 A

Answer: d

Explanation: We draw the Norton equivalent of the left side of xx’ and source transformed right side of yy’.

network-theory-exam-questions-answers-q10a

Norton equivalent as seen from terminal yy’ is

V yy’ = V N =\(\displaystyle\frac{\frac{4}{24} + \frac{8}{8}}{\frac{1}{24} + \frac{1}{8}}\) = 5V

= \(\frac{0.167+1}{0.04167+0.125}\) = 7 V

∴ R N = || 8

= \(\frac{24×8}{24+8}\) = 6 Ω

∴ I N = \(\frac{V_N}{R_N} = \frac{7}{6}\) = 1.167 A.

12. In the figure given below, the power loss in 1 Ω resistor using Norton’s Theorem is ________

network-theory-exam-questions-answers-q12

a) 9.76 W

b) 9.26 W

c) 10.76 W

d) 11.70 W

Answer: b

Explanation: Let us remove the 1 Ω resistor and short x-y.

At Node 1, assuming node potential to be V, \(\frac{V-10}{5}\) + I SC = 5

But I SC = \(\frac{V}{2}\)

∴ \(\frac{V-10}{5} + \frac{V}{2}\) = 5

Or, 0.7 V = 7

That is, V= 10 V

∴ I SC = \(\frac{V}{2}\) = 5 A

To find R int , all constant sources are deactivated. R int = \(\frac{×2}{5+2+2} = \frac{14}{9}\) = 1.56 Ω

R int = 1.56 Ω; I SC = I N = 5A

Here, I = I N \

2 × 1 = 9.26 W.

13. The value of R N from the circuit given below is ________

network-theory-exam-questions-answers-q13

a) 3 Ω

b) 1.2 Ω

c) 5 Ω

d) 12.12 Ω

Answer: d

Explanation: V X = 3\(\frac{V_X}{6}\) + 4

Or, V X = 8 V = V OC

If terminal is short-circuited, V X = 0.

I SC = \(\frac{4}{3+3}\) = 0.66 A

∴ R N = \(\frac{V_{OC}}{I_{SC}}\) = \(\frac{8}{0.66}\) = 12.12 Ω.

14. The current I, as shown in the figure below, is ________

network-theory-exam-questions-answers-q14

a) 3 A

b) 2 A

c) 1 A

d) 0

Answer: c

Explanation: The 3 Ω resistance is an extra element because voltage at node B is independent of the 3 Ω resistance.

I 1 = \

The net current in 2 Ω resistance is I = 1 – I 1

= 2 – 1 = 1 A .

15. While computing the Norton equivalent resistance and the Norton equivalent current, which of the following steps are undertaken?

a) Both the dependent and independent voltage sources are short-circuited and both the dependent and independent current sources are open-circuited

b) Both the dependent and independent voltage sources are open-circuited and both the dependent and independent current sources are short-circuited

c) The dependent voltage source is open-circuited keeping the independent voltage source untouched and the dependent current source is short-circuited keeping the independent current source untouched

d) The dependent voltage source is short-circuited keeping the independent voltage source untouched and the dependent current source is open-circuited keeping the independent current source untouched

Answer: d

Explanation: While computing the Norton equivalent voltage consisting of both dependent and independent sources, we first find the equivalent resistance called the Norton resistance by opening the two terminals. Then while computing the Norton current, we short-circuit the dependent voltage sources keeping the independent voltage sources untouched and open-circuiting the dependent current sources keeping the independent current sources untouched.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Advanced Problems on Superposition Theorem”.

1. The Superposition Theorem is not applicable for _________

a) Power calculation

b) Voltage calculation

c) Current Calculation

d) Both Voltage and Current calculation

Answer: a

Explanation: The Superposition Theorem is not applicable for Power calculation because for power, the calculations involve either the product of voltage and current or the square of current or the square of the voltage thus making them non-linear operations. Hence they cannot be calculated using Superposition Theorem.

2. The current I in the circuit given below is ________

network-theory-questions-answers-advanced-problems-superposition-theorem-q2

a) \

\(\frac{1}{2}\) A

Answer: b

Explanation: Using Superposition Theorem, we can write

I = \(\frac{24}{3+2} – \frac{72}{3+2} – \frac{3I}{3+2} \)

Or, I = 2 – \(\frac{3I}{3+2}\)

Or, I = 2 – \(\frac{3I}{5}\)

∴ I = \(\frac{5}{4}\) A.

3. The value of the voltage V 2 at the node in the circuit given below is __________

network-theory-questions-answers-advanced-problems-superposition-theorem-q3

a) 2.745 V

b) 1.393 V

c) -1.393 V

d) -2.745 V

Answer: c

Explanation: The datum node is the lower branch.

By superposition, the current i is given by, i = 2 \(\frac{7}{15+5+7} + \frac{3}{15+5+7} + 4I \frac{7+15}{15+5+7}\)

Or, I = \(\frac{17}{27} + \frac{88}{27I}\)

∴ The solution for i yields, I = \(\frac{\frac{17}{27}}{1 – \frac{88}{27}} = -\frac{17}{61}\) A

∴ V 2 = 5I = -1.393 V.

4. The value of the voltage V 1 at the node in the circuit given below is __________

network-theory-questions-answers-advanced-problems-superposition-theorem-q3

a) 11.148 V

b) 10.989 V

c) 20.151 V

d) 25.148 V

Answer: a

Explanation: The datum node is the lower branch.

By superposition, the current i is given by, i = 2 \(\frac{7}{15+5+7} + \frac{3}{15+5+7} + 4I \frac{7+15}{15+5+7}\)

Or, I = \(\frac{17}{27} + \frac{88}{27I}\)

∴ The solution for i yields, I = \(\frac{\frac{17}{27}}{1 – \frac{88}{27}} = -\frac{17}{61}\) A

∴ V 1 = V 2 – 15 = 11.148 V.

5. The value of the current I in the circuit given below, is ___________

network-theory-questions-answers-advanced-problems-superposition-theorem-q5

a) 0.7 A

b) 0.5 A

c) 1 A

d) 3.5 A

Answer: a

Explanation: Using Superposition theorem, we can write,

I = \(\frac{30}{6+4+2} + 3\frac{4}{6+4+2} – 8i\frac{6}{6+4+2}\)

Or, I = \(\frac{42}{12}\) – 4I

Or, I = \(\frac{\frac{42}{12}}{1+4}\) = 0.7 A.

6. In superposition theorem when we consider one voltage source, all the other voltage sources are ___________

a) Shorted

b) Removed

c) Undisturbed

d) Opened

Answer: a

Explanation: To determine the contribution of each individual source in case of Superposition Theorem, we short circuit all the other voltage sources and open circuit all the independent current sources .

7. In the figure given below, the value of Resistance R by Superposition Theorem is ___________

network-theory-questions-answers-advanced-problems-superposition-theorem-q7

a) 10

b) 20

c) 30

d) 40

Answer: b

Explanation: Using Superposition theorem, we get,

\(\frac{V_P-100}{10} + \frac{V_P}{10}\) + 2 = 0

Or, 2V P – 100 +20= 0

∴ V P = 80/2 = 40V

∴ R = 20Ω.

8. In the circuit given below, the value of voltage V X using Superposition Theorem is _________

network-theory-questions-answers-advanced-problems-superposition-theorem-q8

a) 5(I O – 3)

b) 2.5(I O – 3)

c) 4.5(3 – I O )

d) 2.5(3 – I O )

Answer: d

Explanation: Using Superposition Theorem the voltage V X is given as,

V X = (3-I O ) + 5V X \(\frac{2}{40+2}\)

Or, V X = \

+ \frac{10}{42}V_X\)

Or, V X = \

\) = 2.5(3 – I O ).

9. In the circuit given below, the value of voltage V O using Superposition Theorem is _________

network-theory-questions-answers-advanced-problems-superposition-theorem-q8

a) −30 – 10I O

b) 30 + 10I O

c) −30 + 10I O

d) +30 – 10I O

Answer: c

Explanation: Using Superposition Theorem the voltage V X is given as,

V X = (3-I O ) + 5V X \(\frac{2}{40+2}\)

Or, V X = \

+ \frac{10}{42}\)V X

Or, V X = \(\frac{\frac{80}{42}}{1-\frac{10}{42}}\)(3 – I O ) = 2.5(3 – I O )

∴ V O = V X – 5V X = −30 + 10I O .

10. The value of the voltage V O in the circuit given below is __________

network-theory-questions-answers-advanced-problems-superposition-theorem-q10

a) 20 V

b) 30 V

c) 40 V

d) 50 V

Answer: b

Explanation: Using Superposition Theorem, the current IB is given by,

I B = \(\frac{70}{4||20+2||10} × \frac{20}{4+20} + \frac{50}{10+4||20||2} × \frac{20||2}{4+20||2} – \frac{2I_B}{20||2+4||10} × \frac{10}{4+10}\)

Or, I B = \(\frac{35}{3} + \frac{25}{18} – \frac{11}{36}\) I B

Now, Solving for I B , we get, I B = \(\frac{\frac{35}{3}+\frac{25}{18}}{1+\frac{11}{36}}\) = 10 A

Now, V O = 70 − 4i b

∴ V O = 30 V.

11. The value of voltage V O in the circuit given below is _____________

network-theory-questions-answers-advanced-problems-superposition-theorem-q11

a) 24 V

b) 48 V

c) 12 V

d) 4 V

Answer: a

Explanation: Using Superposition Theorem, we get,

V ∆ = −0.4V ∆ × 10 + 5 × 10

Solving for V ∆ , we get, V ∆ = \(\frac{5 × 10}{1+0.4×10}\) = 10 V

Again by using Superposition Theorem, we get, I ∆ = \(\frac{10}{5+20} – 0.4V_∆ \frac{20}{5+20}\)

Or, I ∆ = \(\frac{10}{25} – 0.4V_∆ \frac{20}{25} = -\frac{70}{25}\) A

Thus, V O is given by, V O = 10 − 5I ∆

∴ V O = 24 V.

12. In the circuit given below, the value of V in terms of V S and I S using Superposition Theorem is ___________

network-theory-questions-answers-advanced-problems-superposition-theorem-q12

a) V = \

V = \

V = \(\frac{2}{7} I_S – \frac{4}{7} V_S\)

Answer: c

Explanation: By superposition, the current I is given by I = \

∴ V = \(\frac{2}{7} V_S – \frac{4}{7} I_S\).

13. In the circuit given below, the value of V X due to the 10 V source is ____________

network-theory-questions-answers-advanced-problems-superposition-theorem-q13

a) 1 V

b) 2 V

c) 3 V

d) 4 V

Answer: b

Explanation: Due to the effect of the 10 V source, we short the 16 V source and open the 3A and 15 A sources. From the resulting circuit, we can calculate the value of V.

∴ V = \(\frac{10 ×20}{80+20}\) = 2 V.

14. In the following circuit, when R = 0 Ω, the current IR equals to 10 A. The maximum power will be?

network-theory-questions-answers-advanced-problems-superposition-theorem-q14

a) 50 W

b) 100 W

c) 200 W

d) 400 W

Answer: a

Explanation: Using Superposition Theorem, we get, I = 10 A and R = 2 Ω

∴ P max =

2 × 2

= 5 × 5 × 2 = 50 W.

15. The circuit shown in figure has a load equivalent to _________

network-theory-questions-answers-advanced-problems-superposition-theorem-q15

a) \

4 Ω

d) 2 Ω

Answer: b

Explanation: Using Superposition Theorem, in the given circuit, \(\frac{V}{4} + \frac{V-2I}{2}\) = I

Or, \(\frac{3V-4I}{4}\) = I

Or, 3V = 8I

∴ \(\frac{V}{I} = \frac{8}{3}\) Ω.

This set of Network Theory Questions & Answers for Exams focuses on “Advanced Problems on Reciprocity Theorem”.

1. In Reciprocity Theorem, which of the following ratios is considered?

a) Voltage to current

b) Current to current

c) Voltage to voltage

d) No ratio is considered

Answer: a

Explanation: The Reciprocity Theorem states that if an Emf E in one branch produces a current I in a second branch, then if the same emf E is moved from the first to the second branch, it will produce the same current in the first branch, when the Emf E in the first branch is replaced with a short circuit. Therefore the ratio of Voltage to Current is considered in case of Reciprocity Theorem.

2. The Reciprocity Theorem is valid for ___________

a) Non-Linear Time Invariant circuits

b) Linear Time Invariant circuits

c) Non-Linear Time Variant circuits

d) Linear Time Variant circuits

Answer: b

Explanation: A reciprocal network comprises of linear time-invariant bilateral elements. It is applicable to resistors, capacitors, inductors and transformers. However, both dependent and independent sources ate not permissible.

3. In the circuit given below, the current in the 4-ohm resistor is __________

network-theory-questions-answers-advanced-problems-reciprocity-theorem-q3

a) 3.5 A

b) 2.5 A

c) 1.5 A

d) 0.5 A

Answer: c

Explanation: R th = [ || 6] + 12 = 15 Ω

I S = \(\frac{45}{15}\) = 3 A

Now, by current division rule, we get, I = \(\frac{3 × 6}{12} = \frac{18}{12}\) = 1.5 A.

4. The Reciprocity Theorem is applicable for __________

a) Single-source networks

b) Multi-source networks

c) Both Single and Multi-source networks

d) Neither Single nor Multi-source networks

Answer: a

Explanation: According to Reciprocity Theorem, the voltage source and the resulting current source may be interchanged without a change in current. Therefore the theorem is applicable only to single-source networks. It therefore cannot be employed in multi-source networks.

5. In the circuit given below, the current through the 12 Ω resistance is _________

network-theory-questions-answers-advanced-problems-reciprocity-theorem-q5

a) 1.5 A

b) 2.5 A

c) 3.5 A

d) 4.5 A

Answer: a

Explanation: Equivalent resistance, R EQ = [ + 2 + 4] = 10 Ω

I S = \(\frac{45}{10}\) = 4.5 A

Now, by using Current division rule, we get, I = \(\frac{4.5 × 6}{12+6} = \frac{27}{18}\) = 1.5 A.

6. A circuit is given in the figure below. We can infer that ________

network-theory-questions-answers-advanced-problems-reciprocity-theorem-q6

a) The circuit follows Reciprocity Theorem

b) The circuit follows Millman’s Theorem

c) The circuit follows Superposition Theorem

d) The circuit follows Tellegen Theorem

Answer: a

Explanation: Let us consider this circuit,

network-theory-questions-answers-advanced-problems-reciprocity-theorem-q6a

R th = [ || 6] + 12 = 15 Ω

I S = \(\frac{45}{15}\) = 3 A

Now, by current division rule, we get, I 1 = \(\frac{3 × 6}{12} = \frac{18}{12}\) = 1.5 A.

Again, let us consider this circuit,

network-theory-questions-answers-advanced-problems-reciprocity-theorem-q6b

Equivalent resistance,R EQ = [ + 2 + 4] = 10 Ω

I S = \(\frac{45}{10}\) = 4.5 A

Now, by using Current division rule, we get, I 2 = \(\frac{4.5 × 6}{12+6} = \frac{27}{18}\) = 1.5 A.

Since I 1 = I 2 , the circuit follows Reciprocity Theorem.

7. In the circuit given below, the current in the resistance 20 Ω is _________

network-theory-questions-answers-advanced-problems-reciprocity-theorem-q7

a) 8.43 A

b) 5.67 A

c) 1.43 A

d) 2.47 A

Answer: c

Explanation: Equivalent Resistance R EQ = 20 + [30 || )]

= 20 + [30 ||

]

= 20 + [30 || ]

= 20 + [30 || 30]

= 20 + \(\frac{30 × 30}{30+30}\)

= 20 + 15 = 35 Ω

The current drawn by the circuit = \(\frac{200}{35}\) = 5.71 A

Now, by using current division rule, we get, I 2Ω = 1.43 A.

8. In the circuit given below, the value of I is __________

network-theory-questions-answers-advanced-problems-reciprocity-theorem-q8

a) 2.47 A

b) 5.67 A

c) 8.43 A

d) 1.43 A

Answer: d

Explanation: Equivalent Resistance, R EQ = [[ + 20) || 20] + 20]

= \Missing or unrecognized delimiter for \right + 20\right) || 20\Big] + 20\Big]\)

= [[ || 20] + 20]

= [[32 || 20] + 20]

= \Missing or unrecognized delimiter for \right + 20\Big]\)

= [12.31 + 20] = 32.31 Ω

The current drawn by the circuit = \(\frac{200}{32.31}\) = 6.19 A

Now, by using current division rule, we get, I 2Ω = 1.43 A.

9. A circuit is given in the figure below. We can infer that ________

network-theory-questions-answers-advanced-problems-reciprocity-theorem-q7

a) The circuit follows Reciprocity Theorem

b) The circuit follows Millman’s Theorem

c) The circuit follows Superposition Theorem

d) The circuit follows Tellegen Theorem

Answer: a

Explanation: Let us consider this circuit,

network-theory-questions-answers-advanced-problems-reciprocity-theorem-q7

Equivalent Resistance R EQ = 20 + [30 || )] = 20 + [30 ||

]

= 20 + [30 || ]

= 20 + [30 || 30]

= 20 + \(\frac{30 × 30}{30+30}\)

= 20 + 15 = 35 Ω

The current drawn by the circuit = \(\frac{200}{35}\) = 5.71 A

Now, by using current division rule, we get, I 1 = 1.43 A

Again, let us consider this circuit,

network-theory-questions-answers-advanced-problems-reciprocity-theorem-q8

Equivalent Resistance, R EQ = [[ + 20) || 20] + 20]

= \Missing or unrecognized delimiter for \right + 20\right) || 20\Big] + 20\Big]\)

= [[ || 20] + 20]

= [[32 || 20] + 20]

= \Missing or unrecognized delimiter for \right + 20\Big]\)

= [12.31 + 20] = 32.31 Ω

The current drawn by the circuit = \(\frac{200}{32.31}\) = 6.19 A

Now, by using current division rule, we get, I 2 = 1.43 A.

Since I 1 = I 2 , the circuit follows Reciprocity Theorem.

10. In the circuit given below, the current in the 30 Ω resistor is _________

network-theory-questions-answers-advanced-problems-reciprocity-theorem-q10

a) 1 A

b) 2 A

c) 3 A

d) 4 A

Answer: b

Explanation: Equivalent Resistance, R EQ = 20 + [60 || 30]

= 20 + \(\frac{60 × 30}{60+30}\)

= 20 + 20 = 40 Ω

Total current from the source, I = \(\frac{120}{40}\) = 3A

Now, by using current division rule, I 3Ω = \(\frac{3 × 60}{30+60}\) = 2 A.

11. In the circuit given below, the current in the 20 Ω resistor is _________

network-theory-questions-answers-advanced-problems-reciprocity-theorem-q11

a) 5 A

b) 1 A

c) 1.5 A

d) 2 A

Answer: d

Explanation: Equivalent resistance, R EQ = [[20 || 60] + 30]

= \(\Big[\frac{20 × 60}{20+60} + 30\Big]\)

= [15 + 30] = 45 Ω

Total current = \(\frac{120}{45}\) = 2.67 A

Current through the 20Ω resistor is, I 20Ω = \(\frac{2.67 × 60}{60+20}\) = 2 A.

12. A circuit is given in the figure below. We can infer that ________

network-theory-questions-answers-advanced-problems-reciprocity-theorem-q10

a) The circuit follows Reciprocity Theorem

b) The circuit follows Millman’s Theorem

c) The circuit follows Superposition Theorem

d) The circuit follows Tellegen Theorem

Answer: a

Explanation: Let us consider this circuit,

network-theory-questions-answers-advanced-problems-reciprocity-theorem-q10

Equivalent Resistance, Equivalent Resistance, R EQ = 20 + [60 || 30]

= 20 + \(\frac{60 × 30}{60+30}\)

= 20 + 20 = 40 Ω

Total current from the source, I = \(\frac{120}{40}\) = 3A

Now, by using current division rule, I 1 = \(\frac{3 × 60}{30+60}\) = 2 A.

Again, let us consider this circuit,

network-theory-questions-answers-advanced-problems-reciprocity-theorem-q11

Equivalent resistance, R EQ = [[20 || 60] + 30]

= \(\Big[\frac{20 × 60}{20+60} + 30\Big]\)

= [15 + 30] = 45 Ω

Total current = \(\frac{120}{45}\) = 2.67 A

Current through the 20Ω resistor is, I 2 = \(\frac{2.67 × 60}{60+20}\) = 2 A

Since I 1 = I 2 , the circuit follows Reciprocity Theorem.

13. In the circuit given below, the value of load R L , for which maximum power is transferred through it is ___________

network-theory-questions-answers-advanced-problems-reciprocity-theorem-q13

a) 2 Ω

b) 3 Ω

c) 1 Ω

d) 6 Ω

Answer: b

Explanation: I + 0.9 = 10 I

Or, I = 0.1 A

V OC = 3 × 10 I = 30 I

Or, V OC = 3 V

Now, I SC = 10 I = 1 A

R th = 3/1 = 3 Ω.

14. In the circuit given below, the maximum power absorbed by the load resistance R L is ___________

network-theory-questions-answers-advanced-problems-reciprocity-theorem-q14

a) 2200 W

b) 1250 W

c) 1000 W

d) 621 W

Answer: d

Explanation: R L = \

I 1 + 5I 2

And 90 = I 2 = 5I 1

∴ I 1 = 5.5 – 2.75j and I 2 = 4.5 – 2.2j

Total current in R L = I 1 + I 2 = A = 11.15 A

∴ Power absorbed by R L = I 2 R

= 11.15 2 × 5 = 621 W.

15. In the circuit given below, the maximum power delivered to the load is ___________

network-theory-questions-answers-advanced-problems-reciprocity-theorem-q15

a) 3 W

b) 5.2 W

c) 3.2 W

d) 4.2 W

Answer: d

Explanation: Equivalent resistance of the circuit is = [{ || 5} + 10] = = 12.5 Ω

Total current drawn by the circuit is I T = \(\frac{50}{12.5}\) = 4 A

Current in 3 Ω resistor is I 3 = I T × \(\frac{5}{5+5} = \frac{4 × 5}{10}\) = 2 A

V TH = V 3 = 3 × 2 = 6V

R TH = R AB = [ || 3] = 2.1 Ω

For maximum power transfer R L = R TH = 2.1 Ω

∴ Current drawn by R L is I L = \

2 = 4.2 W.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Advanced Problems on Network Theorems – 1”.

1. The temperature coefficient of a metal as the temperature increases will ____________

a) Decreases

b) Increases

c) Remains unchanged

d) Increases and remains same

Answer: a

Explanation: We know that the temperature coefficient is,

Given by, α = \(\frac{α_0}{1 + α_0 t}\)

Since temperature is present at the denominator, so with increase in temperature t, the denominator increases and hence the fraction decreases.

So, temperature coefficient decreases.

2. Given a wire of resistance R Ω. The resistance of a wire of the same material and same weight and double the diameter is ___________

a) 0.5 R

b) 0.25 R

c) 0.125 R

d) 0.0625 R

Answer: d

Explanation: Since diameter is double, area of cross-section is four times and length is one-fourth.

It can be verified by the following equation,

R 2 = \(\frac{\frac{ρl}{4}}{4A}\)

= \(\frac{ρl}{16 A} = \frac{R}{16}\).

3. The star equivalent resistance of 3 resistors having each resistance = 5 Ω is ____________

a) 1.5 Ω

b) 1.67 Ω

c) 3 Ω

d) 4.5 Ω

Answer: b

Explanation: We know that for star connection, R EQ = \(\frac{R X R}{R+R+R}\)

Given R = 5 Ω

So, R EQ = \(\frac{5 X 5}{5+5+5}\)

= \(\frac{25}{15}\) = 1.67 Ω.

4. The charge associated with a bulb rated as 20 W, 200 V and used for 10 minutes is ____________

a) 36 C

b) 60 C

c) 72 C

d) 50 C

Answer: b

Explanation: Charge Q= It

Given I = \(\frac{20}{200}\) = 0.1 A, t = 10 X 60 sec = 600 sec

So, Q = 0.1 X 600 = 60 C.

5. For a series RL circuit having L = 5 H, current = 1 A . The energy stored in magnetic field is ___________

a) 3.6 J

b) 2.5 J

c) 1.5 J

d) 3 J

Answer: b

Explanation: We know that, Energy, E = 0.5 LI 2

Or, E = 0.5 X 5 X 1 2 = 2.5 J.

6. For a practical voltage source, which of the following is correct?

a) Cannot be less than source voltage

b) Cannot be higher than source voltage

c) Is always less than source voltage

d) Is always equal to source voltage

Answer: b

Explanation: A practical voltage source has some resistance. Because of this resistance, some amount of voltage drop occurs across this resistance. Hence, the terminal voltage cannot be higher than source voltage. However, if current is zero, then terminal voltage and source voltage are equal.

7. Consider an electric motor having resistance of 10 Ω, 20 Ω and 30 Ω respectively. The percentage of energy dissipated by 10 Ω earthing plate is ____________

a) More than 50% of total energy

b) Less than 50% of total energy

c) Depends on the materials of the three plates

d) May be more or less than 50% of total energy

Answer: a

Explanation: The parallel combination of 30 Ω and 20 Ω is 12 Ω. Since 12 Ω and 10 Ω are in parallel, the 10 ohm plate draws more than 50% current and dissipates more than 50% energy.

8. Consider a resistive network circuit, having 3 sources of 18 W, 50 W and 98 W respectively and a resistance R. When all source act together the maximum and minimum power is ____________

a) 98 W, 18 W

b) 166 W, 18 W

c) 450 W, 2 W

d) 166 W, 2 W

Answer: c

Explanation: Let us suppose R = 1 Ω

Then, I 1 = 32 A, I 2 = 52 A and I 3 = 72 A

Or, (I 1 + I 2 + I 3 ) 2 R = 2 R = 450 Ω.

9. A current waveform is of the shape of a right angled triangle with period of t = 1 sec. Given a resistance R = 1 Ω. The average power is __________

a) 1 W

b) 0.5 W

c) 0.333 W

d) 0.111 W

Answer: d

Explanation: We know that RMS current is,

I = 1 \

^2 \,dt\)

= \( \int_0^1 t^2 \,dt\)

= \(\frac{1}{3}\) A

Now, power P = I 2 R

= \(\frac{1}{9}\) X 1

= 0.111 W.

10. Given two voltages, V 1 = sin and V 2 = cos . Which of the following is correct?

a) V 1 is leading V 2 by 15°

b) V 1 is leading V 2 by 30°

c) V 2 is leading V 1 by 60°

d) V 2 is leading V 1 by 30°

Answer: c

Explanation: Given that, V 1 = sin and V 2 = cos

Now, V 2 can be written as,

V 2 = sin .

Hence, V 2 is leading V 1 by = 60°.

11. Given two mutually coupled coils have a total inductance of 1500 mH, the self-inductance of each coils if the coefficient of coupling is 0.2 is ____________

a) 325 mH

b) 255 mH

c) 625 mH

d) 550 mH

Answer: c

Explanation: We know that, M = k\(\sqrt{L_1 L_2}\)

Given that, L EQ = 1500 mH and k = 0.2

Again, total inductance = L 1 + L 2 + 2M

Or, 2L + 2kL = 1500 mH

Or, L = 1500

Or, L = 625 mH.

12. For a series RL circuit, the impedance Z = 10 Ω at a frequency of 50 Hz. At 100 Hz the impedance is ___________

a) 10 Ω

b) 20 Ω

c) 1 Ω

d) More than 1 Ω but less than 10 Ω 0 Ω

Answer: d

Explanation: We know that impedance, Z = \( \sqrt{R^2 +

^2}\)

Since frequency is doubled, so \(\frac{1}{ω^2+c^2}\) becomes one-fourth but R2 remains the same. Thus the impedance cannot be exactly measured but we can infer that the resistance is more than 1 Ω but less than 10 Ω.

13. Consider the self-inductances of two coils as 12 H and 5 H. 50 % of one flux links the other. The mutual inductance is ___________

a) 30 H

b) 24 H

c) 9 H

d) 4.5 H

Answer: d

Explanation: We know that, M = k\(\sqrt{L_1 L_2}\)

Given that, L 1 = 12 H, L 2 = 5 H and k = 0.5

So, M = 0.5\(\sqrt{12 X 5}\)

= 0.5\(\sqrt{60}\)

= 0.5 X 7.75 = 3.875 H.

14. In the circuit given below the maximum power that can be transferred from the source voltage is __________

network-theory-questions-answers-advanced-problems-network-theorems-1-q14

a) 1 W

b) 10 W

c) 0.25 W

d) 0.5 W

Answer: c

Explanation: For maximum power transfer to the load resistor R L , R L must be equal to 100Ω.

∴ Maximum power = \(\frac{V^2}{4R_L}\)

= \(\frac{10^2}{4×100} = \frac{100}{400}\) = 0.25 W.

15. In the circuit given below, the value of RL for maximum power transfer is ___________

network-theory-questions-answers-advanced-problems-network-theorems-1-q15

a) 2.4 Ω

b) 2.6 Ω

c) 2.8 Ω

d) 3.0 Ω

Answer: c

Explanation: Using Y-∆ transformation,

network-theory-questions-answers-advanced-problems-network-theorems-1-q15a

R AB = ||

= ||

=\Missing or unrecognized delimiter for \right || \leftMissing or unrecognized delimiter for \right\)

= 4.5 || 3.6

= \(\frac{4.5×3.6}{4.5+3.6}\) = 2.8 Ω.

This set of Advanced Network Theory Questions and Answers focuses on “Advanced Problems on Network Theorems – 2”.

1. A network contains linear resistors and ideal voltage source s. If values of all the resistors are doubled, then voltage across each resistor is __________

a) Halved

b) Doubled

c) Increases by 2 times

d) Remains same

Answer: d

Explanation: V / R ratio is a constant R. If R is doubled then, electric current will become half. So voltage across each resistor is same.

2. A voltage waveform V = 12t 2 is applied across a 1 H inductor for t ≥ 0, with initial electric current through it being zero. The electric current through the inductor for t ≥ 0 is given by __________

a) 12 t

b) 24 t

c) 12 t 3

d) 4 t 3

Answer: d

Explanation: We know that, I = \(\frac{1}{L} \int_0^t V \,dt\)

= 1\(\int_0^t 12 t^2 \,dt\)

= 4 t 3 .

3. The linear circuit element among the following is ___________

a) Capacitor

b) Inductor

c) Resistor

d) Capacitor & Inductor

Answer: c

Explanation: A linear circuit element does not change its value with voltage or current. The resistance is only one among the others does not change its value with voltage or current.

4. In the circuit shown, V C is zero at t=0 s. For t>0, the capacitor current I C , where t is in second, is ___________

advanced-network-theory-questions-answers-q4

a) 0.50 e -25t mA

b) 0.25 e -25t mA

c) 0.50 e -12.5t mA

d) 0.25 e -6.25t mA

Answer: a

Explanation: The capacitor voltage V C = V C – [V C -V C ]e -t/RC

R = 20 || 20 = \(\frac{20×20}{20+20} = \frac{400}{40}\) = 10 kΩ

V C = 10 × \(\frac{20}{20+20}\) = 5 V

Given, V C = 0

∴ V C = 5 – e -t/10×4×10^×10^3

= 5(1 – e -25t )

I C = C\

= 4 × 10 -6 × 5 × 25e -25t

∴ I L = 0.50e -2.5t mA.

5. In the circuit given below, the phasor voltage V AB is _________

advanced-network-theory-questions-answers-q5

a) Zero

b) 5∠30°

c) 12.5∠30°

d) 17∠30°

Answer: d

Explanation: Equivalent impedance = ||

= \(\frac{×}{ + } \)

= \(\frac{25+9}{10}\) = 3.4 Ω

V AB = Current × Impedance

= 5∠30° × 3.4

= 17∠30°.

6. For the circuit given below, the driving point impedance is given by, Z = \(\frac{0.2s}{s^2+0.1s+2} \). The component values are _________

advanced-network-theory-questions-answers-q6

a) L = 5 H, R = 0.5 Ω, C = 0.1 F

b) L = 0.1 H, R = 0.5 Ω, C = 5 F

c) L = 5 H, R = 2 Ω, C = 0.1 F

d) L = 0.1 H, R = 2 Ω, C = 5 F

Answer: d

Explanation: Dividing point impedance = R || sL || \

= \(\frac{0.2s}{s^2+0.1s+2} \)

∴ On comparing, we get L = 0.1 H, R = 2 Ω, C = 5 F.

7. What is the power loss in the 10 Ω resistor in the network shown in the figure below?

advanced-network-theory-questions-answers-q7

a) 15.13 W

b) 11.23 W

c) 16 W

d) 14 W

Answer: a

Explanation: In mesh aef, 8(I 1 – I 3 ) + 3(I 1 – I 2 ) = 15

Or, 11 I 1 – 3 I 2 – 8I 3 = 15

In mesh efd, 5(I 2 – I 3 ) + 2I 2 + 3(I 2 – I 1 ) = 0

Or, -3 I 1 + 10I 2 – 5I 3 = 0

In mesh abcde, 10I 3 + 5(I 3 – I 2 ) + 8(I 3 – I 1 ) = 0

Or, -8I 1 – 5I 2 +23I 3 = 0

Thus loop equations are,

11 I 1 – 3 I 2 – 8I 3 = 15

-3 I 1 + 10I 2 – 5I 3 = 0

-8I 1 – 5I 2 + 23I 3 = 0

Solving by Cramer’s rule, I 3 = current through the 10Ω resistor = 1.23 A

∴ Current through 10 Ω resistor = 1.23 A

Power loss = \

2 ×10 = 15.13 W.

8. The switch S is the circuit shown in the figure is ideal. If the switch is repeatedly closed for 1 ms and opened for 1 ms, the average value of i is ____________

advanced-network-theory-questions-answers-q8

a) 0.25 mA

b) 0.35 mA

c) 0.5 mA

d) 1 mA

Answer: a

Explanation: Since i = \(\frac{5}{10 × 10^{-3}}\) = 0.5 × 10 3 = 0.5 mA

As the switch is repeatedly close, then i will be a square wave.

So average value of electric current is

= 0.25 mA.

9. In the circuit given below the value of resistance, R eq is ___________

advanced-network-theory-questions-answers-q9

a) 10 Ω

b) 11.86 Ω

c) 11.18 Ω

d) 25 Ω

Answer: c

Explanation: The circuit is as shown in figure below.

advanced-network-theory-questions-answers-q9a

R eq = 5 + \(\frac{10

}{10 + 5 + R_{eq}}\)

Or, \(R_{eq}^2 + 15R_{eq}\) = 5R eq + 75 + 10R eq + 50

Or, R eq = \(\sqrt{125}\) = 11.18 Ω.

10. A particular electric current is made up of two component: a 10 A and a sine wave of peak value 14.14 A. The average value of electric current is __________

a) 0

b) 24.14 A

c) 10 A

d) 14.14 A

Answer: c

Explanation: Average dc electric current = 10 A

Average ac electric current = 0 A as it is alternating in nature.

Average electric current = 10 + 0 = 10 A.

11. Given that, R 1 = 36 Ω and R 2 = 75 Ω, each having tolerance of ±5% are connected in series. The value of resultant resistance is ___________

a) 111 ± 0 Ω

b) 111 ± 2.77 Ω

c) 111 ± 5.55 Ω

d) 111 ± 7.23 Ω

Answer: c

Explanation: R 1 = 36 ± 5% = 36 ± 1.8 Ω

R 2 = 75 ± 5% = 75 ± 3.75 Ω

∴ R 1 + R 2 = 111 ± 5.55 Ω.

12. In the circuit of figure below a charge of 600 C is delivered to 100 V source in a 1 minute. The value of V 1 is ___________

advanced-network-theory-questions-answers-q12

a) 30 V

b) 60 V

c) 120 V

d) 240 V

Answer: d

Explanation: In order for 600 C charges to be delivered to 100 V source, the electric current must be anti-clockwise.

\(i = \frac{dQ}{dt} = \frac{600}{60}\) = 10A

Applying KVL we get

V 1 + 60 – 100 = 10 × 20 ⇒ V 1 = 240 V.

13. The energy required to charge a 10 μF capacitor to 100 V is ____________

a) 0.01 J

b) 0.05 J

c) 5 X 10 -9 J

d) 10 X 10 -9 J

Answer: b

Explanation: E = \(\frac{1}{2} CV^2\)

= 5 X 10 -6 X 100 2

= 0.05 J.

14. Among the following, the active element of electrical circuit is ____________

a) Voltage source

b) Current source

c) Resistance

d) Voltage and current source both

Answer: d

Explanation: We know that active elements are the ones that are used to drive the circuit. They also cause the electric current to flow through the circuit or the voltage drop across the element. Here only the voltage and current source are the ones satisfying the above conditions.

15. In the circuit given below, the source current is _________

advanced-network-theory-questions-answers-q15

a) 2 A

b) 3 A

c) 0.54 A

d) 1.5 A

Answer: c

Explanation: In the given figure, R X = R Y = R Z = \(\frac{5×5}{5+5+5}\) = 1.67 Ω

Here, R = [(R X + 2) || (R Y + 3)] + R X

= + R Z

= \(\frac{3.67 × 4.67}{3.67 + 4.67}\) + 1.67

= \(\frac{17.1389}{8.34}\) + 1.67

= 3.725 Ω

∴ I = \(\frac{V}{R} = \frac{2}{3.725}\) = 0.54 A.

This set of Network Theory Questions & Answers for Exams focuses on “Advanced Problems on Millman’s Theorem and Maximum Power Transfer Theorem”.

1. The basic elements of an electric circuit are _____________

a) R, L and C

b) Voltage

c) Current

d) Voltage as well as current

Answer: a

Explanation: The elements which show their behaviour only when excited are called as basic circuit elements. Here resistance, inductance and capacitance show their behaviour only when excited. Hence they are the basic elements of an electric circuit.

2. In the circuit given below, the value of the maximum power transferred through R L is ___________

network-theory-questions-answers-advanced-problems-millmans-theorem-maximum-power-transfer-theorem-q2

a) 0.75 W

b) 1.5 W

c) 2.25 W

d) 1.125 W

Answer: a

Explanation: I + 0.9 = 10 I

Or, I = 0.1 A

V OC = 3 × 10 I = 30 I

Or, V OC = 3 V

Now, I SC = 10 I = 1 A

R th = 3/1 = 3 Ω

V th = V OC = 3 V

R L = 3 Ω

P max = \(\frac{3^2}{4×3}\) = 0.75 W.

3. The energy stored in the magnetic field at a solenoid 100 cm long and 10 cm diameter wound with 1000 turns of wire carrying an electric current of 10 A, is ___________

a) 1.49 J

b) 0.49 J

c) 0.1 J

d) 1 J

Answer: b

Explanation: L = \(\frac{N^2 μ_0 A}{l}\)

= \(\frac{10^6.4π.10^{-7}.\frac{π}{4}.

}{1}\)

= \(\frac{π^2 X 10^{-3}}{1}\)

Energy = 0.5 LI 2

= 0.49 J.

4. The resistance of a strip of copper of rectangular cross section is 2 Ω. A metal of resistivity twice that of a copper is coated on its upper surface to a thickness equal to that of copper strip. The resistance of composite strip will be _________

a) \

6 Ω

Answer: b

Explanation: Given that copper and coated metal strip have resistance of 2 ohms respectively. These two strips are connected in parallel.

Hence, the resistance of the composite strip = \(\frac{2 X 4}{2 + 4}\)

= \(\frac{8}{6} = \frac{4}{3}\) Ω.

5. In the circuit shown below what is the value of R L for which maximum power is transferred to R L ?

network-theory-questions-answers-advanced-problems-millmans-theorem-maximum-power-transfer-theorem-q5

a) 2.4 Ω

b) \

4 Ω

d) 6 Ω

Answer: c

Explanation: Maximum power is transferred to R L when the load resistance equals the Thevenin resistance of the circuit.

R L = R TH = \(\frac{V_{OC}}{I_{SC}} \)

Due to open-circuit, V OC = 100 V; I SC = I 1 + I 2

Applying KVL in lower loop, 100 – 8I 1 = 0

Or, I 1 = \(\frac{100}{8} = \frac{25}{2}\)

And V X = -4I 1 = -4 × \(\frac{25}{2}\) = -50V

KVL in upper loop, 100 + V X – 4I 2 = 0

I 2 = \(\frac{100-50}{4} = \frac{25}{2}\)

Hence, I SC = I 1 + I 2 = \(\frac{25}{2} + \frac{25}{2}\) = 25

R TH = \(\frac{V_{OC}}{I_{SC}} = \frac{100}{25}\) = 4 Ω

R L = R TH = 4 Ω.

6. A 3 V DC supply with an internal resistance of 2 Ω supplies a passive non-linear resistance characterized by the relation V NL = \Missing open brace for subscript 1 W

b) 1.5 W

c) 2.5 W

d) 3 W

Answer: a

Explanation: 3 = 2I + I 2

∴ I = 1 A; V NL = 1V

∴ Power dissipated in R NL = 1 × 1 = 1 W.

7. The two windings of a transformer have an inductor of 2 H each. If mutual inductor between them is also 2 H, then which of the following is correct?

a) Turns ratio of the transformer is also 2

b) Transformer is an ideal transformer

c) It is a perfect transformer

d) It is a perfect as well as an ideal transformer

Answer: c

Explanation: We know that, K = \(\frac{M}{\sqrt{L_1 L_2}}\)

= \(\frac{2}{\sqrt{2 X 2}}\) = 1

Hence, it is a perfect transformer.

8. In the circuit given below, what is the amount of maximum power transfer to R?

network-theory-questions-answers-advanced-problems-millmans-theorem-maximum-power-transfer-theorem-q8

a) 56 W

b) 76 W

c) 60 W

d) 66 W

Answer: d

Explanation: Drop across V 1Ω = 5 × 1 = 5V

Also, \(\frac{V-V_{1Ω}}{10} + \frac{V-20-V_{1Ω}}{2} + \frac{V-V_{OC}}{5}\) = 2

Or, 0.1 V – 0.1V 1Ω + 0.5V – 10 – 0.5V 1Ω + 0.2 – 0.2V OC = 2

Or, 0.8V – 0.6V 1Ω = 12 + 0.2V OC

Or, 0.8 V – 0.2V OC = 12 +3=15 (Putting V 1Ω = 5)

Again, \(\frac{V_{OC}-V}{5}\) + 2 = 5

Or, 0.2V OC – 0.2V = 3

Again, R TH = { + 1} + 5

+ 5 = 7.67 Ω

Following the theorem of maximum power transfer, R = R TH = 7.67 Ω

And P MAX = \(\frac{V_{OC}^2}{4R} = \frac{45^2}{4×7.67}\) = 66 W.

9. In the circuit given below, the value of R L for which it absorbs maximum power is ___________

network-theory-questions-answers-advanced-problems-millmans-theorem-maximum-power-transfer-theorem-q9

a) \

350.38 Ω

d) \(\frac{4}{9}\) kΩ

Answer: c

Explanation: 5 = 200I – 50 × 2I

Or, I = \(\frac{5}{100}\) = 0.05 A

V OC = 100 × 3I + 200 × I = 25 V

V 1 = \(\frac{\frac{5}{50}}{\frac{1}{50} + \frac{1}{200} + \frac{1}{100}} \)

= \(\frac{0.1}{0.02+0.005+0.01} \)

= 2.85 V

I = \(\frac{2.85}{100}\) = 0.0142 A = 14.2 mA

I SC = \(\frac{2.85}{100}\) + 3 × 0.0142 = 0.07135 A

∴ R TH = \(\frac{V_{OC}}{I_{SC}} = \frac{25}{0.07135}\) = 350.38 Ω.

10. The form factor of sinusoidal alternating electric current is ___________

a) 0

b) 1

c) 1.11

d) 1.15

Answer: c

Explanation: We know that for alternating electric current form factor is defined as the ratio of rms value and average value of alternating current.

Now the rms value of alternating electric current = 0.07 × maximum value of alternating current.

Average value of alternating electric current = 0.637 × maximum value of alternating current.

∴ Form factor = \(\frac{0.707}{0.637}\) = 1.11.

11. The average power delivered to the 6 Ω load in the circuit of figure below is ___________

network-theory-questions-answers-advanced-problems-millmans-theorem-maximum-power-transfer-theorem-q11

a) 8 W

b) 76.68 W

c) 625 kW

d) 2.50 kW

Answer: b

Explanation: I 2 = \(\frac{V_2}{6}\), I 1 = \(\frac{I_2}{5} = \frac{V_2}{30}\)

V 1 = 5V 2

50 = 400(I 1 – 0.04V 2 ) + V 1

Or, V 2 = 21.45 V

∴ P L = \(\frac{V_2^2}{6} \)

= \(\frac{21.45^2}{6} \)

= \(\frac{460.1025}{6}\) = 76.68 W.

12. The rms value of the sine wave is 100 A. Its peak value is ____________

a) 70.7 A

b) 141.4 A

c) 150 A

d) 282.8 A

Answer: b

Explanation: We know that for sinusoidal alternating electric current the peak factor or amplitude factor can be expressed the ratio of maximum or peak value and rms value of alternating current.

So the peak value = rms value of alternating electric current × peak factor of alternating electric current = 100 × 1.414 = 141.4 A.

13. Potential of earth is – 50 V. If the potential difference between anode and cathode is measured as 150 V, actual voltage on anode is __________

a) 0 V

b) 100 V

c) 200 V

d) 250 V

Answer: c

Explanation: Given that, potential difference between anode and cathode is measured as 150 V and potential of earth is – 50 V.

So, actual voltage on anode, V = 150 –

= 150 + 50

= 200 V.

14. An alternating voltage V = 150 sint is supplied to a device which offers a resistance of 20 Ω in forward direction of electric current while preventing the flow of electric current in reverse direction. The form factor is ___________

a) 0.57

b) 0.318

c) 1.414

d) 1.57

Answer: d

Explanation: From the voltage equation, we can get V m = 150 V and I m = 150 / 20 = 7 A

RMS value of the current, I rms = I m / 2 = 7/2 = 3.5 A

Average value of the current, I avg = I m / π = 2.228 A

Form factor = I rms / I avg = 3.5 / 2.228 = 1.57.

15. A coil of 300 turns is wound on a non-magnetic core having a mean circumference of 300 mm and a cross sectional area of 300 mm 2 . The inductor of the coil corresponding to a magnetizing electric current of 3 A will be?

a) 37.68 μH

b) 47.68 μH

c) 113.04 μH

d) 120.58 μH

Answer: c

Explanation: Inductance of the coil, L = \(\frac{μ_0 n^2 A}{l}\)

= \(\frac{4π X 10^{-7} X 300 X 300 X 300 X 10^{-6}}{300 X 10^{-3}}\)

= 113.04 μH.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Angular Relation of a Sine Wave”.

1. The response of a second order system is?

a) pulse

b) saw tooth

c) square

d) sinusoid

Answer: d

Explanation: The response of a second order system is sinusoid. Any periodic waveform can be written in terms of sinusoidal function according to Fourier transform.

2. If a function f is periodic, then?

a) f = f

b) f = f

c) f = f

d) f = f

Answer: a

Explanation: If a function f is periodic, then f = f in general we can say that a function f is periodic, then f = f for all integer values of n.

3. The period of a function is measured as?

a) zero crossing of one cycle to zero crossing of next cycle

b) positive peak of one cycle to positive peak of next cycle

c) negative peak of one cycle to negative peak of next cycle

d) all of the mentioned

Answer: d

Explanation: The period of a function is measured as zero crossing of one cycle to zero crossing of next cycle or positive peak of one cycle to positive peak of next cycle or negative peak of one cycle to negative peak of next cycle.

4. The number of cycles a wave completes in one second is called?

a) time period

b) frequency

c) energy

d) wavelength

Answer: b

Explanation: The number of cycles a wave completes in one second is called frequency. And the time period is the inverse of frequency.

5. The relation between frequency and time period is?

a) f=1/T

b) f=T

c) f=1/T 2

d) f=1/T 3

Answer: a

Explanation: The relation between frequency and time period is f=1/T. So the frequency and time period are inversely proportional to each other.

6. The period of a sine wave is 40ms. What is the frequency?

a) 25

b) 50

c) 75

d) 100

Answer: a

Explanation: We know that the frequency and time period are inversely proportional to each other. So f = 1/T = 1/40ms = 25Hz.

7. The frequency of a sine wave is 30Hz. What is its period?

a) 3333ms

b) 333.3ms

c) 33.33ms

d) 3.333ms

Answer: c

Explanation: We know that the frequency and time period are inversely proportional to each other. So T=1/f =1/30=0.03333s=33.33ms.

8. A sine wave completes half cycle in ____ radians.

a) π/2

b) π

c) π/4

d) 2π

Answer: b

Explanation: The period of a sine wave is 360⁰ or 2π radians. So, a sine wave takes 90⁰ or π/2 radians to complete a half cycle.

9. A sine wave completes quarter cycle in ____ radians.

a) 2π

b) π

c) π/2

d) π/4

Answer: d

Explanation: The period of a sine wave is 360⁰ or 2π radians. So, a sine wave takes 45⁰ or π/4 radians to complete a quarter cycle.

10. A sine wave completes full cycle in ____ radians.

a) π

b) 2π

c) π/4

d) π/2

Answer: b

Explanation: The period of a sine wave is 360⁰ or 2π radians. So, a sine wave takes 90⁰ or 2π radians to complete a full cycle.

This set of Network Theory Questions and Answers for Experienced people focuses on “Voltage and Current Values of a Sine Wave”.

1. Find the instantaneous value of the sine value at 90⁰ point having an amplitude 10V and time period 360⁰.

a) 5

b) 10

c) 15

d) 20

Answer: b

Explanation: The equation for sine wave A is v = 10sinωt. The value at 90⁰ in this wave is v = 10sin90⁰ = 10V.

2. Find the instantaneous value at 90⁰ point of the sine wave if the wave is shifted by 45⁰.

a) 5.66

b) 6.66

c) 7.66

d) 8.66

Answer: a

Explanation: The equation for sine wave A is v = 10sinωt. The equation for sine wave A is v = 8sin. ωt = π/2. v = 8sin = 8sin45⁰=8 = 5.66V.

3. The value of the sine wave at some particular instant is called?

a) peak value

b) peak to peak value

c) instantaneous value

d) average value

Answer: c

Explanation: The value of the sine wave at some particular instant is called instantaneous value. This value is different at different points along the waveform.

4. The maximum value of the wave during a positive half cycle or maximum value of the wave during negative cycle is called?

a) instantaneous value

b) peak value

c) peak to peak value

d) average value

Answer: b

Explanation: The maximum value of the wave during positive half cycle or maximum value of the wave during negative cycle is called peak value. Since the values of these two are equal in magnitude, a sine wave is characterized by a single peak value.

5. The total area under the complete curve divided by the distance of the curve is called?

a) peak to peak value

b) RMS value

c) average value

d) effective value

Answer: c

Explanation: The total area under the complete curve divided by the distance of the curve is called average value. The average value of a sine wave over on e complete cycle is always zero.

6. The value from positive to negative peak of the sine wave is called?

a) effective value

b) average value

c) peak value

d) peak to peak value

Answer: d

Explanation: The value from positive to negative peak of the sine wave is called peak to peak value of a sine wave.

7. The RMS value of sine wave is?

a) 0.707V p

b) 0.607V p

c) 0.807V p

d) 0.907V p

Answer: a

Explanation: The root mean square value of a sine wave is a measure of the heating effect of the wave. The RMS value of sine wave is 0.707V p .

8. A wire is carrying a direct current of 20A and a sinusoidal alternating current of peak value 20A. Find the rms value of the resultant current.

a) 24

b) 24.5

c) 25

d) 25.5

Answer: b

Explanation: Direct current = 20A, sinusoidal alternating current of peak value = 20A. The rms value of the combined wave = √(20 2 +20 2 /2) = 24.5A.

9. The peak factor of the sinusoidal waveform is?

a) 4

b) 2

c) 1.414

d) 8

Answer: c

Explanation: The peak factor of any waveform is defined as the ratio of the peak value of the wave to the rms value of the wave. Peak factor = V p /(V rms ) = V p /(V p /√2)=√2=1.414.

10. The form factor of the sinusoidal waveform is?

a) 1.11

b) 2.22

c) 3.33

d) 4.44

Answer: a

Explanation: Form factor of any waveform is defined as the ratio of the rms value to the average value of the wave. Form factor = V rms /V av = (V p /√2)/0.637V p =1.11.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Phase Relation in a Pure Resistor”.

1. The phase difference between voltage and current in case of resistor is?

a) in phase

b) out of phase

c) 45⁰ out of phase

d) 90⁰ out of phase

Answer: a

Explanation: The phase difference between voltage and current in case of resistor is in phase. The amplitudes of the waveform may differ according to the value of resistance.

2. In the case of inductor, the voltage?

a) is in phase with the current

b) lags behind the current by 90⁰

c) leads the current by 90⁰

d) is out of phase with the current

Answer: c

Explanation: In the case of inductor, the voltage and current are out of phase. So, the voltage leads the current by 90⁰.

3. For inductor, what is the current?

a) is out of phase with the current

b) leads the current by 90⁰

c) is in phase with the current

d) lags behind the current by 90⁰

Answer: d

Explanation: For inductor, the voltage and current are out of phase. So, the current lags behind the current by 90⁰.

4. The value of inductance reactance is?

a) R

b) ωL

c) 1/ωL

d) ωC

Answer: b

Explanation: The value of inductance reactance is ωL. Hence a pure inductor has an impedance whose value is ωL.

5. The current in the pure capacitor?

a) lags behind the voltage by 90⁰

b) is in phase with the voltage

c) lags behind the voltage by 45⁰

d) leads the voltage by 90⁰

Answer: d

Explanation: In a capacitor, there exists a phase difference between current and voltage. The current in pure capacitor leads the voltage by 90⁰.

6. In case of pure capacitor, the voltage?

a) leads the voltage by 90⁰

b) lags behind the voltage by 45⁰

c) lags behind the voltage by 90⁰

d) is in phase with the voltage

Answer: c

Explanation: In a capacitor, there exists a phase difference between current and voltage. In case of pure capacitor, the voltage lags behind the voltage by 90⁰.

7. The impedance value of a pure capacitor is?

a) ωC

b) 1/ ωC

c) ωL

d) R

Answer: b

Explanation: The impedance value of a pure capacitor is 1/ ωC. And the impedance of the capacitor is called capacitive reactance.

8. One sine wave has a positive peak at 75⁰, and another has a positive peak at 100⁰. How much is first sine wave shifted in phase from the 0⁰ reference?

a) leads reference angle by 10⁰

b) lags reference angle by 10⁰

c) leads reference angle by 15⁰

d) lags reference angle by 15⁰

Answer: c

Explanation: As first sine wave has a positive peak at 75⁰, and another has a positive peak at 100⁰. First sine wave leads reference angle by 15⁰ from the 0⁰ reference.

9. One sine wave has a positive peak at 75⁰, and another has a positive peak at 100⁰. How much is second sine wave shifted in phase from the 0⁰ reference?

a) lags reference angle by 15⁰

b) leads reference angle by 15⁰

c) lags reference angle by 10⁰

d) leads reference angle by 10⁰

Answer: a

Explanation: As first sine wave has a positive peak at 75⁰, and another has a positive peak at 100⁰. Second sine wave lags reference angle by 15⁰ from the 0⁰ reference.

10. One sine wave has a positive peak at 75⁰, and another has a positive peak at 100⁰. What is the phase angle between two sine waves?

a) 10⁰

b) 15⁰

c) 20⁰

d) 25⁰

Answer: d

Explanation: As first sine wave has a positive peak at 75⁰, and another has a positive peak at 100⁰. The phase angle between two sine waves mentioned above is 25⁰.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Impedence Diagram”.

1. Impedance is a complex quantity having the real part as _______ and the imaginary part as ______

a) resistance, resistance

b) resistance, reactance

c) reactance, resistance

d) reactance, reactance

Answer: b

Explanation: Almost all electric circuits offer impedance to the flow of current. Impedance is a complex quantity having the real part as resistance and the imaginary part as reactance.

2. The voltage function v in the circuit shown below is?

network-theory-questions-answers-impedence-diagram-q2

a) v = V m e -tjω

b) v = V m e tjω

c) v = e tjω

d) v = e -tjω

Answer: b

Explanation: The voltage function v in the circuit is a complex function and is given by v = V m e tjω =Vm.

3. The current i in the circuit shown below is?

network-theory-questions-answers-impedence-diagram-q2

a) i=(V m /)e tjω

b) i=(V m ) e tjω

c) i=(V m ) e tjω

d) i=(V m /) e tjω

Answer: d

Explanation: i=I m e tjω . V m e tjω =RI m e tjω +L d/dt (I m e tjω ). V m e tjω =RI m e tjω +L I m e tjω . I m = V m /. i=(V m /) e tjω .

4. The impedance of the circuit shown below is?

network-theory-questions-answers-impedence-diagram-q2

a) R + jωL

b) R – jωL

c) R + 1/jωL

d) R – 1/jωL

Answer: a

Explanation: Impedance is defined as he ratio of the voltage to current function. The impedance of the circuit Z= V m e tjω /((V m /) e tjω )=R+jωL.

5. What is the magnitude of the impedance of the following circuit?

network-theory-questions-answers-impedence-diagram-q2

a) √

b) √

c)√(R 2 + 2 )

d) √(R 2 - 2 )

Answer: c

Explanation: Complex impedance is the total opposition offered by the circuit elements to ac current and can be displayed on the complex plane. The magnitude of the impedance of the circuit is √(R 2 + 2 ).

6. The phase angle between current and voltage in the circuit shown below is?

network-theory-questions-answers-impedence-diagram-q2

a) tan -1 ⁡ωL/R

b) tan -1 ⁡ωR/

c) tan -1 ⁡R/ωL

d) tan -1 ⁡L/ωR

Answer: a

Explanation: The angle between impedance and reactance is the phase angle between the current and voltage applied to the circuit. θ=tan -1 ⁡ωL/R

7. The voltage function v in the circuit shown below is?

network-theory-questions-answers-impedence-diagram-q7

a) v = e -tjω

b) v = e tjω

c) v = V m e tjω

d) v = V m e -tjω

Answer: c

Explanation: If we consider the RC series circuit and apply the complex function v to the circuit then voltage function is given by v = V m e tjω .

8. The impedance of the circuit shown below is?

network-theory-questions-answers-impedence-diagram-q7

a) R + jωC

b) R – jωC

c) R + 1/jωC

d) R – 1/jωC

Answer: c

Explanation: The impedance of the circuit shown above is R + 1/jωC. Here the impedance Z consists resistance which is real part and capacitive reactance which is imaginary part of the impedance.

9. The magnitude of the impedance of the circuit shown below is?

network-theory-questions-answers-impedence-diagram-q7

a) √

b) √

c) √(R 2 + 2 )

d) √(R 2 - 2 )

Answer: c

Explanation: The magnitude of the impedance of the circuit shown above is √(R 2 + 2 ). Here the impedance is the vector sum of the resistance and the capacitive reactance.

10. The angle between resistance and impedance in the circuit shown below.

network-theory-questions-answers-impedence-diagram-q7

a) tan -1 1/ωRC

b) tan -1 ⁡C/ωR

c) tan -1 ⁡R/ωC

d) tan -1 ⁡ωRC

Answer: a

Explanation: The angle between resistance and impedance in the circuit shown above tan -1 ⁡1/ωRC. The angle between resistance and impedance is the phase angle between the applied voltage and current in the circuit.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Series Circuits”.

1. If we apply a sinusoidal input to RL circuit, the current in the circuit is __________ and the voltage across the elements is _______________

a) square, square

b) square, sinusoid

c) sinusoid, square

d) sinusoid, sinusoid

Answer: d

Explanation: If we apply a sinusoidal input to RL circuit, the current in the circuit is square and the voltage across the elements is sinusoid. In the analysis of the RL series circuit, we can find the impedance, current, phase angle and voltage drops.

2. The circuit shown below consists of a 1kΩ resistor connected in series with a 50mH coil, a 10V rms, 10 KHz signal is applied. Find impedance Z in rectangular form.

network-theory-questions-answers-series-circuits-q2

a) Ω

b) Ω

c) Ω

d) Ω

Answer: c

Explanation: Inductive Reactance X L = ωL = 2πfL = (10 4 )(50×10 -3 ) = 3140Ω. In rectangular form, total impedance Z = Ω.

3. Find the current I in the circuit shown below.

network-theory-questions-answers-series-circuits-q2

a) 3.03

b) 30.3

c) 303

d) 0.303

Answer: a

Explanation: Total impedance Z = Ω. Magnitude = 3295.4Ω

Current I=V s /Z = 10/3295.4=3.03mA.

4. Find the phase angle θ in the circuit shown below.

network-theory-questions-answers-series-circuits-q2

a) 62.33⁰

b) 72.33⁰

c) 82.33⁰

d) 92.33⁰

Answer: b

Explanation: Phase angle θ = tan -1 ⁡(X L /R). The values of X L , R are X L = 3140Ω and R = 1000Ω. On substituting the values in the equation, phase angle θ =tan -1 ⁡=72.33⁰.

5. In the circuit shown below, find the voltage across resistance.

network-theory-questions-answers-series-circuits-q2

a) 0.303

b) 303

c) 3.03

d) 30.3

Answer: c

Explanation: Voltage across resistance = IR. The values of I = 3.03 mA and R = 10000Ω. On substituting the values in the equation, the voltage across resistance = 3.03 x 10 -3 ×1000 = 3.03V.

6. In the circuit shown below, find voltage across inductive reactance.

network-theory-questions-answers-series-circuits-q2

a) 9.5

b) 10

c) 9

d) 10.5

Answer: a

Explanation: Voltage across inductor = IX L . The values of I = 3.03 mA and X L = 10000Ω. On substituting the values in the equation, the voltage across inductor = 3.03×10 -3 ×1000 = 9.51V.

7. Determine the source voltage if voltage across resistance is 70V and the voltage across inductor is 20V as shown in the figure.

network-theory-questions-answers-series-circuits-q7

a) 71

b) 72

c) 73

d) 74

Answer: c

Explanation: If voltage across resistance is 70V and the voltage across inductor is 20V, source voltage V s =√(70 2 +20 2 ) = 72.8≅73V.

8. Find the phase angle in the circuit shown below.

network-theory-questions-answers-series-circuits-q7

a) 15

b) 16

c) 17

d) 18

Answer: b

Explanation: The phase angle is the angle between the source voltage and the current. Phase angle θ=tan -1 (V L /V R ). The values of V L = 20V and V R = 70V. On substituting the values in the equation, phase angle in the circuit = tan -1 =15.94 o ≅ 16 o .

9. An AC voltage source supplies a 500Hz, 10V rms signal to a 2kΩ resistor in series with a 0.1µF capacitor as shown in the following figure. Find the total impedance.

network-theory-questions-answers-series-circuits-q9

a) 3750.6Ω

b) 3760.6Ω

c) 3780.6Ω

d) 3790.6Ω

Answer: b

Explanation: The capacitive reactance X C = 1/2πfC = 1/(6.28×500×0.1×10 -6 ))=3184.7Ω. In rectangular form, X = Ω. Magnitude = 3760.6Ω.

10. Determine the phase angle in the circuit shown below.

network-theory-questions-answers-series-circuits-q9

a) 58

b) 68

c) -58

d) -68

Answer: c

Explanation: The phase angle in the circuit is phase angle θ = tan -1⁡ (-X C /R) =tan -1 ⁡/2000)=-57.87 o ≅-58 o .

11. Find the current I in the circuit shown below.

network-theory-questions-answers-series-circuits-q9

a) 2.66

b) 3.66

c) 4.66

d) 5.66

Answer: a

Explanation: The term current is the ratio of voltage to the impedance. The current I in the circuit is current I = V S / Z = 10/3760.6 = 2.66mA

12. Find the voltage across the capacitor in the circuit shown below.

network-theory-questions-answers-series-circuits-q9

a) 7

b) 7.5

c) 8

d) 8.5

Answer: d

Explanation: The voltage across the capacitor in the circuit is capacitor voltage = 2.66×10 -3 ×3184.7 =8.47V.

13. Determine the voltage across the resistor in the circuit shown below.

network-theory-questions-answers-series-circuits-q9

a) 3

b) 4

c) 5

d) 6

Answer: c

Explanation: The voltage across the resistor in the circuit resistive voltage = 2.66×10 -3 ×3184.7 = 5.32V.

14. In the circuit shown below determine the total impedance.

network-theory-questions-answers-series-circuits-q14

a) 161

b) 162

c) 163

d) 164

Answer: b

Explanation: Reactance across capacitor = 1/(6.28×50×10×10 -6 ) = 318.5Ω.

Reactance across inductor = 6.28×0.5×50=157Ω. In rectangular form, Z = Ω = Ω. Magnitude=161.8Ω.

15. Find the current in the circuit shown below.

network-theory-questions-answers-series-circuits-q14

a) 0.1

b) 0.2

c) 0.3

d) 0.4

Answer: c

Explanation: The term current is the ratio of voltage to the impedance. The current in the circuit is current I=V S /Z = 50/161.8 = 0.3A.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Parallel Circuits”.

1. A signal generator supplies a sine wave of 20V, 5 kHz to the circuit as shown. Determine total current IT.

network-theory-questions-answers-parallel-circuits-q1

a) 0.21∠33⁰

b) 0.22∠33⁰

c) 0.23∠33⁰

d) 0.24∠33⁰

Answer: d

Explanation: Capacitive reactance = 1/(6.28×5×10 3 ×0.2×10 -6 )=159.2Ω.Current in the resistance branch I R =V S /R=20/100 = 0.2A.Current in capacitive branch I C =V S /X C = 20/159.2 = 0.126A.Total current I T = (I R +j I C ) A = A. In polar form, I T = 0.24∠33⁰ A.

2. Find the phase angle in the circuit shown below.

network-theory-questions-answers-parallel-circuits-q1

a) 31⁰

b) 32⁰

c) 33⁰

d) 34⁰

Answer: c

Explanation: We obtained I T = 0.24∠33⁰ A, the phase angle between applied voltage and current is 33⁰. The phase angle between the applied voltage and total current is 33⁰.

3. Determine the total impedance in the circuit.

network-theory-questions-answers-parallel-circuits-q1

a) 73.3∠33⁰

b) 83.3∠-33⁰

c) 83.3∠33⁰

d) 73.3∠-33⁰

Answer: b

Explanation: Z = V S /I T = 20∠0⁰ / 0.24∠33⁰ = 83.3∠-33⁰Ω. The phase angle indicates that the total line current is 0.24 A and leads the voltage by 33⁰.

4. A 50Ω resistor is connected in parallel with an inductive reactance of 30Ω. A 20V signal is applied to the circuit. Find the line current in the circuit.

network-theory-questions-answers-parallel-circuits-q4

a) 0.77∠-58.8⁰

b) 0.77∠58.8⁰

c) 0.88∠-58.8⁰

d) 0.88∠58.8⁰

Answer: a

Explanation: Since the voltage across each element is the same as the applied voltage, the current in resistive branch is I R = V S /R = 20∠0⁰/50∠0⁰=0.4A. Current in the inductive branch is I L = V S /X L = 20∠0⁰/30∠90⁰= 0.66∠-90⁰. Total current is I T = 0.4-j0.66 = 0.77∠-58.8⁰.

5. Determine total impedance in the circuit shown below.

network-theory-questions-answers-parallel-circuits-q4

a) 25∠-58.8⁰

b) 25∠-58.8⁰.

c) 26∠-58.8⁰.

d) 26∠58.8⁰.

Answer: c

Explanation: The current lags behind the voltage by 58.8⁰. Total impedance Z = V S /I T = 20∠0⁰ / 0.77∠-58.8⁰ = 25.97∠-58.8⁰Ω. So the total impedance in the circuit shown is 25.97∠-58.8⁰Ω.

6. Determine Z in the figure shown below.

network-theory-questions-answers-parallel-circuits-q6

a) 26∠-20.5⁰

b) 26∠20.5⁰

c) 25∠-20.5⁰

d) 25∠20.5⁰

Answer: b

Explanation: First the inductive reactance is calculated. X L =6.28 x 50 x 0.1 = 31.42Ω. In figure the 10Ω resistance is in series with the parallel combination of 20Ω and j31.42Ω. Z T = 10 + /=24.23+j9.06=25.87∠20.5⁰.

7. Find I T in the figure shown below.

network-theory-questions-answers-parallel-circuits-q6

a) 0.66∠-20.5⁰

b) 0.66∠20.5⁰

c) 0.77∠20.5⁰

d) 0.77∠-20.5⁰

Answer: d

Explanation: The current lags behind the applied voltage by 20.5⁰. Total current I T = V S /Z T = 20/25.87∠20.5 o =0.77∠-20.5 o .

8. Find the phase angle θ in the circuit shown below.

network-theory-questions-answers-parallel-circuits-q6

a) 20.5⁰

b) 20⁰

c) 19.5⁰

d) 19⁰

Answer: a

Explanation: As I T = 0.77∠-20.5 o , the phase angle θ between the voltage and the current in the circuit is 20.5⁰.

9. Find the impedance in the circuit shown below.

network-theory-questions-answers-parallel-circuits-q9

a) 25

b) 26

c) 27

d) 28

Answer: c

Explanation: Capacitive reactance X C = 1/2πfC = 1/(6.28×50×100×10 -6 )=31.83Ω. Capacitive susceptance B C = 1/X C = 1/31.83 = 0.031S. Conductance G=1/R = 1/50 = 0.02S. Total admittance Y=√(G 2 +B c 2 )=√(0.02 2 +0.031 2 )=0.037S. Total impedance Z = 1/Y = 1/0.037 = 27.02Ω.

10. Determine the phase angle in the circuit shown below.

network-theory-questions-answers-parallel-circuits-q9

a) 56⁰

b) 56.5⁰

c) 57.5⁰

d) 57⁰

Answer: c

Explanation: Phase angle θ=tan -1 (R/X c ). Resistance R = 50Ω and capacitive reactance X c = 31.83Ω. So the phase angle in the circuit = tan -1 =57.52⁰.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Instantaneous Power”.

1. In purely resistive circuit, energy delivered by source is ____________ by resistance.

a) dissipated in the form of heat

b) stored as electric field

c) stored as magnetic field

d) returned to source

Answer: a

Explanation: In purely resistive circuit, energy delivered by source is dissipated in the form of heat by resistance and is not stored as either electric field or magnetic field.

2. In inductor, the energy delivered by source is ____________ by inductor.

a) stored as magnetic field

b) dissipated in the form of heat

c) returned to source

d) stored as electric field

Answer: a

Explanation: In inductor, the energy delivered by source is stored as magnetic field by inductor and is not dissipated in the form of heat or stored as electric field.

3. In capacitor, the energy delivered by source is ____________ by capacitor.

a) returned to source

b) dissipated in the form of heat

c) stored as electric field

d) stored as magnetic field

Answer: c

Explanation: In capacitor, the energy delivered by source is stored as electric field by capacitor and is not stored as magnetic field or dissipated in the form of heat.

4. If there is complex impedance in a circuit, part of energy is ____________ by reactive part and part of its energy is ____________ by the resistance.

a) alternately stored and returned, alternately stored and returned

b) alternately stored and returned, dissipated

c) dissipated, alternately stored and returned

d) dissipated, dissipated

Answer: b

Explanation: If there is complex impedance in a circuit, part of energy is alternately stored and returned by reactive part and part of its energy is dissipated by the resistance. The amount of energy dissipated is determined by relative values of resistance and reactance.

5. The equation of instantaneous power is?

a) P = (V m I m /2)+sin⁡θ)

b) P = (V m I m /2)+cos⁡θ)

c) P = (V m I m /2)+cos⁡θ)

d) P = (V m I m /2)+sin⁡θ)

Answer: c

Explanation: The equation of instantaneous power is P = (V m I m /2)+cos⁡θ). It consists of two parts. One is a fixed part and the other is time varying which has frequency twice that of the voltage or current wave forms.

6. The time varying part in the equation of instantaneous power has frequency ________________ that of the frequency of voltage or current wave forms.

a) equal to

b) twice

c) thrice

d) four times

Answer: b

Explanation: The time varying part in the equation of instantaneous power has a frequency twice that of voltage or current wave forms and the other part is a fixed part.

7. Instantaneous power is negative, when the polarities of voltage and current are of __________

a) opposite sign

b) same sign

c) voltage is zero

d) current is zero

Answer: a

Explanation: Instantaneous power is negative, when voltage and current have opposite sign that is if voltage is positive, the current is negative and if current is positive, the voltage is negative.

8. In P equation, if θ=0, then P =?

a) (V m I m /2)

b) (V m I m /2)

c) (V m I m /2)

d) (V m I m )

Answer: d

Explanation: In P equation, if θ=0⁰, then P =(V m I m /2). The power wave has a frequency twice that of the voltage or current. Here the average value of power is V m I m /2.

9. The average value of power if θ=0⁰ is?

a) V m I m /2

b) V m I m /2

c) V m I m /4

d) V m I m /8

Answer: b

Explanation: The average value of power if θ=0⁰ is V m I m /2. So, average power = V m I m /2 at θ=0⁰. When phase angle is increased the negative portion of the power cycle increases and lesser power is dissipated.

10. At θ=π/2, positive portion is __________ negative portion in power cycle.

a) greater than

b) less than

c) equal to

d) greater than or equal to

Answer: c

Explanation: At θ=π/2, the area under positive portion is equal to the area under negative portion in power cycle. At this instant the power dissipated in the circuit is zero.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Average Power”.

1. The equation of the average power (P avg ) is?

a) (V m I m /2)cosθ

b) (V m I m /2)sinθ

c) V m I m cosθ

d) V m I m sinθ

Answer: a

Explanation: To find the average value of any power function we have to take a particular time interval from t1 to t2, by integrating the function we get the average power. The equation of the average power (P avg ) is P avg = (V m I m /2)cosθ.

2. Average power (P avg )=?

a) V eff I m cosθ

b) V eff I eff cosθ

c) V m I m cosθ

d) V m I eff cosθ

Answer: b

Explanation: To get average power we have to take the product of the effective values of both voltage and current multiplied by cosine of the phase angle between the voltage and current. The expression of average power is Average power (P avg ) = V eff I eff cosθ

3. In case of purely resistive circuit, the average power is?

a) V m I m

b) V m I m /2

c) V m I m /4

d) V m I m /8

Answer: b

Explanation: In case of purely resistive circuit, the phase angle between the voltage and current is zero that is θ=0⁰. Hence the average power = V m I m /2.

4. In case of purely capacitive circuit, average power = ____ and θ = _____

a) 0, 0⁰

b) 1, 0⁰

c) 1, 90⁰

d) 0, 90⁰

Answer: d

Explanation: In case of purely capacitive circuit, the phase angle between the voltage and current is zero that is θ=90⁰. Hence the average power = 0.

5. In case of purely inductive circuit, average power = ____ and θ = ______

a) 0, 90⁰

b) 1, 90⁰

c) 1, 0⁰

d) 0, 0⁰

Answer: a

Explanation: In case of purely inductive circuit, the phase angle between the voltage and current is zero that is θ=90⁰. Hence the average power = 0.

6. If a circuit has complex impedance, the average power is ______

a) power stored in inductor only

b) power stored in capacitor only

c) power dissipated in resistor only

d) power stored in inductor and power dissipated in resistor

Answer: c

Explanation: If a circuit has complex impedance, the average power is power dissipated in resistor only and is not stored in capacitor or inductor.

7. A voltage v = 100sinωt is applied to a circuit. The current flowing through the circuit is i = 15sin⁰. Find the effective value of voltage.

a) 70

b) 71

c) 72

d) 73

Answer: b

Explanation: The expression of effective value of voltage is V eff = V m /√2. Given V m = 100. On substituting the value in the equation we get effective value of voltage = 100/√2 = 71V.

8. A voltage v = 100sinωt is applied to a circuit. The current flowing through the circuit is i = 15sin⁰. Find the effective value of current.

a) 9

b) 10

c) 11

d) 12

Answer: c

Explanation: The expression of effective value of current is I eff = I m /√2. Given I m = 15. On substituting the value in the equation we get effective value of current = 15/√2 = 11V.

9. A voltage v = 100sinωt is applied to a circuit. The current flowing through the circuit is i = 15sin⁰. Determine the average power delivered to the circuit.

a) 620

b) 630

c) 640

d) 650

Answer: d

Explanation: The expression of average power delivered to the circuit is P avg = V eff I eff cosθ, θ = 30⁰. We have V eff = 71, I eff = 11. So the average power delivered to the circuit P avg = 71 x 11 x cos 30⁰ = 650W.

10. Determine the average power delivered to the circuit consisting of an impedance Z = 5+j8 when the current flowing through the circuit is I = 5∠30⁰.

a) 61.5

b) 62.5

c) 63.5

d) 64.5

Answer: b

Explanation: The expression of the average power delivered to the circuit is P avg = I m 2 R/2. Given I m = 5, R = 5. So the average power delivered to the circuit = 5 2 ×5/2 = 62.5W.

This set of Network Theory Interview Questions and Answers for Experienced people focuses on “Apparent Power and Power Factor”.

1. The highest power factor will be?

a) 1

b) 2

c) 3

d) 4

Answer: a

Explanation: The power factor is useful in determining the useful power transferred to a load. The highest power factor will be 1.

2. If power factor = 1, then the current to the load is ______ with the voltage across it.

a) out of phase

b) in phase

c) 90⁰ out of phase

d) 45⁰ out of phase

Answer: b

Explanation: If power factor = 1, then the current to the load is in phase with the voltage across it because the expression of power factor is power factor = cosθ.

3. In case of resistive load, what is the power factor?

a) 4

b) 3

c) 2

d) 1

Answer: d

Explanation: In case of resistive load, the power factor = 1 as the current to the load is in phase with the voltage across it.

4. If power factor = 0, then the current to a load is ______ with the voltage.

a) in phase

b) out of phase

c) 45⁰ out of phase

d) 90⁰ out of phase

Answer: d

Explanation: If the power factor = 0, then the current to a load is 90⁰ out of phase with the voltage and it happens in case of reactive load.

5. For reactive load, the power factor is equal to?

a) 0

b) 1

c) 2

d) 3

Answer: a

Explanation: For reactive load, the power factor is equal to 0. Power factor = 0 when current to a load is 90⁰ out of phase with the voltage.

6. Average power is also called?

a) apparent power

b) reactive power

c) true power

d) instantaneous power

Answer: c

Explanation: The average power is expressed in watts. It means the useful power transferred from the source to the load, which is also called true power. Average power is also called true power.

7. If we apply a sinusoidal voltage to a circuit, the product of voltage and current is?

a) true power

b) apparent power

c) average power

d) reactive power

Answer: b

Explanation: If we apply a sinusoidal voltage to a circuit, the product of voltage and current is apparent power. The apparent power is expressed in volt amperes or simply VA.

8. The expression of apparent power (P app ) is?

a) V m I m

b) V m I eff

c) V eff I eff

d) V eff I m

Answer: c

Explanation: In case of sinusoidal voltage applied to the circuit, the product of voltage and the current is not the true power or average power and it is apparent power. The expression of apparent power (P app ) is P app = V eff I eff .

9. The power factor=?

a) sinθ

b) cosθ

c) tanθ

d) secθ

Answer: b

Explanation: The expression of power factor is power factor= cosθ. As the phase angle between the voltage and the current increases the power factor decreases.

10. The power factor is the ratio of ________ power to the ______ power.

a) average, apparent

b) apparent, reactive

c) reactive, average

d) apparent, average

Answer: a

Explanation: The power factor is the ratio of average power to the apparent power. Power factor =/. Power factor is also defined as the factor with which the volt amperes are to be multiplied to get true power in the circuit.

11. The power factor is called leading power factor in case of ____ circuits.

a) LC

b) RC

c) RL

d) RLC

Answer: b

Explanation: The power factor is called leading power factor in case of RC circuits and not in RLC circuits and RL circuits and LC circuits.

12. The term lagging power factor is used in which circuits?

a) RLC

b) RC

c) RL

d) LC

Answer: c

Explanation: The term lagging power factor is used in RL circuits and not in RLC circuits and RC circuits and LC circuits.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Reactive Power”.

1. The reactive power equation (P r ) is?

a) I eff 2 sin2

b) I eff 2 cos2

c) I eff 2 sin

d) I eff 2 cos

Answer: a

Explanation: If we consider a circuit consisting of a pure inductor, the power in the inductor is reactive power and the reactive power equation (P r ) is P r =I eff 2 sin2.

2. Reactive power is expressed in?

a) Watts

b) Volt Amperes Reactive

c) Volt Ampere

d) No units

Answer: b

Explanation: Reactive power is expressed in Volt Amperes Reactive and power is expressed in watts and apparent power is expressed in Volt Ampere .

3. The expression of reactive power is?

a) V eff I m sinθ

b) V m I m sinθ

c) V eff I eff sinθ

d) V m I eff sinθ

Answer: c

Explanation: The expression of reactive power (P r ) is V eff I eff sinθ Volt Amperes Reactive . Reactive power = V eff I eff sinθ Volt Amperes Reactive .

4. The power factor is the ratio of ________ power to the ______ power.

a) average, apparent

b) apparent, reactive

c) reactive, average

d) apparent, average

Answer: a

Explanation: The power factor is the ratio of average power to the apparent power. Power factor = /.

5. The expression of true power (P true ) is?

a) P a sinθ

b) P a cosθ

c) P a tanθ

d) P a secθ

Answer: b

Explanation: True power is the product of the apparent power and cosθ. The expression of true power (P true ) is P a cosθ. True power = P a cosθ.

6. The average power (P avg ) is expressed as?

a) P a secθ

b) P a tanθ

c) P a cosθ

d) P a sinθ

Answer: c

Explanation: The average power is the product of the apparent power and cosθ. The average power (P avg ) is expressed as P a cosθ. Average power = P a cosθ.

7. The equation of reactive power is?

a) P a cosθ

b) P a secθ

c) P a sinθ

d) P a tanθ

Answer: c

Explanation: The reactive power is the product of the apparent power and sinθ. The equation of reactive power is P a sinθ. Reactive power = P a sinθ.

8. A sinusoidal voltage v = 50sinωt is applied to a series RL circuit. The current in the circuit is given by I = 25sin ⁰. Determine the apparent power .

a) 620

b) 625

c) 630

d) 635

Answer: c

Explanation: The expression of apparent power is P app = V eff I eff = (V m /√2)×(I m /√2). On substituting the values V m = 50, I m = 25, we get apparent power = /2 = 625VA.

9. A sinusoidal voltage v = 50sinωt is applied to a series RL circuit. The current in the circuit is given by I = 25sin ⁰. Find the power factor.

a) 0.4

b) 0.5

c) 0.6

d) 0.7

Answer: c

Explanation: In sinusoidal sources the power factor is the cosine of the phase angle between the voltage and the current. The expression of power factor = cosθ = cos53⁰ = 0.6.

10. A sinusoidal voltage v = 50sinωt is applied to a series RL circuit. The current in the circuit is given by I = 25sin ⁰. Determine the average power.

a) 365

b) 370

c) 375

d) 380

Answer: c

Explanation: The average power, P avg = V eff I eff cosθ. We know the values of V eff , I eff are V eff = 625 and I eff – 0.6. So the average power = 625 x 0.6 = 375W.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Series Resonance”.

1. The circuit is said to be in resonance if the current is ____ with the applied voltage.

a) in phase

b) out of phase

c) 45⁰ out of phase

d) 90⁰ out of phase

Answer: a

Explanation: The circuit is said to be in resonance if the current is in phase with the applied voltage and not if the current is out of phase with the applied voltage. The study of resonance is very useful particularly in the area of communications.

2. In a series resonance circuit, series resonance occurs when?

a) X L = 1

b) X C = 1

c) X L = X C

d) X L = -X C

Answer: c

Explanation: In a series resonance circuit, series resonance occurs when capacitive reactance is equal to the inductive reactance that is X L = X C .

3. As X L = X C in a series resonance circuit, the impedance is_________

a) purely capacitive

b) purely inductive

c) purely resistive

d) capacitive and inductive

Answer: c

Explanation: As X L = X C in a series resonance circuit, the impedance is purely resistive. In a series RLC circuit the current lags behind or leads the applied voltage depending on the values of X L and X C .

4. At resonant frequency, the voltage across capacitor is _______ the voltage across inductor.

a) greater than

b) less than

c) greater than or equal to

d) equal to

Answer: d

Explanation: At resonant frequency, the voltage across capacitor is equal to the voltage across inductor. If one of the parameters of the series RLC circuit is varied in such a way that the current in the circuit is in phase with the applied voltage, then the circuit is said to be in resonance.

5. In series RLC circuit, the voltage across capacitor and inductor are ______ with each other.

a) in phase

b) 180⁰ out of phase

c) 90⁰ out of phase

d) 45⁰ out of phase

Answer: b

Explanation: In series RLC circuit, the voltage across capacitor and inductor are 180⁰ out of phase with each other. The frequency at which the resonance occurs is called resonant frequency.

6. The voltage across the LC combination in a series RLC circuit is?

a) 0

b) 1

c) 2

d) 3

Answer: a

Explanation: Since the voltage across capacitor and inductor are 180⁰ out of phase with each other, the voltage across the LC combination in a series RLC circuit is 0V.

7. The expression of resonant frequency in a series resonant circuit is?

a) 1/

b) 1/

c) 2π√LC

d) 1/

Answer: d

Explanation: The expression of resonant frequency is resonant frequency = 1/. In a series RLC circuit resonance may be produced by varying the frequency, keeping L and C constant.

8. For the circuit shown in figure determine the capacitive reactance at resonance.

network-theory-questions-answers-series-resonance-q8

a) 15

b) 20

c) 25

d) 30

Answer: c

Explanation: We know at resonance, capacitive reactance is equal to the inductive reactance that is X L = X C . Given inductive reactance X L = 25. On substituting in the equation we get X C = 25Ω.

9. What is the value of the impedance at resonance in the circuit shown below?

network-theory-questions-answers-series-resonance-q8

a) 25

b) 50

c) 75

d) 100

Answer: b

Explanation: We know that at resonance the value of impedance at resonance. So Z = R. Given R = 50Ω. On substituting in the equation we get Z = 50Ω.

10. Determine the resonant frequency for the circuit shown below.

network-theory-questions-answers-series-resonance-q10

a) 2.25

b) 22.5

c) 225

d) 2250

Answer: a

Explanation: The expression of resonant frequency is resonant frequency f r = 1/. Given L = 0.5mH and C = 10uF. On substituting in the equation we get resonant frequency f r = 1/(2π√(10×10 -6 )×0.5×10 -3 )) = 2.25kHz.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Bandwidth of an RLC Circuit”.

1. The expression of power (P 1 ) at lower half power frequency is?

a) (I 2 max R)/8

b) (I 2 max R)/4

c) (I 2 max R)/2

d) I 2 max R

Answer: c

Explanation: The upper and lower cut-off frequencies are sometimes called the half-power frequencies,. At these frequencies the power from the source is half of the power delivered at the resonant frequency. The expression of power (P 1 ) at lower half power frequency is P 1 = (I 2 max R)/2.

2. At upper half power frequency, the expression for power (P 2 ) is?

a) I 2 max R

b) (I 2 max R)/2

c) (I 2 max R)/4

d) (I 2 max R)/8

Answer: b

Explanation: At upper half power frequency, the expression for power (P 2 ) is P 2 = (I 2 max R)/2. The response curve is also called the selectivity curve of the circuit.

3. Determine the resonant frequency for the specifications: R = 10Ω, L = 0.1H, C = 10µF.

a) 157

b) 158

c) 159

d) 160

Answer: c

Explanation: The frequency at which the resonance occurs is called resonant frequency. The expression of the resonant frequency is given by f r = 1/. On substituting the given values we get resonant frequency = 1/(2π√(0.1×10×10 -6 ))=159.2 Hz.

4. The expression for lower half power frequency is?

a) (-R+√(x 2 +4LC))/4πL

b) (–R-√(x 2 +4LC))/4πL

c) (R-√(x 2 +4LC))/4πL

d) (R+√(x 2 +4LC))/4πL

Answer: a

Explanation: Selectivity indicates how well a resonant circuit responds to a certain frequency and eliminates all other frequencies. At lower power frequency, X C > X L (1/2πf 1 C)-2πf 1 L=R. f 1 = (-R+√(x 2 +4LC))/4πL.

5. The expression for upper half power frequency is?

a) (R+√(x 2 +4LC))/4πL

b) (R-√(x 2 +4LC))/4πL

c) (–R-√(x 2 +4LC))/4πL

d) (-R+√(x 2 +4LC))/4πL

Answer: a

Explanation: The narrower the bandwidth the greater the selectivity. At upper half power frequency, X C < X L => -(1/2πf 2 C)+2πf 2 L=R. f 2 = (R+√(x 2 +4LC))/4πL.

6. The expression for bandwidth is?

a) R/πL

b) R/2πL

c) R/4πL

d) R/8πL

Answer: b

Explanation: The bandwidth of any system is the range of frequencies for which the current or output voltage is equal to 70.7% of its value at the resonant frequency. The expression of bandwidth is BW = f 2 – f 1 = R/2πL.

7. In series circuits, the expression for quality factor is?

a) f r

b) BW

c) f r /BW

d) BW/f r

Answer: c

Explanation: The quality factor is the ratio of the reactive power in the inductor or capacitor to the true power in the resistance in series with the coil or capacitor. The expression of quality factor is Q = f r /BW.

8. In a series circuit having resistance and inductance, the quality factor is?

a) ωL/R

b) R/ωL

c) ωL

d) R

Answer: a

Explanation: Quality factor Q = ωL/R. A higher value of circuit Q results in smaller bandwidth and a lower value of Q causes a larger bandwidth.

9. If a series circuit contains resistor and capacitor, the expression for quality factor is?

a) C

b) ωRC

c) ωC

d) 1/ωRC

Answer: d

Explanation: The expression of quality factor is Q = 1/ωRC. The ratio of voltage across either L or C to the voltage applied the resonance can be defined as magnification.

10. The quality factor of the coil for a series circuit having R = 10Ω, L = 0.1H, C = 10µF.

a) 1

b) 5

c) 10

d) 15

Answer: c

Explanation: The resonant frequency is given by f r = 1/=1/(2π√(0.1×10×10 -6 ))=159.2 Hz. The relation between quality factor, resonant frequency and bandwidth is Q = f r /BW = 2πf r L/R = /10=10.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Parallel Resonance”.

1. For the circuit shown below, determine its resonant frequency.

network-theory-questions-answers-parallel-resonance-q1

a) 6.12

b) 7.12

c) 8.12

d) 9.12

Answer: b

Explanation: The resonant frequency of the circuit is f r = 1/. Given L = 5H and C = 100uf. On substituting the given values in the equation we get resonant frequency = 1/(2π√(5×100×10 -6 )) = 7.12 Hz.

2. Find the quality factor of the following circuit.

network-theory-questions-answers-parallel-resonance-q1

a) 2.24

b) 3.34

c) 4.44

d) 5.54

Answer: a

Explanation: The quality factor of the circuit is Q = X L /R = 2πf r L/R. Given f = 7.12 Hz and L = 5H and R = 100. On substituting the given values in the equation we get the quality factor = /100 = 2.24.

3. Find the bandwidth of the circuit shown below.

network-theory-questions-answers-parallel-resonance-q1

a) 1

b) 2

c) 3

d) 4

Answer: c

Explanation: The bandwidth of the circuit is BW = f r /Q. we obtained f r = 7.12 Hz and Q = 2.24. On substituting the given values in the equation we get the bandwidth = 7.12/2.24 = 3.178Hz.

4. The magnification in resonance considering the voltage across inductor is?

a) V/V L

b) V L /V

c) V x V L

d) V L

Answer: b

Explanation: The ratio of voltage across inductor to the voltage applied at resonance can be defined as magnification. The magnification in resonance considering the voltage across inductor is Q = V L /V.

5. Considering the voltage across the capacitor, the magnification in resonance is?

a) V C

b) V x V C

c) V C /V

d) V/V C

Answer: c

Explanation: The ratio of voltage across capacitor to the voltage applied at resonance can be defined as magnification. Considering the voltage across the capacitor, the magnification in resonance is Q = V C /V.

6. The value of ω r in parallel resonant circuit is?

a) 1/

b) 1/√LC

c) 1/

d) 1/

Answer: b

Explanation: Basically parallel resonance occurs when X L = X L . The frequency at which the resonance occurs is called the resonant frequency. The value of ω r in parallel resonant circuit is ω r = 1/√LC.

7. The expression of resonant frequency for parallel resonant circuit is?

a) 1/

b) 1/

c) 1/

d) 1/√LC

Answer: a

Explanation: The condition for resonance occurs when X L = X L . The expression of resonant frequency for parallel resonant circuit is f r = 1/.

8. Find the resonant frequency in the ideal parallel LC circuit shown in the figure.

network-theory-questions-answers-parallel-resonance-q8

a) 7.118

b) 71.18

c) 711.8

d) 7118

Answer: d

Explanation: The expression for resonant frequency is f r = 1/. Given L = 50mH and C = 0.01uF. On substituting the given values in the equation we get the resonant frequency = 1/(2π√(50×10 -3 )×0.01×10 -6 ))=7117.6 Hz.

9. If the value of Q of the circuit is high, then its effect on bandwidth is?

a) large bandwidth

b) small bandwidth

c) no effect on bandwidth

d) first increases and then decreases

Answer: b

Explanation: If the value of Q of the circuit is high, then small bandwidth because bandwidth is inversely proportional to the quality factor.

10. If in a circuit, if Q value is decreased then it will cause?

a) small bandwidth

b) no effect on bandwidth

c) first increases and then decreases

d) large bandwidth

Answer: d

Explanation: If in a circuit, if the Q value is decreased then bandwidth increases and the bandwidth do not decrease.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Resonant Frequency for a Tank Circuit”.

1. For the tank circuit shown below, find the resonant frequency.

network-theory-questions-answers-tank-circuit-q1

a) 157.35

b) 158.35

c) 159.35

d) 160.35

Answer: b

Explanation: The parallel resonant circuit is generally called a tank circuit because of the fact that the circuit stores energy in the magnetic field of the coil and in the electric field of the capacitor. The resonant frequency f r = √-(R 2 /L 2 )).

2. The expression of ωr in a parallel resonant circuit is?

a) 1/

b) 1/√LC

c) 1/

d) 1/

Answer: b

Explanation: The stored energy is transferred back and forth between the capacitor and coil and vice-versa. The expression of ω r in parallel resonant circuit is ω r = 1/√LC.

3. The expression of bandwidth for the parallel resonant circuit is?

a) 1/RC

b) RC

c) 1/R

d) 1/C

Answer: a

Explanation: The expression of bandwidth for parallel resonant circuit is BW = 1/RC. the circuit is said to be in a resonant condition when the susceptance part of admittance is zero.

4. The quality factor in case of parallel resonant circuit is?

a) C

b) ω r RC

c) ω r C

d) 1/ω r RC

Answer: d

Explanation: The quality factor in case of parallel resonant circuit is Q = 1/ω r RC. The impedance of parallel resonant circuit is maximum at the resonant frequency and decreases at lower and higher frequencies.

5. The quality factor is the product of 2π and the ratio of ______ to _________

a) maximum energy stored, energy dissipated per cycle

b) energy dissipated per cycle, maximum energy stored

c) maximum energy stored per cycle, energy dissipated

d) energy dissipated, maximum energy stored per cycle

Answer: a

Explanation: At low frequencies X L is very small and X C is very large so the total impedance is essentially inductive. The quality factor is the product of 2π and the ratio of maximum energy stored to energy dissipated per cycle.

6. The maximum energy stored in a capacitor is?

a) CV 2

b) CV 2 /2

c) CV 2 /4

d) CV 2 /8

Answer: b

Explanation: The maximum energy stored in a capacitor is CV 2 /2. Maximum energy = CV 2 /2. As frequency increases the impedance also increases and the inductive reactance dominates until the resonant frequency is reached.

7. The expression of quality factor is?

a) I L /I

b) I/I L

c) IL

d) I

Answer: a

Explanation: The expression of quality factor is I L /I. Quality factor = I L /I. At the the point X L = X C , the impedance is at its maximum.

8. The quality factor is defined as?

a) I

b) I C

c) I/I C

d) I C /I

Answer: d

Explanation: The quality factor is defined as I C /I. Quality factor = I C /I. As the frequency goes above resonance capacitive reactance dominates and impedance decreases.

9. In the circuit shown in the figure, an inductance of 0.1H having a Q of 5 is in parallel with a capacitor. Determine the value coil resistance of at a resonant frequency of 500 rad/sec.

network-theory-questions-answers-tank-circuit-q9

a) 10

b) 20

c) 30

d) 40

Answer: a

Explanation: Quality factor Q = ω r L/R. L = 0.1H, Q = 5, ω r = 500 rad/sec. On solving, R = 10Ω. While plotting the voltage and current variation with frequency, at resonant frequency, the current is maximum.

10. Find the value of capacitance µ in the circuit shown below.

network-theory-questions-answers-tank-circuit-q9

a) 10

b) 20

c) 30

d) 40

Answer: d

Explanation: ω 2 r = 1/LC. L = 0.1H, ωr = 500 rad/sec. On solving, C = 40 µF. In order to tune a parallel circuit to a lower frequency the capacitance must be increased.

This set of Network Theory Questions & Answers for Exams focuses on “Problems of Series Resonance Involving Quality Factor”.

1. A circuit tuned to a frequency of 1.5 MHz and having an effective capacitance of 150 pF. In this circuit, the current falls to 70.7 % of its resonant value. The frequency deviates from the resonant frequency by 5 kHz. Q factor is?

a) 50

b) 100

c) 150

d) 200

Answer: c

Explanation: Q = \(\frac{ω}{ω1 – ω2} = \frac{f}{f2-f1}\)

Here, f = 1.5 × 10 6 Hz

f1 = (1.5 × 10 6 – 5 × 10 3 )

f2 = (1.5 × 10 6 + 5 × 10 3 )

So, f2 –f1 = 10 × 10 3 Hz

∴ Q = \(\frac{1.5 × 10^6}{10 × 10^3}\) = 150.

2. A circuit tuned to a frequency of 1.5 MHz and having an effective capacitance of 150 pF. In this circuit, the current falls to 70.7 % of its resonant value. The frequency deviates from the resonant frequency by 5 kHz. Effective resistance of the circuit is?

a) 2 Ω

b) 3 Ω

c) 5.5 Ω

d) 4.7 Ω

Answer: d

Explanation: R = \(\frac{f2-f1}{2πf^2 L}\)

Here, f = 1.5 × 10 6 Hz

f1 = (1.5 × 10 6 – 5 × 10 3 )

f2 = (1.5 × 10 6 + 5 × 10 3 )

So, f2 – f1 = 10 × 10 3 Hz

∴ R = 4.7 Ω.

3. Consider a circuit consisting of two capacitors C 1 and C 2 . Let R be the resistance and L be the inductance which are connected in series. Let Q 1 and Q 2 be the quality factor for the two capacitors. While measuring the Q value by the Series Connection method, the value of the Q factor is?

a) Q = \

Q = \

Q = \(\frac{

C_1 C_2}{Q_1 C_1-Q_2 C_2}\)

Answer: a

Explanation: ωL = \(\frac{1}{ωC}\) and Q 1 = \(\frac{ωL}{R} = \frac{1}{ωC_1 R}\)

X S = \(\frac{C_1 – C_2}{ωC_1 C_2}\), R S = \(\frac{Q_1 C_1 – Q_2 C_2}{ωC_1 C_2 Q_1 Q_2}\)

Q X = \(\frac{X_S}{R_S} = \frac{

Q_1 Q_2}{Q_1 C_1-Q_2 C_2}\).

4. Consider a circuit consisting of two capacitors C 1 and C 2 . Let R be the resistance and L be the inductance which are connected in series. Let Q 1 and Q 2 be the quality factor for the two capacitors. While measuring the Q value by the Parallel Connection method, the value of the Q factor is?

a) Q = \

Q = \

Q = \(\frac{

C_1 C_2}{Q_1 C_1-Q_2 C_2}\)

Answer: b

Explanation: \(\frac{1}{R_P} = \frac{ωC_1}{Q_2} – \frac{1}{RQ_1^2}\), X P = \(\frac{1}{ω

}\)

Q = \(\frac{

Q_1 Q_2}{Q_1 C_1-Q_2 C_2}\).

5. A 50 Hz voltage is measured with a moving iron voltmeter and a rectifier type AC voltmeter connected in parallel. If the meter readings are V a and V b respectively. Then the form factor may be estimated as?

a) \

\(\frac{πV_a}{V_b}\)

Answer: b

Explanation: Form factor of the wave = \(\frac{RMS \,value}{Mean \,value}\)

Moving iron instrument will show rms value. Rectifier voltmeter is calibrated to read rms value of sinusoidal voltage that is, with form factor of 1.11.

∴ Mean value of the applied voltage = \(\frac{V_b}{1.11}\)

∴ Form factor = \(\frac{V_a}{V_b/1.11} = \frac{1.11V_a}{V_b}\).

6. For the resonant circuit given below, the value of the quality factor of the circuit is __________

network-theory-questions-answers-problems-series-resonance-involving-quality-factor-q6

a) 5.6

b) 7.1

c) 8.912

d) 12.6

Answer: b

Explanation: f = \(\frac{1}{2π\sqrt{LC}} \)

= \(\frac{1}{6.28\sqrt{

}}\)

= \(\frac{1}{6.28\sqrt{2 × 10^{-15}}}\)

= \(\frac{1}{888 × 10^{-9}}\) = 1.13 MHz

Inductive Reactance, X L = 2πfL = (1.13 × 10 6 )(10 × 10 -6 )

= 70.96 Ω

∴ Q = \(\frac{X_L}{R} = \frac{70.96}{10}\) = 7.1.

7. For the series resonant circuit shown below, the value of the resonant frequency is _________

network-theory-questions-answers-problems-series-resonance-involving-quality-factor-q7

a) 36.84 kHz

b) 40.19 kHz

c) 25.28 kHz

d) 15.9 kHz

Answer: d

Explanation: Resonant Frequency, F R = \(\frac{1}{2π\sqrt{LC}} \)

= \(\frac{1}{6.28\sqrt{

}}\)

= \(\frac{1}{6.28\sqrt{10^{-10}}}\)

= \(\frac{1}{6.28 × 10^{-5}}\) = 15.9 kHz.

8. For the series resonant circuit given below, the value of the quality factor is ___________

network-theory-questions-answers-problems-series-resonance-involving-quality-factor-q7

a) 15

b) 36

c) 25

d) 10

Answer: d

Explanation: Resonant Frequency, F R = \(\frac{1}{2π\sqrt{LC}} \)

= \(\frac{1}{6.28\sqrt{

}}\)

= \(\frac{1}{6.28\sqrt{10^{-10}}}\)

= \(\frac{1}{6.28 × 10^{-5}}\) = 15.9 kHz.

Inductive Reactance, X L = 2πfL = (15.92 × 10 3 )(5 × 10 -6 )

= 0.5 kΩ

∴ Quality factor Q = \(\frac{X_L}{R} = \frac{0.5 kΩ}{50}\) = 10.

9. For the series resonant circuit given below, the bandwidth is ____________

network-theory-questions-answers-problems-series-resonance-involving-quality-factor-q7

a) ±351 Hz

b) ±796 Hz

c) ±531 Hz

d) ±225 Hz

Answer: b

Explanation: Resonant Frequency, F R = \(\frac{1}{2π\sqrt{LC}} \)

= \(\frac{1}{6.28\sqrt{

}}\)

= \(\frac{1}{6.28\sqrt{10^{-10}}}\)

= \(\frac{1}{6.28 × 10^{-5}}\) = 15.9 kHz.

Inductive Reactance, X L = 2πfL = (15.92 × 10 3 )(5 × 10 -6 )

= 0.5 kΩ

∴ Quality factor Q = \(\frac{X_L}{R} = \frac{0.5 kΩ}{50}\) = 10

∴ ∆F = \(\frac{f}{Q} = \frac{15.9 kHz}{10}\) = 1.592 kHz

∴ Bandwidth = \(\frac{∆F}{2}\) = ±796 Hz.

10. For the series circuit given below, the value of the cut-off frequencies are ____________

network-theory-questions-answers-problems-series-resonance-involving-quality-factor-q7

a) 78.235 kHz; 16.215 kHz

b) 13.135 kHz; 81.531 kHz

c) 16.716 kHz; 15.124 kHz

d) 50.561 kHz; 18.686 kHz

Answer: c

Explanation: Resonant Frequency, F R = \(\frac{1}{2π\sqrt{LC}} \)

= \(\frac{1}{6.28\sqrt{

}}\)

= \(\frac{1}{6.28\sqrt{10^{-10}}}\)

= \(\frac{1}{6.28 × 10^{-5}}\) = 15.9 kHz.

Inductive Reactance, X L = 2πfL = (15.92 × 10 3 )(5 × 10 -6 )

= 0.5 kΩ

∴ Quality factor Q = \(\frac{X_L}{R} = \frac{0.5 kΩ}{50}\) = 10

∴ ∆F = \(\frac{f}{Q} = \frac{15.9 kHz}{10}\) = 1.592 kHz

∴ Bandwidth = \(\frac{∆F}{2}\) = ±796 Hz.

Therefore, f 2 = f + \(\frac{∆F}{2}\) = 16.716 kHz and f 1 = f – \(\frac{∆F}{2}\) = 15.124 kHz.

11. In a series resonance type BPF, C = 1.8 pf, L = 25 mH, R F = 52 Ω and R L = 9 kΩ. The Resonant frequency f is __________

network-theory-questions-answers-problems-series-resonance-involving-quality-factor-q11

a) 75.1 kHz

b) 751 kHz

c) 575 kHz

d) 57.5 kHz

Answer: b

Explanation: f = \(\frac{1}{2π\sqrt{LC}} = \frac{1}{2π\sqrt{0.025×1.8×10^{-12}}}\)

= \(\frac{1}{2π} × \frac{10^6}{\sqrt{0.45}}\)

= \(\frac{1}{2π} × \frac{10^6}{0.67}\)

= 751000

∴ f = 751 kHz.

12. In a series resonance type BPF, C = 1.8 pf, L = 25 mH, R F = 52 Ω and R L = 9 kΩ. The Bandwidth is ___________

network-theory-questions-answers-problems-series-resonance-involving-quality-factor-q12

a) 331 kHz

b) 575 kHz

c) 331 Hz

d) 575 Hz

Answer: c

Explanation: R = \(\frac{R_f×R_L}{R_f+R_L}\) ≈ 52 Ω

∴ Q factor = \(\frac{ω_r L}{R} = \frac{2π×751×25×10^{-3}}{52} \)

= \(\frac{π×37580}{52}\) = 2270

Now, B W = \(\frac{f_r}{Q} = \frac{751 × 10^3}{2270}\)

∴ B W = 331 Hz.

13. A 50 μF capacitor, when connected in series with a coil having a resistance of 40Ω, resonates at 1000 Hz. The inductance of the coil for the resonant circuit is ____________

a) 2.5 mH

b) 1.2 mH

c) 0.5 mH

d) 0.1 mH

Answer: c

Explanation: At resonance, X L = X C

Or, 2πfL = \(\frac{1}{2πfC} \)

∴ L = \(\frac{1}{4π^2 f^2 C} \)

= \(\frac{1}{4π^2 × ^2 × 50 × 10^{-6}} \)

= \(\frac{1}{39.46×50} \)

= 0.0005

So, L = 0.5 mH.

14. A coil has been designed for a high Quality Factor performance. The voltage is rated at and a specified frequency. If the frequency of operation is increased 10 times and the coil is operated at the same rated voltage. The new value of Q factor and the active power P will be?

a) P is doubled and Q is halved

b) P is halved and Q is doubled

c) P remains constant and Q is doubled

d) P decreases 100 times and Q is increased 10 times

Answer: d

Explanation: ω 2 L = 10 ω 1 LR will remain constant

∴ Q 2 = \(\frac{10 ω_1 L}{R}\) = 10 Q 1

That is Q is increased 10 times.

Now, I 1 = \(\frac{V}{ω_1 L} \)

For a high Q coil, ωL >> R,

I 2 = \(\frac{V}{10 ω_1 L} = \frac{I_1}{10}\)

∴ P 2 = R \

^2 = \frac{P_1}{100}\)

Thus, P decreases 100 times and Q is increased 10 times.

15. A 50 Hz, bar primary CT has a secondary with 800 turns. The secondary supplies 7 A current into a purely resistive burden of 2 Ω. The magnetizing ampere-turns are 300. The phase angle is?

a) 3.1°

b) 85.4°

c) 94.6°

d) 175.4°

Answer: a

Explanation: Secondary burden is purely resistive and the resistance of burden is equal to the resistance of the secondary winding; the resistance of secondary winding = 1Ω. The voltage induced in secondary × resistance of secondary winding = 7 × 2 = 14V. Secondary power factor is unity as the load is purely resistive. The loss component of no-load current is to be neglected i.e. I e = 0. I M = 300 A.

Secondary winding current I S = 7 A

Reflected secondary winding current = n I S = 5600 A

∴ tan θ = \(\frac{I_M}{nI_S} \). So, θ = 3.1°.

This set of Network Theory Questions & Answers for Exams focuses on “Problems of Series Resonance Involving Frequency Response of Series Resonant Circuit”.

1. A series RLC circuit has R = 50 Ω, L = 0.01 H and C = 0.04 × 10 -6 F. System voltage is 100 V. The frequency, at which the maximum voltage appears across the capacitor, is?

a) 5937.81 Hz

b) 7370 Hz

c) 7937.81 Hz

d) 981 Hz

Answer: c

Explanation: Frequency f c at which V C is maximum f c = \(\frac{1}{2π}\Big[\frac{1}{0.01×0.01×10^{-6}} – \frac{50×50}{2×0.01×0.01}\Big]^{0.5}\)

So, f c = 7937.81 Hz.

2. A series RLC circuit has R = 50 Ω, L = 0.01 H and C = 0.04 × 10 -6 F. System voltage is 100 V. At this frequency, the circuit impedance is?

a) 50 – j2.5 Ω

b) 50 – j2.8 Ω

c) 50 + j2.5 Ω

d) 50 + j2.8 Ω

Answer: a

Explanation: Z = R + j

= 50 + j \Missing or unrecognized delimiter for \right\) = 50 – j2.5 Ω.

3. Given, R = 10 Ω, L = 100 mH and C = 10 μF. The values of ω 0 , ω 1 and ω 2 respectively are?

network-theory-questions-answers-resonance-q3

a) ω 0 = 1000 rad/s, ω 1 = 951.25 rad/s, ω 2 = 1075.54 rad/s

b) ω 0 = 1000 rad/s, ω 1 = 975.25 rad/s, ω 2 = 1051.25 rad/s

c) ω 0 = 1000 rad/s, ω 1 = 951.25 rad/s, ω 2 = 1051.25 rad/s

d) ω 0 = 1000 rad/s, ω 1 = 825.30 rad/s, ω 2 = 1075.54 rad/s

Answer: c

Explanation: ω 0 = \(\frac{1}{\sqrt{100×10^{-3}×10×10^{-6}}}\) = 1000 rad/s

Now, ω 1 = \(\frac{-R + \sqrt{R^2 + \frac{4L}{C}}}{2L} = \frac{-10 + \frac{\sqrt{100 + 4 × 100 × 10^{-3}}}{10×10^{-6}}}{2×100×10^{-3}}\) = 951.25 rad/s

And, ω 2 = \(\frac{+R+ \sqrt{R^2+ 4L/C}}{2L}\) = 1051.25 rad/s.

4. Given, R = 10 Ω, L = 100 mH and C = 10 μF. Selectivity is?

network-theory-questions-answers-resonance-q3

a) 10

b) 1.2

c) 0.15

d) 0.1

Answer: d

Explanation: Selectivity = \(\frac{1}{Q}\)

Q = \(\frac{ωL}{R} = \frac{1000×100×10^{-3}}{10}\)

∴ Q = 10

So, Selectivity = 0.1.

5. Two series resonant filters are shown below. Let the 3 dB bandwidth of filter 1 be B 1 and that of filter 2 be B 2 . The value of \(\frac{B_1}{B_2}\) is ____________

network-theory-questions-answers-resonance-q5

a) 0.25

b) 1

c) 0.5

d) 0.75

Answer: a

Explanation: For series resonant circuit, 3dB bandwidth is \(\frac{R}{L}\)

B 1 = \(\frac{R}{L_1}\)

B 2 = \(\frac{R}{L_2} = \frac{4R}{L_1}\)

Hence, \(\frac{B_1}{B_2}\) = 0.25.

6. A 440 V, 50 HZ AC source supplies a series LCR circuit with a capacitor and a coil. If the coil has 100 mΩ resistance and 15 mH inductance, then at a resonance frequency of 50 Hz, the half power frequencies of the circuit are?

a) 50.53 Hz, 49.57 Hz

b) 52.12 HZ, 49.8 Hz

c) 55.02 Hz, 48.95 Hz

d) 50 HZ, 49 Hz

Answer: a

Explanation: Bandwidth, BW = \(\frac{f_o}{Q} = \frac{50}{47.115}\) = 1.061 Hz

f 2 , higher half power frequency = f 0 + \(\frac{BW}{2}\)

∴ f 2 = 50 + \(\frac{1.061}{2}\) =50.53 Hz

f 1 , lower half power frequency = f 0 – \(\frac{BW}{2}\)

∴ f 1 = 100 – \(\frac{1.59}{2}\) = 49.47 Hz.

7. At what frequency will the current lead the voltage by 30° in a series circuit with R = 10Ω and C = 25 μF?

a) 11.4 kHz

b) 4.5 kHz

c) 1.1 kHz

d) 24.74 kHz

Answer: a

Explanation: From the impedance diagram, 10 – jX C = Z∠-30°

∴ – X C = 10 tan = -5.77 Ω

∴ X C = 5.77 Ω

Then, X C = \(\frac{1}{2πfC}\) or, f = \(\frac{1}{2πX_C C}\) = 1103 HZ = 1.1 kHz.

8. A 440 V, 50 HZ AC source supplies a series LCR circuit with a capacitor and a coil. If the coil has 100 mΩ resistance and 15 mH inductance, then at a resonance frequency of 50 Hz, what is the value of the capacitor?

a) 676 μF

b) 575 μF

c) 591 μF

d) 610 μF

Answer: a

Explanation: We know that f o = \(\frac{1}{2π\sqrt{LC}}\)

Or, 50 = \(\frac{1}{2π\sqrt{5×10^{-3}×C}}\)

Or, C = \(\frac{1}{4π^2×50^2×15×10^{-3}}\) ≈ 676 μF.

9. A 440 V, 50 HZ AC source supplies a series LCR circuit with a capacitor and a coil. If the coil has 100 mΩ resistance and 15 mH inductance, then at a resonance frequency of 50 Hz, the Q factor of the circuit is?

a) 82.63

b) 47.115

c) 27.38

d) 92.38

Answer: b

Explanation: We know that, Q = \(\frac{ωL}{R}\)

Or, Q = \(\frac{2π×50×15×10^{-3}}{100×10^{-3}}\) = 47.115.

10. For the circuit given below, if the frequency of the source is 50 Hz, then a value of t o which results in a transient free response is?

network-theory-questions-answers-resonance-q10

a) 0

b) 1.78 ms

c) 7.23 ms

d) 9.21 ms

Answer: b

Explanation: For the ideal case, transient response will die out with time constant.

T = \(\frac{L}{R} = \frac{0.01}{5}\) = 0.002 s = 2 ms

Practically, T will be less than 2 ms.

11. In the circuit given below, the magnitudes of V L and V C are twice that of V K . Calculate the inductance of the coil, given that f = 50.50 Hz.

network-theory-questions-answers-resonance-q11

a) 6.41 mH

b) 5.30 mH

c) 3.18 mH

d) 2.31 mH

Answer: c

Explanation: V L = V C = 2 VR

∴ Q = \(\frac{V_L}{V_R}\) = 2

But we know, Q = \(\frac{ωL}{R} = \frac{1}{ωCR}\)

∴ 2 = \(\frac{2πf × L}{5}\)

Or, L = 3.18 mH.

12. A circuit with a resistor, inductor and capacitor in series is resonant with frequency f Hz. If all the component values are now doubled, then the new resonant frequency will be?

a) 2 f

b) f

c) \

\(\frac{f}{2}\)

Answer: d

Explanation: Resonance frequency = \(\frac{1}{2π\sqrt{LC}} = \frac{1}{2π2\sqrt{LC}}\)

Hence, \(\frac{f}{f_o} = \frac{\frac{1}{2π\sqrt{LC}}}{\frac{1}{2π2\sqrt{LC}}}\) = 2

∴ f = 2f o .

13. A series RLC circuit has a resonance frequency of 1 kHz and a quality factor Q = 50. If R and L are doubled and C is kept same, the new Q of the circuit is?

a) 25.52

b) 35.35

c) 45.45

d) 20.02

Answer: b

Explanation: Quality factor Q of the series RLC circuit is given by, Q = \(\frac{1}{R} \sqrt{\frac{L}{C}}\)

Q new = \(\frac{1}{2R} \sqrt{\frac{2L}{C}} = \frac{1}{2} × \frac{1}{R} \sqrt{\frac{2L}{C}} = \frac{1}{2} × \sqrt{2} × Q\) = 35.35.

14. In a series RLC circuit R = 10 kΩ, L = 0.5 H and C = \

4.089 × 10 4 Hz

b) \

4.089π × 10 4 Hz

d) \(\frac{1}{π}\) × 10 4 Hz

Answer: b

Explanation: Resonance frequency = \(\frac{1}{2π\sqrt{LC}}\)

= \(\frac{1}{2π\sqrt{\frac{1}{250} × 10^{-6} × 0.5}} = \frac{11111.1}{π}\)Hz.

15. In a series RLC circuit for lower frequency and for higher frequency, power factors are respectively ______________

a) Leading, Lagging

b) Lagging, Leading

c) Independent of Frequency

d) Same in both cases

Answer: a

Explanation: A Leading power factor means that the current in the circuit leads the applied voltage. This condition occurs in capacitive circuits. On the other hand, a lagging power factor indicates that the current lags the voltage and this condition happens in an inductive circuit.

This set of Network Theory Questions & Answers for Exams focuses on “Problems of Parallel Resonance Involving Quality Factor”.

1. For the parallel resonant circuit shown below, the value of the resonant frequency is _________

network-theory-questions-answers-problems-parallel-resonance-involving-quality-factor-q1

a) 1.25 MHz

b) 2.5 MHz

c) 5 MHz

d) 1.5 MHz

Answer: a

Explanation: Resonant Frequency, F R = \(\frac{1}{2π\sqrt{LC}} \)

= \(\frac{1}{6.28\sqrt{

}}\)

= \(\frac{1}{6.28\sqrt{1.621×10^{-14}}}\)

= \(\frac{1}{799×10^{-9}}\) = 1.25 MHz.

2. For the parallel resonant circuit given below, the value of the inductive and capacitive reactance is _________

network-theory-questions-answers-problems-parallel-resonance-involving-quality-factor-q1

a) X L = 654.289 Ω; X C = 458.216 Ω

b) X L = 985.457 Ω; X C = 875.245 Ω

c) X L = 785.394 Ω; X C = 785.417 Ω

d) X L = 125.354 Ω; X C = 657.215 Ω

Answer: c

Explanation: Resonant Frequency, f = \(\frac{1}{2π\sqrt{LC}} \)

= \(\frac{1}{6.28\sqrt{

}}\)

= \(\frac{1}{6.28\sqrt{1.621×10^{-14}}}\)

= \(\frac{1}{799×10^{-9}}\) = 1.25 MHz.

Inductive Reactance, X L = 2πfL = (1.25×10 6 )(100×10 -6 )

= 785.394 Ω

Capacitive Reactance, X C = \(\frac{1}{2πfC} = \frac{1}{

}\)

= \(\frac{1}{1.273×10^{-3}}\) = 785.417 Ω.

3. For the parallel resonant circuit given below, the current through the capacitor and inductor are _________

network-theory-questions-answers-problems-parallel-resonance-involving-quality-factor-q1

a) I C = 10.892 mA; I L = 12.732 mA

b) I C = 12.732 mA; I L = 10.892 mA

c) I C = 10.892 mA; I L = 10.892 mA

d) I C = 12.732 mA; I L = 12.732 mA

Answer: d

Explanation: Resonant Frequency, f = \(\frac{1}{2π\sqrt{LC}} \)

= \(\frac{1}{6.28\sqrt{

}}\)

= \(\frac{1}{6.28\sqrt{1.621×10^{-14}}}\)

= \(\frac{1}{799×10^{-9}}\) = 1.25 MHz.

Inductive Reactance, X L = 2πfL = (1.25×10 6 )(100×10 -6 )

= 785.394 Ω

Capacitive Reactance, X C = \(\frac{1}{2πfC} = \frac{1}{

}\)

= \(\frac{1}{1.273×10^{-3}}\) = 785.417 Ω.

I C = \(\frac{V_A}{X_C} \)

= \(\frac{10}{785.417} \) = 12.732 mA

I L = \(\frac{V_A}{X_L} \)

= \(\frac{10}{785.394} \) = 12.732 mA.

4. For the parallel resonant circuit given below, the value of the equivalent impedance of the circuit is ________

network-theory-questions-answers-problems-parallel-resonance-involving-quality-factor-q1

a) 56.48 kΩ

b) 78.58 kΩ

c) 89.12 kΩ

d) 26.35 kΩ

Answer: b

Explanation: f = \(\frac{1}{2π\sqrt{LC}} \)

= \(\frac{1}{6.28\sqrt{

}}\)

= \(\frac{1}{6.28\sqrt{1.621×10^{-14}}}\)

= \(\frac{1}{799×10^{-9}}\) = 1.25 MHz.

Inductive Reactance, X L = 2πfL = (1.25×10 6 )(100×10 -6 )

= 785.394 Ω

Q = \(\frac{X_L}{R} = \frac{785.394}{7.85}\) = 100.05

∴ Z EQ = QX L = = 78.58 kΩ.

5. For the parallel resonant circuit, the bandwidth of the circuit is ____________

network-theory-questions-answers-problems-parallel-resonance-involving-quality-factor-q1

a) ±6.25 kHz

b) ±8.56 kHz

c) ±10.35 kHz

d) ±6.37 kHz

Answer: a

Explanation: Resonant Frequency, f = \(\frac{1}{2π\sqrt{LC}} \)

= \(\frac{1}{6.28\sqrt{

}}\)

= \(\frac{1}{6.28\sqrt{1.621×10^{-14}}}\)

= \(\frac{1}{799×10^{-9}}\) = 1.25 MHz.

Inductive Reactance, X L = 2πfL = (1.25×10 6 )(100×10 -6 )

= 785.394 Ω

Q = \(\frac{X_L}{R} = \frac{785.394}{7.85}\) = 100.05

∴ ∆F = \(\frac{f}{Q} = \frac{1.25 × 10^6}{100.05}\) = 12.5 kHz

Hence, bandwidth = \(\frac{∆F}{2}\) = 6.25 kHz.

6. For the parallel resonant circuit given below, the cut-off frequencies are ____________

network-theory-questions-answers-problems-parallel-resonance-involving-quality-factor-q1

a) ∆f 1 = 2.389 MHz; ∆f 2 = 441.124 MHz

b) ∆f 1 = 1.256 MHz; ∆f 2 = 1.244 MHz

c) ∆f 1 = 5.658 MHz; ∆f 2 = 6.282 MHz

d) ∆f 1 = 3.656 MHz; ∆f 2 = 8.596 MHz

Answer: b

Explanation: Resonant Frequency, f = \(\frac{1}{2π\sqrt{LC}} \)

= \(\frac{1}{6.28\sqrt{

}}\)

= \(\frac{1}{6.28\sqrt{1.621×10^{-14}}}\)

= \(\frac{1}{799×10^{-9}}\) = 1.25 MHz.

Inductive Reactance, X L = 2πfL = (1.25×10 6 )(100×10 -6 )

= 785.394 Ω

Q = \(\frac{X_L}{R} = \frac{785.394}{7.85}\) = 100.05

∴ ∆F = \(\frac{f}{Q} = \frac{1.25 × 10^6}{100.05}\) = 12.5 kHz

Hence, bandwidth = \(\frac{∆F}{2}\) = 6.25 kHz

∴ ∆f 1 = f + \(\frac{∆F}{2}\) = 1.25 MHz + 6.25 kHz = 1.256 MHz

∴ ∆f 1 = f – \(\frac{∆F}{2}\) = 1.25 MHz – 6.25 kHz = 1.244 MHz.

7. For the series resonant circuit shown below, the value of the resonant frequency is _________

network-theory-questions-answers-problems-parallel-resonance-involving-quality-factor-q7

a) 10.262 kHz

b) 44.631 kHz

c) 50.288 kHz

d) 73.412 kHz

Answer: d

Explanation: Resonant Frequency, F R = \(\frac{1}{2π\sqrt{LC}} \)

= \(\frac{1}{6.28\sqrt{

}}\)

= \(\frac{1}{6.28\sqrt{4.7×10^{-12}}}\)

= \(\frac{1}{1.362×10^{-5}}\) = 73.412 kHz.

8. For the series resonant circuit given below, the value of the inductive and capacitive reactance is _________

network-theory-questions-answers-problems-parallel-resonance-involving-quality-factor-q7

a) X L = 5.826 kΩ; X C = 5.826 kΩ

b) X L = 2.168 kΩ; X C = 2.168 kΩ

c) X L = 6.282 kΩ; X C = 6.282 kΩ

d) X L = 10.682 kΩ; X C = 10.682 kΩ

Answer: c

Explanation: Resonant Frequency, \(\frac{1}{2π\sqrt{LC}} \)

= \(\frac{1}{6.28\sqrt{

}}\)

= \(\frac{1}{6.28\sqrt{4.7×10^{-12}}}\)

= \(\frac{1}{1.362×10^{-5}}\) = 73.412 kHz

Inductive Reactance, X L = 2πfL = (73.142 × 10 3 )(4.7 × 10 -6 )

= 2.168 kΩ

Capacitive Reactance, X C = \(\frac{1}{2πfC} = \frac{1}{

}\)

= \(\frac{1}{4.613×10^{-4}}\) = 2.168 kΩ.

9. For the series resonant circuit given below, the value of the equivalent impedance of the circuit is ________

network-theory-questions-answers-problems-parallel-resonance-involving-quality-factor-q7

a) 55 Ω

b) 47 Ω

c) 64 Ω

d) 10 Ω

Answer: b

Explanation: Resonant Frequency, \(\frac{1}{2π\sqrt{LC}} \)

= \(\frac{1}{6.28\sqrt{

}}\)

= \(\frac{1}{6.28\sqrt{4.7×10^{-12}}}\)

= \(\frac{1}{1.362×10^{-5}}\) = 73.412 kHz

Inductive Reactance, X L = 2πfL = (73.142 × 10 3 )(4.7 × 10 -6 )

= 2.168 kΩ

Capacitive Reactance, X C = \(\frac{1}{2πfC} = \frac{1}{

}\)

= \(\frac{1}{4.613×10^{-4}}\) = 2.168 kΩ

We see that, X C = X L are equal, along with being 180° out of phase.

Hence the net reactance is zero and the total impedance equal to the resistor.

∴ Z EQ = R = 47 Ω.

10. For the series resonant circuit given below, the value of the total current flowing through the circuit is ____________

network-theory-questions-answers-problems-parallel-resonance-involving-quality-factor-q7

a) 7.521 mA

b) 6.327 mA

c) 2.168 mA

d) 9.136 mA

Answer: a

Explanation: Resonant Frequency, \(\frac{1}{2π\sqrt{LC}} \)

= \(\frac{1}{6.28\sqrt{

}}\)

= \(\frac{1}{6.28\sqrt{4.7×10^{-12}}}\)

= \(\frac{1}{1.362×10^{-5}}\) = 73.412 kHz

Inductive Reactance, X L = 2πfL = (73.142 × 10 3 )(4.7 × 10 -6 )

= 2.168 kΩ

Capacitive Reactance, X C = \(\frac{1}{2πfC} = \frac{1}{

}\)

= \(\frac{1}{4.613×10^{-4}}\) = 2.168 kΩ

Z EQ = R = 47 Ω

I T = \(\frac{V_{in}}{Z_{EQ}} = \frac{V_{in}}{R} = \frac{0.3535}{47}\) = 7.521 mA.

11. For the series circuit given below, the value of the voltage across the capacitor and inductor are _____________

network-theory-questions-answers-problems-parallel-resonance-involving-quality-factor-q7

a) V C = 16.306 V; V L = 16.306 V

b) V C = 11.268 V; V L = 11.268 V

c) V C = 16.306 V; V L = 16.306 V

d) V C = 14.441 V; V L = 14.441 V

Answer: c

Explanation: Resonant Frequency, \(\frac{1}{2π\sqrt{LC}} \)

= \(\frac{1}{6.28\sqrt{

}}\)

= \(\frac{1}{6.28\sqrt{4.7×10^{-12}}}\)

= \(\frac{1}{1.362×10^{-5}}\) = 73.412 kHz

Inductive Reactance, X L = 2πfL = (73.142 × 10 3 )(4.7 × 10 -6 )

= 2.168 kΩ

Capacitive Reactance, X C = \(\frac{1}{2πfC} = \frac{1}{

}\)

= \(\frac{1}{4.613×10^{-4}}\) = 2.168 kΩ

Z EQ = R = 47 Ω

I T = \(\frac{V_{in}}{Z_{EQ}} = \frac{V_{in}}{R} = \frac{0.3535}{47}\) = 7.521 mA

∴ Voltage across the capacitor, V C = X C I T = = 16.306 V

∴ Voltage across the inductor, V L = X L I T = = 16.306 V.

12. For the series resonant circuit given below, the value of the quality factor is ___________

network-theory-questions-answers-problems-parallel-resonance-involving-quality-factor-q7

a) 35.156

b) 56.118

c) 50.294

d) 46.128

Answer: d

Explanation: Resonant Frequency, \(\frac{1}{2π\sqrt{LC}} \)

= \(\frac{1}{6.28\sqrt{

}}\)

= \(\frac{1}{6.28\sqrt{4.7×10^{-12}}}\)

= \(\frac{1}{1.362×10^{-5}}\) = 73.412 kHz

Inductive Reactance, X L = 2πfL = (73.142 × 10 3 )(4.7 × 10 -6 )

= 2.168 kΩ

Quality factor Q = \(\frac{X_L}{R} = \frac{2.168 kΩ}{47}\) = 46.128.

13. For a parallel RLC circuit, the incorrect statement among the following is _____________

a) The bandwidth of the circuit decreases if R is increased

b) The bandwidth of the circuit remains same if L is increased

c) At resonance, input impedance is a real quantity

d) At resonance, the magnitude of input impedance attains its minimum value

Answer: d

Explanation: BW = 1/RC

It is clear that the bandwidth of a parallel RLC circuit is independent of L and decreases if R is increased.

At resonance, imaginary part of input impedance is zero. Hence, at resonance input impedance is a real quantity.

In parallel RLC circuit, the admittance is minimum, at resonance. Hence the magnitude of input impedance attains its maximum value at resonance.

14. A circuit excited by voltage V has a resistance R which is in series with an inductor and capacitor, which are connected in parallel. The voltage across the resistor at the resonant frequency is ___________

a) 0

b) \

Answer: a

Explanation: Dynamic resistance of the tank circuit, Z DY = L/(R L C)

But given that R L = 0

So, Z DY = L/(0X C ) = ∞

Therefore current through circuit, I = \(\frac{V}{∞}\) = 0

∴ V D = 0.

15. For the circuit given below, the nature of the circuit is ____________

network-theory-questions-answers-problems-parallel-resonance-involving-quality-factor-q7

a) Inductive

b) Capacitive

c) Resistive

d) Both inductive as well as capacitive

Answer: c

Explanation: θ = 0° since X L and X C are cancelling, which means at resonance the circuit is purely resistive.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Advanced Problems on Resonance”.

1. For a series RLC circuit, excited by voltage V = 20 sinωt. The values of R = 2 Ω, L = 1mH and C = 0.4 µF. The resonant frequency of the circuit is ______________

a) 20 krad/s

b) 30 krad/s

c) 40 krad/s

d) 50 krad/s

Answer: d

Explanation: We know that the resonant frequency w 0 = 1/ 0.5

Given R = 2 Ω, L = 1mH and C = 0.4 µF.

So, w 0 = \(\frac{1}{\sqrt{10^{-3}×0.4×10^{-6}}}\)

Or, w 0 = 50 krad/s.

2. For a series RLC circuit, excited by voltage V = 20 sinωt. The values of R = 2 Ω, L = 1mH and C = 0.4 µF. The quality factor and bandwidth of the circuit is ______________

a) 25, 2

b) 30, 5

c) 40, 10

d) 50, 2

Answer: a

Explanation: We know that, the bandwidth is, B = w 2 – w 1 = 2 krad/s

Also, B = \(\frac{R}{L} = \frac{2}{10^{-3}}\) = 2 krad/s

Also, the quality factor is given by, Q = \(\frac{w_0}{B}\)

Or, Q = \(\frac{w_0}{B} = \frac{50}{2}\) = 25.

3. An ideal transformer is rated as 2400/120 V, 9.6 kVA and has 50 turns on the secondary side. The turns ratio is ____________

a) 0.04

b) 0.05

c) 0.06

d) 0.07

Answer: b

Explanation: The given transformer is a step down transformer.

So, V 1 = 2400 V

But it is greater than V 2 which is equal to 120V

So, the turns ratio, n = \(\frac{V_2}{V_1} = \frac{120}{2400}\) = 0.05.

4. In the circuit shown in the figure, if ω = 40 rad/s, then Z in is _____________

network-theory-questions-answers-advanced-problems-resonance-q4

a) 4.77 – j1.15 Ω

b) 2.96 – j0.807 Ω

c) 2.26 – j3.48 Ω

d) 2.26 – j3.48 Ω

Answer: b

Explanation: Z = Z 11 + \

+ \frac{40^2

^2}{4+j\frac{1}{2}}\)

= 2 + j + \

+ \

+ \(\frac{400-j }{416}\)

= 2.96 – j0.807 Ω.

5. For the circuit shown in the figure below, the value of R for critical damping will be?

network-theory-questions-answers-advanced-problems-resonance-q5

a) 10.5 Ω

b) 8.57 Ω

c) 3.5 Ω

d) 3 Ω

Answer: b

Explanation: For a parallel RLC circuit, \(\frac{d^2 V}{dt^2} + \frac{1}{RC} \frac{dV}{dt} + \frac{V}{LC}\) = 0

Solving the 2nd order differential equation, we get, 2rω n = \(\frac{1}{RC}\) and ω n = \(\frac{1}{\sqrt{LC}}\)

As, r=1,

∴ R = \(\frac{1}{2} \sqrt{\frac{L}{C}} \)

= \(\frac{1}{2} \sqrt{\frac{7}{\frac{1}{42}}} = \frac{1}{2}\) × 17.15 = 8.57 Ω.

6. The circuit shown in the figure below will act as an ideal source with respect to terminal A and B when the frequency is ___________

network-theory-questions-answers-advanced-problems-resonance-q6

a) Zero

b) 1 rad/s

c) 4 rad/s

d) 16 rad/s

Answer: c

Explanation: Z t = \

= \

= ∞

Or, 1 – LCω 2 = 0

Or, ω = \(\frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{\frac{1}{16}×1}}\) = 4 rad/s.

7. An ideal capacitor is charged to a voltage V and connected at t=0 across an ideal inductor L. If we take ω = \

b) V cos

c) V sin

d) Ve -ωt cos

Answer: b

Explanation: Voltage across the capacitor will discharge through inductor unless the voltage across the capacitor becomes zero. Now, the inductor will start charging the capacitor. Voltage across the capacitor will be decreasing from V and will be periodic. But it will not be decaying with time, since both L and C are ideal.

∴ Voltage across the capacitor at time t>0 is given by V cos .

8. The RL circuit in the below figure is fed from a constant magnitude, variable frequency sinusoidal voltage source V IN . At 50 Hz, the R and L elements each have a voltage drop V rms . If the frequency of the source is changed to 25 Hz, the new voltage drop across R is ___________

network-theory-questions-answers-advanced-problems-resonance-q8

a) \(\sqrt{\frac{5}{8}}\) V rms

b) \(\sqrt{\frac{2}{3}}\) V rms

c) \(\sqrt{\frac{8}{5}}\) V rms

d) \(\sqrt{2}\) V rms

Answer: c

Explanation: At 50 Hz, V R = V L and R = ω L

Also V rms = \(\frac{V_{in}}{\sqrt{2}} \)

At 25 Hz, I in = \(\frac{V_{in}}{\sqrt{R^2+\frac{ω^2 L^2}{4}}}\)

= \(\frac{V_{in}}{\sqrt{ω^2 L^2 + \frac{ω^2 L^2}{4}}} = \frac{2V_{in}}{\sqrt{5} ωL}\)

So, V new = \(\frac{2V_{in}}{\sqrt{5} ωL} × R = \frac{2V_{in}}{\sqrt{5}}\)

= \(\frac{2\sqrt{2} V_{rms}}{\sqrt{5}} = \sqrt{\frac{8}{5}}\) V rms .

9. In a series circuit with R = 10Ω and C = 25 μF, at what frequency will the current lead the voltage by 30°?

a) 11.4 kHz

b) 4.5 kHz

c) 1.1 kHz

d) 24.74 kHz

Answer: a

Explanation: From the impedance diagram, 10 -jX C = Z∠-30°

∴ – X C = 10 tan = -5.77 Ω

∴ X C = 5.77 Ω

Then, X C = \(\frac{1}{2πfC}\) or, f = \(\frac{1}{2πX_C C} \)

= \(\frac{1}{2π×5.77×25×10^{-6}}\)

= \(\frac{10^6}{906.18}\)

= 1103 HZ = 1.1 kHz.

10. The frequency for the given circuit when the circuit is in a state of resonance will be?

network-theory-questions-answers-advanced-problems-resonance-q10

a) 1 rad/s

b) 2 rad/s

c) 3 rad/s

d) 4 rad/s

Answer: c

Explanation: Z = jωL + \

At resonance, imaginary part must be zero.

0.1 ω – \(\frac{ω}{ 1+ ω^2}\) = 0

Or, 0.1 = \(\frac{1}{1+ω^2}\)

Or, ω 2 + 1 = 10

∴ ω 2 = 9

Or, ω = 3 rad/s.

11. The time constant for the given circuit under resonance condition will be?

network-theory-questions-answers-advanced-problems-resonance-q11

a) \

4 s

d) 9 s

Answer: c

Explanation: For finding the time constant, we neglect current source as an open circuit.

Hence, the circuit becomes:

network-theory-questions-answers-advanced-problems-resonance-q11a

So, from the above circuit, the Time constant = R eq C eq = 6 × \(\frac{2}{3}\) = 4 s.

12. A 50 μF capacitor, when connected in series with a coil having a resistance of 40Ω, resonates at 1000 Hz. The circuit is in resonating condition. The voltage across the capacitor is __________

a) 8.04 V

b) 7.96 V

c) 5.68 V

d) 1.25 V

Answer: c

Explanation: At resonance, | I | = \(\frac{V}{Z} = \frac{V}{R} \)

Given that, R = 40Ω and V = 100 V

∴ | I | = \(\frac{100}{40}\) = 2.5 A

Power loss of the coil = I 2 R = 2.5 2 × 40 = 250 W

Also, V C = IX C = 2.5 × \(\frac{1}{2πfC}\) and X C = \(\frac{1}{2πfC}\)

= 2.5 × \(\frac{1}{2π×1000×50×10^{-6}}\)

= 2.5 × \(\frac{100}{31.41}\) = 5.68 V.

13. An ideal transformer is rated as 2400/120 V, 9.6 kVA and has 50 turns on the secondary side. The number of turns on the primary side is ____________

a) 1000

b) 2000

c) 3000

d) 4000

Answer: a

Explanation: Turns ratio, n = \(\frac{N_2}{N_1}\)

Now, the turns ratio, n = \(\frac{V_2}{V_1} = \frac{120}{2400}\) = 0.05

Also, n = \(\frac{50}{N_1}\)

So, N 1 = \(\frac{50}{0.05}\) = 1000 turns.

14. A 50 μF capacitor, when connected in series with a coil having resistance of 40Ω, resonates at 1000 Hz. The circuit current if the applied voltage is 100 V for the resonant circuit is ___________

a) 2.5 A

b) 3.5 A

c) 4.5 A

d) 0.5 A

Answer: a

Explanation: We know that the magnitude of current |I| for a resonating circuit is given by,

| I | = \(\frac{V}{Z} = \frac{V}{R} \)

Given that, R = 40Ω and V = 100 V

∴ | I | = \(\frac{100}{40}\) = 2.5 A.

15. A 50 μF capacitor, when connected in series with a coil having resistance of 40Ω, resonates at 1000 Hz. The circuit is in resonating condition. The voltage across the coil is __________

a) 100.31 V

b) 200.31 V

c) 300.31 V

d) 400.31 V

Answer: a

Explanation: At resonance, | I | = \(\frac{V}{Z} = \frac{V}{R} \)

Given that, R = 40Ω and V = 100 V

∴ | I | = \(\frac{100}{40}\) = 2.5 A

X L = ωL = 2π × 1000 × 0.5 × 10 -3

= 2π × 0.5 = 3.14 Ω

∴ V COIL = I Z

= 2.5 \(\sqrt{40^2 + 3.14^2}\)

= 2.5 × 40.122 = 100.31 V.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Dot Convention in Magnetically Coupled Circuits”.

1. A series RLC circuit has a resonance frequency of 1 kHz and a quality factor Q = 100. If each of R, L and C is doubled from its original value, the new Q of the circuit is _____________

a) 25

b) 50

c) 100

d) 200

Answer: b

Explanation: Q = \(\frac{f_0}{BW}\)

And f 0 = 1/2π 0.5

BW = R/L

Or, Q = \

^{0.5}\)

When R, L and C are doubled, Q’ = 50.

2. In the circuit given below, the input impedance Z IN of the circuit is _________

network-theory-questions-answers-dot-convention-magnetically-coupled-circuits-q2

a) 0.52 – j4.30 Ω

b) 0.52 + j15.70 Ω

c) 64.73 + j17.77 Ω

d) 0.3 – j33.66 Ω

Answer: c

Explanation: Z IN = || (Z A )

Z A = j10 + \(\frac{12^2}{}\)

= 0.49 + j5.82

Z IN = \(\frac{}{}\)

= 64.73 + j17.77 Ω.

3. The switch in the circuit shown was on position X for a long time. The switch is then moved to position Y at time t=0. The current I for t>0 is ________

network-theory-questions-answers-dot-convention-magnetically-coupled-circuits-q3

a) 0.2e -125t u mA

b) 20e -1250t u mA

c) 0.2e -1250t u mA

d) 20e -1000t u mA

Answer: c

Explanation: C EQ = \(\frac{0.8 × 0.2}{0.8+0.2}\) = 0.16

V C (t=0 – ) = 100 V

At t≥0,

The discharging current I = \(\frac{V_O}{R} e^{-\frac{t}{RC}}\)

= \(\frac{100}{5000} e^{- \frac{t}{5×10^3×0.16×10^{-6}}}\)

= 0.2e -1250t u mA.

4. In the circuit shown, the voltage source supplies power which is _____________

network-theory-questions-answers-dot-convention-magnetically-coupled-circuits-q4

a) Zero

b) 5 W

c) 10 W

d) 100 W

Answer: a

Explanation: Let the current supplied by a voltage source.

Applying KVL in outer loop,

10 – × – × 2 = 0

10 – 2 – 2 = 0

Or, I = 0

∴ Power VI = 0.

5. In the circuit shown below the current I for t≥0 + is ___________

network-theory-questions-answers-dot-convention-magnetically-coupled-circuits-q5

a) 0.5-0.125e -1000t A

b) 1.5-0.125e -1000t A

c) 0.5-0.5e -1000t A

d) 0.375e -1000t A

Answer: a

Explanation: I = \(\frac{1.5}{3}\) = 0.5

L EQ = 15 mH

R EQ = 5+10 = 15Ω

L = \

A – e -t = 0.5 – e -1000t

= 0.5 e -1000t

= 0.5 – 0.125 e -1000t

I = 0.5-0.125e -1000t .

6. Initial voltage on capacitor VO as marked |V O | = 5 V, V S = 8 u , where u is the unit step. The voltage marked V at t=0 + is _____________

network-theory-questions-answers-dot-convention-magnetically-coupled-circuits-q6

a) 1 V

b) -1 V

c) \

–\(\frac{13}{3}\) V

Answer: c

Explanation: Applying voltage divider method, we get,

I = \(\frac{V}{R_{EQ}} \)

= \(\frac{8}{1+1||1} \)

= \(\frac{8}{1+\frac{1}{2}} = \frac{16}{3}\) A

I 1 = \(\frac{16}{3} × \frac{1}{2} = \frac{8}{3}\) A

And \(I’_2 = \frac{V}{R_{EQ}} = \frac{5}{1+1||1}\)

= \(\frac{5}{1+\frac{1}{2}} = \frac{10}{3}\) A

Now, \(I’_1 = I’_2 × \frac{1}{1+1} \)

= \(\frac{10}{3} × \frac{1}{2} = \frac{5}{3}\) A

Hence, the net current in 1Ω resistance = I1 + \(I’_1\)

= \(\frac{8}{3} + \frac{5}{3} = \frac{13}{3}\) A

∴ Voltage drop across 1Ω = \(\frac{13}{3} × 1 = \frac{13}{3}\) V.

7. For a unit step signal u , the response is V 1 = (1-e -3t ) for t>0. If a signal 3u + δ is applied, the response will be ?

a) (3-6e -3t )u

b) (3-3e -3t )u

c) 3u

d) (3+3e -3t )u

Answer: c

Explanation: For u = 1, t>0

V 1 = (1-e -3t )

Or, V 1 = \Missing or unrecognized delimiter for \right = \frac{3}{s}\)

And T = \

Response, H = R T = \

= \frac{3}{s}\)

Or, h = 3 u .

8. In the circuit given below, for time t<0, S 1 remained closed and S 2 open., S 1 is initially opened and S 2 is initially closed. If the voltage V 2 across the capacitor C 2 at t=0 is zero, the voltage across the capacitor combination at t=0 + will be ____________

network-theory-questions-answers-dot-convention-magnetically-coupled-circuits-q8

a) 1 V

b) 2 V

c) 1.5 V

d) 3 V

Answer: d

Explanation: When S 1 is closed and S 2 is open,

V C1 (0 – ) = V C1 (0 + ) = 3V

When S 1 is opened and S 2 is closed, V C2 (0 + ) = V C2 (0 + ) = 3V.

9. In the circuit given below, the switch S 1 is initially closed and S 2 is opened. The inductor L carries a current of 10A and the capacitor is charged to 10 V with polarities as shown. The current through C during t=0 + is _____________

network-theory-questions-answers-dot-convention-magnetically-coupled-circuits-q9

a) 55 A

b) 5.5 A

c) 45 A

d) 4.5 A

Answer: d

Explanation: By KCL, we get,

\(\frac{V_L}{10} – 10 + \frac{V_L-10}{10}\) = 0

Hence, 2 V L = 110

∴ V L = 55 V

Or, I C = \(\frac{55-10}{10}\) = 4.5 A.

10. In the circuit given below, the switch S 1 is initially closed and S 2 is opened. The inductor L carries a current of 10A and the capacitor is charged to 10 V with polarities as shown. The current through L during t=0 + is _____________

network-theory-questions-answers-dot-convention-magnetically-coupled-circuits-q10

a) 55 A

b) 5.5 A

c) 45 A

d) 4.5 A

Answer: a

Explanation: By KCL, we get,

\(\frac{V_L}{10} – 10 + \frac{V_L-10}{10}\) = 0

Hence, 2 V L = 110

∴ V L = 55 V.

11. An ideal capacitor is charged to a voltage V O and connected at t=0 across an ideal inductor L. If ω = \Missing open brace for subscript V O

b) V O cos

c) V O sin

d) V O e -ωt cos

Answer: b

Explanation: Voltage across capacitor will discharge through inductor up to voltage across the capacitor becomes zero. Now, inductor will start charging capacitor.

Voltage across capacitor will be decreasing from V O and periodic and is not decaying since both L and C is ideal.

∴ Voltage across the capacitor at time t>0 is V O cos.

12. In the figure given below, what is the RMS value of the periodic waveform?

network-theory-questions-answers-dot-convention-magnetically-coupled-circuits-q12

a) 2\ 6\ \

1.5 A

Answer: a

Explanation: The rms value for any waveform is = \

^2 dt + \int_{\frac{T}{2}}^T 6^2 dt]^{1/2}\)

= \

]^{\frac{1}{2}}\)

= \([6+18]^{\frac{1}{2}} = \sqrt{24} = 2\sqrt{6}\) A.

13. In the circuit given below, the capacitor initially has a charge of 10 C. The current in the circuit at t=1 sec after the switch S is closed will be ___________

network-theory-questions-answers-dot-convention-magnetically-coupled-circuits-q13

a) 14.7 A

b) 18.5 A

c) 40 A

d) 50 A

Answer: a

Explanation: Using KVL, 100 = R\(\frac{dq}{dt} + \frac{q}{C} \)

Or, 100 C = RC\(\frac{dq}{dt}\) + q

Now, \(\int_{q_o}^q \frac{dq}{100C-q} = \frac{1}{RC} ∫_0^t dt\)

Or, 100C – q = (100C – q o ) e -t/RC

I = \(\frac{dq}{dt} = \frac{

}{RC} e^{-1/1}\)

= 40e -1 = 14.7 A.

14. In the circuit given below, the switch is closed at time t=0. The voltage across the inductance just at t=0 + is ____________

network-theory-questions-answers-dot-convention-magnetically-coupled-circuits-q14

a) 2 V

b) 4 V

c) -6 V

d) 8 V

Answer: b

Explanation: A t=0 + ,

I (0 + ) = \(\frac{10}{4||4+3} = \frac{10}{5}\) = 2A

∴ I 2(0 + ) = \(\frac{2}{2}\) = 1A

V L(0 + ) = 1 × 4 = 4 V.

15. A rectangular voltage wave of magnitude A and duration B is applied to a series combination of resistance R and capacitance C. The voltage developed across the capacitor is ____________

a) A[1 – exp

]

b) \

d) A exp

Answer: a

Explanation: V C = \(\frac{1}{C} ∫Idt\)

= \(\frac{1}{C} ∫_0^B \frac{A}{R} e^{-\frac{t}{RC}}\) dt

V C = A[1 – exp

]

Hence, maximum voltage = V [1 – exp

This set of Network Theory Multiple Choice Questions & Answers focuses on “Problems Involving Dot Conventions”.

1. The current through an electrical conductor is 1A when the temperature of the conductor 0°C and 0.7 A when the temperature is 100°C. The current when the temperature of the conductor is 1200°C is ___________

a) 0.08 A

b) 0.16 A

c) 0.32 A

d) 0.64 A

Answer: b

Explanation: \(\frac{1}{0.7} = \frac{R_O }{R_O}\)

= 1 + α100

∴α = 0.0043 per °C

Current at 1200 °C is given by, \(\frac{1}{I} = \frac{R_O }{R_O}\)

= 1 + α1200

= 1 + 0.0043 × 1200 = 6.16

∴ I = \(\frac{1}{6.16}\) = 0.16 A.

2. In the circuit given below, the equivalent capacitance is ____________

network-theory-questions-answers-problems-involving-dot-conventions-q2

a) 1.625 F

b) 1.583 F

c) 0.583 F

d) 0.615 F

Answer: d

Explanation: C CB = \Missing or unrecognized delimiter for \right\) + C 5 = 1.5 F

Now, C AB =\Missing or unrecognized delimiter for \right\) + C 6 = 1.6 F

C XY = \(\frac{C_{AB} × C_4}{C_{AB} + C_4}\) = 0.615 F.

3. In the circuit given below, the equivalent capacitance is ______________

network-theory-questions-answers-problems-involving-dot-conventions-q3

a) 3.5 μF

b) 1.2 μF

c) 2.4 μF

d) 4.05 μF

Answer: d

Explanation: The 2.5 μF capacitor is in parallel with 1 μF capacitor and this combination is in series with 1.5 μF.

Hence, C 1 = \(\frac{1.5}{1.5+2.5+1}\)

= \(\frac{5.25}{5}\) = 1.05

Now, C 1 is in parallel with the 3 μF capacitor.

∴ C EQ = 1.05 + 3 = 4.05 μF.

4. In the circuit given below, the voltage across A and B is?

network-theory-questions-answers-problems-involving-dot-conventions-q4

a) 13.04 V

b) 17.84 V

c) 12 V

d) 10.96 V

Answer: b

Explanation: Loop current I 1 = \(\frac{6}{10}\) = 0.6 A

I 2 = \(\frac{12}{14}\) = 0.86 A

V AB = + 12 +

= 2.4 + 12 + 3.44

= 17.84 V.

5. In the figure given below, the voltage source provides the circuit with a voltage V. The number of non-planar graph of independent loop equations is ______________

network-theory-questions-answers-problems-involving-dot-conventions-q5

a) 8

b) 12

c) 7

d) 5

Answer: d

Explanation: The total number of independent loop equations are given by L = B – N + 1 where,

L = number of loop equations

B = number of branches = 12

N = number of nodes = 8

∴ L = 12 – 8 + 1 = 5.

6. When a DC voltage is applied to an inductor, the current through it is found to build up in accordance with I = 20(1-e -50t ). After the lapse of 0.02 s, the voltage is equal to 2 V. What is the value of inductance?

a) 2 mH

b) 5.43 mH

c) 1.54 mH

d) 0.74 mH

Answer: b

Explanation: V L = L\(\frac{dI}{dt}\)

Where, I = 20(1-e -50t )

Therefore, V L = L\(\frac{d 20

}{dt}\)

= L × 20 × 50e -50t

At t = 0.02 s, V L = 2 V

∴ L = \(\frac{2}{20 × 50 × e^{-50×0.02}}\)

= 5.43 μH.

7. An air capacitor of capacitance 0.005 μF is connected to a direct voltage of 500 V. It is disconnected and then immersed in oil with a relative permittivity of 2.5. The energy after immersion is?

a) 275 μJ

b) 250 μJ

c) 225 μJ

d) 625 μJ

Answer: b

Explanation: E = \(\frac{1}{2}\) CV 2

Or, C = \

= 2.5 × 0.005 × 10 -6

∴ C NEW = 12.5 × 10 -9 F

Now, q = CV = 0.005 × 10 -6 × 500 = 2.5 × 10 -6

V NEW = \(\frac{q} {C_{NEW}}\)

= \(\frac{2.5 × 10^{-6}}{12.5 × 10^{-9}}\) V NEW = 200

E = \(\frac{1}{2}\) CV 2

= \(\frac{1}{2}\) × 12.5 × 10 -9 × 2

= 250 μJ.

8. The resistance of copper motor winding at t=20°C is 3.42 Ω. After extended operation at full load, the motor windings measures 4.22 Ω. If the temperature coefficient is 0.0426, what is the rise in temperature?

a) 60°C

b) 45.2°C

c) 72.9°C

d) 10.16°C

Answer: d

Explanation: Given that, R 1 = 3.42 Ω

T 1 = 20° and α = 0.0426

R 2 = 4.22 Ω

Now, \(\frac{R_1}{1 + αT_1} = \frac{R_2}{1 + αT_2}\)

Or, \(\frac{3.42}{1 + 0.0426 × 20} = \frac{4.22}{1 + 0.0426 T_2}\)

∴ Rise in temperature = T 2 – T 1

= 30.16 – 20

= 10.16°C.

9. A capacitor of capacitance 50 μF is connected in parallel to another capacitor of capacitance 100 μF. They are connected across a time-varying voltage source. At a particular time, the current supplied by the source is 5 A. The magnitude of instantaneous current through the capacitor of capacitance 50 μF is?

a) 1.57 A

b) 1.87 A

c) 1.67 A

d) 2.83 A

Answer: c

Explanation: As the capacitors are in parallel, then the voltage V is given by,

V = \(\frac{1}{C_1} \int I_1 \,dt \)

∴ I 1 = C 1 \

Also, I 1 + I 2 = 5 A …………………………..

Solving and , we get, I 1 = 1.67 A.

10. A capacitor of capacitance 50 μF is connected in parallel to another capacitor of capacitance 100 μF. They are connected across a time-varying voltage source. At a particular time, the current supplied by the source is 5 A. The magnitude of instantaneous current through the capacitor of capacitance 100 μF is?

a) 2.33 A

b) 3.33 A

c) 1.33 A

d) 4.33 A

Answer: b

Explanation: As the capacitors are in parallel, then the voltage V is given by,

V = \(\frac{1}{C_2} \int I_2 \,dt \)

∴ I 2 = C 2 \

Also, I 1 + I 2 = 5 A …………………………..

Solving and , we get, I 2 = 3.33 A.

11. In the circuit given below, the resonant frequency is _______________

network-theory-questions-answers-problems-involving-dot-conventions-q11

a) \

\(\frac{1}{\sqrt{2}2π}\) Hz

Answer: c

Explanation: I EQ = L 1 + L 2 + 2M

L EQ = 1 + 2 + 2 × \(\frac{1}{2}\) = 4 H

∴ F O = \(\frac{1}{2π\sqrt{LC}} \)

= \(\frac{1}{2π\sqrt{4 × 1}} \)

= \(\frac{1}{4π}\) Hz.

12. In a series resonant circuit, V C = 300 V, V L = 300 V and V R = 100 V. What is the value of the source voltage?

a) Zero

b) 100 V

c) 350 V

d) 200 V

Answer: b

Explanation: As V C and V L are equal, then X C is equal to X L and both the voltages are then cancelled out.

That is V S = V R

∴ V S = 100 V.

13. For the circuit given below, what is the value of the Q factor for the inductor?

network-theory-questions-answers-problems-involving-dot-conventions-q13

a) 4.74

b) 4.472

c) 4.358

d) 4.853

Answer: c

Explanation: Q IN = \(\sqrt{\frac{L}{CR^2} – 1}\)

= \(\sqrt{\frac{1}{2 × 5^2 × 10^{-3} – 1}}\)

= \(\sqrt{\frac{1}{50 × 10^{-3} – 1}}\)

= \(\sqrt{19}\) = 4.358.

14. In the circuit given below, the value of the voltage source E is _______________

network-theory-questions-answers-problems-involving-dot-conventions-q14

a) -65 V

b) 40 V

c) -60 V

d) 65 V

Answer: a

Explanation: Going from 10 V to 0 V, we get,

network-theory-questions-answers-problems-involving-dot-conventions-q14

50 + 10 + E + 5 = 0

Or, E + 65 = 0

Hence, E = -65 V.

15. In the circuit given below, bulb X uses 48 W when lit, bulb Y uses 22 W when lit and bulb Z uses 14.4 W when lit. The number of additional bulbs in parallel to this circuit, that would be required to below the fuse is _______________

network-theory-questions-answers-problems-involving-dot-conventions-q15

a) 4

b) 5

c) 6

d) 7

Answer: a

Explanation: I X = \(\frac{48}{12}\) = 4 A

I Y = \(\frac{22}{12}\) = 1.8 A

I Z = \

= 20 – 7 = 13

∴ Number of additional bulbs required = \(\frac{13}{3}\) = 4.33

So, 4 additional bulbs are required.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Problems Involving Coupling Coefficient”.

1. If the 3-phase balanced source in the given figure delivers 1500 W at a leading power factor of 0.844, then the value of Z L is approximately __________

network-theory-questions-answers-problems-involving-coupling-coefficient-q1

a) 90∠32.44°

b) 80∠32.44°

c) 80∠-32.44°

d) 90∠-32.44°

Answer: d

Explanation: 3V P I P cosθ = 1500

Or, 3\

\) cos θ = 1500

Or, Z L = \

= ∠-32.44°.

2. An RLC series circuit has a resistance R of 20Ω and a current which lags behind the applied voltage by 45°. If the voltage across the inductor is twice the voltage across the capacitor, the value of inductive resistance is ____________

a) 10 Ω

b) 20 Ω

c) 40 Ω

d) 60 Ω

Answer: c

Explanation: Z = 20 + j20

V = V R = j (V L – V C )

Given, V L = 2 V C

Or, Z L = 2 Z C

Or, Z L – Z C = 20

Or, 2 Z C – Z C = 20

Or, Z C = 20 Ω

Or, Z L = 2Z C = 40 Ω.

3. In the Wheatstone bridge shown below, if the resistance in each arm is increased by 0.05%, then the value of V out will be?

network-theory-questions-answers-problems-involving-coupling-coefficient-q3

a) 50 mV

b) Zero

c) 5mV

d) 0.1mV

Answer: b

Explanation: In Wheatstone bridge, balance condition is

R 1 R 3 = R 2 R 4

Here, R 1 = 5, R 2 = 10, R 3 = 16, R 4 = 8

And when the Wheatstone bridge is balanced then, at V out voltage will be Zero.

4. The voltage drop across a standard resistor of 0.2 Ω is balanced at 83 cm. The magnitude of the current, if the standard cell emf of 1.53 V is balanced at 42 m is ___________

a) 13.04 A

b) 10 A

c) 14.95 A

d) 12.56 A

Answer: c

Explanation: Voltage drop per unit length = \(\frac{1.53}{42}\) = 0.036 V/cm

Voltage drop across 83 cm length = 0.036 × 83 = 2.99 V

∴ Current through resistor, I = \(\frac{2.99}{0.2}\) = 14.95 A.

5. The readings of polar type potentiometer are

I = 12.4∠27.5°

V = 31.5∠38.4°

Then, Reactance of the coil will be?

a) 2.51 Ω

b) 2.56 Ω

c) 2.54 Ω

d) 2.59 Ω

Answer: c

Explanation: Here, V = 31.5∠38.4°

I = 12.4∠27.5°

Z = \(\frac{31.5∠38.4°}{12.4∠27.5°}\) = 2.54∠10.9°

But Z = R + jX = 2.49 + j0.48

∴ Reactance X = 2.54 Ω.

6. The simultaneous applications of signals x and y to the horizontal and vertical plates respectively, of an oscilloscope, produce a vertical figure of 8 displays. If P and Q are constants and x = P sin , then y is equal to _________

a) Q sin

b) Q sin

c) Q sin

d) Q sin

Answer: b

Explanation: \

= Q sin .

7. A resistor of 10 kΩ with a tolerance of 5% is connected in parallel with 5 kΩ resistors of 10% tolerance. The tolerance limit is __________

a) 9%

b) 12.4%

c) 8.33%

d) 7.87%

Answer: c

Explanation: Here, R 1 and R 2 are in parallel.

Then, \(\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}\)

Or, R = \(\frac{50}{15}\) kΩ

∴ \(\frac{△R}{R} = \frac{△R_1}{R_1^2} + \frac{△R_2}{R_2^2}\)

And △R 1 = 0.5 × 10 3 , △R 2 = 0.5 × 10 3

∴ \(\frac{△R}{R} = \frac{10 × 10^3}{3 × 10 × 10^3} × \frac{0.5 × 10^3}{10 × 10^3} + \frac{10}{3} × \frac{10^3}{5 × 10^3} × \frac{0.5 × 10^3}{5 × 10^3}\)

= \(\frac{0.5}{30} + \frac{1}{15} = \frac{2.5}{30}\) = 8.33%.

8. A 200/1 Current Transformer is wound with 400 turns on the secondary on a toroidal core. When it carries a current of 180 A on the primary, the ratio is found to be -0.5%. If the number of secondary turns is reduced by 1, the new ratio error will be?

a) 0.0

b) -0.5

c) -1.0

d) -2.0

Answer: c

Explanation: Turn compensation only alters ratio error n=400

Ratio error = -0.5% = – \(\frac{0.5}{100}\) × 400 = -2

So, Actual ratio = R = n+1 = 401

Nominal Ratio K N = \(\frac{400}{1}\) = 400

Now, if the number of turns are reduced by one, n = 399, R = 400

Ratio error = \(\frac{K_N-R}{R} = \frac{200-200}{200}\) = 0.

9. In the Two wattmeter method of measuring power in a balanced three-phase circuit, one wattmeter shows zero and the other positive maximum. The load power factor is ___________

a) 0

b) 0.5

c) 0.866

d) 1.0

Answer: b

Explanation: The load power factor is = 0.5. Since at this power factor one wattmeter shows zero and the other shows a positive maximum value of power.

10. The meter constant of a single-phase, 230 V induction watt-meter is 600 rev/kW-h. The speed of the meter disc for a current of 15 A at 0.8 power factor lagging will be?

a) 30.3 rpm

b) 25.02 rpm

c) 27.6 rpm

d) 33.1 rpm

Answer: c

Explanation: Meter constant = \(\frac{Number\, of\, revolution}{Energy} = \frac{600 × 230 × 15 × 0.8}{1000}\) = 1656

∴ Speed in rpm = \(\frac{1656}{60}\) = 27.6 rpm.

11. In the figure given below, a 220 V 50 Hz supplies a 3-phase balanced source. The pressure Coil and Current Coil of a watt-meter are connected to the load as shown. The watt-meter reading is _________

network-theory-questions-answers-problems-involving-coupling-coefficient-q11

a) Zero

b) 1600 W

c) 242 W

d) 400 W

Answer: c

Explanation: Watt-meter reading = Current through CC × Voltage across PC × cos .

I BR = I CC = \(\frac{220∠120°}{100°}\) = 2.2∠120°

V YB = V PC = 220∠-120°

w = 2.2∠120° × 220∠-120° × cos 240° = – 242 W.

12. In the Owen’s bridge shown in below figure, Z 1 = 200∠60°, Z 2 = 400∠-90°, Z 3 = 300∠0°, Z 4 = 400∠30°. Then,

network-theory-questions-answers-problems-involving-coupling-coefficient-q12

a) Bridge is balanced with given impedance values

b) Bridge can be balanced, if Z 4 = 600∠60°

c) Bridge can be balanced, if Z 3 = 400∠0°

d) Bridge cannot be balanced with the given configuration

Answer: d

Explanation: For Bridge to be balanced, the product of impedances of the opposite arm should be equal in magnitude as well as phase angle. Here Z 3 Z 2 ≠ Z 1 Z 4 for whatever chosen value. Therefore the Bridge cannot be balanced.

13. In Maxwell’s capacitance bridge for calculating unknown inductance, the various values at balance are, R 1 = 300 Ω, R 2 = 700 Ω, R 3 = 1500 Ω, C 4 = 0.8 μF. The values of R 1 , L 1 and Q factor, if the frequency is 1100 Hz are ____________

a) 240 Ω, 0.12 H, 3.14

b) 140 Ω, 0.168 H, 8.29

c) 140 Ω, 0.12 H, 5.92

d) 240 Ω, 0.36 H, 8.29

Answer: b

Explanation: From Maxwell’s capacitance, we have

R 1 = \(\frac{R_2 R_3}{R_4} = \frac{300 ×700}{1500}\) = 140 Ω

L 1 = R 2 R 3 C 4

= 300 × 700 × 0.8 × 10 -6 = 0.168 H

∴ Q = \(\frac{ωL_1}{R_1}\)

= \(\frac{2 × π × 1100 × 0.168}{140}\) = 8.29.

14. In the figure below, the values of the resistance R 1 and inductance L 1 of a coil are to be calculated after the bridge is balanced. The values are _________________

network-theory-questions-answers-problems-involving-coupling-coefficient-q14

a) 375 Ω and 75 mH

b) 75 Ω and 150 mH

c) 37.5 Ω and 75 mH

d) 75 Ω and 75 mH

Answer: a

Explanation: Applying the usual balance condition relation,

Z 1 Z 4 = Z 2 Z 3

We have, (R 1 + jL 1 ω) \(\frac{R_4/jωC_4}{R_4+1/jωC_4}\) = R 2 R 3

Or, R 1 R 4 + jL 1 ωR 4 = R 2 R 3 + j R 2 R 3 R 4 C 4 ω

∴ R 1 = 2000 × \(\frac{750}{4000}\) = 375 Ω

∴ L 1 = 2000 × 750 × 0.5 × 10 -6 = 75 mH.

15. The four arms of an AC bridge network are as follows:

Arm AB: unknown impedance

Arm BC: standard capacitor C 2 of 1000pf

Arm CD: a non-inductive resistance of R of 100 Ω in parallel to a capacitor of 0.01 μF

Arm DA: a non-inductive resistance of 1000 Ω

The supply frequency is 50 Hz and connected across terminals B and D. If the bridge is balanced with the above value, determine the value of unknown Impedance.

a) 10 kΩ

b) 100 kΩ

c) 250 kΩ

d) 20 kΩ

Answer: a

Explanation: For the balance conditions,

Z 1 Z 3 = Z 2 Z 4

1000 × \

\frac{100}{1 + j100 × ω × 0.01 × 10^{-6}}\)

Or, \

\leftMissing or unrecognized delimiter for \right\)

Or, \(\frac{- j 10^{10}}{ω}\) – 10 4 = R + jX

Comparing the real part, we get,

R = 10 kΩ.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Advanced Problems on Magnetically Coupled Circuits – 1”.

1. For the circuit given below, the effective inductance of the circuit across the terminal AB is ___________

network-theory-questions-answers-advanced-problems-magnetically-coupled-circuits-1-q1

a) 9 H

b) 21 H

c) 11 H

d) 6 H

Answer: c

Explanation: L EFF across AB = L 1 + L 2 + L 3 – 2M 12 – 2M 13 + 2M 23

= 11.

2. Two coils are having self-inductance of 5 mH and 10 mH and a mutual inductance of 0.5 mH in a differential connection. The equivalent inductance of the combination is ___________

a) 14 mH

b) 5.85 mH

c) 6 mH

d) 6.15 mH

Answer: a

Explanation: When 2 coils are connected in series, then effective inductance,

L EFF = L 1 + L 2 ± 2M

For this case, L EFF = L 1 + L 2 – 2M

= 5 + 10 – 2 × 0.5

= 15 – 1

= 14 mH.

3. When two coupled coils of equal self-inductance are connected in series in one way the net inductance is 20 mH and when they are connected in the other way, the net inductance is 12 mH. The maximum value of net inductance when they are connected in parallel is __________

a) 2 mH

b) 5 mH

c) 4 mH

d) 6 mH

Answer: b

Explanation: L EFF = L 1 + L 2 ± 2M

Now, L 1 = L 2 = L

∴ L EFF = 2L ± 2M

Or, 2L + 2M = 20

Or, 2L – 2M = 12

∴ L = 8, M = 2

Now, to get maximum value in parallel connection,

L MAX = \(\frac{}{2L+2M}\)

= \(\frac{}{}\)

= \(\frac{100}{20}\) = 5 mH.

4. For 2 coupled inductors L A and L B , their mutual inductance M LALB satisfies ____________

a) M LALB = \

M LALB > \

M LALB ≤ \(\sqrt{L_A L_B}\)

Answer: d

Explanation: M = K \(\sqrt{L_A L_B}\)

Or, K = \(\frac{M}{\sqrt{L_A L_B}}\)

∴ K≤1

∴ \(\frac{M}{\sqrt{L_A L_B}}\) ≤ 1

Or, M ≤ \(\frac{M}{\sqrt{L_A L_B}}\).

5. The impedance seen by the source in the circuit is given by __________

network-theory-questions-answers-advanced-problems-magnetically-coupled-circuits-1-q5

a) Ω

b) Ω

c) Ω

d) Ω

Answer: c

Explanation: Z 1 = 10∠30° ×

Z 1 = Ω

Total impedance= +

= Ω.

6. For the circuit given below, the inductance measured across the terminals 1 and 2 was 15 H with open terminals 3 and 4. It was 30 H when terminals 3 and 4 were short-circuited. Both the inductors are having inductances 2H. The coefficient of coupling is ______________

network-theory-questions-answers-advanced-problems-magnetically-coupled-circuits-1-q6

a) 1

b) 0.707

c) 0.5

d) Inderminate due to insufficient data

Answer: d

Explanation: When 2 coils are connected in series, then effective inductance,

L EFF = L 1 + L 2 ± 2M

For this case, L EFF = L 1 + L 2 – 2M

However, L 1 + L 2 + 2M or L 1 + L 2 – 2M cannot be determined.

Hence, data is insufficient to calculate the value of coupling coefficient .

7. In the circuit given below, the resonant frequency is ________________

network-theory-questions-answers-advanced-problems-magnetically-coupled-circuits-1-q7

a) \

\(\frac{1}{π\sqrt{2}}\) Hz

Answer: b

Explanation: L EFF = L 1 + L 2 + 2M

= 2 + 2 + 2 × 1

Or, L EFF = 6 H

At resonance, ωL = \(\frac{1}{ωC}\)

Or, ω = \(\frac{1}{\sqrt{L_{EFF} C}} = \frac{1}{\sqrt{12}}\)

Or, I = \(\frac{1}{4π\sqrt{3}}\) Hz.

8. The maximum value of mutual inductance of 2 inductively coupled coils with self inductance L A = 49 mH and L B = 81 mH is ______________

a) 130 mH

b) 63 mH

c) 32 mH

d) 3969 mH

Answer: b

Explanation: M ≤ K\

Or, M max = K\(\sqrt{L_A L_B}\)

Or, M max = \(\sqrt{L_A L_B}\)

= \(\sqrt{40 ×81}\) = 63 mH.

9. In the circuit shown below, I 1 = 4 sin2t A and I 2 = 0. The value of V 1 is _______________

network-theory-questions-answers-advanced-problems-magnetically-coupled-circuits-1-q9

a) -16 cos2t V

b) 16 cos2t V

c) 4 cos2t V

d) -4 cos2t V

Answer: b

Explanation: V 1 = \(\frac{2dI_1}{dt} + \frac{1dI_2}{dt}\)

∴ V 1 = \(\frac{2dI_1}{dt}\) = 16 cos2t V.

10. In the circuit shown below, I 1 = 4 sin2t A and I 2 = 0. The value of V 2 is _______________

network-theory-questions-answers-advanced-problems-magnetically-coupled-circuits-1-q9

a) 2 cos2t V

b) -2 cos2t V

c) 8 cos 2t V

d) -8 cos2t V

Answer: c

Explanation: V 2 = \(\frac{dI_2}{dt} + \frac{dI_1}{dt}\)

Or, V 2 = \(\frac{dI_1}{dt}\)

∴ V 2 = 8 cos2t V.

11. A parallel resonant circuit has a midband admittance of 20×10 -3 S, quality factor of 60 and a resonance frequency of 200 k rad/s. The value of R is _____________

a) 50 Ω

b) 56.57 Ω

c) 80 Ω

d) 28.28 Ω

Answer: a

Explanation: At midband frequency, Z = R, Y = \(\frac{1}{R}\)

Or, R = \(\frac{1}{20 × 10^{-3}}\) = 50 Ω.

12. A circuit resonates at 1 MHz It also has a Q of 100. Bandwidth between half power points is _____________

a) 10 kHz

b) 100 kHz

c) 10 Hz

d) 100 Hz

Answer: a

Explanation: We know that, Q = \(\frac{f}{∆f}\)

Or, ∆f = \(\frac{f}{Q}\)

= \(\frac{10^6}{100}\) = 10 kHz.

13. A series RLC circuit has R = 10 Ω, |X L | = 20 Ω and |X C | = 20 Ω is connected across an AC supply of 200 V rms . The RMS voltage across the capacitor is ____________

a) 200∠-90° V

b) 200∠90° V

c) 400∠90° V

d) 400∠-90° V

Answer: d

Explanation: Q = \(\frac{|X_L|}{R} = \frac{|X_C|}{R}\)

So, Q = \(\frac{20}{10}\) = 2

Rms voltage across capacitor, V Crms = QV∠-90°

Or, V Crms = 2 X 200∠-90°

Or, V Crms = 400∠-90° V.

14. For the circuit given below, the value of input frequency which is required to cause a gain equal to 1.5 is _____________

network-theory-questions-answers-advanced-problems-magnetically-coupled-circuits-1-q14

a) 20 rad/s

b) 20 Hz

c) 10 rad/s

d) No such value exists

Answer: d

Explanation: H = \(\frac{V_O}{V_i}\)

= \(\frac{1}{1+jωRC}\)

∴ Gain = \(\frac{1}{\sqrt{1 + ω^2 RC}}\)

For any value of ω, R and C gain is ≤ 1.

15. Two resistances 100 ± 5Ω and 150 ± 15Ω are connected in series. If the error is specified as standard deviations, the resultant error will be ________________

a) ±10 Ω

b) ±10.6 Ω

c) ±15.8 Ω

d) ±20 Ω

Answer: c

Explanation: Given, R 1 = 100 ± 5 Ω

R 2 = 150 ± 15 Ω

Now, R = R 1 + R 2

The probable errors in this case, R = ±

^{0.5}\) = ± 15.8 Ω.

This set of Tricky Network Theory Questions and Answers focuses on “Advanced Problems on Magnetically Coupled Circuits – 2”.

1. In the circuit given below, the resonant frequency is ____________

tricky-network-theory-questions-answers-q1

a) \

\(\frac{1}{2π\sqrt{6}}\) Hz

Answer: d

Explanation: f = \(\frac{1}{2π\sqrt{L_{EQ} C}} \)

Here, L EQ = 2 + 2 + 2 × 1 = 6

So, C = 1 F

F R = \(\frac{1}{2π\sqrt{6 × 1}} = \frac{1}{2π\sqrt{6}}\) Hz.

2. A coil is designed for high Q performance at a rated voltage and a specified frequency. If the frequency is made twice the original and the coil is operated at the same rated voltage, then the Q factor will be affected as ____________

a) Q is halved

b) Q remains unchanged

c) Q is doubled

d) Q increases or decreases but magnitude cannot be measured

Answer: c

Explanation: ω 2 L = 2ω 1 L

∴ Q 2 = \(\frac{2ω_1 L}{R}\) = 2Q 1

∴ Q is doubled.

3. A coil is designed for high Q performance at a rated voltage and a specified frequency. If the frequency is made twice the original and the coil is operated at the same rated voltage, then the active power P will be affected as ____________

a) P is halved

b) P remains unchanged

c) P is doubled

d) P decreases 4 times

Answer: d

Explanation: I 1 = \(\frac{V}{\sqrt{R^2+ ω_1^2 L^2}} = \frac{V}{ω_1 L}\)

For a high current coil, ωL >> R

I 2 = \(\frac{V_1}{2ω_1 L} = \frac{I_1}{2}\)

∴ P 2 = R

2 = \(\frac{P_1}{4}\)

Therefore, P decreases 4 times.

4. In the figure given below, the time constant of the circuit is ______________

tricky-network-theory-questions-answers-q4

a) 2RC

b) 3RC

c) \

\(\frac{2RC}{3}\)

Answer: d

Explanation: The simplified circuit is:

tricky-network-theory-questions-answers-q4a

Resistance faced by C with the source shorted,

R EQ = \(\frac{R × 2R}{3R} = \frac{2R}{3}\)

Time constant of the circuit, τ = R EQ × C

= \(\frac{2R}{3} × C = \frac{2}{3}\) RC.

5. The effective inductance of the circuit across the terminals A, B is _______________

tricky-network-theory-questions-answers-q5

a) 9 H

b) 21 H

c) 11 H

d) 6 H

Answer: c

Explanation: Effective inductance across AB terminals

= L 1 + L 2 + L 3 – 2M 12 – 2M 13 + 2M 23

= 4 + 5 + 6 – 2 – 2 + 2

= 15 + 4 – 2 – 6 = 11 H.

6. The inductance of a certain moving- iron ammeter is expressed as L = 10 + 3θ – \(\frac{θ^2}{4}\) μH, where θ is the deflection in radian from the zero position. The control spring torque is 25 × 10 -6 Nm/rad. If the meter is carrying a current of 5 A, the deflection is ____________

a) 2.4

b) 2.0

c) 1.2

d) 1.0

Answer: c

Explanation: At equilibrium,

Kθ = \(\frac{1}{2} I^2 \frac{dL}{dθ}\)

(25 × 10 -6 ) θ = \

× 10^{-6}\)

∴ 2 θ + \(\frac{θ}{2}\) = 3

Or, θ = 1.2.

7. A 50 Hz voltage is measured with a moving iron voltmeter and a rectifier type AC voltmeter connected in parallel. If the meter readings are V A and V B respectively. Then the form factor may be estimated as?

a) \

\(\frac{πV_A}{V_B}\)

Answer: b

Explanation: Form factor of the wave = \(\frac{RMS \,value}{Mean \,value}\)

Moving iron instrument will show rms value. Rectifier voltmeter is calibrated to read rms value of sinusoidal voltage that is, with form factor of 1.11.

∴ Mean value of the applied voltage = \(\frac{V_B}{1.11}\)

∴ Form factor = \(\frac{V_A}{V_B/1.11} = \frac{1.11V_A}{V_B}\).

8. A , 50 Hz current transformer has a primary bar. The secondary has a pure resistance of 1 Ω. It also draws a current of 5 A. The magnetic core requires 350 AT for magnetization. Find the percentage ratio error.

a) 10.56

b) -28.57

c) 11.80

d) -11.80

Answer: b

Explanation: I m = 350/1 =350 A

I p = \

^2 +

^2)^{0.5}\) = 490.05

Or, n = \(\frac{350}{7}\) = 50

∴ R = \(\frac{I_P}{I_S} = \frac{490.05}{7}\) = 70

∴ Percentage ratio error = \(\frac{50-70}{70}\) × 100 = -28.57%.

9. The CT supplies current to the current coil of a wattmeter power factor meter, energy meter and, an ammeter. These are connected as?

a) All coils in parallel

b) All coils in series

c) Series-parallel connection with two in each arm

d) Series-parallel connection with one in each arm

Answer: b

Explanation: Since the CT supplies the current to the current coil of a wattmeter, therefore the coils are connected in series so that the current remains the same. If they were connected in parallel then the voltages would have been same but the currents would not be same and thus efficiency would decrease.

10. A current of [2 + \ + 2\] is measured with a thermocouple type, 5A full scale, class 1 meter. The meter reading would lie in the range?

a) 5 A ± 1 %

b) A ± 1%

c) 3 A ± 1.7 %

d) 2 A ± 2.5 %

Answer: c

Explanation: I = [2 + \ + 2\]

Thermocouple measure the rms value of current.

I rms = \Missing or unrecognized delimiter for \right^2 + \leftMissing or unrecognized delimiter for \right^2\Big]^{1/2} = \sqrt{9}\) = 3 A ± 1.7%.

11. The average power absorbed by an impedance Z = 30 – j 70 Ω when a voltage V = 120∠0° is applied is _____________

a) 35

b) 37.24

c) 45

d) 50.25

Answer: b

Explanation: The current through the impedance is given by,

I = \(\frac{V}{Z} = \frac{120∠0°}{30-j70}\)

= \(\frac{120∠0°}{76.16∠-66.8°}\)

= 1.576∠66.8° A

The average power is, P = 0.5V m I m cos (θ v – θ i )

= 0.5 cos

= 37.24 W.

12. A moving iron ammeter produces a full-scale torque of 240 μN-m with a deflection of 120° at a current of 10 A. the rate of change of self-inductance of the instrument at full scale is?

a) 2.0 μH/rad

b) 4.8 μH/rad

c) 12.0 μH/rad

d) 114.6 μH/rad

Answer: b

Explanation: At full scale position, \(\frac{1}{2} I^2 \frac{dL}{dθ}\) = T C

\(\frac{1}{2} 10^2 \frac{dL}{dθ}\) = 240 × 10 -6

∴ \(\frac{dL}{dθ}\) = 4.8 μH/rad.

13. The relation between the Q factor of a coil measured by the Q Meter and the actual Q of the coil is _________

a) Equal to

b) Same but somewhat lesser than

c) Same but somewhat higher than

d) Not equal to

Answer: b

Explanation: The Q factor measured by the Q meter cannot be exactly equal to the actual Q of the coil because of the presence of errors. Also, it is not practically possible for the value to be higher than the actual one. But the value is somewhat lesser and almost equal to the actual value.

14. Consider a circuit consisting of two capacitors C 1 and C 2 . Let R be the resistance and L be the inductance which are connected in series. Let Q 1 and Q 2 be the quality factor for the two capacitors. While measuring the Q value by the Series Connection method, the value of the Q factor is?

a) Q = \

Q = \

Q = \(\frac{

C_1 C_2}{Q_1 C_1-Q_2 C_2}\)

Answer: a

Explanation: ωL = \(\frac{1}{ωC}\) and Q 1 = \(\frac{ωL}{R} = \frac{1}{ωC_1 R}\)

X S = \(\frac{C_1 – C_2}{ωC_1 C_2}\), R S = \(\frac{Q_1 C_1 – Q_2 C_2}{ωC_1 C_2 Q_1 Q_2}\)

Q X = \(\frac{X_S}{R_S} = \frac{

Q_1 Q_2}{Q_1 C_1-Q_2 C_2}\).

15. The meter constant of a single-phase, 230 V induction watt-meter is 600 rev/kW-h. The speed of the meter disc for a current of 15 A at 0.8 power factor lagging will be?

a) 30.3 rpm

b) 25.02 rpm

c) 27.6 rpm

d) 33.1 rpm

Answer: c

Explanation: Meter constant = \(\frac{Number \,of \,revolution}{Energy} = \frac{600 × 230 × 15 × 0.8}{1000}\) = 1656

∴ Speed in rpm = \(\frac{1656}{60}\) = 27.6 rpm.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Polyphase System”.

1. The power generated by a machine increases _____________ percent from single phase to two phase.

a) 40.4

b) 41.4

c) 42.4

d) 43.4

Answer: b

Explanation: In an ac system it is possible to connect two or more number of individual circuits to a common poly phase source. The power generated by machine increases 41.4 percent from single phase to two phase. So, percentage increase = 41.4%.

2. The percentage of power increased from single phase to three phase is?

a) 50

b) 100

c) 150

d) 200

Answer: a

Explanation: The percentage of power increased from single phase to three phase is 50. So, percentage increase = 50%. Beyond three phase the maximum possible increase is only seven percent but the complications are many.

3. When the power factor is __________ the power becomes zero 100 times a second in a 50Hz supply.

a) 0

b) 1

c) 2

d) 3

Answer: b

Explanation: The power in a single phase circuit is pulsating. When the power factor is one, the power becomes zero 100 times a second in a 50Hz supply. Therefore single phase motors have a pulsating torque.

4. Which motors are called self-starting motors?

a) single phase

b) two phase

c) three phase

d) four phase

Answer: c

Explanation: Three phase motors are more easily started than single phase motors. Three phase motors are called self-starting motors. Single phase or two phase or four phase motors are not called self-starting motors.

5. In three phase system, the three voltages differ in phase by __________electrical degrees from each other in a particular sequence.

a) 30

b) 60

c) 90

d) 120

Answer: d

Explanation: In general a three phase system of voltages is merely a combination of three single phase systems of voltages. In three phase system, the three voltages differ in phase by 120⁰ from each other in a particular sequence.

6. In a two phase generator, the armature has two distinct windings that are displaced __________ apart.

a) 45⁰

b) 90⁰

c) 135⁰

d) 180⁰

Answer: b

Explanation: Single phase voltages and currents are generated by single phase generators and the armature of such generator has only one winding or one set of coils. In a two phase generator, the armature has two distinct windings that are displaced 90⁰ apart. θ= 90⁰.

7. In three phase system at any given instant, the algebraic sum of three voltages must be?

a) 0

b) 1

c) 2

d) 3

Answer: a

Explanation: For three phase alternator the three voltages are of same magnitude and frequency. In three phase system at any given instant, the algebraic sum of three voltages must be zero. Algebraic sum of three voltages = 0.

8. Phase sequence depends on the _________________

a) field

b) rotation of the field

c) armature

d) rotation of the armature

Answer: b

Explanation: The sequence of voltages in the three phases undergo changes one after the other abd this is called phase sequence. Phase sequence depends on the rotation of the field not on rotation of armature or on field or on armature.

9. If RR ‘ , YY ‘ and BB ‘ constitutes three phase sequence if V ‘ RR = V m sinωt its corresponding field magnets are in clockwise direction, then V ‘ YY =?

a) V m sinωt

b) V m sin⁰

c) V m sin⁰

d) V m sin⁰

Answer: c

Explanation: If the field system is rotated in the anticlockwise direction, then the sequence of voltages in the three phases are in order RBY. The value of V ‘ YY is V ‘ YY = V m sin⁰.

10. If RR ‘ , YY ‘ and BB ‘ constitutes three phase sequence if V ‘ RR = V m sinωt its corresponding field magnets are in clockwise direction, then the value of V ‘ BB is?

a) V m sin⁰

b) V m sin⁰

c) V m sin⁰

d) V m sinωt

Answer: a

Explanation: The value of V ‘ BB is V ‘ BB = V m sin⁰. There are only two possible phase sequences; they are RBY and RYB.

This set of Network Theory Quiz focuses on “Inter Connection of Three-Phase Sources and Loads”.

1. In a three phase alternator, there are __________ independent phase windings or coils.

a) 1

b) 2

c) 3

d) 4

Answer: c

Explanation: In a three phase alternator, there are 3 independent phase windings or coils. So, 3 independent phase windings or coils. The end connections of the three sets of the coils may be brought out of the machine to form three separate single phase sources to feed three individual circuits.

2. Each coil in three phase alternator has ________________ number of terminals.

a) 2

b) 4

c) 6

d) 8

Answer: a

Explanation: Each coil in three phase alternator has 2 number of terminals, viz. start and finish. So, 2 number of terminals. the coils are inter connected to form a wye or delta connected three phase system to achieve economy and reduce the number of conductors and thereby the complexity of the circuit.

3. In wye or star connection _____________ of the three phases are joined together within the alternator.

a) similar ends

b) opposite ends

c) one similar end, two opposite ends

d) one opposite end, two opposite ends

Answer: a

Explanation: In wye or star connection, similar ends of the three phases are joined together within the alternator. The common terminal so formed is referred to as the neutral point or neutral terminal.

4. The voltage between __________ and ___________ is called phase voltage.

a) line and line

b) line and reference

c) neutral point and reference

d) line and neutral point

Answer: d

Explanation: In a three phase four wire star connected system, the terminals R, Y and B are called the line terminals of the source. The voltage between line and neutral point is called phase voltage. And the voltage between line and line is called line voltage.

5. The voltage between ______________ is called line voltage.

a) line and neutral point

b) line and reference

c) line and line

d) neutral point and reference

Answer: c

Explanation: The voltage between line and line is called line voltage. And the voltage between line and neutral point is called phase voltage. The currents flowing through the phases are called the phase currents, while those flowing in the lines are called the line currents.

6. Figure below represents three phases of an alternator. The phase voltage for the star connection among the options given below is?

network-theory-questions-answers-quiz-q6

a) V RY

b) V RN

c) V YB

d) V BR

Answer: b

Explanation: If the neutral wire is not available for external connection, the system is called a three phase,three wire star connected system. Phase voltage = V RN . And V RY , V YB and V BR are not phase voltages.

7. In the figure shown below, what will be the line voltage?

network-theory-questions-answers-quiz-q6

a) V BR

b) V BN

c) V RN

d) V YN

Answer: a

Explanation: The star connected system formed will supply equal line voltages displaced 120 from one another and acting simultaneously in the cicruit like three independent single phase sources in the same frame of a three phase alternator. Line voltage = V BR . And V RN , V YN and V BN are not line voltages.

8. In the Delta or Mesh connection, there will be __________ number of common terminals.

a) 1

b) 2

c) 3

d) 0

Answer: d

Explanation: The three line conductors are taken from the three junctions of the mesh or delta connection to feed the three phase load. This constitutes a three phase, three wire, delta connected system. In the Delta or Mesh connection, there will be zero number of common terminals. Number of common terminals = 0.

9. The relation between line voltage and phase voltage in Delta or Mesh connection is?

a) V phase > V line

b) V phase < V line

c) V phase = V line

d) V phase >= V line

Answer: c

Explanation: When the sources are connected in delta, loads can be connected only across the three line terminal. The relation between line voltage and phase voltage in Delta or Mesh connection is V phase = V line .

10. Which of the following voltage is a phase voltage in the delta connection?

a) V RN

b) V BR

c) V YN

d) V BN

Answer: b

Explanation: A balanced three phase source is one in which the three individual sources have equal magnitude with 120 phase differences. V BR is a phase voltage in delta connection. And V RN , V YN and V BN are not phase voltages.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Star to Delta and Delta to Star Transformation”.

1. If a resistor Z R is connected between R and N, Z BR between R and B, Z RY between R and Y and Z BY between B and Y form a delta connection, then after transformation to star, the impedance at R is?

a) (Z BR Z BY )/(Z RY +Z BY +Z BR )

b) (Z RY Z BR )/(Z RY +Z BY +Z BR )

c) (Z RY Z BY )/(Z RY +Z BY +Z BR )

d) (Z RY )/(Z RY +Z BY +Z BR )

Answer: b

Explanation: After transformation to star, the impedance at R is

(Z RY Z BR )/(Z RY +Z BY +Z BR ).

2. If a resistor Z R is connected between R and N, Z BR between R and B, Z RY between R and Y and Z BY between B and Y form a delta connection, then after transformation to star, the impedance at Y is?

a) (Z RY )/(Z RY +Z BY +Z BR )

b) (Z BY )/(Z RY +Z BY +Z BR )

c) (Z RY Z BY )/(Z RY +Z BY +Z BR )

d) (Z RY Z BR )/(Z RY +Z BY +Z BR )

Answer: c

Explanation: After transformation to star, the impedance at Y is

(Z RY Z BY )/(Z RY +Z BY +Z BR ).

3. If a resistor Z R is connected between R and N, Z BR between R and B, Z RY between R and Y and Z BY between B and Y form a delta connection, then after transformation to star, the impedance at B is?

a) (Z BR Z BY )/(Z RY +Z BY +Z BR )

b) (Z RY Z BY )/(Z RY +Z BY +Z BR )

c) (Z BY )/(Z RY +Z BY +Z BR )

d) (Z BR )/(Z RY +Z BY +Z BR )

Answer: a

Explanation: After transformation to star, the impedance at Y is

(Z BR Z BY )/(Z RY +Z BY +Z BR ).

4. If the resistors of star connected system are Z R , Z Y , Z B then the impedance Z RY in delta connected system will be?

a) (Z R Z Y + Z Y Z B + Z B Z R )/Z B

b) (Z R Z Y + Z Y Z B + Z B Z R )/Z Y

c) (Z R Z Y + Z Y Z B + Z B Z R )/Z R

d) (Z R Z Y + Z Y Z B + Z B Z R )/(Z R +Z Y )

Answer: a

Explanation: After transformation to delta, the impedance Z RY in delta connected system will be (Z R Z Y + Z Y Z B + Z B Z R )/Z B .

5. If the resistors of star connected system are Z R , Z Y , Z B then the impedance Z BY in delta connected system will be?

a) (Z R Z Y + Z Y Z B + Z B Z R )/(Z B +Z Y )

b) (Z R Z Y + Z Y Z B + Z B Z R )/Z B

c) (Z R Z Y + Z Y Z B + Z B Z R )/Z Y

d) (Z R Z Y + Z Y Z B + Z B Z R )/Z R

Answer: d

Explanation: After transformation to delta, the impedance Z BY in delta connected system will be (Z R Z Y + Z Y Z B + Z B Z R )/Z R .

6. If the resistors of star connected system are Z R , Z Y , Z B then the impedance Z BR in delta connected system will be?

a) (Z R Z Y + Z Y Z B + Z B Z R )/Z Y

b) (Z R Z Y + Z Y Z B + Z B Z R )/ R

c) (Z R Z Y + Z Y Z B + Z B Z R )/Z B

d) (Z R Z Y + Z Y Z B + Z B Z R )/(Z B +Z R )

Answer: a

Explanation: After transformation to delta, the impedance Z BR in delta connected system will be (Z R Z Y + Z Y Z B + Z B Z R )/Z Y .

7. A symmetrical three-phase, three-wire 440V supply is connected to star-connected load. The impedances in each branch are Z R = Ω, Z Y = Ω, Z B = Ω. Find Z RY .

a) Ω

b) Ω

c) Ω

d) Ω

Answer: c

Explanation: Z RY = (Z R Z Y + Z Y Z B + Z B Z R )/Z B

= Ω.

8. A symmetrical three-phase, three-wire 440V supply is connected to star-connected load. The impedances in each branch are Z R = Ω, Z Y = Ω, Z B = Ω. Find Z BY .

a) Ω

b) Ω

c) Ω

d) Ω

Answer: b

Explanation: Z BY = (Z R Z Y + Z Y Z B + Z B Z R )/Z R

= Ω.

9. A symmetrical three-phase, three-wire 440V supply is connected to star-connected load. The impedances in each branch are Z R = Ω, Z Y = Ω, Z B = Ω. Find Z BR .

a) Ω

b) Ω

c) Ω

d) Ω

Answer: b

Explanation: Z BR = (Z R Z Y + Z Y Z B + Z B Z R )/Z Y

= Ω.

10. If a star connected system has equal impedances Z 1 , then after converting into delta connected system having equal impedances Z 2 , then?

a) Z 2 = Z 1

b) Z 2 = 2Z 1

c) Z 2 = 3Z 1

d) Z 2 = 4Z 1

Answer: c

Explanation: If a star connected system has equal impedances Z 1 , then after converting into delta connected system having equal impedances Z 2 , then Z 2 = 3Z 1 .

This set of Network Theory Multiple Choice Questions & Answers focuses on “Voltage, Current and Power in a Star Connected System”.

1. In star connected system, V RY is equal to?

a) V YR

b) -V YR

c) 2V YR

d) 3V YR

Answer: b

Explanation: The voltage available between any pair of terminals is called the line voltage. The double script notation is purposefully used to represent voltages and currents in poly phase circuits. In star connected system, V RY = – V YR .

2. In three phase system, the line voltage V RY is equal to?

a) phasor sum of V RN and V NY

b) phasor difference of V RN and V NY

c) phasor sum of V RN and V NY

d) algebraic sum of V RN and V NY

Answer: a

Explanation: In three phase system, the line voltage V RY is equal to the phasor sum of V RN and V NY which is also equal to the phasor difference of V RN and V YN .

3. The relation between the lengths of the phasors V RN and – V YN is?

a) |V RN | > – |V YN |

b) |V RN | < – |V YN |

c) |V RN | = – |V YN |

d) |V RN | >= – |V YN |

Answer: c

Explanation: The voltage V RY is found by compounding V RN and V YN reversed. The relation between the lengths of the phasors V RN and – V YN is |V RN | = – |V YN |.

4. In a star connected system, the phasors V RN , V YN are ____ apart.

a) 15⁰

b) 30⁰

c) 45⁰

d) 60⁰

Answer: d

Explanation: In a star connected system, the phasors V RN , V YN are separated by θ = 60⁰. To subtract V YN from V RN , we reverse the phase V YN and find its phasor sum with V RN .

5. The relation between V RY , Vph in a star connected system is?

a) V RY = V ph

b) V RY = √3V ph

c) V RY = 3√3V ph

d) V RY = 3V ph

Answer: b

Explanation: The two phasors V YN and V BN are equal in length and are 60⁰apart. The relation between V RY , V ph in a star connected system is V RY = √3V ph .

6. In a star connected system, the relation between V YB , V ph is?

a) V YB = V ph

b) V YB = 3√3V ph

c) V YB = 3V ph

d) V YB = √3V ph

Answer: d

Explanation: In a star connected system, the relation between V YB , V ph is V YB = √3V ph . The line voltage V YB is equal to the phasor difference of V YN and V BN and is equal to √3V ph .

7. The voltages, V BR ,V ph are related in star connected system is?

a) V BR = 3V ph

b) V BR = 3√3V ph

c) V BR = √3V ph

d) V BR = V ph

Answer: c

Explanation: The voltages, V BR , V ph in star connected system are related as V BR = √3V ph . The line voltage V YB is equal to the phasor difference of V BN and V RN and is equal to √3V ph .

8. A symmetrical star connected system has V RN = 230∠0⁰. The phase sequence is RYB. Find V RY .

network-theory-questions-answers-voltage-current-power-star-q8

a) 398.37∠30⁰

b) 398.37∠-30⁰

c) 398.37∠90⁰

d) 398.37∠-90⁰

Answer: a

Explanation: Since the system is a balanced system, all the phase voltages are equal in magnitude but displaced by 120⁰. V RN = 230∠0⁰V. V RY = √3×230∠(0 o +30 o )V=398.37∠30 o V.

9. A symmetrical star connected system has V RN = 230∠0⁰. The phase sequence is RYB. Find V YB .

network-theory-questions-answers-voltage-current-power-star-q8

a) 398.37∠-30⁰

b) 398.37∠210⁰

c) 398.37∠90⁰

d) 398.37∠-90⁰

Answer: d

Explanation: Corresponding line voltages are equal to √3 times the phase voltages and are 30⁰ ahead of the respective phase voltages. V YN = 230∠-120⁰V. V YB = √3×230∠(-120 o +30 o )V=398.37∠-90⁰V.

10. A symmetrical star connected system has V RN = 230∠0⁰. The phase sequence is RYB. Find V BR .

network-theory-questions-answers-voltage-current-power-star-q8

a) 398.37∠210⁰

b) 398.37∠-210⁰

c) 398.37∠120⁰

d) 398.37∠-120⁰

Answer: b

Explanation: All the line voltages are equal in magnitude and are displaced by 120⁰. V BN = 230∠-240⁰V. V BR = √3×230∠(-240 o +30 o )V=398.37∠-210 o V.

This set of Network Theory MCQs focuses on “Voltage, Current and Power in a Delta Connected System”.

1. In a delta connected system, the voltage across the terminals R and Y is 400∠0⁰. Calculate the line voltage V RY . Assume R RY phase sequence.

a) 400∠0⁰

b) 400∠120⁰

c) 400∠-120⁰

d) 400∠240⁰

Answer: a

Explanation: In a balanced delta-connected system we know |V RY | = |V Ph |, and it is displaced by 120⁰, therefore the line voltage V RY is V RY = 400∠0⁰V.

2. In a delta connected system, the voltage across the terminals R and Y is 400∠0⁰. Find the line voltage V YB .

a) 400∠120⁰

b) 400∠-120⁰

c) 400∠240⁰

d) 400∠-240⁰

Answer: b

Explanation: As |V YB | = |V Ph |, and is displaced by 120⁰, therefore the line voltage V YB is V YB = 400∠-120⁰V. A balanced three phase, three wire, delta connected system is referred to as mesh connection because it forms a closed circuit.

3. In a delta connected system, the voltage across the terminals R and Y is 400∠0⁰. Find the line voltage V BR .

a) 400∠240⁰

b) 400∠120⁰

c) 400∠-240⁰

d) 400∠-120⁰

Answer: c

Explanation: We know, |V BR | = |V Ph |, and is displaced by 120⁰, therefore the line voltage V BR is V BR = 400∠-240⁰V. Delta connection is so called because the three branches in the circuit can be arranged in the shape of delta.

4. In delta-connected system, the currents I R , I Y , I B are equal in magnitude and they are displaced by _____ from one another.

a) 0⁰

b) 60⁰

c) 90⁰

d) 120⁰

Answer: d

Explanation: In delta-connected system, the currents I R , I Y , I B are equal in magnitude and they are displaced by 120⁰ from one another. From the manner of interconnection of the three phases in the circuit, it may appear that the three phase are short circuited among themselves.

5. In a delta-connected system, the currents I R = I B = I Y =?

a) I Ph

b) 2I Ph

c) 3I Ph

d) 4I Ph

Answer: a

Explanation: In a delta-connected system, the currents I R = I B = I Y = I Ph . Since the system is balanced, the sum of the three voltages round the closed mesh is zero; consequently no current can flow around the mesh when the terminals are open.

6. The relation between I L and I Ph is in a delta connected system is?

a) I L = I Ph

b) I L = √3 I Ph

c) I L = 3 I Ph

d) I L = 3√3I Ph

Answer: b

Explanation: The relation between I L and I Ph is in a delta connected system is I L = √3 I Ph . The arrows placed alongside the voltages of the three phases indicate that the terminals are positive during their positive half cycles.

7. The line currents are ________ behind respective phase currents in a delta connected system.

a) 120⁰

b) 90⁰

c) 60⁰

d) 30⁰

Answer: d

Explanation: In a delta connected system, all the line currents are equal in magnitude but displaced by 120⁰ from one another and the line currents are 30⁰ behind the respective phase currents.

8. In a delta connected system, the expression of power is?

a) V L I L cosφ W

b) √3 V L I L cosφ W

c) 3V L I L cosφ W

d) 3√3V L I L cosφ W

Answer: b

Explanation: The total power in the delta circuit is the sum of the powers in the three phases. In a delta connected system, the expression of power is P = √3V L I L cosφ W.

9. A balanced delta-connected load of Ω per phase is connected to a balanced three-phase 440V supply. The phase current is 10A. Find the total active power.

a) 7.26W

b) 726W

c) 7260W

d) 72.6W

Answer: c

Explanation: Z Ph = √(2 2 +3 2 ) = 3.6∠56.3⁰Ω. cosφ = R Ph /Z Ph = 2/3.6 = 0.55. I L = √3× I Ph = 17.32A. Active power = √3 V L I L cosφ = √3×440×17.32×0.55= 7259.78W.

10. A balanced delta-connected load of Ω per phase is connected to a balanced three-phase 440V supply. The phase current is 10A. Find the apparent power.

a) 10955.67 VAR

b) 10.95567 VAR

c) 109.5567 VAR

d) 1.095567 VAR

Answer: a

Explanation: Sinφ = 0.83. Reactive power = √3 V L I L sinφ. V L = 440V, I L = 17.32A. On substituting we get reactive power = √3 x 440 x 17.32 = 10955.67 VAR.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Three-Phase Balanced Circuits”.

1. In a balanced three-phase system-delta load, if we assume the line voltage is V RY = V∠0⁰ as a reference phasor. Then the source voltage V YB is?

a) V∠0⁰

b) V∠-120⁰

c) V∠120⁰

d) V∠240⁰

Answer: b

Explanation: As the line voltage V RY = V∠0⁰ is taken as a reference phasor. Then the source voltage V YB is V∠-120⁰.

2. In a balanced three-phase system-delta load, if we assume the line voltage is V RY = V∠0⁰ as a reference phasor. Then the source voltage V BR is?

a) V∠120⁰

b) V∠240⁰

c) V∠-240⁰

d) V∠-120⁰

Answer: c

Explanation: As the line voltage V RY = V∠0⁰ is taken as a reference phasor. Then the source voltage V BR is V∠-240⁰.

3. In a delta-connected load, the relation between line voltage and the phase voltage is?

a) line voltage > phase voltage

b) line voltage < phase voltage

c) line voltage = phase voltage

d) line voltage >= phase voltage

Answer: c

Explanation: In a delta-connected load, the relation between line voltage and the phase voltage is line voltage = phase voltage.

4. If the load impedance is Z∠Ø, the current (I R ) is?

a) ∠-Ø

b) ∠Ø

c) ∠90-Ø

d) ∠-90+Ø

Answer: a

Explanation: As the load impedance is Z∠Ø, the current flows in the three load impedances and the current flowing in the R impedance is I R = V BR ∠0⁰/Z∠Ø = ∠-Ø.

5. If the load impedance is Z∠Ø, the expression obtained for current (I Y ) is?

a) ∠-120+Ø

b) ∠120-Ø

c) ∠120+Ø

d) ∠-120-Ø

Answer: d

Explanation: As the load impedance is Z∠Ø, the current flows in the three load impedances and the current flowing in the Y impedance is I Y = V YB ∠120⁰/Z∠Ø = ∠-120-Ø.

6. If the load impedance is Z∠Ø, the expression obtained for current (I B ) is?

a) ∠-240+Ø

b) ∠-240-Ø

c) ∠240-Ø

d) ∠240+Ø

Answer: b

Explanation: As the load impedance is Z∠Ø, the current flows in the three load impedances and the current flowing in the B impedance is I B = V BR ∠240⁰/Z∠Ø = ∠-240-Ø.

7. A three-phase balanced delta connected load of Ω is connected across a 400V, 3 – Ø balanced supply. Determine the phase current I R . Assume the phase sequence to be R YB .

a) 44.74∠-63.4⁰A

b) 44.74∠63.4⁰A

c) 45.74∠-63.4⁰A

d) 45.74∠63.4⁰A

Answer: a

Explanation: Taking the line voltage V RY = V∠0⁰ as a reference V RY = 400∠0⁰V, V YB = 400∠-120⁰V and V BR = 400∠-240⁰V. Impedance per phase = Ω = 8.94∠63.4⁰Ω. Phase current I R = (400∠0 o )/(8.94∠63.4 o )= 44.74∠-63.4⁰A.

8. A three-phase balanced delta connected load of Ω is connected across a 400V, 3 – Ø balanced supply. Determine the phase current I Y .

a) 44.74∠183.4⁰A

b) 45.74∠183.4⁰A

c) 44.74∠183.4⁰A

d) 45.74∠-183.4⁰A

Answer: c

Explanation: Taking the line voltage V RY = V∠0⁰ as a reference V RY = 400∠0⁰V, V YB = 400∠-120⁰V and V BR = 400∠-240⁰V. Impedance per phase = Ω = 8.94∠63.4⁰Ω. Phase current I Y = (400∠120 o )/(8.94∠63.4 o )= 44.74∠-183.4⁰A.

9. A three-phase balanced delta connected load of Ω is connected across a 400V, 3 – Ø balanced supply. Determine the phase current I B .

a) 44.74∠303.4⁰A

b) 44.74∠-303.4⁰A

c) 45.74∠303.4⁰A

d) 45.74∠-303.4⁰A

Answer: b

Explanation: Taking the line voltage V RY = V∠0⁰ as a reference V RY = 400∠0⁰V, V YB = 400∠-120⁰V and V BR = 400∠-240⁰V. Impedance per phase = Ω = 8.94∠63.4⁰Ω. Phase current I B = (400∠240 o )/(8.94∠63.4 o ) = 44.74∠-303.4⁰A.

10. Determine the power drawn by the load.

a) 21

b) 22

c) 23

d) 24

Answer: d

Explanation: Power is defined as the product of voltage and current. So the power drawn by the load is P = 3V Ph I Ph cosØ = 24kW.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Three-Phase Unbalanced Circuits”.

1. If the system is a three-wire system, the currents flowing towards the load in the three lines must add to ___ at any given instant.

a) 1

b) 2

c) 3

d) zero

Answer: d

Explanation: If the system is a three-wire system, the currents flowing towards the load in the three lines must add to zero at any given instant.

2. The three impedances Z 1 = 20∠30⁰Ω, Z 2 = 40∠60⁰Ω, Z 3 = 10∠-90⁰Ω are delta-connected to a 400V, 3 – Ø system. Determine the phase current I R .

a) A

b) A

c) A

d) A

Answer: a

Explanation: Taking V RY = V∠0⁰ as a reference phasor, and assuming RYB phase sequence, we have V RY = 400∠0⁰V Z 1 = 20∠30⁰Ω = Ω I R = (400∠0 o )/(20∠30 o ) = A.

3. The three impedances Z 1 = 20∠30⁰Ω, Z 2 = 40∠60⁰Ω, Z 3 = 10∠-90⁰Ω are delta-connected to a 400V, 3 – Ø system. Find the phase current I Y .

a) A

b) A

c) A

d) A

Answer: c

Explanation: The voltage V YB is V YB = 400∠-120⁰V. The impedance Z 2 is Z 2 = 40∠60⁰Ω => I Y = (400∠-120 o )/(40∠60 o )=A.

4. The three impedances Z 1 = 20∠30⁰Ω, Z 2 = 40∠60⁰Ω, Z 3 = 10∠-90⁰Ω are delta-connected to a 400V, 3 – Ø system. Find the phase current I B .

a) A

b) A

c) A

d) A

Answer: d

Explanation: The voltage V BR is V BR = 400∠-240⁰V. The impedance Z 3 is Z 3 = 10∠-90⁰Ω => I B = (400∠240 o )/(10∠-90 o )=A.

5. The three impedances Z 1 = 20∠30⁰Ω, Z 2 = 40∠60⁰Ω, Z 3 = 10∠-90⁰Ω are delta-connected to a 400V, 3 – Ø system. Find the line current I 1 .

a) A

b) A

c) A

d) A

Answer: c

Explanation: The line current I 1 is the difference of I R and I B . So the line current I 1 is I 1 = I R – I B = A.

6. The three impedances Z 1 = 20∠30⁰Ω, Z 2 = 40∠60⁰Ω, Z 3 = 10∠-90⁰Ω are delta-connected to a 400V, 3 – Ø system. Find the line current I 2 .

a) A

b) A

c) A

d) A

Answer: a

Explanation: The line current I 2 is the difference of I Y and I R . So the line current I 2 is I 2 = I Y – I R = A.

7. The three impedances Z 1 = 20∠30⁰Ω, Z 2 = 40∠60⁰Ω, Z 3 = 10∠-90⁰Ω are delta-connected to a 400V, 3 – Ø system. Find the line current I 3 .

a) A

b) A

c) A

d) A

Answer: c

Explanation: The line current I 3 is the difference of I B and I Y . So the line current I 3 is I 3 = I B – I Y = A.

8. The three impedances Z 1 = 20∠30⁰Ω, Z 2 = 40∠60⁰Ω, Z 3 = 10∠-90⁰Ω are delta-connected to a 400V, 3 – Ø system. Find the power in the R phase.

a) 6628

b) 6728

c) 6828

d) 6928

Answer: d

Explanation: The term power is defined as the product of square of current and the impedance. So the power in the R phase = 20 2 x 17.32 = 6928W.

9. The three impedances Z 1 = 20∠30⁰Ω, Z 2 = 40∠60⁰Ω, Z 3 = 10∠-90⁰Ω are delta-connected to a 400V, 3 – Ø system. Find the power in the Y phase.

a) 1000

b) 2000

c) 3000

d) 4000

Answer: b

Explanation: The term power is defined as the product of square of current and the impedance. So the power in the Y phase = 10 2 x 20 = 2000W.

10. The three impedances Z 1 = 20∠30⁰Ω, Z 2 = 40∠60⁰Ω, Z 3 = 10∠-90⁰Ω are delta-connected to a 400V, 3 – Ø system. Find the power in the B phase.

a) 0

b) 1

c) 3

d) 2

Answer: a

Explanation: The term power is defined as the product of square of current and the impedance. So the power in the B phase = 40 2 x 0 = 0W.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Power Measurement in Three-Phase Circuits”.

1. The wattmeter method is used to measure power in a three-phase load. The wattmeter readings are 400W and -35W. Calculate the total active power.

a) 360

b) 365

c) 370

d) 375

Answer: b

Explanation: Wattmeters are generally used to measure power in the circuits. Total active power = W 1 + W 2 = 400 + =365W.

2. The wattmeter method is used to measure power in a three-phase load. The wattmeter readings are 400W and -35W. Find the power factor.

a) 0.43

b) 0.53

c) 0.63

d) 0.73

Answer: a

Explanation: We know tanØ = √3((W R – W Y )/(W R + W Y )) => tanØ = √3 )/)=2.064 => Ø = 64.15⁰. Power factor = 0.43.

3. The wattmeter method is used to measure power in a three-phase load. The wattmeter readings are 400W and -35W. Find the reactive power.

a) 751.44

b) 752.44

c) 753.44

d) 754.44

Answer: c

Explanation: Reactive power = √3V L I L sinØ. We know that W R – W Y = 400-) = 435 = V L I L sinØ. Reactive power = √3 x 435 = 753.44 VAR.

4. The input power to a three-phase load is 10kW at 0.8 Pf. Two watt meters are connected to measure the power. Find the reading of higher reading wattmeter.

a) 7.165

b) 6.165

c) 6.165

d) 4.165

Answer: a

Explanation: W R + W Y = 10kW. Ø = cos -1 0.8=36.8 o => tanØ = 0.75 = √3 (W R -W Y )/(W R +W Y )=(W R -W Y )/10. W R -W Y =4.33kW. W R +W Y =10kW. W R =7.165kW.

5. The input power to a three-phase load is 10kW at 0.8 Pf. Two watt meters are connected to measure the power. Find the reading of lower reading wattmeter.

a) 1.835

b) 2.835

c) 3.835

d) 4.835

Answer: b

Explanation: W R + W Y = 10kW. Ø = cos -1 0.8=36.8 o => tanØ = 0.75 = √3 (W R -W Y )/(W R +W Y )=(W R -W Y )/10. W R -W Y =4.33kW. W R +W Y =10kW. W Y =2.835kW.

6. The readings of the two watt meters used to measure power in a capacitive load are -3000W and 8000W respectively. Calculate the input power. Assume RYB sequence.

a) 5

b) 50

c) 500

d) 5000

Answer: d

Explanation: Toatal power is the sum of the power in R and power in Y. So total power = W R +W Y = -3000+8000 = 5000W

7. The readings of the two watt meters used to measure power in a capacitive load are -3000W and 8000W respectively. Find the power factor.

a) 0.25

b) 0.5

c) 0.75

d) 1

Answer: a

Explanation: As the load is capacitive, the wattmeter connected in the leading phase gives less value. W R =-3000. W Y =8000. tanØ = √3 )/5000=3.81 => Ø = 75.29⁰ => cosØ = 0.25.

8. The wattmeter reading while measuring the reactive power with wattmeter is?

a) V L I L secØ

b) V L I L sinØ

c) V L I L tanØ

d) V L I L cosØ

Answer: b

Explanation: The wattmeter reading while measuring the reactive power with wattmeter is wattmeter reading = V L I L sinØ VAR.

9. The total reactive power in the load while measuring the reactive power with wattmeter is?

a) √3V L I L cosØ

b) √3V L I L tanØ

c) √3V L I L sinØ

d) √3 V L I L secØ

Answer: c

Explanation: To obtain the reactive power, wattmeter reading is to be multiplied by √3. Total reactive power = √3V L I L sinØ.

10. A single wattmeter is connected to measure the reactive power of a three-phase, three-wire balanced load. The line current is 17A and the line voltage is 440V. Calculate the power factor of the load if the reading of the wattmeter is 4488 VAR.

a) 0.6

b) 0.8

c) 1

d) 1.2

Answer: b

Explanation: Wattmeter reading = V L I L sinØ => 4488 = 440 x 17sinØ => sinØ = 0.6. Power factor = cosØ = 0.8.

This set of Network Theory Multiple Choice Questions & Answers focuses on “DC Response of an R-L Circuit”.

1. The expression of current in R-L circuit is?

a) i=t))

b) i=-t))

c) i=-t))

d) i=t))

Answer: d

Explanation: The expression of current in R-L circuit is i = -exp⁡t). On solving we get i = t)).

2. The steady state part in the expression of current in the R-L circuit is?

a) t))

b) t))

c) V/R

d) R/V

Answer: c

Explanation: The steady state part in the expression of current in the R-L circuit is steady state part = V/R. When the switch S is closed, the response reaches a steady state value after a time interval.

3. In the expression of current in the R-L circuit the transient part is?

a) R/V

b) t))

c) t))

d) V/R

Answer: b

Explanation: The expression of current in the R-L circuit has the transient part as

t)). The transition period is defined as the time taken for the current to reach its final or steady state value from its initial value.

4. The value of the time constant in the R-L circuit is?

a) L/R

b) R/L

c) R

d) L

Answer: a

Explanation: The time constant of a function e -t is the time at which the exponent of e is unity where e is the base of the natural logarithms. The term L/R is called the time constant and is denoted by ‘τ’.

5. After how many time constants, the transient part reaches more than 99 percent of its final value?

a) 2

b) 3

c) 4

d) 5

Answer: d

Explanation: After five time constants, the transient part of the response reaches more than 99 percent of its final value.

6. A series R-L circuit with R = 30Ω and L = 15H has a constant voltage V = 60V applied at t = 0 as shown in the figure. Determine the current in the circuit at t = 0 + .

network-theory-questions-answers-dc-response-rl-q6

a) 1

b) 2

c) 3

d) 0

Answer: d

Explanation: Since the inductor never allows sudden changes in currents. At t = 0 + that just after the initial state the current in the circuit is zero.

7. The expression of current obtained from the circuit in terms of differentiation from the circuit shown below?

network-theory-questions-answers-dc-response-rl-q6

a) di/dt+i=4

b) di/dt+2i=0

c) di/dt+2i=4

d) di/dt-2i=4

Answer: c

Explanation: Let the i be the current flowing through the circuit. By applying Kirchhoff’s voltage law, we get 15 di/dt+30i=60 => di/dt+2i=4.

8. The expression of current from the circuit shown below is?

network-theory-questions-answers-dc-response-rl-q6

a) i=2(1-e -2t )A

b) i=2(1+e -2t )A

c) i=2(1+e 2t )A

d) i=2(1+e 2t )A

Answer: a

Explanation: At t = 0 + the current in the circuit is zero. Therefore at t = 0 + , i = 0 => 0 = c + 2 =>c = -2. Substituting the value of ‘c’ in the current equation, we have i = 2(1-e -2t )A.

9. The expression of voltage across resistor in the circuit shown below is?

network-theory-questions-answers-dc-response-rl-q6

a) V R = 60(1+e 2t )V

b) V R = 60(1-e -2t )V

c) V R = 60(1-e 2t )V

d) V R = 60(1+e -2t )V

Answer: b

Explanation: Voltage across the resistor V R = iR. On substituting the expression of current we get voltage across resistor = (2(1-e -2t ))×30=60(1-e -2t )V.

10. Determine the voltage across the inductor in the circuit shown below is?

network-theory-questions-answers-dc-response-rl-q6

a) V L = 60(-e -2t )V

b) V L = 60(e 2t )V

c) V L = 60(e -2t )V

d) V L = 60(-e 2t )V

Answer: c

Explanation: Voltage across the inductor V L = Ldi/dt. On substituting the expression of current we get voltage across the inductor = 15×(2(1-e -2t )))=60(e -2t )V.

This set of Network Theory Multiple Choice Questions & Answers focuses on “DC Response of an R-C Circuit”.

1. The current in the R-L circuit at a time t = 0 + is?

a) V/R

b) R/V

c) V

d) R

Answer: a

Explanation: The capacitor never allows sudden changes in voltage, it will act as a short circuit at t = 0 + . So the current in the circuit at t = 0 + is V/R.

2. The expression of current in R-C circuit is?

a) i=exp⁡

b) i=exp⁡

c) i=-exp

d) i=-exp⁡

Answer: b

Explanation: The particular solution of the current equation is zero. So the expression of current in R-C circuit is i=exp⁡.

3. In an R-C circuit, when the switch is closed, the response ____________

a) do not vary with time

b) decays with time

c) rises with time

d) first increases and then decreases

Answer: b

Explanation: In a R-C circuit, when the switch is closed, the response decays with time that is the response V/R decreases with increase in time.

4. The time constant of an R-C circuit is?

a) RC

b) R/C

c) R

d) C

Answer: a

Explanation: The time constant of an R-C circuit is RC and it is denoted by τ and the value of τ in dc response of R-C circuit is RC sec.

5. After how many time constants, the transient part reaches more than 99 percent of its final value?

a) 2

b) 3

c) 4

d) 5

Answer: d

Explanation: After five time constants, the transient part of the response reaches more than 99 percent of its final value.

6.A series R-C circuit consists of resistor of 10 and capacitor of 0.1F as shown in the figure. A constant voltage of 20V is applied to the circuit at t = 0. What is the current in the circuit at t = 0?

network-theory-questions-answers-dc-response-rc-q6

a) 1

b) 2

c) 3

d) 4

Answer: b

Explanation: At t = 0, switch S is closed. Since the capacitor does not allow sudden changes in voltage, the current in the circuit is i = V/R = 20/10 = 2A. At t = 0, i = 2A.

7. The expression of current obtained from the circuit in terms of differentiation from the circuit shown below?

network-theory-questions-answers-dc-response-rc-q6

a) di/dt+i=1

b) di/dt+i=2

c) di/dt+i=3

d) di/dt+i=0

Answer: d

Explanation: By applying Kirchhoff’s law, we get

network-theory-questions-answers-dc-response-rc-q7

Differentiating with respect to t, we get 10 di/dt+i/0.1=0 => di/dt+i=0.

8. The current equation in the circuit shown below is?

network-theory-questions-answers-dc-response-rc-q6

a) i=2(e -2t )A

b) i=2(e 2t )A

c) i=2(-e -2t )A

d) i=2(-e 2t )A

Answer: a

Explanation: At t = 0, switch S is closed. Since the capacitor does not allow sudden changes in voltage, the current in the circuit is i = V/R = 20/10 = 2A. At t = 0, i = 2A. The current equation is i=2(e -2t )A.

9. The expression of voltage across resistor in the circuit shown below is?

network-theory-questions-answers-dc-response-rc-q6

a) V R = 20(e t )V

b) V R = 20(-e -t )V

c) V R = 20(-e t )V

d) V R = 20(e -t )V

Answer: d

Explanation: The expression of voltage across resistor in the circuit is V R = iR =(2(e -t ))×10=20(e -t )V.

10. Determine the voltage across the capacitor in the circuit shown below is?

network-theory-questions-answers-dc-response-rc-q6

a) V C = 60(1-e -t )V

b) V C = 60(1+e t )V

c) V C = 60(1-e t )V

d) V C = 60(1+e -t )V

Answer: a

Explanation: The expression of voltage across capacitor in the circuit V C = V(1-e -t/RC ) = 20(1-e -t )V.

This set of Network Theory Multiple Choice Questions & Answers focuses on “DC Response of an R-L-C Circuit”.

1. For an R-L-C circuit, we get [D – (K 1 + K 2 )][D – (K 1 – K 2 )] i = 0. If K 2 is positive, then the curve will be?

a) damped

b) over damped

c) under damped

d) critically damped

Answer: b

Explanation: For an R-L-C circuit, we get [D – (K 1 + K 2 )][D – (K 1 – K 2 )] i = 0. If K 2 is positive, then the curve will be over damped response.

2. If the roots of an equation are real and unequal, then the response will be?

a) critically damped

b) under damped

c) over damped

d) damped

Answer: c

Explanation: If the roots of an equation are real and unequal, then the response will be over damped response. Over damped response of a system is defined as the system returns to equilibrium without oscillating.

3. If the roots of an equation are complex conjugate, then the response will be?

a) over damped

b) critically damped

c) damped

d) under damped

Answer: d

Explanation: If the roots of an equation are complex conjugate, then the response will be under damped response. Damping is an influence within or upon an oscillatory system that has the effect of reducing, restricting or preventing its oscillations.

4. If the roots of an equation are real and equal, then the response will be?

a) over damped

b) damped

c) critically damped

d) under damped

Answer: c

Explanation: If the roots of an equation are real and equal, then the response will be critically damped response. For a critically damped system, the system returns to equilibrium as quickly as possible without oscillating.

5. The circuit shown in the figure consists of resistance, capacitance and inductance in series with a 100V source when the switch is closed at t = 0. Find the equation obtained from the circuit in terms of current.

a) 100 = 20i + 0.05 \

100 = 20i – 0.05 \

100 = 20i + 0.05 \

100 = 20i – 0.05 \(\frac{di}{dt} – \frac{1}{20 \times 10^{-6}} \int idt\)

Answer: a

Explanation: At t = 0, switch S is closed when the 100V source is applied to the circuit and results in the following differential equation.

100 = 20i + 0.05 \(\frac{di}{dt} + \frac{1}{20 \times 10^{-6}} \int idt\)

6. Replacing the differentiation with D 1 , D 2 in the equation 100 = 20i + 0.05 \(\frac{di}{dt} + \frac{1}{20 \times 10^{-6}} \int idt\). Find the values of D 1 , D 2 .

a) 200±j979.8

b) -200±j979.8

c) 100±j979.8

d) -100±j979.8

Answer: b

Explanation: Let the roots of the characteristic equation are denoted by D 1 , D 2 . So on differentiating the equation 100 = 20i + 0.05 \(\frac{di}{dt} + \frac{1}{20 \times 10^{-6}} \int idt\), we get D 1 = -200+j979.8, D 2 = -200-j979.8.

7. The expression of current from the circuit shown below.

a) i=e -200t [c 1 cos979.8t+c 2 979.8t]A

b) i=e 200t [c 1 cos979.8t-c 2 979.8t]A

c) i=e -200t [c 1 cos979.8t-c 2 979.8t]A

d) i=e 200t [c 1 cos979.8t+c 2 979.8t]A

Answer: a

Explanation: The expression of current from the circuit will be i = e K 1 t [c 1 cosK 1 t + c 2 sinK 2 t]. So, i=e -200t [c 1 cos979.8t+c 2 979.8t]A.

8. At time t = 0, the value of current in the circuit shown below.

a) 1

b) 2

c) 3

d) 0

Answer: d

Explanation: At t = 0 that is initially the current flowing through the circuit is zero that is i = 0. So, i = 0.

9. The voltage across the inductor at t = 0 in the circuit shown below.

a) 50

b) 100

c) 150

d) 200

Answer: b

Explanation: At t = 0, that is initially the voltage across the inductor is 100V. => V = 100V. So we can write Ldi/dt = 100.

10. The current equation obtained from the circuit shown below is?

a) i=e -200t A

b) i=e -200t A

c) i=e -200t A

d) i=e -200t A

Answer: b

Explanation: On solving the values of c 1 , c 2 are obtained as c 1 = 0, c 2 = 2.04. So, the current equation is i=e -200t A.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Sinusoidal Response of an R-L Circuit”.

1. In the sinusoidal response of R-L circuit, the complementary function of the solution of i is?

a) i c = ce -t

b) i c = ce t

c) i c = ce -t

d) i c = ce t

Answer: a

Explanation: From the R-L circuit, we get the characteristic equation as i=V/L cos⁡. The complementary function of the solution i is i c = ce -t .

2. The particular current obtained from the solution of i in the sinusoidal response of R-L circuit is?

a) i p = V/√(R 2 + 2 ) cos⁡(ωt+θ+tan -1 )

b) i p = V/√(R 2 + 2 ) cos⁡(ωt+θ-tan -1 )

c) i p = V/√(R 2 + 2 ) cos⁡(ωt-θ+tan -1 )

d) i p = V/√(R 2 + 2 ) cos⁡(ωt-θ+tan -1 )

Answer: b

Explanation: The characteristic equation consists of two parts, viz. complementary function and particular integral. The particular integral is i p = V/√(R 2 + 2 ) cos⁡(ωt+θ-tan -1 ).

3. The value of ‘c’ in complementary function of ‘i’ is?

a) c = -V/√(R 2 + 2 ) cos⁡(θ+tan -1 )

b) c = -V/√(R 2 + 2 ) cos⁡(θ-tan -1 )

c) c = V/√(R 2 + 2 ) cos⁡(θ+tan -1 )

d) c = V/√(R 2 + 2 ) cos⁡(θ-tan -1 )

Answer: b

Explanation: Since the inductor does not allow sudden changes in currents, at t = 0, i = 0. So, c = -V/√(R 2 + 2 ) cos⁡(θ-tan -1 ).

4. The complete solution of the current in the sinusoidal response of R-L circuit is?

a) i = e -t [V/√(R 2 + 2 ) cos⁡(θ-tan -1 )⁡)]+V/√(R 2 + 2 ) cos⁡(ωt+θ-tan -1 )⁡)

b) i = e -t [-V/√(R 2 + 2 ) cos⁡(θ-tan -1 ))]-V/√(R 2 + 2 ) cos⁡(ωt+θ-tan -1 )⁡)

c) i = e -t [V/√(R 2 + 2 ) cos⁡(θ-tan -1 )⁡)]-V/√(R 2 + 2 ) cos⁡(ωt+θ-tan -1 )⁡)

d) i = e -t [-V/√(R 2 + 2 ) cos⁡(θ-tan -1 )⁡)]+V/√(R 2 + 2 ) cos⁡(ωt+θ-tan -1 )⁡)

Answer: d

Explanation: The complete solution for the current becomes i = e -t [-V/√(R 2 + 2 ) cos⁡(θ-tan -1 )⁡)]+V/√(R 2 + 2 )cos⁡(ωt+θ-tan -1 )⁡).

5. In the circuit shown below, the switch is closed at t = 0, applied voltage is v = 100cos , resistance R = 20Ω and inductance L = 0.1H. The complementary function of the solution of ‘i’ is?

network-theory-questions-answers-sinusoidal-response-rl-q5

a) i c = ce -100t

b) i c = ce 100t

c) i c = ce -200t

d) i c = ce 200t

Answer: c

Explanation: By applying Kirchhoff’s voltage law to the circuit, we have 20i+0.1di/dt=100cos⁡(10 3 t+π/2) => i=1000cos⁡. The complementary function is i c = ce -200t .

6. In the circuit shown below, the switch is closed at t = 0, applied voltage is v = 100cos , resistance R = 20Ω and inductance L = 0.1H. The particular integral of the solution of ‘i p ’ is?

network-theory-questions-answers-sinusoidal-response-rl-q5

a) i p = 0.98cos⁡(1000t+π/2-78.6 o )

b) i p = 0.98cos⁡(1000t-π/2-78.6 o )

c) i p = 0.98cos⁡(1000t-π/2+78.6 o )

d) i p = 0.98cos⁡(1000t+π/2+78.6 o )

Answer: a

Explanation: Assuming particular integral as i p = A cos + B sin. We get i p = V/√(R 2 + 2 ) cos⁡(ωt+θ-tan -1 ) where ω = 1000 rad/sec, V = 100V, θ = π/2, L = 0.1H, R = 20Ω. On substituting, we get i p = 0.98cos⁡(1000t+π/2-78.6 o ).

7. In the circuit shown below, the switch is closed at t = 0, applied voltage is v = 100cos , resistance R = 20Ω and inductance L = 0.1H. The complete solution of ‘i’ is?

network-theory-questions-answers-sinusoidal-response-rl-q5

a) i = ce -200t + 0.98cos⁡(1000t-π/2-78.6 o )

b) i = ce -200t + 0.98cos⁡(1000t+π/2-78.6 o )

c) i = ce -200t + 0.98cos⁡(1000t+π/2+78.6 o )

d) i = ce -200t + 0.98cos⁡(1000t-π/2+78.6 o )

Answer: b

Explanation: The complete solution for the current is the sum of the complementary function and the particular integral. The complete solution for the current becomes i = ce -200t + 0.98cos⁡(1000t+π/2-78.6 o ).

8. The current flowing through the circuit at t = 0 in the circuit shown below is?

network-theory-questions-answers-sinusoidal-response-rl-q5

a) 1

b) 2

c) 3

d) 0

Answer: d

Explanation: At t = 0 that is initially the current flowing through the circuit is zero that is i = 0. So, i = 0.

9. In the circuit shown below, the switch is closed at t = 0, applied voltage is v = 100cos , resistance R = 20Ω and inductance L = 0.1H. The value of c in the complementary function of ‘i’ is?

network-theory-questions-answers-sinusoidal-response-rl-q5

a) c = -0.98cos⁡(π/2-78.6 o )

b) c = -0.98cos⁡(π/2+78.6 o )

c) c = 0.98cos⁡(π/2+78.6 o )

d) c = 0.98cos⁡(π/2-78.6 o )

Answer: a

Explanation: At t = 0, the current flowing through the circuit is zero. Placing i = 0 in the current equation we get c = -0.98cos⁡(π/2-78.6 o ).

10. In the circuit shown below, the switch is closed at t = 0, applied voltage is v = 100cos , resistance R = 20Ω and inductance L = 0.1H. The complete solution of ‘i’ is?

network-theory-questions-answers-sinusoidal-response-rl-q5

a) i = [-0.98 cos⁡(π/2-78.6 o )] exp⁡+0.98cos⁡(1000t+π/2-78.6 o )

b) i = [-0.98 cos⁡(π/2-78.6 o )] exp⁡-0.98cos⁡(1000t+π/2-78.6 o )

c) i = [0.98 cos⁡(π/2-78.6 o )] exp⁡-0.98cos⁡(1000t+π/2-78.6 o )

d) i = [0.98 cos⁡(π/2-78.6 o )] exp⁡+0.98cos⁡(1000t+π/2-78.6 o )

Answer: a

Explanation: The complete solution for the current is the sum of the complementary function and the particular integral.

So, i = [-0.98 cos⁡(π/2-78.6 o )] exp⁡+0.98cos⁡(1000t+π/2-78.6 o ).

This set of Network Theory online test focuses on “Sinusoidal Response of an R-C Circuit”.

1. In the sinusoidal response of R-C circuit, the complementary function of the solution of i is?

a) i c = ce -t/RC

b) i c = ce t/RC

c) i c = ce -t/RC

d) i c = ce t/RC

Answer: a

Explanation: From the R-c circuit, we get the characteristic equation as i=-Vω/R sin⁡. The complementary function of the solution i is i c = ce -t/RC .

2. The particular current obtained from the solution of i in the sinusoidal response of R-C circuit is?

a) i p = V/√(R 2 + 2 ) cos⁡(ωt+θ+tan -1 )

b) i p = -V/√(R 2 + 2 ) cos⁡(ωt+θ-tan -1 )

c) i p = V/√(R 2 + 2 ) cos⁡(ωt+θ-tan -1 )

d) i p = -V/√(R 2 + 2 ) cos⁡(ωt+θ+tan -1 )

Answer: a

Explanation: The characteristic equation consists of two parts, viz. complementary function and particular integral. The particular integral is i p = V/√(R 2 + 2 ) cos⁡(ωt+θ+tan -1 ).

3. The value of ‘c’ in complementary function of ‘i’ is?

a) c = V/R cosθ+V/√(R 2 +) 2 ) cos⁡(θ+tan -1 )

b) c = V/R cosθ+V/√(R 2 +) 2 ) cos⁡(θ-tan -1 )

c) c = V/R cosθ-V/√(R 2 +) 2 ) cos⁡(θ-tan -1 )

d) c = V/R cosθ-V/√(R 2 +) 2 ) cos⁡(θ+tan -1 )

Answer: d

Explanation: Since the capacitor does not allow sudden changes in voltages, at t = 0, i = V/R cosθ. So, c = V/R cosθ-V/√(R 2 +) 2 ) cos⁡(θ+tan -1 ).

4. The complete solution of the current in the sinusoidal response of R-C circuit is?

a) i = e -t/RC [V/R cosθ+V/√(R 2 +) 2 ) cos⁡(θ+tan -1 )+V/√(R 2 + 2 ) cos⁡(ωt+θ+tan -1 ]

b) i = e -t/RC [V/R cosθ-V/√(R 2 + 2 ) cos⁡(θ+tan -1 )-V/√(R 2 + 2 ) cos⁡(ωt+θ+tan -1 ]

c) i = e -t/RC [V/R cosθ+V/√(R 2 + 2 ) cos⁡(θ+tan -1 )-V/√(R 2 + 2 ) cos⁡(ωt+θ+tan -1 ]

d) i = e -t/RC [V/R cosθ-V/√(R 2 +) 2 ) cos⁡(θ+tan -1 )+V/√(R 2 + 2 ) cos⁡(ωt+θ+tan -1 ]

Answer: d

Explanation: The complete solution for the current becomes i = e -t/RC [V/R cosθ-V/√(R 2 +) 2 ) cos⁡(θ+tan -1 )+V/√(R 2 + 2 ) cos⁡(ωt+θ+tan -1 ).

5. In the circuit shown below, the switch is closed at t = 0, applied voltage is v = 50cos , resistance R = 10Ω and capacitance C = 1µF. The complementary function of the solution of ‘i’ is?

network-theory-questions-answers-online-test-q5

a) i c = c exp (-t/10 -10 )

b) i c = c exp(-t/10 10 )

c) i c = c exp (-t/10 -5 )

d) i c = c exp (-t/10 5 )

Answer: c

Explanation: By applying Kirchhoff’s voltage law to the circuit, we have

network-theory-questions-answers-online-test-q5a

(D+1/10 -5 ) )i=-500sin⁡. The complementary function is i c = c exp (-t/10 -5 ).

6. In the circuit shown below, the switch is closed at t = 0, applied voltage is v = 50cos , resistance R = 10Ω and capacitance C = 1µF. The particular integral of the solution of ‘i p ’ is?

network-theory-questions-answers-online-test-q5

a) i p = (4.99×10 -3 ) cos⁡(100t+π/4-89.94 o )

b) i p = (4.99×10 -3 ) cos⁡(100t-π/4-89.94 o )

c) i p = (4.99×10 -3 ) cos⁡(100t-π/4+89.94 o )

d) i p = (4.99×10 -3 ) cos⁡(100t+π/4+89.94 o )

Answer: d

Explanation: Assuming particular integral as i p = A cos + B sin

we get i p = V/√(R 2 + 2 ) cos⁡(ωt+θ-tan -1 )

where ω = 1000 rad/sec, θ = π/4, C = 1µF, R = 10Ω. On substituting, we get i p = (4.99×10 -3 ) cos⁡(100t+π/4+89.94 o ).

7. In the circuit shown below, the switch is closed at t = 0, applied voltage is v = 50cos , resistance R = 10Ω and capacitance C = 1µF. The current flowing in the circuit at t = 0 is?

network-theory-questions-answers-online-test-q5

a) 1.53

b) 2.53

c) 3.53

d) 4.53

Answer: c

Explanation: At t = 0 that is initially current flowing through the circuit is i = V/R cosθ = cos = 3.53A.

8. In the circuit shown below, the switch is closed at t = 0, applied voltage is v = 50cos , resistance R = 10Ω and capacitance C = 1µF. The complete solution of ‘i’ is?

network-theory-questions-answers-online-test-q5

a) i = c exp (-t/10 -5 ) – (4.99×10 -3 ) cos⁡(100t+π/2+89.94 o )

b) i = c exp (-t/10 -5 ) + (4.99×10 -3 ) cos⁡(100t+π/2+89.94 o )

c) i = -c exp(-t/10 -5 ) + (4.99×10 -3 ) cos⁡(100t+π/2+89.94 o )

d) i = -c exp(-t/10 -5 ) – (4.99×10 -3 ) cos⁡(100t+π/2+89.94 o )

Answer: b

Explanation: The complete solution for the current is the sum of the complementary function and the particular integral. The complete solution for the current becomes i = c exp (-t/10 -5 ) + (4.99×10 -3 ) cos⁡(100t+π/2+89.94 o ).

9. In the circuit shown below, the switch is closed at t = 0, applied voltage is v = 50cos , resistance R = 10Ω and capacitance C = 1µF. The value of c in the complementary function of ‘i’ is?

network-theory-questions-answers-online-test-q5

a) c = (3.53-4.99×10 -3 ) cos⁡(π/4+89.94 o )

b) c = (3.53+4.99×10 -3 ) cos⁡(π/4+89.94 o )

c) c = (3.53+4.99×10 -3 ) cos⁡(π/4-89.94 o )

d) c = (3.53-4.99×10 -3 ) cos⁡(π/4-89.94 o )

Answer: a

Explanation: At t = 0, the current flowing through the circuit is 3.53A. So, c = (3.53-4.99×10 -3 ) cos⁡(π/4+89.94 o ).

10. In the circuit shown below, the switch is closed at t = 0, applied voltage is v = 50cos , resistance R = 10Ω and capacitance C = 1µF. The complete solution of ‘i’ is?

network-theory-questions-answers-online-test-q5

a) i = [(3.53-4.99×10 -3 )cos⁡(π/4+89.94 o )] exp⁡+4.99×10 -3 ) cos⁡(100t+π/2+89.94 o )

b) i = [(3.53+4.99×10 -3 )cos⁡(π/4+89.94 o )] exp⁡+4.99×10 -3 ) cos⁡(100t+π/2+89.94 o )

c) i = [(3.53+4.99×10 -3 )cos⁡(π/4+89.94 o )] exp⁡-4.99×10 -3 ) cos⁡(100t+π/2+89.94 o )

d) i = [(3.53-4.99×10 -3 )cos⁡(π/4+89.94 o )] exp⁡-4.99×10 -3 ) cos⁡(100t+π/2+89.94 o )

Answer: a

Explanation: The complete solution for the current is the sum of the complementary function and the particular integral. So, i = [(3.53-4.99×10 -3 )cos⁡(π/4+89.94 o )] exp⁡+4.99×10 -3 ) cos⁡(100t+π/2+89.94 o ).

This set of Network Theory Multiple Choice Questions & Answers focuses on “Sinusoidal Response of an R-L-C Circuit”.

1. The particular current obtained from the solution of i in the sinusoidal response of R-L-C circuit is?

a) i p = V/√(R 2 + 2 ) cos⁡(ωt+θ+tan -1 )⁡/R))

b) i p = V/√(R 2 + 2 ) cos⁡(ωt+θ+tan -1 )⁡/R))

c) i p = V/√(R 2 + 2 ) cos⁡(ωt+θ+tan -1 )⁡/R))

d) i p = V/√(R 2 + 2 ) cos⁡(ωt+θ+tan -1 )⁡/R))

Answer: b

Explanation: The characteristic equation consists of two parts, viz. complementary function and particular integral. The particular integral is i p = V/√(R 2 + 2 ) cos⁡(ωt+θ+tan -1 )⁡/R)).

2. In the sinusoidal response of R-L-C circuit, the complementary function of the solution of i is?

a) i c = c 1 e (K 1 +K 2 )t + c 1 e (K 1 -K 2 )t

b) i c = c 1 e (K 1 -K 2 )t + c 1 e (K 1 -K 2 )t

c) i c = c 1 e (K 1 +K 2 )t + c 1 e (K 2 -K 1 )t

d) i c = c 1 e (K 1 +K 2 )t + c 1 e (K 1 +K 2 )t

Answer: a

Explanation: From the R-L circuit, we get the characteristic equation as

(D 2 +R/L D+1/LC)=0. The complementary function of the solution i is i c = c 1 e (K 1 +K 2 )t + c 1 e (K 1 -K 2 )t .

3. The complete solution of the current in the sinusoidal response of R-L-C circuit is?

a) i = c 1 e (K 1 +K 2 )t + c 1 e (K 1 -K 2 )t – V/√(R 2 + 2 ) cos⁡(ωt+θ+tan -1 )⁡/R))

b) i = c 1 e (K 1 +K 2 )t + c 1 e (K 1 -K 2 )t – V/√(R 2 + 2 ) cos⁡(ωt+θ-tan -1 )⁡/R))

c) i = c 1 e (K 1 +K 2 )t + c 1 e (K 1 -K 2 )t + V/√(R 2 + 2 ) cos⁡(ωt+θ+tan -1 )⁡/R))

d) i = c 1 e (K 1 +K 2 )t + c 1 e (K 1 -K 2 )t + V/√(R 2 + 2 ) cos⁡(ωt+θ-tan -1 )⁡/R))

Answer: c

Explanation: The complete solution for the current becomes i = c 1 e (K 1 +K 2 )t + c 1 e (K 1 -K 2 )t + V/√(R 2 + 2 ) cos⁡(ωt+θ+tan -1 )⁡/R)).

4. In the circuit shown below, the switch is closed at t = 0. Applied voltage is v = 400cos . Resistance R = 15Ω, inductance L = 0.2H and capacitance = 3 µF. Find the roots of the characteristic equation.

network-theory-questions-answers-sinusoidal-response-rlc-q4

a) -38.5±j1290

b) 38.5±j1290

c) 37.5±j1290

d) -37.5±j1290

Answer: d

Explanation: By applying Kirchhoff’s voltage law to the circuit,

network-theory-questions-answers-sinusoidal-response-rlc-q4a

On differentiating the above equation and on solving, we get roots of the characteristic equation as -37.5±j1290.

5. In the circuit shown below, the switch is closed at t = 0. Applied voltage is v = 400cos . Resistance R = 15Ω, inductance L = 0.2H and capacitance = 3 µF. Find the complementary current.

network-theory-questions-answers-sinusoidal-response-rlc-q4

a) i c = e -37.5t (c 1 cos1290t + c 2 sin1290t)

b) i c = e -37.5t (c 1 cos1290t – c 2 sin1290t)

c) i c = e 37.5t (c 1 cos1290t – c 2 sin1290t)

d) i c = e 37.5t (c 1 cos1290t + c 2 sin1290t)

Answer: a

Explanation: The roots of the charactesistic equation are D 1 = -37.5+j1290 and D 2 = -37.5-j1290. The complementary current obtained is i c = e -37.5t (c 1 cos1290t + c 2 sin1290t).

6. In the circuit shown below, the switch is closed at t = 0. Applied voltage is v = 400cos . Resistance R = 15Ω, inductance L = 0.2H and capacitance = 3 µF. Find the particular solution.

network-theory-questions-answers-sinusoidal-response-rlc-q4

a) i p = 0.6cos⁰

b) i p = 0.6cos⁰

c) i p = 0.7cos⁰

d) i p = 0.7cos⁰

Answer: d

Explanation: Particular solution is i p = V/√(R 2 + 2 ) cos⁡(ωt+θ+tan -1 )⁡/R)). i p = 0.7cos⁰.

7. In the circuit shown below, the switch is closed at t = 0. Applied voltage is v = 400cos . Resistance R = 15Ω, inductance L = 0.2H and capacitance = 3 µF. Find the complete solution of current.

network-theory-questions-answers-sinusoidal-response-rlc-q4

a) i = e -37.5t (c 1 cos1290t + c 2 sin1290t) + 0.7cos⁰

b) i = e -37.5t (c 1 cos1290t + c 2 sin1290t) + 0.7cos⁰

c) i = e -37.5t (c 1 cos1290t + c 2 sin1290t) – 0.7cos⁰

d) i = e -37.5t (c 1 cos1290t + c 2 sin1290t) – 0.7cos⁰

Answer: a

Explanation: The complete solution is the sum of the complementary function and the particular integral. So i = e -37.5t (c 1 cos1290t + c 2 sin1290t) + 0.7cos⁰.

8. The value of the c 1 in the following equation is?

i = e -37.5t (c 1 cos1290t + c 2 sin1290t) + 0.7cos⁰.

a) -0.5

b) 0.5

c) 0.6

d) -0.6

Answer: b

Explanation: At t = 0 that is initially the current flowing through the circuit is zero that is i = 0. So, c 1 = -0.71cos ⁰ = 0.49.

9. The value of the c 2 in the following equation is?

i = e -37.5t (c 1 cos1290t + c 2 sin1290t) + 0.7cos⁰.

a) 2.3

b) -2.3

c) 1.3

d) -1.3

Answer: c

Explanation: Differentiating the current equation, we have di/dt = e -37.5t (-1290c 1 sin1290t + 1290c 2 cos1290t) – 37.5e -37.5t (c 1 cos1290t+c 2 sin1290t) – 0.71x500sin(500t+45 o +88.5 o ). At t = 0, di/dt = 1414. On solving, we get c 2 = 1.31.

10. The complete solution of current obtained by substituting the values of c 1 and c 2 in the following equation is?

i = e -37.5t (c 1 cos1290t + c 2 sin1290t) + 0.7cos⁰.

a) i = e -37.5t + 0.7cos⁰

b) i = e -37.5t – 0.7cos⁰

c) i = e -37.5t – 0.7cos⁰

d) i = e -37.5t + 0.7cos⁰

Answer: d

Explanation: The complete solution is the sum of the complementary function and the particular integral. So i = e -37.5t + 0.7cos⁰.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Definition of the Laplace Transform”.

1. The Laplace transform of a function f is?

a) \

e -st

b) \

e -st

c) \

e st

d) \

e st

Answer: a

Explanation: The Laplace transform is a powerful analytical technique that is widely used to study the behavior of linear, lumped parameter circuits. L) = F

network-theory-questions-answers-definition-laplace-q1a

2. Laplace transform changes the ____ domain function to the _____ domain function.

a) time, time

b) time, frequency

c) frequency, time

d) frequency, frequency

Answer: b

Explanation: Laplace transform changes the time domain function f to the frequency domain function F. Similarly Laplace transformation converts frequency domain function F to the time domain function f.

3. In the bilateral Laplace transform, the lower limit is?

a) 0

b) 1

c) ∞

d) – ∞

Answer: d

Explanation: If the lower limit is 0, then the transform is referred to as one-sided or unilateral Laplace transform. In the two-sided or bilateral Laplace transform, the lower limit is – ∞.

4. The unit step is not defined at t =?

a) 0

b) 1

c) 2

d) 3

Answer: a

Explanation: If k is 1, the function is defined as unit step function. And the unit step is not defined at t = 0.

5. The total period of the function shown in the figure is 4 sec and the amplitude is 10. Find the function f 1 from t = 0 to 1 in terms of unit step function.

network-theory-questions-answers-definition-laplace-q5

a) 10t [u – u ]

b) 10t [u + u ]

c) 10t [u + u ]

d) 10t [u – u ]

Answer: d

Explanation: The function shown in the figure is made up of linear segments with break points at 0, 1, 3 and 4 seconds. From the graph, f 1 = 10t for 0 < t < 1. In terms of unit step function, f 1 = 10t [u – u ].

6. Find the function f 2 from the time t = 1 to 3 sec.

a) [u +u ]

b) [u – u ]

c) [u + u ]

d) [u – u ]

Answer: b

Explanation: From the graph, f 2 = -10t + 20 for 1 < t < 3. In terms of unit step function, f 2 = [u – u ]. This function turn off at t = 1, turn off at t = 3.

7. Find the function f 3 from the time t = 3 to 4 sec.

a) [u – u ]

b) [u – u ]

c) [u + u ]

d) [u + u ]

Answer: a

Explanation: From the graph, f 3 = 20t – 40 for 3 < t < 4. In terms of unit step function, f 3 = [u – u ]. This function turn off at t = 3, turn off at t = 4.

8. Find the expression of f in the graph shown below.

network-theory-questions-answers-definition-laplace-q5

a) 10t [u – u ] – [u – u ] + [u – u ]

b) 10t [u – u ] – [u – u ] – [u – u ]

c) 10t [u – u ] + [u – u ] + [u – u ]

d) 10t [u – u ] + [u – u ] – [u – u ]

Answer: c

Explanation: We use the step function to initiate and terminate these linear segments at the proper times. The expression of f is f = 10t [u – u ] + [u – u ] + [u – u ].

9. In the graph shown below, find the expression f .

network-theory-questions-answers-definition-laplace-q9

a) 2t

b) 3t

c) 4t

d) 5t

Answer: c

Explanation: The waveform shown in the figure starts at t = 0 and ends at t = 5 sec. The equation for the above waveform is f = 4t.

10. Find the function f in terms of unit step function in the graph shown below.

network-theory-questions-answers-definition-laplace-q9

a) 4t [u – u ]

b) 4t [u + u ]

c) 4t [u – u ]

d) 4t [u + u ]

Answer: c

Explanation: The waveform shown in the figure starts at t = 0 and ends at t = 5 sec. In terms of unit step function the waveform can be expressed as f = 4t [u – u ].

This set of Network Theory Multiple Choice Questions & Answers focuses on “Operational Transforms”.

1. The Laplace transform of kf is?

a) F

b) kF

c) F/k

d) k 2 F

Answer: b

Explanation: Operational transforms indicate how mathematical operations performed in either f or F are converted into the opposite domain. Linearity property states that L ) = kF .

2. The Laplace transform of f 1 + f 2 is?

a) F 1 + F 2

b) F 1 – F 2

c) F 1 – 2F 2

d) F 1 + 2F 2

Answer: a

Explanation: Addition or subtraction in time domain translates into addition or subtraction in frequency domain. L (f 1 + f 2 ) = F 1 + F 2 .

3. Find the Laplace transform of the function f = 4t 3 + t 2 – 6t + 7.

a) 24/s 4 + 2/s 3 + 6/s 2 + 7/s

b) 24/s 4 – 2/s 3 – 6/s 2 + 7/s

c) 24/s 4 + 2/s 3 – 6/s 2 + 7/s

d) 24/s 4 – 2/s 3 + 6/s 2 + 7/s

Answer: c

Explanation: L (4t 3 + T 2 -6t +7) = 4L (t 3 ) + L(t 2 )-6L + 7L = 4×3!/s 4 + 2!/s 3 – 6 /(s 2 )+71/s = 24/s 4 + 2/s 3 -6/s 2 + 7/s.

4. Find the Laplace transform of the function f = cos2t.

a) (2s 2 +4)/2s(s 2 -4)

b) (2s 2 -4)/2s(s 2 -4)

c) (2s 2 -4)/2s(s 2 +4)

d) (2s 2 +4)/2s(s 2 +4)

Answer: d

Explanation: The Laplace transform of the function f = cos2t is L = L/2) = L+L = 1/2[L+L] = (2s 2 +4)/2s(s 2 +4).

5. Find the Laplace transform of the function f = 3t 4 – 2t 3 + 4e -3t – 2sin5t + 3cos2t.

a) 72/s 5 – 12/s 4 + 4/+10/(s 2 +25)+3s/(s 2 +4)

b) 72/s 5 – 12/s 4 + 4/-10/(s 2 +25)+3s/(s 2 +4)

c) 72/s 5 – 12/s 4 – 4/+10/(s 2 +25)+3s/(s 2 +4)

d) 72/s 5 – 12/s 4 – 4/-10/(s 2 +25)+3s/(s 2 +4)

Answer: b

Explanation: L (3t 4 -2t 3 +4e -3t – 2sin5t +3cos2t) = 3 L (t 4 )-2L (t 3 )+4L (e -3t )-2L + 3L = 72/s 5 ) -12/s 4 +4/-10/(s 2 +25)+3s/(s 2 +4).

6. Find the Laplace transform of e at sinbt.

a) b/ 2 +b 2 )

b) b/ 2 +b 2 )

c) b/ 2 -b 2 )

d) b/ 2 -b 2 )

Answer: a

Explanation: The Laplace transform of sinbt is L=b/(s 2 +b 2 ). So the Laplace transform of e at sinbt is L sinbt)=b/ 2 +b 2 ).

7. Find the Laplace transform of 2 e t .

a) 2/ 3 – 2/ 2 + 4/

b) 2/ 3 – 2/ 2 – 4/

c) 2/ 3 + 2/ 2 + 4/

d) 2/ 3 + 2/ 2 – 4/

Answer: c

Explanation: The Laplace transform of t 2 +2t+4 is L(t 2 +2t+4)=2/ 3 + 2/ 2 +4/s. So the Laplace transform of 2 e t is L 2 e t ) = 2/ 3 + 2/ 2 + 4/.

8. Find the Laplace transform of ramp function r = t.

a) 1/s

b) 1/s 2

c) 1/s 3

d) 1/s 4

Answer: b

Explanation: We know

network-theory-questions-answers-operational-transforms-q8

9.Find the Laplace transform of the function f = tsin2t.

a) 4s/(s 2 +4) 2

b) -4s/(s 2 +4) 2

c) -4s/(s 2 -4) 2

d) 4s/(s 2 -4) 2

Answer: a

Explanation: The Laplace transform of the function of sin2t is L=2/(s 2 +4). So the Laplace transform of the function f = tsin2t is L = -d/ds [2/(s 2 +4)] = 4s/(s 2 +4) 2 .

10.If u = 1 for t >= 0 and u = 0 for t < 0, determine the Laplace transform of [u – u ].

a) 1/s(1+e )

b) 1/s(1-e )

c) 1/s(1+e as )

d) 1/s(1-e as )

Answer: b

Explanation: As u = 1 for t >= 0 and u = 0 for t < 0, the Laplace transform of [u – u ] is L[u – u ] = 1/s-e 1/s = 1/s (1-e ).

This set of Network Theory Multiple Choice Questions & Answers focuses on “Inverse Transforms”.

1. For the function F = (s 2 +s+1)/s, after splitting this function into partial fractions, the co-efficient of the term 1/s is?

a) 1/5

b) 1/10

c) 1/15

d) 1/20

Answer: c

Explanation: To obtain the constant A, multiply the given equation with and putting s = 0. The co-efficient of the term 1/s is A = sF |s=0 = (s 2 +s+1)/|s=0 =1/15.

2. For the function F = (s 2 +s+1)/s, after splitting this function into partial fractions, the co-efficient of 1/ is?

a) 1.1

b) 2.1

c) 3.1

d) 4.1

Answer: b

Explanation: To obtain the constant B, multiply the given equation with and putting s = -5. B = F |s=-5 = (s 2 +s+1)/s|s=-5 = 2.1.

3. For the function F = (s 2 +s+1)/s, after splitting this function into partial fractions, co-efficient of 1/ is?

a) -1.17

b) 1.17

c) -2.27

d) 2.27

Answer: a

Explanation: To obtain the constant C, multiply the given equation with and putting s = -3. C = F |s = -3 = (s 2 +s+1)/s|s=-3 = -1.17.

4. For the function F = (s 2 +s+1)/s, after splitting this function into partial fractions. Find the partial fraction expansion of the function.

a) 1/15s-2.1/+1.17/

b) 1/15s-2.1/-1.17/

c) 1/15s+2.1/+1.17/

d) 1/15s+2.1/-1.17/

Answer: d

Explanation: The values of A, B,C are A = 1/15, B = 2.1, C = -1.17. Partial fraction expansion of the function is (s 2 +s+1)/s =1/15s+2.1/-1.17/.

5. For the function F = /s(s 2 +2s+5), after splitting this function into the partial fractions, 1/s co-efficient is?

a) 1

b) 2

c) 3

d) 4

Answer: a

Explanation: To obtain the constant A, multiply the given equation with and putting s = 0. A= sF|s=0 = /((s 2 +2s+5))=1.

6. For the function F = /s(s 2 +2s+5), after splitting this function into the partial fractions, the co-efficient of 1/ is?

a) 1/2

b) -1/2

c) 1/4

d) -1/4

Answer: b

Explanation: To obtain the constant B, multiply the given equation with and putting s = -1+j2. B = F |s = = /s|s=-1+j2 = -1/2.

7. For the function F = /s(s 2 +2s+5), after splitting this function into the partial fractions, determine the co-efficient of 1/?

a) -1/4

b) 1/4

c) -1/2

d) 1/2

Answer: c

Explanation: To obtain the constant B*, multiply the given equation with and putting s = -1-j2. B* = F |s = -1 – j2 = /s|s=-1-j2 = -1/2.

8. For the function F = /s(s 2 +2s+5), after splitting this function into the partial fractions. What is the expression of F after splitting into partial fractions is?

a) 1/s-1/2 -1/2

b) 1/s+1/2 -1/2

c) 1/s+1/2 +1/2

d) 1/s-1/2 +1/2

Answer: a

Explanation: The expression of F after splitting into partial fractions is /s(s 2 +2s+5) = 1/s-1/2 -1/2.

9. For the function F = /s(s 2 +2s+5), after splitting this function into the partial fractions. What is the inverse transform of F is?

a) 1+ 1/2 e t -1/2 e t

b) 1+ 1/2 e t +1/2 e t

c) 1- 1/2 e t -1/2 e t

d) 1- 1/2 e t +1/2 e t

Answer: c

Explanation: The inverse transform F is f, f = L-1) = L-1 – 1/2) = 1-1/2 e t -1/2 e t .

10. The inverse transform of the function k/ is?

a) ke -at u

b) ke at u

c) ke -at u

d) ke at u

Answer: b

Explanation: The inverse transform of the function k/ is k. The inverse transform of the function k/ is ke at u. k/ <—–> ke at u.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Circuit Elements in the S-Domain”.

1. The resistance element __________ while going from the time domain to frequency domain.

a) does not change

b) increases

c) decreases

d) increases exponentially

Answer: a

Explanation: The s-domain equivalent circuit of a resistor is simply resistance of R ohms that carries a current I ampere seconds and has a terminal voltage V volts-seconds. The resistance element does not change while going from the time domain to the frequency domain.

2. The relation between current and voltage in the case of inductor is?

a) v=Ldt/di

b) v=Ldi/dt

c) v=dt/di

d) v=di/dt

Answer: b

Explanation: Consider an inductor with an initial current I o . The time domain relation between current and voltage is v=Ldi/dt.

3. The s-domain equivalent of the inductor reduces to an inductor with impedance?

a) L

b) sL

c) s 2 L

d) s 3 L

Answer: b

Explanation: If the initial energy stored in the inductor is zero, the equivalent circuit of the inductor reduces to an inductor with impedance sL ohms.

4. The voltage and current in a capacitor are related as?

a) i=Cdt/dv

b) v=Cdv/dt

c) i=Cdv/dt

d) v=Cdt/dv

Answer: c

Explanation: Consider an initially charged capacitor and the initial voltage on the capacitor is V o . The voltage current relation in the time domain is i=Cdv/dt.

5. The s-domain equivalent of the capacitor reduces to a capacitor with impedance?

a) sC

b) C

c) 1/C

d) 1/sC

Answer: d

Explanation: The s-domain equivalent of the capacitor can be derived for the charged capacitor and it reduces to an capacitor with impedance 1/sC.

6. From the circuit shown below, find the value of current in the loop.

network-theory-questions-answers-circuit-elements-s-domain-q6

a) /

b) /

c) /

d) /

Answer: a

Explanation: Applying Kirchhoff’s law around the loop, we have V/s=1/sC I+RI. Solving above equation yields I=CV/=/.

7. After taking the inverse transform of current in the circuit shown below, the value of current is?

network-theory-questions-answers-circuit-elements-s-domain-q6

a) i=e -t/R

b) i=e -t/RC

c) i=e -t/RC

d) i=e -t/R

Answer: c

Explanation: We had assumed the capacitor is initially charged to V o volts. By taking the inverse transform of the current, we get i= e -t/RC .

8. The voltage across the resistor in the circuit shown below is?

network-theory-questions-answers-circuit-elements-s-domain-q6

a) Ve t/R

b) Ve -t/RC

c) Ve -t/R

d) Ve t/RC

Answer: b

Explanation: We can determine the voltage v by simply applying the ohm’s law from the circuit. And applying the Ohm’s law from the circuit v = Ri = Ve -t/RC .

9. The voltage across the resistor in the parallel circuit shown is?

network-theory-questions-answers-circuit-elements-s-domain-q9

a) V/

b) V/

c) V/

d) V/

Answer: c

Explanation: The given circuit is converted to parallel equivalent circuit. By taking the node equation, we get v/R+sCv=CV. Solving the above equation, v=V/.

10. Taking the inverse transform of the voltage across the resistor in the circuit shown below is?

network-theory-questions-answers-circuit-elements-s-domain-q9

a) Ve -t/τ

b) Ve t/τ

c) Ve tτ

d) Ve -tτ

Answer: a

Explanation: By taking the inverse transform, we get v=Ve -t/RC =Ve -t/τ , where τ is the time constant and τ = RC. And v is the voltage across the resistor.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Transfer Function”.

1. The transfer function of a system having the input as X and output as Y is?

a) Y/X

b) Y * X

c) Y + X

d) Y – X

Answer: a

Explanation: The transfer function is defined as the s-domain ratio of the laplace transfrom of the output to the laplace transfrom of the input. The transfer function of a system having the input as X and output as Y is H = Y/X.

2. In the circuit shown below, if current is defined as the response signal of the circuit, then determine the transfer function.

network-theory-questions-answers-transfer-function-q2

a) H=C/(S 2 LC+RCS+1)

b) H=SC/(S 2 LC-RCS+1)

c) H=SC/(S 2 LC+RCS+1)

d) H=SC/(S 2 LC+RCS-1)

Answer: c

Explanation: If the current is defined as the response signal of the circuit, then the transfer function H = I/V = 1/ = sC/(s 2 Lc+RCs+1) where I corresponds to the output Y and V corresponds to the input X.

3. In the circuit shown below, if voltage across the capacitor is defined as the output signal of the circuit, then the transfer function is?

network-theory-questions-answers-transfer-function-q2

a) H=1/(S 2 LC-RCS+1)

b) H=1/(S 2 LC+RCS+1)

c) H=1/(S 2 LC+RCS-1)

d) H=1/(S 2 LC-RCS-1)

Answer: b

Explanation: If the voltage across the capacitor is defined as the output signal of the circuit, the transfer function is H = V o /V = /=1/(S 2 LC+RCS+1).

4. Let us assume x = A cos, then the Laplace transform of x is?

a) X=AØØ/(S 2 -ω 2 )

b) X=AØØ/(S 2 +ω 2 )

c) X=AØØ/(S 2 -ω 2 )

d) X=AØØ/(S 2 +ω 2 )

Answer: d

Explanation: We use the transfer function to relate the study state response to the excitation source. And we had assumed that x = A cos. On expanding and taking the laplace transform we get X = AcosØs/(s 2 +ω 2 )-AsinØs/(s 2 +ω 2 ) = AØØ/(S 2 +ω 2 ).

5. Let us assume x = A cos, what is the s-domain expression?

a) Y=H AØØ/(S 2 -ω 2 )

b) Y=H AØØ/(S 2 +ω 2 )

c) Y=H AØØ/(S 2 +ω 2 )

d) Y=H AØØ/(S 2 -ω 2 )

Answer: c

Explanation: We had the equation Y = HX to find the steady state solution of y. The s-domain expression for the response for the input is Y=H AØØ/(S 2 +ω 2 ).

6. Let us assume x = A cos, on taking the partial fractions for the response we get?

a) Y=k 1 /+(k 1 ‘ )/+Σterms generated by the poles of H

b) Y=k 1 /+(k 1 ‘ )/+Σterms generated by the poles of H

c) Y=k 1 /+(k 1 ‘ )/+Σterms generated by the poles of H

d) Y=k 1 /+(k 1 ‘ )/+Σterms generated by the poles of H

Answer: a

Explanation: By taking partial fractions, Y=k 1 /+(k 1 ‘ )/+Σterms generated by the poles of H. The first two terms result from the complex conjugate poles of the deriving source.

7. Let us assume x = A cos, what is the value of k 1 ?

a) 1/2 HAe jØ

b) HAe -jØ

c) HAe jØ

d) 1/2 HAe -jØ

Answer: d

Explanation: The first two terms on the right hand side of Y determine the steady state response. k 1 =H ØØ)/|s=jω = 1/2 HAe -jØ .

8. The relation between H and θ is?

a) H=e -jθ

b) H=|H|e -jθ

c) H=|H|e jθ

d) H=e jθ

Answer: c

Explanation: In general, H is a complex quantity, thus H = |H|e jθ where |H| is the magnitude and the phase angle is θ of the transfer function vary with frequency ω.

9. Let us assume x = A cos, what is the value of k 1 by considering θ is?

a) |H|e j[θ +Ø]

b) A/2|H|e j[θ +Ø]

c) |H|e -j[θ +Ø]

d) A/2 |H|e -j[θ +Ø]

Answer: b

Explanation: The expression of k 1 becomes K 1 = A/2|H|e j[θ +Ø] . We obtain the steady state solution for y by taking the inverse transform ignoring the terms generated by the poles of H.

10. Let us assume x = A cos, What is the final steady state solution for y ?

a) A|H|cos⁡[ωt+Ø+ θ ]

b) A|H|cos⁡[ωt-Ø+ θ ]

c) A|H|cos⁡[ωt-Ø- θ ]

d) A|H|cos⁡[ωt+Ø- θ ]

Answer: a

Explanation: We obtain the steady state solution for y by taking the inverse transform of Y ignoring the terms generated by the poles of H . Thus y ss = A|H|cos⁡[ωt+Ø+ θ ] which indicates how to use the transfer function to find the steady state response of a circuit.

This set of Network Theory Multiple Choice Questions & Answers focuses on “The Impulse Function in Circuit Analysis”.

1. For the circuit shown below, find the voltage across the capacitor C 1 at the time the switch is closed.

network-theory-questions-answers-impulse-function-q1

a) 0

b) V/4

c) V/2

d) V

Answer: d

Explanation: We use two different cicuits to illustrate how an impulse function can be created with a switching operation. The capacitor is charged to an initial voltage of V at the time the switch is closed.

2. The charge on capacitor C 2 in the circuit shown below is?

network-theory-questions-answers-impulse-function-q1

a) 0

b) 1

c) 2

d) 3

Answer: a

Explanation: In the circuit, the initial charge on C 2 is zero. So, charge on capacitor C 2 = zero. Capacitor circuit and series inductor circuit are two different cicuits to illustrate how an impulse function can be created with a switching operation.

3. The current in the circuit shown below is?

network-theory-questions-answers-impulse-function-q1

a) /(s+1/C e )

b) /(s+1/RC e )

c) /(s-1/RC e )

d) /(s-1/C e )

Answer: b

Explanation: The current flowing through the circuit is given by I = /(R+1/sC 1 +1/sC 2 ) = /(s+1/RC e ) where the equivalent capacitance C 1 C 2 /(C 1 +C 2 ) and is replaced by C e .

4. By taking the inverse transform of the current /(s+1/RC e ), the value of the current is?

network-theory-questions-answers-impulse-function-q1

a) V/R e t/C e

b) V/R e t/RC e

c) V/R e -t/RC e

d) V/R e -t/C e

Answer: c

Explanation: By taking the inverse transform of the current, we obtain i = V/R e -t/RC e which indicates that as R decreases, the initial current increases and the time constant decreases.

5. Consider the circuit shown below. On applying the Kirchhoff’s current law, the equation will be?

network-theory-questions-answers-impulse-function-q5

a) V/++30])/=0

b) V/++30])/=0

c) V/++30])/=0

d) V/++30])/=0

Answer: c

Explanation: The current in the 3H inductor at t = 0 is 10A and the current in 2H inductor at t = 0 is zero. Applying Kirchhoff’s current law, we get V/++30])/=0.

6. The value of the voltage V in the circuit shown below is?

network-theory-questions-answers-impulse-function-q5

a) 40/s – 12/

b) 40/s – 12/

c) 40/s + 12/

d) 40/s + 12/

Answer: d

Explanation: Solving for V yields V=40/s + 12/. The current in the 3H inductor at t = 0 is 10A and the current in 2H inductor at t = 0 is zero.

7. The value of the voltage V after taking the partial fractions in the equation V/++30])/=0 is?

network-theory-questions-answers-impulse-function-q5

a) 12 + 60/s + 10/

b) 12 – 60/s + 10/

c) 12 – 60/s – 10/

d) 12 + 60/s – 10/

Answer: a

Explanation: By taking the partial fractions we get V = 60/s-20/+12+30/ and on solving the equation we get V = 12+60/s+10/.

8. Determine the voltage V after taking the inverse transform ).

network-theory-questions-answers-impulse-function-q5

a) 12δ-(60-10e -5t )u

b) 12δ+(60+10e -5t )u

c) 12δ-(60+10e -5t )u

d) 12δ+(60-10e -5t )u

Answer: b

Explanation: By taking inverse transform of V = 12+60/s+10/ we have v=12δ+(60+10e -5t )u volts and we have to derive the expression for the current when t > 0.

9. The current equation for the circuit shown below is?

network-theory-questions-answers-impulse-function-q5

a) I=4/s-2/

b) I=4/s-2/

c) I=4/s+2/

d) I=4/s+2/

Answer: c

Explanation: After the switch has been opened, the current in L 1 is the same as the current in L 1 . The current equation is I=/. On solving we get I = 4/s-2/.

10. The value of the current after taking the inverse transform of the current is?

network-theory-questions-answers-impulse-function-q5

a) (4-2e 5t )u

b) (4-2e -5t )u

c) (4+2e 5t )u

d) (4-2e -5t )u

Answer: d

Explanation: By taking the inverse transform of I = 4/s-2/ gives i = (4-2e -5t )u. Before the switch is opened,the current in L 1 is 10A and the current in L 2 is 0A.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Problems Involving Standard Time Functions”.

1. A voltage waveform V = 12t 2 is applied across a 1 H inductor for t≥0, with initial current through it being 0. The current through the inductor for t greater than 0 is ___________

a) 12t

b) 24t

c) 12t 3

d) 4t 3

Answer: d

Explanation: We know that, I = \

= 12t 2

∴ I = \(\int_0^t 12t^2 \,dt\)

= \(\big[\frac{12t^3}{3}\big]_0^t\) = 4t 3 A.

2. In the circuit given below, Z 1 = 10∠-60°, Z 2 = 10∠60°, Z 3 = 50∠53.13°. The Thevenin impedance as seen from A-B is _____________

network-theory-questions-answers-problems-involving-standard-time-functions-q2

a) 56.5∠45°

b) 60∠30°

c) 70∠30°

d) 34.4∠65°

Answer: a

Explanation: Z TH = Z A-B = Z 1 || Z 2 + Z 3

= \(\frac{Z_1 × Z_2}{Z_1 + Z_2} + Z_3\)

= \(\frac{10∠-60° × 10∠60°}{10∠-60° + 10∠60°}\) + 50∠53.13°

= 56.66∠45°.

3. In the circuit given below, the value of Z which would cause parallel resonance at 500 Hz is ___________

network-theory-questions-answers-problems-involving-standard-time-functions-q3

a) 125 mH

b) 34.20 μF

c) 2 μF

d) 0.05 μF

Answer: d

Explanation: At resonance the circuit should have unity power factor

Hence, Z should be capacitive.

Now, \(\frac{1}{jLω} + \frac{1}{1/jCω}\) = 0

Or, \(\frac{-1}{jLω}\) + jCω = 0

∴ C = \(\frac{1}{L × ω^2} = \frac{1}{2 × ^2}\)

= 0.05 μF.

4. In the circuit given below, the current source is 1∠0° A, R = 1 Ω, the impedances are Z C = -j Ω and Z L = 2j Ω. The Thevenin equivalent looking across A-B is __________

network-theory-questions-answers-problems-involving-standard-time-functions-q4

a) \ Ω

b) 2∠45° V, Ω

c) 2∠45° V, Ω

d) \ Ω

Answer: d

Explanation: Z TH = 1 + = Ω

∴ V TH = 1 = \(\sqrt{2}\)∠45° V.

5. A periodic rectangular signal X has the waveform as shown below. The frequency of the fifth harmonic of its spectrum is ______________

network-theory-questions-answers-problems-involving-standard-time-functions-q5

a) 40 Hz

b) 200 Hz

c) 250 Hz

d) 1250 Hz

Answer: d

Explanation: Periodic time = 4 ms = 4 × 10 -3

Fundamental frequency = \(\frac{10^3}{4}\) = 250 Hz

∴ Frequency of the fifth harmonic = 250 × 5 = 1250 Hz.

6. A voltage having the waveform of a sine curve is applied across a capacitor. When the frequency of the voltage is increased, what happens to the current through the capacitor?

a) Increases

b) Decreases

c) Remains same

d) Is zero

Answer: a

Explanation: The current through the capacitor is given by,

I C = ωCV cos .

As the frequency is increased, I C also increases.

7. An inductive coil connected to a 200 V, 50 Hz AC supply with 10 A current flowing through it dissipates 1000 W. Which of the following will have least value?

a) Resistance R

b) Reactance X L

c) Impedance Z

d) Reactance X C

Answer: a

Explanation: Power dissipated is given by,

I 2 R = 1000 W

Hence, R = \(\frac{1000}{100}\)

Or, R = 10 Ω

Impedance Z = \(\frac{V}{I} = \frac{200}{10}\)

Or, Z = 20 Ω

Reactance X L = \(\sqrt{Z^2 – R^2}\)

= \(\sqrt{20^2 – 10^2}\)

∴ X L = 17.3 Ω.

8. An inductive coil connected to a 200 V, 50 Hz AC supply with 10 A current flowing through it dissipates 1000 W. The value of the impedance Z and Reactance X L are respectively ____________

a) 10 Ω and 15 Ω

b) 20 Ω and 17.3 Ω

c) 17.3 Ω and 20 Ω

d) 15 Ω and 10 Ω

Answer: b

Explanation: Power dissipated is given by,

I 2 R = 1000 W

Hence, R = \(\frac{1000}{100}\)

Or, R = 10 Ω

Impedance Z = \(\frac{V}{I} = \frac{200}{10}\)

Or, Z = 20 Ω

Reactance X L = \(\sqrt{Z^2 – R^2}\)

= \(\sqrt{20^2 – 10^2}\)

∴ X L = 17.3 Ω.

9. A relay coil having voltage of magnitude 210 V and frequency 50 Hz is connected to a 210 V, 50 Hz supply. If it has resistance of 50 Ω and an inductance of 0.2 H, the apparent power is _____________

a) 549.39 VA

b) 275.6 VA

c) 157 VA

d) 187 VA

Answer: a

Explanation: Z= 50 + j = 50 + 62.8

S = \(\frac{|V|^2}{Z} = \frac{210^2}{50 – 62.8}\)

Apparent Power |S| = \(\frac{210^2}{\sqrt{50^2 + 62.8^2}}\)

= 549.39 VA.

10. In the circuit given below, the value of Z L for maximum power to be transferred is _____________

network-theory-questions-answers-problems-involving-standard-time-functions-q10

a) R

b) R + jωL

c) R – jωL

d) jωL

Answer: c

Explanation: The value of load for maximum power transfer is given by the complex conjugate of Z AB

Z AB = R + jX L

= R + jωL

∴ Z L for maximum power transfer is given by Z L = R – jωL.

11. A 10 V is connected across a load whose V-I characteristics is given by 7I = V 2 + 2V. The internal resistance of the battery is of magnitude 1Ω. The current delivered by the battery is ____________

a) 6 A

b) 5 A

c) 7 A

d) 8 A

Answer: b

Explanation: 7I = V 2 + 2V ………………….

Now, V = 10 – 1 ×I

Putting the value of V in eqt , we get,

&I = 2 + 2

Or, I = 100 + I2 – 20I + 20 – 2I

Or, I2 – 29I + 120 = 0………………….

∴ I = \

∴ I = 5 A.

12. A current I given by I = – 8 + 6\) A is passed through three meters. The respective readings will be?

a) 8, 6 and 10

b) 8, 6 and 8

c) – 8, 10 and 10

d) -8, 2 and 2

Answer: c

Explanation: PMMC instrument reads only DC value and since it is a centre zero type, so it will give – 8 values.

So, rms = \(\sqrt{8^2 + \left

^2\right)}\) = 10 A

Moving iron also reads rms value, so its reading will also be 10 A.

13. A rectifier type AC voltmeter consists of a series resistance R, an ideal full-wave rectifier bridge and a PMMC instrument. A DC current of 1 mA produces a full-scale deflection of the instrument. The resistance of the instrument is 100 Ω. A voltage of 100 V is applied to the input terminals. The value of R required is?

a) 63.56 Ω

b) 89.83 Ω

c) 89.83 kΩ

d) 141.3 kΩ

Answer: c

Explanation: V OAverage = 0.636 × \(\sqrt{2}\)V rms = 0.8993 V rms

The deflection with AC is 0.8993 times that with DC for the same value of voltage V

S AC = 0.8993 S DC

S DC of a rectifier type instrument is \(\frac{1}{I_{fs}}\) where I fs is the current required to produce full scale deflection, I fs = 1 mA; R m = 100 Ω; S DC = 10 3 Ω/V

S AC = 0.8993 × 1000 = 899.3 Ω/V. Resistance of multiplier R S = S AC V – R m – 2R d , where R d is the resistance of diode, for ideal diode R d = 0

∴ R S = 899.3 × 100 – 100 = 89.83 kΩ.

14. The discharge of a capacitor through a ballistic galvanometer produces a damped frequency of 0.125 Hz and successive swings of 120, 96 and 76.8 mm. The logarithmic decrement is?

a) 0.225

b) 0.565

c) 0.484

d) 0.7887

Answer: a

Explanation: Logarithmic decrement, δ = ln \(\frac{x_1}{x_2} = ln \frac{120}{96}\) = 0.225

Now, δ is related to damping ratio K as, K = \(\frac{1}{\left

^2)^{0.5}\right)}\)

∴ K = 0.0357.

15. 12, 1 H inductors are used as edge to form a cube. The inductance between two diagonally opposite corners of the cube is ____________

a) \

2 H

d) \(\frac{3}{2}\) H

Answer: a

Explanation: V AB = \(\frac{1}{3}L + \frac{1}{6} L + \frac{1}{3}L\)

V AB = IL \([\frac{5}{6}] \)

Now, \(\frac{V_{AB}}{I} = \frac{5L}{6}\)

Or, L AB = \(\frac{5L}{6}\)

Here, L = 1Ω

∴ L AB = \(\frac{5}{6}\) H.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Laplace Transform of a Periodic Function”.

1. For a network having 1 Ω resistor and 1 F capacitor in series, the impedance Z is ____________

a) \

\(\frac{s}{s+2}\)

Answer: a

Explanation: We know that the impedance, Z is given by,

Z = resistance + 1/capacitor

= 1 + \(\frac{1}{s}\)

= \(\frac{s+1}{s}\).

2. A system function H = \Missing open brace for subscript 5, 10

b) 5, 4

c) 10, 4

d) 10, 10

Answer: c

Explanation: Given that, H = \(\frac{25}{s^2+4s+100}\)

So, we can infer that, ω n = 0.5

= 10

And hence, ω r ≃ 10

So, resonant frequency = 10 rad/sec

And Bandwidth = 10 rad/sec.

3. Given a series RLC circuit. The impedance Z of the circuit will be?

a) \

\(\frac{s^2+10s+100}{s}\)

Answer: d

Explanation: The impedance Z = R + sL + \(\frac{1}{sC}\)

= \(\frac{s^2 LC+RCs+1}{sC}\)

This is similar to \(\frac{s^2+10s+100}{s}\).

4. A system function H = \

= u and value of system is 0 for t<0. Then v is ___________

a) 1 – e -4t

b) e -4t

c) e 4t

d) 1 + e -4t

Answer: b

Explanation: Given that, V = \

= e -4t .

5. A system is at rest for t < 0. It is given by \

sin⁡. If steady state is reached at t = 0, then the value of angle A is ___________

a) 0°

b) 45°

c) -45°

d) ∞

Answer: b

Explanation: Given that, \

sin⁡

Or, y.s + 2y = \(\frac{ω}{s^2+4}\)

Or, s + 2 = 0

Or, j ω + 2 = 0

Or, j2 + 2 = 0

Or, 1 + j = 0

Or, tan -1 = 45°.

6. The value of 2 [u – u ] + u ) at t = 3 sec is ____________

a) 0

b) 4

c) ∞

d) 1

Answer: b

Explanation: Putting t=3 in the given equation, we get,

2[u – u ] [u + u ]

= 2 [1 – 0] [1 + 1]

= 4.

7. Barlett’s Bisection Theorem is applicable to ___________

a) Unsymmetrical networks

b) Symmetrical networks

c) Both unsymmetrical and symmetrical networks

d) Neither to unsymmetrical nor to symmetrical networks

Answer: b

Explanation: A symmetrical network can be split into two halves.

So the z parameters of the network are symmetrical as well as reciprocal of each other. Hence Barlett’s Bisection Theorem is applicable to Symmetrical networks.

8. The values of z 11 and z 21 for a T circuit having resistances 20 Ω each is _____________

a) 40, 20

b) 40, 60

c) 60, 40

d) 40, 40

Answer: a

Explanation: To determine the values of z 11 and z 21 , we apply a voltage source V 1 to the input port and leave the output port open.

Thus, z 11 = \(\frac{V_1}{I_1} = \frac{ I_1}{I_1}\)

= 40 Ω

Now, z 21 is the input impedance at port 1.

So, z 21 = \(\frac{V_2}{I_1}\) = 20 Ω.

9. The values of z 12 and z 22 for a T circuit having resistances 20 Ω each is _____________

a) 40, 20

b) 40, 60

c) 20, 40

d) 40, 40

Answer: c

Explanation: To find z 12 and z 22 , we apply a voltage source V 2 to the output port and leave the input port open.

Thus, z 12 = \(\frac{V_1}{I_2} = \frac{ I_2}{I_2}\)

= 20 Ω

Now, z 22 is the input impedance at port 1.

So, z 22 = \(\frac{V_2}{I_2} = \frac{ I_2}{I_2}\) = 40 Ω.

10. If y = 120e 10x , then the relation is _________

a) Dynamic

b) Static

c) Memory

d) Memoryless but not static

Answer: b

Explanation: Given relation, y = 120 10x .

The system represented by the above relation is static because the present output of the system as well as memoryless as its present output does not depend on its past input. It is not a dynamic system since the value of the system increases exponentially.

11. The z parameters form a matrix of the form ___________

a) [z 11 z 12 ; z 21 z 22 ]

b) [z 11 z 12 ; z 22 z 21 ]

c) [z 12 z 11 ; z 21 z 22 ]

d) [z 11 z 22 ; z 12 z 21 ]

Answer: a

Explanation: Z parameters are also called as the impedance parameters.

There are 4 main types of Z-parameters, z 11 , z 12 , z 21 , z 22

They are arranged in the form of a matrix given by [z 11 z 12 ; z 21 z 22 ].

12. The Laplace transform of the function e -2t cos + 5e -2t sin is ____________

a) \

\(\frac{-15}{^2+9}\)

Answer: c

Explanation: L {e -2t cos + 5e -2t sin} = \(\frac{}{^2+9} + \frac{5 3}{^2+9}\)

= \(\frac{+15}{^2+9}\).

13. The Laplace transform of the function 6e 5t cos – e 7t is ______________

a) \

\(\frac{6}{^2+4} + \frac{1}{s-7}\)

Answer: a

Explanation: L {6e 5t cos – e 7t } = \(\frac{6}{^2+4} – \frac{1}{s-7}\).

14. The Laplace transform of the function cosh 2 is ____________

a) \

\(\frac{s^2+2}{s

}\)

Answer: b

Explanation: L ((\(\frac{1}{2}\)(e t – e -t )) 2 )

= L \Missing or unrecognized delimiter for \right\)

= \(\frac{1}{4} \frac{1}{s-2} + \frac{1}{2} \frac{1}{s} + \frac{1}{4} \frac{1}{s+2}\)

= \(\frac{s^2-2}{s

}\).

15. Given a system function H = \

is a unit step, then V in the steady state is ___________

a) \

d) ∞

Answer: a

Explanation: V = \

= \frac{}{s^2}\)

At steady state, sV = \(\frac{0+4}{^2}\)

= \(\frac{4}{9}\).

This set of Network Theory Multiple Choice Questions & Answers focuses on “Problems on Initial and Final Value Theorem”.

1. What is the steady state value of F , if it is known that F = \

Cannot be determined

Answer: b

Explanation: From the equation of F, we can infer that, a simple pole is at origin and all other poles are having negative real part.

∴ F = lim s→0 s F

= lim s→0 \(\frac{2s}{s}\)

= \(\frac{2}{}\)

= \(\frac{2}{6} = \frac{1}{3}\).

2. What is the steady state value of F , if it is known that F = \

b) –\

Cannot be determined

Answer: d

Explanation: The steady state value of this Laplace transform is cannot be determined since; F has a pole s = 1. Hence the answer is that it cannot be determined.

3. What is the steady state value of F , if it is known that F = \

Cannot be determined

c) 0

d) \(\frac{1}{8}\)

Answer: c

Explanation: The steady state value of F exists since all poles of the given Laplace transform have negative real part.

∴F = lim s→0 s F

= lim s→0 \(\frac{s}{^2 }\)

= 0.

4. What is the steady state value of F , if it is known that F = \

-5

b) 5

c) 10

d) Cannot be determined

Answer: d

Explanation: The steady state value of this Laplace transform is cannot be determined since; F is having two poles on the imaginary axis . Hence the answer is that it cannot be determined.

5. What is the steady state value of F , if it is known that F = b/), where a>0?

a) \

d) Cannot be determined

Answer: a

Explanation: F = lim s→0 s F

= lim s→0 \(\frac{sb}{s}\)

= lim s→0 \(\frac{b}{}\)

= \(\frac{b}{a}\).

6. The inverse Laplace transform of F = \Missing open brace for subscript -2e -2t + 2e -t

b) 2e -2t + 2e -t

c) -2e -2t – 2e -t

d) 2e -t + e -2t

Answer: a

Explanation: s 2 + 3s + 2 =

Now, F = \

F | s=-2

= \(\frac{2}{s+1}\)| s=-2 = -2

And, B = F | s=-1

= \(\frac{2}{s+2}\)| s=-1 = 2

∴ F = \

= L -1 {F}

= -2e -2t + 2e -t for t≥0.

7. The inverse Laplace transform of F = \Missing open brace for subscript 2e -k u

b) 2e -k u

c) 2e k u

d) 2e k u

Answer: b

Explanation: Let G = \

= L -1 {G} = 2e -ct

∴ F = L -1 {G e -bs }

= 2e -k u .

8. The inverse Laplace transform of F = \Missing open brace for subscript 3 sin – \

b) 3 sin + \

c) 3 cos + \

d) 3 cos – \

Answer: c

Explanation: F = L -1 {F}

= L -1 \(\Big\{\frac{3s}{s^2+7} + \frac{5}{s^2+7}\Big\}\)

= L -1 \

+ \

9. The inverse Laplace transform of F = \Missing open brace for subscript \

\(\frac{5}{6}e^{3t} – \frac{5}{6e^{-3t}}\)

Answer: d

Explanation: F = L -1 {F}

= L -1 {\(\frac{5}{s^2-9}\)}

= L -1 \

\(\frac{5}{}|_{s=-3} = -\frac{5}{6}\)

The inverse Laplace transform is \(\frac{5}{6}e^{3t} – \frac{5}{6e^{-3t}}\).

10. The inverse Laplace transform of F = \Missing open brace for subscript {0.5 + 0.5e -2 – e - } u

b) {0.5 + 0.5e -2 – e - } u

c) {0.5 – 0.5e -2 – e - } u

d) 0.5 + 0.5e -2t – e -t )

Answer: b

Explanation: Let G = \

= G e -3s

G = L -1 {G}

= L -1 \

= 0.5 + 0.5e -2t -e -t

The inverse Laplace transform is F = {0.5 + 0.5e -2 – e - } u .

11. Given a sinusoidal voltage that has a peak to peak value of 100 V. The RMS value of the sinusoidal voltage is ___________

a) 50 V

b) 70.7 V

c) 35.35 V

d) 141.41 V

Answer: c

Explanation: Given that,

Peak value = 50 V

We know that, RMS value = \(\frac{Peak \,Value}{\sqrt{2}}\)

= \(\frac{50}{\sqrt{2}}\) = 35.35 V.

12. The Laplace transform of F = sincos is ______________

a) \

\(\frac{1}{2s} + \frac{s}{2

}\)

Answer: a

Explanation: Using the Trigonometric Identity,

We get, sin cos = \

)

∴L {sin cos } = \(\frac{4}{2

}\).

13. Convolution of step signal 100 times that is 100 convolution operations. The Laplace transform is ______________

a) \

d) s 100

Answer: a

Explanation: n times = u * u * …… * u

Laplace transform of the above function = \(\frac{1}{s^n}\), where n is number of convolutions.

∴ Laplace transform for 100 convolutions = \(\frac{1}{s^{100}}\).

14. For the circuit given below, the Time-constant is __________

network-theory-questions-answers-problems-initial-final-value-theorem-q14

a) 1.5

b) 1.25

c) 2.5

d) 2.25

Answer: b

Explanation: We know that,

Time constant is given by R eq .C

The equivalent resistance is given by,

R eq = R || R

= \(\frac{R*R}{R+R}\)

= \(\frac{R}{2}\)

= 5 Ω

So, Time-constant = \(\frac{5X0.5}{2}\)

= 1.25.

15. In a dual slop integrating type digital voltmeter, the first integrating is carried out for 50 periods of the supply frequency of 50 Hz. If the reference voltage used is 10 V, the total conversion time for an input of 40 V is?

a) 3 s

b) 2 s

c) 4 s

d) 1 s

Answer: c

Explanation: In a dual slope integrating digital voltmeter,

\) V in = V ref

Where, t 1 = first integration time = 50 × \(\frac{1}{50}\) = 1

But V in = 40 V and V ref = 10 V

∴ t 2 = \(\frac{V_{in} t_1}{V_{ref}}\) = 4 s.

This set of Tough Network Theory Questions and Answers focuses on “Calculation of I, di/dt, d2I/dt2 in Circuits Involving both Capacitor and Inductor”.

1. A coil of inductance 10 H, resistance 40 Ω is connected as shown in the figure. The switch S connected with point 1 for a very long time, is moved to point 2 at t=0. If, at t=0 + , the voltage across the coil is 120 V, the value of the resistance R is __________

tough-network-theory-questions-answers-q1

a) Zero

b) 20 Ω

c) 40 Ω

d) 60 Ω

Answer: a

Explanation: I L at 0 – = \

∴ R = 0.

2. A coil of inductance 10 H, resistance 40 Ω is connected as shown in figure. The switch S connected with point 1 for a very long time, is moved to point 2 at t=0. For the value of R obtained, the time taken for 95 % of the stored energy dissipated is ________

tough-network-theory-questions-answers-q1

a) 0.10 s

b) 0.15 s

c) 0.50 s

d) 1.0 s

Answer: c

Explanation: For source free circuit,

I = I o \

= 0.05 = 2 × \(e^{-\frac{60}{10} t}\)

Or, t = 0.61 ≈ 0.5 s.

3. In the circuit shown below steady state is obtained before the switch closes at t = 0. The switch is closed for 1.5 s and is then opened. At t = 1 s, V will be?

tough-network-theory-questions-answers-q3

a) – 3.24 V

b) 1.97 V

c) 5.03 V

d) 13.24 V

Answer: c

Explanation: V (0 – ) = 10 V = V (0 + )

For 0<t≤1.5 s, τ = 4 × 0.05 = 0.2 s

V OC = 5 V

V = 5 + \(e^{\frac{-t}{0.2}}\) = 5 + 5e -5t

V = 5.03 V.

4. The circuit shown below is at steady state before the switch closes at t=0. The switch is closed for 1.5 s and is then opened. At t = 2 s, V will be?

tough-network-theory-questions-answers-q3

a) 5.12 V

b) 6.43 V

c) 8.57 V

d) 9.88 V

Answer: c

Explanation: V = 5.002 V

For t > 1.5 s, τ = 8 × 0.05 = 0.4

V = 10+ \(e^{\frac{-}{0.4}}\) = 10 – 5e -2.5

For t≥1.5 s, V = 8.57 V.

5. In the circuit given below, the voltage across capacitor when switch is closed at t = ∞ is ____________

tough-network-theory-questions-answers-q5

a) 50 V

b) 20 V

c) 30 V

d) 7.5 V

Answer: a

Explanation: From the figure, we can infer that,

Voltage across capacitor = voltage across 10 Ω resistance.

Now, voltage across the 10 Ω resistance = \(\frac{100}{10+10}\) X 10

= \(\frac{100}{20}\).10

= 50 V.

6. In the circuit given below, the current source is 1 A, voltage source is 5 V, R 1 = R 2 = R 3 = 1 Ω, L 1 = L 2 = L 3 = 1 H, C 1 = C 2 = 1 F. The current through R 3 is _________

tough-network-theory-questions-answers-q6

a) 1 A

b) 5 A

c) 6 A

d) 8 A

Answer: b

Explanation: At steady state, the circuit becomes,

tough-network-theory-questions-answers-q6a

∴ The current through R 3 = \(\frac{5}{1}\) = 5 A.

7. In the circuit given below, the capacitor is initially having a charge of 10 C. 1 second after the switch is closed, the current in the circuit is ________

tough-network-theory-questions-answers-q7

a) 14.7 A

b) 18.5 A

c) 40.0 A

d) 50.0 A

Answer: a

Explanation: Using KVL, 100 = R\(\frac{dq}{dt} + \frac{q}{C}\)

100 C = RC\(\frac{dq}{dt}\) + q

Or, \(\int_{q_o}^q \frac{dq}{100C-q} = \frac{1}{RC} \int_0^t \,dt\)

100C – q = (100C – q o )e -t/RC

I = \(\frac{dq}{dt} = \frac{

}{RC} e^{-1/1}\)

∴ e -t/RC = 40e -1 = 14.7 A.

8. In the circuit given below, the current source is 1 A, voltage source is 5 V, R 1 = R 2 = R 3 = 1 Ω, L 1 = L 2 = L 3 = 1 H, C 1 = C 2 = 1 F. The current through the voltage source V is _________

tough-network-theory-questions-answers-q6

a) 1 A

b) 3 A

c) 2 A

d) 4 A

Answer: d

Explanation: At steady state, the circuit becomes,

tough-network-theory-questions-answers-q6a

∴ The current through the voltage source V = 5 – 1 = 4 A.

9. In the circuit given below, the voltage across R by mesh analysis is _________

tough-network-theory-questions-answers-q9

a) 1.59 V

b) 1 V

c) 2.54 V

d) 1.54 V

Answer: d

Explanation: In loop 1, by KVL, I 1 + I 2 + 0I 3 = 5

Or, 12 I 1 – 2I 2 = 5

In loop 2, -2I 1 + I 2 + I 3 = 0

Or, -2I 1 + 34 I 2 – 2I 3 = 0

In loop 3, 0I 1 + – 2I 2 + I 3 = 10

Or, 0I 1 – 2I 2 + 12I 3 = 10

Now, the voltage across R is V R = (I 2 – I 3 ) R

Now, I 2 = \(\frac{360}{4800} = \frac{3}{40}\)

Now, I 3 = \(\frac{4060}{4800} = \frac{203}{240}\) A

Therefore, V R = (I 2 – I 3 ) R = \(\Big[\frac{3}{40} – \frac{203}{240}\Big] 2\) = 1.54 V.

10. In a dual slop integrating type digital voltmeter, the first integrating is carried out for 10 periods of the supply frequency of 50 Hz. If the reference voltage used is 2 V, the total conversion time for an input of 1 V is _________

a) 0.01 s

b) 0.05 s

c) 0.1 s

d) 1 s

Answer: c

Explanation: In a dual slope integrating digital voltmeter,

\) V in = V ref

Where, t 1 = first integration time = 10 × \(\frac{1}{50}\) = 0.25

But V in = 1 V and V ref = 2 V

∴ t 2 = \(\frac{V_{in} t_1}{V_{ref}}\) = 0.1 s.

11. A rectifier type AC voltmeter consists of a series resistance R, an ideal full-wave rectifier bridge and a PMMC instrument. The internal resistance and a full-scale deflection produced by a DC current are 100 Ω and 1 mA respectively. A voltage of 100 V is applied to the input terminals. The value of R required is _________

a) 63.56 Ω

b) 89.83 Ω

c) 89.83 kΩ

d) 141.3 kΩ

Answer: c

Explanation: V OAverage = 0.636 × \(\sqrt{2}\)V rms = 0.8993 V rms

The deflection with AC is 0.8993 times that with DC for the same value of voltage V

S AC = 0.8993 S DC

S DC of a rectifier type instrument is \(\frac{1}{I_{fs}}\) where I fs is the current required to produce full scale deflection, I fs = 1 mA; R m = 100 Ω; S DC = 10 3 Ω/V

S AC = 0.8993 × 1000 = 899.3 Ω/V. Resistance of multiplier R S = S AC V – R m – 2R d , where R d is the resistance of diode, for ideal diode R d = 0

∴ R S = 899.3 × 100 – 100 = 89.83 kΩ.

12. A 1 μF capacitor is connected across a 50 V battery. The battery is kept closed for a long time. The circuit current and voltage across capacitor is __________

a) 0.5 A and 0 V

b) 0 A and 50 V

c) 20 A and 5 V

d) 0.05 A and 5 V

Answer: b

Explanation: We know that, when the capacitor is fully charged, it acts as an open circuit.

That is when the capacitor is fully charged.

So, the circuit current and voltage across capacitor are 0 A and 50 V respectively.

13. In the circuit given below, the switch is closed at t = 0. At t = 0 + the current through C is ___________

tough-network-theory-questions-answers-q5

a) 2 A

b) 3 A

c) 4 A

d) 5 A

Answer: d

Explanation: We know that,

When the capacitor is fully charged, it acts as a short circuit.

The equivalent resistance R eq = Ω

= 20 Ω

Given voltage = 100 V

So, current through C = \(\frac{100}{20}\) A = 5 A.

14. A digital Multimeter reads 10 V when fed with a triangular wave, which is symmetric about the time-axis. If the input is same, the rms reading meter will read?

a) \

–\

\(\frac{10}{\sqrt{3}}\)

Answer: d

Explanation: For triangular wave Average value = \(\frac{V_m}{3}\), rms value = \(\frac{V_m}{\sqrt{3}}\)

∴ \(\frac{V_m}{3}\) = 10 V or, V m = 30 V.

15. A resistance R is measured using the connection shown in the below figure.

tough-network-theory-questions-answers-q15

The current measured is 10 A on ranges 100A and the voltage measured is 125 V on 150 V range. The scales of the ammeter and voltmeter are uniform. The total number of scale divisions of the ammeter is 100 and that of the voltmeter is 150. The scale division can be distinguished. The constructional error of the ammeter is ± 0.3% and that of voltmeter±0.4%. The resistance of the ammeter is 0.25 Ω. The value of R is _________

a) 12.75 Ω

b) 12.0 Ω

c) 12.25 Ω

d) 12.5 Ω

Answer: c

Explanation: Percentage error in ammeter = ± \(\frac{1}{10×100}\) × 100 = ± 0.1%

Percentage error in voltmeter= ±\(\frac{1}{10 ×150}\) × 100 = ± 0.067%

So, δI = ± 0.3 ± 0.1 = ± 0.4%

δV = ± 0.4 ± 0.067 = ± 0.467%

R = \(\frac{V}{I}\)

So, error = ± δV ± δI = ± 0.867

Measured value of resistance = R m = \(\frac{125}{10}\) = 12.5

∴ True value = R m

= 12.25 Ω.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Advanced Problems on Application of Laplace Transform – 1”.

1. The resistance of a 230 V, 100 W lamp is ____________

a) 529 Ω

b) 2300 Ω

c) 5290 Ω

d) 23 Ω

Answer: a

Explanation: P = \(\frac{V^2}{R}\)

Or, R = \(\frac{V^2}{P}\)

Or, R = \(\frac{230 X 230}{100}\)

= 529 Ω.

2. A network has two branches in parallel. One branch contains impedance Z a and the other branch has impedance Z b . If it is fed from an AC voltage V of frequency f, the current through Z a depends, on which of the following?

a) V, Z a , Z b

b) V, Z a

c) Z a , Z b

d) V, Z a , f

Answer: d

Explanation: We know that, in parallel branches, the current through any branch depends only on V , Z and f only.

So, the current through Z a depends on V, Z a , and f.

3. Two coils having self-inductances of 1 mH and 2 mH are mutually coupled. The maximum possible mutual inductance is ___________

a) 1.414 mH

b) 2 mH

c) 1 mH

d) 5.5 mH

Answer: a

Explanation: We know that, maximum mutual inductance = L 1 L 2

Given that, L 1 = 1 mH and L 2 = 2 mH

So, maximum mutual inductance = 0.5 = \(\sqrt{2}\)

= 1.414 mH.

4. A constant k band pass filter has a pass band from 100 to 400 Hz. The resonant frequency of series and shunt arms is ____________

a) 300 Hz

b) 250 Hz

c) 200 Hz

d) 150 Hz

Answer: c

Explanation: We know that,

Resonant frequency of arms in constant k band pass filter = [(f c1 )(f c2 )] 0.5

Given that, f c1 = 100 Hz and f c2 = 400 Hz

So, Resonant frequency = \(\sqrt{100 X 400}\)

= 200 Hz.

5. An RLC series circuit has Q = 100 and ω 0 = 20 rad/sec. The bandwidth is ____________

a) 0.2 rad/sec

b) 2 rad/sec

c) 20 rad/sec

d) 2000 rad/sec

Answer: a

Explanation: We know that,

Bandwidth = \(\frac{ω_0}{Q}\)

Given that, ω 0 = 20 rad/sec and Q = 100

So, Bandwidth = \(\frac{20}{100}\)

= 0.2 rad/sec.

6. In an unloaded transformer, the flux limiting the primary is 10 mWb and secondary is 30 mWb. The coefficient of coupling is ____________

a) 1

b) 0.1

c) 0.33

d) 0.67

Answer: c

Explanation: Φ 11 = Φ 1

Or, φ 11 = φ 1 – φ 2

Or, φ 11 = 30 – 10 = 20 mWb

Now, 20 mWb = 30m

Or, 0.67 = 1 – k

Or, k = 0.33.

7. Poles and zeros of a driving point function of a network are simple and alternate on jω axis. The network consists of ___________

a) R and C

b) L and C

c) R and L

d) R, L and C

Answer: b

Explanation: In network having only L and C, poles and zeros of driving point function are simple and alternate on jω axis.

So, the network consists of L and C.

8. In a two terminals network the open circuit voltage measured at the given terminals is 110 V and short circuit currents at the same terminals 10 A. If a load of 50 Ω resistance is connected at the terminals, load current is ___________

a) 1.8 A

b) 1.25 A

c) 6 A

d) 6.25 A

Answer: a

Explanation: R TH = \(\frac{110}{10}\) = 11 Ω

So, I = \(\frac{110}{50+11}\) = 1.8 A.

9. A coil has resistance R and inductance L. At ω = ∞ the phase angle between voltage and current is _____________

a) 0°

b) 180°

c) 45°

d) 90°

Answer: d

Explanation: We know that,

When ω = ∞, X L = ωL = ∞.

Therefore, θ = tan -1 \(\frac{ωL}{R}\) = 90°.

10. An RLC series circuit has R = 7.07 Ω, L = .707 H and C = 7.07 μF. At Half power frequencies the circuit impedance is ___________

a) 7.07 Ω

b) 10 Ω

c) 14.14 Ω

d) 20 Ω

Answer: b

Explanation: We know that,

At half power frequency the circuit impedance is 2 times resistance

Given that, resistance = 7.07 Ω

So, circuit impedance = 2 X 7.07

= 14.14 Ω.

11. Two similar coils have self-inductance of 20 mH each. Coefficient of coupling is 0.4. The mutual inductance M is ______________

a) 2.5 mH

b) 8 mH

c) 7 mH

d) 1 mH

Answer: b

Explanation: We know that, the mutual inductance M is given by,

M = k\(\sqrt{L_1 L_2}\)

Given that, k = 0.4, L 1 = 20 mH and L 2 = 20 mH

So, M = 0.4\(\sqrt{20 X 20}\)

= 8 mH.

12. For any given signal, average power in its 8 harmonic components as 50 mW each and fundamental component also has 50 mV power. Then, average power in the periodic signal is _______________

a) 750

b) 400

c) 100

d) 50

Answer: b

Explanation: We know that according to Parseval’s relation, average power is equal to the sum of the average powers in all of its harmonic components.

∴ P avg = 50×8 = 400.

13. The Current Transformer supplies current to the current coil of a power factor meter, energy meter and, an ammeter. These are connected as?

a) All coils in parallel

b) All coils in series

c) Series-parallel connection with two in each arm

d) Series-parallel connection with one in each arm

Answer: b

Explanation: Since the CT supplies the current to the current coil, therefore the coils are connected in series so that the current remains the same. If they were connected in parallel then the voltages would have been same but the currents would not be same and thus efficiency would decrease.

14. A CRO probe has an impedance of 500 kΩ in parallel with a capacitance of 10 pF. The probe is used to measure the voltage between P and Q as shown in the figure. The measured voltage will be?

network-theory-questions-answers-advanced-problems-application-laplace-transform-1-q14

a) 3.53 V

b) 3.47 V

c) 5.54 V

d) 7.00 V

Answer: b

Explanation: X C = \(\frac{1}{jCω} = \frac{-j}{2 × 3.14 × 100 × 10^3 × 10 × 10^{-12}}\)

Applying KCL at node,

\(\frac{V_a-10}{100} + \frac{V_a}{100} + \frac{V_a}{500} + \frac{V_a}{-j159}\)

∴ V a = 4.37∠-15.95°.

15. M = 2, 0≤t<4;

t 2 , t≥4;

The Laplace transform of W is ___________

a) \Missing or unrecognized delimiter for \right\)

b) \Missing or unrecognized delimiter for \right\)

c) \Missing or unrecognized delimiter for \right\)

d) \Missing or unrecognized delimiter for \right\)

Answer: d

Explanation: M = 2 + u (t 2 – 2)

L {2 + u (t 2 – 2)} = \(\frac{2}{s}\) + L {u (t 2 – 2)}

= \(\frac{2}{s} + e^{-4s}\) L { 2 -2}

= \(\frac{2}{s} + e^{-4s}\) L {t 2 + 8t + 14}

= \Missing or unrecognized delimiter for \right\).

This set of Network Theory Objective Questions & Answers focuses on “Advanced Problems on Application of Laplace Transform – 2”.

1. A capacitor of 110 V, 50 Hz is needed for AC supply. The peak voltage of the capacitor should be ____________

a) 110 V

b) 460 V

c) 220 V

d) 230 V

Answer: c

Explanation: We know that,

Peak voltage rating = 2 .

Given that, rms voltage rating = 110 V

So, Peak voltage rating = 110 X 2 V

= 220 V.

2. Given I = 10[1 + sin ]. The RMS value of I is ____________

a) 10

b) 5

c) \ 105

Answer: c

Explanation: Given that, I = 10[1 + sin ]

Now, RMS value = \(\sqrt{10^2 + \frac{10^2}{\sqrt{2}}}\)

= \(\sqrt{100+50}\)

= \(\sqrt{150}\).

3. A two branch parallel circuit has a 50 Ω resistance and 10 H inductance in one branch and a 1 μF capacitor in the second branch. It is fed from 220 V ac supply, at resonance, the input impedance of the circuit is _____________

a) 447.2 Ω

b) 500 Ω

c) 235.48 Ω

d) 325.64 Ω

Answer: a

Explanation: We know that,

For a parallel resonance circuit impedance = \([\frac{L}{RC}]^{0.5}\)

Given, R = 50 Ω, L = 10 H and C = 1 μF

= \([\frac{10}{50*1}]^{0.5}\)

So, impedance = 447.2 Ω.

4. The impedance matrices of two, two-port network are given by [3 2; 2 3] and [15 5; 5 25]. The impedance matrix of the resulting two-port network when the two networks are connected in series is ____________

a) [3 5; 2 25]

b) [18 7; 7 28]

c) [15 2; 5 3]

d) Indeterminate

Answer: b

Explanation: Given that two impedance matrices are [3 2; 2 3] and [15 5; 5 25]. Here, the resulting impedance is the sum of the two given impedances.

So resulting impedance = [18 7; 7 28].

5. The electrical energy required to heat a bucket of water to a certain temperature is 10 kWh. If heat losses, are 10%, the energy input is ____________

a) 2.67 kWh

b) 3 kWh

c) 2.5 kWh

d) 3.5 kWh

Answer: a

Explanation: Given that, 10% of input is lost.

So, 0.90 Input = 10

Or, Input = \(\frac{10}{0.9}\) = 11.11 kWh.

6. The current rating of a cable depends on ___________

a) Length of the cable

b) Diameter of the cable

c) Both length and diameter of the cable

d) Neither length nor diameter of the cable

Answer: b

Explanation: We know that Current rating depends only on the area of cross-section. Since the area of cross section is a function of radius, which in turn is a function of diameter, so the current rating depends only on the Diameter of the cable.

7. In an AC circuit, the maximum and minimum values of power factor can be ___________

a) 2 and 0

b) 1 and 0

c) 0 and -1

d) 1 and -1

Answer: b

Explanation: We know that, power factor is maximum and equal to 1 for a purely resistive load. Power factor is minimum and equal to zero for a purely reactive load.

Hence, in an AC circuit, the maximum and minimum values of power factor are 1 and 0 respectively.

8. In the circuit given below, the series circuit shown in figure, for series resonance, the value of the coupling coefficient, K will be?

network-theory-objective-questions-answers-q8

a) 0.25

b) 0.5

c) 0.1

d) 1

Answer: a

Explanation: Mutual inductance w m = ω k \(\sqrt{L_1 L_2}\)

Or, m = k.2.8 j 2

= 4k.j

At resonance, Z = 20 – j10 + j2 + j6 + 2.4kj – 2j + 8kj = 0

Or, -2 = -8k

Or, k = \(\frac{1}{4}\) = 0.25.

9. In an RC series circuit R = 100 Ω and X C = 10 Ω. Which of the following is possible?

a) The current and voltage are in phase

b) The current leads the voltage by about 6°

c) The current leads the voltage by about 84°

d) The current lags the voltage by about 6°

Answer: b

Explanation: In RC circuit the current leads the voltage.

Or, θ = tan -1 \(\frac{10}{100}\)

This is nearly equal to 6°

Hence, the current lags the voltage by about 6°.

10. The impedance of an RC series circuit is 12 Ω at f = 50 Hz. At f = 200 Hz, the impedance will be?

a) More than 12

b) Less than 3

c) More than 3 Ω but less than 12 Ω

d) More than 12 Ω but less than 24 Ω

Answer: c

Explanation: We know that the impedance Z, is given by

Z = R 2 + \(X_C^2\)

When f is made four times, X C becomes one-fourth but R remains the same.

So, the impedance is more than 3 Ω but less than 12 Ω.

11. A 3 phase balanced supply feeds a 3-phase unbalanced load. Power supplied to the load can be measured by ___________

a) 2 Wattmeter and 1 Wattmeter

b) 2 Wattmeter and 3 Wattmeter

c) 2 Wattmeter and 2 Wattmeter

d) Only 3 Wattmeter

Answer: b

Explanation: We know that, a minimum of 2 wattmeters is required to measure a 3-phase power. Also, power can be measured by using only one wattmeter in each phase. Hence, power supplied to the load can be measured by 2 Wattmeter and 3 Wattmeter methods.

12. A series RC circuit has R = 15 Ω and C = 1 μF. The current in the circuit is 5 sin 10t. The applied voltage is _____________

a) 218200 cos

b) 218200 sin

c) 218200 sin

d) 218200 cos

Answer: b

Explanation: Given that, R = 15Ω, X C = \(\frac{1}{ωC}\)

= \(\frac{10^6}{10 X 1}\) = 10 5 Ω

Now, θ = tan -1 \(\frac{X_C}{R}\) = 89.99°

We know that, impedance Z is given by,

Z = \(\sqrt{R^2 + X_C^2}\) = 10 2 kΩ.

Hence, V = 100000.sin

Or, V = 218200 sin .

13. A capacitor stores 0.15C at 5 V. Its capacitance is ____________

a) 0.75 F

b) 0.75 μF

c) 0.03 F

d) 0.03 μF

Answer: c

Explanation: We know that, Q = CV

Given, V = 5 V, Q = 0.15 C

Hence, 0.15 = C

Or, C = 0.03 F.

14. For a transmission line, open circuit and short circuit impedances are 10 Ω and 20 Ω. Then characteristic impedance is ____________

a) 100 Ω

b) 50 Ω

c) 25 Ω

d) 200 Ω

Answer: d

Explanation: We know that, the characteristic impedance Z0 is given by,

Z 0 = Z oc Z sc

Given that, open circuit impedance, Z oc = 10 Ω and short circuit impedance, Z sc = 20 Ω.

So, Z 0 = Ω

= 200 Ω.

15. A circuit excited by voltage V has a resistance R which is in series with an inductor and a capacitor connected in parallel. The voltage across the resistor at the resonant frequency is ___________

a) 0

b) \

Answer: a

Explanation: Dynamic resistance of the tank circuit, Z DY = \(\frac{L}{R_LC}\)

But given that R L = 0

So, Z DY = \(\frac{L}{0XC}\) = ∞

Therefore current through the circuit, I = \(\frac{V}{∞}\) = 0

∴ V D = 0.

This set of Network Theory Multiple Choice Questions & Answers focuses on “The Concept of Complex Frequency”.

1. The solution of differential equations for networks is of the form?

a) i=K n e ⁡(s n t)

b) i=K n e⁡ (-s n t)

c) i=-K n e⁡ (-s n t)

d) i=-K n e⁡ (s n t)

Answer: a

Explanation: The solution of differential equations for networks is of the form

i=K n e ⁡(s n t) where S n is a complex number which is a root of the characteristic equation.

2. The real part of the complex frequency is called?

a) radian frequency

b) neper frequency

c) sampling frequency

d) angular frequency

Answer: b

Explanation: The complex number consists of two parts, the real part and the imaginary part. The real part of the complex frequency is called neper frequency.

3. The imaginary part of the complex frequency is called?

a) angular frequency

b) sampling frequency

c) neper frequency

d) radian frequency

Answer: d

Explanation: The complex number consists of two parts, the real part of the complex frequency is called radian frequency. The radian frequency is expressed in radian/sec and is related to the frequency or the periodic time.

4. The ratio of transform voltage to the transform current is defined as _________ of the resistor.

a) transform voltage

b) transform current

c) transform impedance

d) transform admittance

Answer: c

Explanation: Transform impedance of the resistor is defined as the ratio of transform voltage to the transform current and is expressed as Z R = V R /I R = R.

5. The ratio of transform current to the transform voltage is defined as ________ of the resistor.

a) transform admittance

b) transform impedance

c) transform current

d) transform voltage

Answer: a

Explanation: Transform admittance of the resistor is defined as the ratio of transform current to the transform voltage and it is also defined as the reciprocal of transform impedance. Y R = I R /V R = G.

6. The transform impedance of the inductor is?

a) L

b) 1/L

c) sL

d) 1/sL

Answer: c

Explanation: Considering the sum of the transform voltage and the initial current voltage as V 1 we have the transform impedance of the inductor. The transform impedance of the inductor is Z L = V 1 /I L = sL.

7. The transform admittance of the inductor is?

a) 1/sL

b) sL

c) 1/L

d) L

Answer: a

Explanation: The transform admittance of the inductor is Y L = I 1 /V L = 1/sL where I 1 is the total transform current through the inductor L.

8. The equivalent transform circuit contains an admittance of value ____ and equivalent transform current source.

a) 1/L

b) 1/sL

c) L

d) sL

Answer: b

Explanation: The time domain representation of inductor L has initial current i L (0 + ). The equivalent transform circuit contains an admittance of value 1/sL and equivalent transform current source.

9. The transform impedance of the capacitor is?

a) C

b) 1/C

c) sC

d) 1/sC

Answer: d

Explanation: The transform impedance of the capacitor is the ratio of the transform voltage V 1 to the transform current I C and is Z C = 1/Cs.

10. The transform admittance of the capacitor is?

a) 1/sC

b) sC

c) 1/C

d) C

Answer: b

Explanation: The transform admittance of the capacitor is the ratio of transform current I 1 to transform voltage V C and is Y C = sC.

This set of Network Theory online quiz focuses on “Series and Parallel Combination of Elements”.

1. In the circuit shown below, switch K is moved from position to position 2 at time t = 0. At time t = 0-, the current in the inductor is I 0 and the voltage at the capacitor is V 0 . The inductor is represented by a transform impedance _________ in series with a voltage source __________

network-theory-questions-answers-online-quiz-q1

a) Ls, L V 0

b) Ls, LI 0

c) 1/Ls, LI 0

d) 1/Ls, L V 0

Answer: a

Explanation: The inductor has an initial current I 0 . It is represented by a transform impedance Ls in series with a voltage source L V 0 .

2. In the circuit shown below, the capacitor is replaced by a transform impedance of __________ with an initial voltage ___________

network-theory-questions-answers-online-quiz-q1

a) 1/Cs, V 0 /S

b) 1/Cs, I 0 /S

c) Cs, I 0 /S

d) Cs, V 0 /S

Answer: a

Explanation: The capacitor has an initial voltage V 0 across it. It is represented by a transform impedance of 1/Cs with an initial voltage V 0 /S.

3. The value of the total voltage after replacing the inductor and capacitor is?

network-theory-questions-answers-online-quiz-q1

a) V 1 -LI 0 -V 0 /S

b) V 1 +LI 0 -V 0 /S

c) V 1 +LI 0 +V 0 /S

d) V 1 -LI 0 +V 0 /S

Answer: b

Explanation: The current I is given as the total transform voltage in the circuit divided by the total transform impedance. The value of the total voltage after replacing the inductor and capacitor is V = V 1 +LI 0 -V 0 /S.

4. The value of the total impedance after replacing the inductor and capacitor is?

network-theory-questions-answers-online-quiz-q1

a) R-LS-1/CS

b) R-LS+1/CS

c) R+LS+1/CS

d) R+LS-1/CS

Answer: c

Explanation: The value of the total impedance after replacing the inductor and capacitor is Z = R+LS+1/CS. By knowing the V and Z we can calculate I as I is given as the total transform voltage in the circuit divided by the total transform impedance.

5. The current flowing in the following circuit is?

network-theory-questions-answers-online-quiz-q1

a) (V 1 -LI 0 -V 0 /S)/

b) (V 1 -LI 0 +V 0 /S)/

c) (V 1 +LI 0 +V 0 /S)/

d) (V 1 +LI 0 -V 0 /S)/

Answer: d

Explanation: The current I is given as the total transform voltage in the circuit divided by the total transform impedance. The current flowing in the circuit is I = V/I = (V 1 +LI 0 -V 0 /S)/.

6. Obtain the admittance of the last two elements in the parallel combination after transformation in the circuit shown below.

network-theory-questions-answers-online-quiz-q6

a) 1+s

b) 2+s

c) 3+s

d) 4+s

Answer: d

Explanation: The term admittance is defined as the inverse of impedance. The admittance of capacitor is 1/s and the admittance of resistor is 1/4 mho. So the admittance of the last two elements in the parallel combination is Y 1 = 4 + s.

7. The impedance of the last two elements in the parallel combination after transformation in the circuit shown below is?

network-theory-questions-answers-online-quiz-q6

a) 1/

b) 1/

c) 1/

d) 1/

Answer: a

Explanation: The impedance of resistor is 4Ω and the impedance of capacitor is s. So the impedance of the last two elements in the parallel combination is Z 1 = 1/.

8. The series combination of the last elements after replacing 1/s and 1/4Ω with 1/ is?

network-theory-questions-answers-online-quiz-q6

a) /2s

b) /2s

c) /2s

d) /2s

Answer: c

Explanation: We got the impedance of last two elements in parallel combination as Z 1 = 1/ and now the impedance of capacitor is 1/2s. So the series combination of the last elements is Z 2 = 1/2s+1/ = /2s.

9. Determine the admittance parallel combination of the last elements after replacing with /2s is?

network-theory-questions-answers-online-quiz-q6

a) (4s 2 -19s+4)/

b) (4s 2 +19s-4)/

c) (4s 2 +19s-4)/

d) (4s 2 +19s+4)/

Answer: d

Explanation: The term admittance is defined as the inverse of the term impedance. As the impedance is Z 2 = 1/2s+1/=/2s, the admittance parallel combination of the last elements is Y 2 = 1/2+2s/=(4s 2 +19s+4)/.

10. Obtain the transform impedance of the network shown below.

network-theory-questions-answers-online-quiz-q6

a) /(4s 2 +19s-4)

b) /(4s 2 +19s+4)

c) /(4s 2 -19s+4)

d) /(4s 2 +19s+4)

Answer: b

Explanation: The term impedance is the inverse of the term admittance. We got admittance as Y 2 = (4s 2 +19s+4)/. So the transform impedance of the network is

Z = 1/Y 2 = /(4s 2 +19s+4).

This set of Network Theory Multiple Choice Questions & Answers focuses on “Network Function for the One-Port and Two-Port”.

1. The ratio of voltage transform at first port to the voltage transform at the second port is called?

a) Voltage transfer ratio

b) Current transfer ratio

c) Transfer impedance

d) Transfer admittance

Answer: a

Explanation: Voltage transfer ratio is the ratio of voltage transform at first port to the voltage transform at the second port and is denoted by G. G 21 = V 2 /V 1 G 12 = V 1 /V 2 .

2. The ratio of the current transform at one port to current transform at other port is called?

a) Transfer admittance

b) Transfer impedance

c) Current transfer ratio

d) Voltage transfer ratio

Answer: c

Explanation: Current transfer ratio is the ratio of the current transform at one port to current transform at other port and is denoted by α. α 12 = I 1 /I 2 α 21 = I 2 /I 1 .

3. The ratio of voltage transform at the first port to the current transform at the second port is called?

a) Voltage transfer ratio

b) Transfer admittance

c) Current transfer ratio

d) Transfer impedance

Answer: d

Explanation: Transfer impedance is the ratio of voltage transform at first port to the current transform at the second port and is denoted by Z. Z 21 = V 2 /I 1 Z 12 = V 1 /I 2 .

4. For the network shown in the figure, find the driving point impedance.

network-theory-questions-answers-network-function-one-port-q4

a) (s 2 -2s+1)/s

b) (s 2 +2s+1)/s

c) (s 2 -2s-1)/s

d) (s 2 +2s-1)/s

Answer: b

Explanation: Applying Kirchoff’s law at port 1, Z=V/I, where V is applied at port 1 and I is current flowinmg through the network. Then Z=V/I = 2+S+1/S = (s 2 +2s+1)/s.

5. Obtain the transfer function G 21 in the circuit shown below.

network-theory-questions-answers-network-function-one-port-q5

a) /s

b) s+1

c) s

d) s/

Answer: d

Explanation: Applying Kirchhoff’s law V 1 = 2 I 1 + 2 sI 1 V 2 = I 1 X 2s Hence G 21 = V 2 /V 1 = 2 s/=s/.

6. Determine the transfer function Z 21 in the circuit shown below.

network-theory-questions-answers-network-function-one-port-q5

a) s

b) 2 s

c) 3 s

d) 4 s

Answer: b

Explanation: The transfer function Z 21 is Z 21 = V 2 /I 1 . V 2 = I 1 X 2s. V 2 /I 1 =2s. On substituting Z 21 = 2s.

7. Find the driving point impedance Z 11 in the circuit shown below.

network-theory-questions-answers-network-function-one-port-q5

a) 2

c) 2

Answer: c

Explanation: The driving point impedance Z 11 is Z 11 =V 1 /I 1 . V 1 = 2 I 1 + 2 sI 1 => V 1 = I 1 => V 1 /I 1 = 2. On substituting Z 11 = 2.

8. Obtain the transfer function G 21 in the circuit shown below.

network-theory-questions-answers-network-function-one-port-q8

a) /

b) /

c) /

d) /

Answer: d

Explanation: From the circuit, the parallel combination of resistance and capacitance can be combined into equivalent in impedance. Z eq = 1/=2/. Applying Kirchhoff’s laws, we have V 2 = 2 I 1 => V 1 = I 1 [2/+2] = I 1 [/] The transfer function G 21 = V 2 /V 1 = 2 I 1 //)I 1 = /.

9. Obtain the transfer function Z 21 in the circuit shown below.

network-theory-questions-answers-network-function-one-port-q8

a) 1

b) 2

c) 3

d) 4

Answer: b

Explanation: The transfer function Z 21 is Z 21 = V 2 /I 1 . V 2 = 2 I 1 => V 2 /I 1 = 2. On substituting Z 21 = 2.

10. Determine the driving point impedance Z 11 in the circuit shown below.

network-theory-questions-answers-network-function-one-port-q8

a) /

b) /

c) /

d) /

Answer: d

Explanation: The driving point impedance Z 11 is Z 11 = V 1 /I 1 . V 1 = I 1 )+2) = I 1 /) => V 1 /I 1 = /). On substituting we get Z 11 = /.

This set of Basic Network Theory Questions and Answers focuses on “Poles and Zeros of Network Functions”.

1. The coefficients of the polynomials P and Q in the network function N are ________ for passive network.

a) real and positive

b) real and negative

c) complex and positive

d) complex and negative

Answer: a

Explanation: The coefficients of the polynomials P and Q in the network function N are real and positive for passive network. On factorising the network function we obtain the poles and zeros.

2. The scale factor is denoted by the letter?

a) G

b) H

c) I

d) J

Answer: b

Explanation: The scale factor is denoted by the letter ‘H’ and its value is equal to the ratio of a o to b o .

3. The zeros in the transfer function are denoted by?

a) 3

b) 2

c) 1

d) 0

Answer: d

Explanation: The roots of the equation P = 0 are zeros of the transfer function. The zeros in the transfer function are denoted by ‘o’.

4. The poles in the transfer function are denoted by?

a) x

b) y

c) z

d) w

Answer: a

Explanation: The roots of the equation Q = 0 are poles of the transfer function. The poles in the transfer function are denoted by ‘x’.

5. The network function N becomes _________ when s is equal to anyone of the zeros.

a) 1

b) 2

c) 0

d) ∞

Answer: c

Explanation: The network function N becomes zero when s in the transfer function is equal to anyone of the zeros as the network function is completely defined by its poles and zeros.

6. The N becomes ________ when s is equal to any of the poles.

a) ∞

b) 0

c) 1

d) 2

Answer: a

Explanation: The network function is completely defined by its poles and zeros and the network function N becomes infinite when s in the transfer function is equal to anyone of the poles.

7. If the poles or zeros are not repeated, then the function is said to be having __________ poles or ________ zeros.

a) simple, multiple

b) multiple, simple

c) simple, simple

d) multiple, multiple

Answer: c

Explanation: If the poles or zeros are not repeated, then the function is said to be having simple poles or simple zeros and the network function is said to be stable when the real parts of the poles and zeros are negative.

8. If the poles or zeros are repeated, then the function is said to be having __________ poles or ________ zeros.

a) multiple, multiple

b) simple, simple

c) multiple, simple

d) simple, multiple

Answer: a

Explanation: If there are repeated poles or zeros, then function is said to be having multiple poles or multiple zeros and the network function is stable if the poles and zeros lie within the left half of the s-plane.

9. If the number of zeros are greater than the number of poles , then there will be _________ number of zeros at s = ∞.

a) n

b) m

c) n-m

d) n+m

Answer: c

Explanation: If the number of zeros are greater than the number of poles , then there will be number of zeros at s = ∞ and to obtain zeros at s = ∞ the condition is n>m.

10. If the number of poles are greater than the number of zeros , then there will be _________ number of zeros at s = ∞.

a) m+n

b) m-n

c) m

d) n

Answer: b

Explanation: If the number of poles are greater than the number of zeros , then there will be number of zeros at s = ∞ and to obtain poles at s = ∞ the condition is m>n.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Properties of Driving Point Functions”.

1. The driving point function is the ratio of polynomials in s. Polynomials are obtained from the __________ of the elements and their combinations.

a) transform voltage

b) transform current

c) transform impedance

d) transform admittance

Answer: c

Explanation: The driving point function is the ratio of polynomials in s. Polynomials are obtained from the transform impedance of the elements and their combinations and if the zeros and poles are not repeated then the poles or zeros are said to be distinct or simple.

2. The pole is that finite value of S for which N becomes __________

a) 0

b) 1

c) 2

d) ∞

Answer: d

Explanation: The quantities P 1 , P 2 … P m are called poles of N if N = ∞ at those points. The pole is that finite value of S for which N becomes infinity.

3. A function N is said to have a pole at infinity if the function N has a pole at S = ?

a) ∞

b) 2

c) 0

d) 1

Answer: c

Explanation: A function N is said to have a pole at infinity, if the function N has a pole at S = infinity. A zero or pole is said to be of multiplicity ‘r’ if r or r is a factor of P or Q.

4. The number of zeros including zeros at infinity is __________ the number of poles including poles at infinity.

a) greater than

b) equal to

c) less than

d) greater than or equal to

Answer: b

Explanation: The number of zeros including zeros at infinity is equal to the number of poles including poles at infinity and it cannot be greater than or less than the number of poles including poles at infinity.

5. The poles of driving point impedance are those frequencies corresponding to ___________ conditions.

a) short circuit

b) voltage source

c) open circuit

d) current source

Answer: c

Explanation: A zero of N is a zero of V, it signifies a short circuit. Similarly, a pole of Z is a zero of I. The poles of driving point impedance are those frequencies corresponding to open circuit conditions.

6. The zeros of driving point impedance are those frequencies corresponding to ___________ conditions.

a) current source

b) open circuit

c) voltage source

d) short circuit

Answer: d

Explanation: The zeros of driving point impedance are those frequencies corresponding to short circuit conditions as a pole of Z is a zero of I and zero of N is a zero of V, it signifies a short circuit.

7. In the driving point admittance function, a zero of Y means a _______ of I .

a) 1

b) 2

c) 3

d) zero

Answer: d

Explanation: In the driving point admittance function, a zero of Y means a zero of I i.e., the open circuit condition as the driving point admittance function is the ratio of I to V.

8. In the driving point admittance function, a pole of Y means a _______ of V .

a) zero

b) 1

c) 2

d) 3

Answer: a

Explanation: The driving point admittance function Y = I/V. In the driving point admittance function, a pole of Y means a zero of V i.e., the short circuit condition.

9. The real part of all zeros and poles must be?

a) positive or zero

b) negative or zero

c) positive

d) negative

Answer: b

Explanation: The real part of all zeros and poles must be negative or zero. But the poles or zeros should not be positive because if they are positive, then they will lie in the right-half of the s-plane.

10. Poles or zeros lying on the jω axis must be?

a) complex

b) at least one complex pole

c) at least one complex zero

d) simple

Answer: d

Explanation: Poles or zeros lying on the jω axis must be simple because on jω axis the imaginary part of poles or zeros will be zero.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Properties of Transfer Functions”.

1. The coefficients of numerator polynomial and the denominator polynomial in a transfer function must be?

a) real

b) complex

c) at least one real coefficient

d) at least one complex coefficient

Answer: a

Explanation: The coefficients of P, the numerator polynomial and of Q, the denominator polynomial in a transfer function must be real. Therefore all poles and zeros if complex must occur in conjugate pairs.

2. In a transfer function, the degree of numerator polynomial is ___________ than the degree of the denominator polynomial.

a) greater than

b) less than

c) equal to

d) less than or equal to

Answer: d

Explanation: In a transfer function, the degree of numerator polynomial is less than or equal to than the degree of the denominator polynomial. And the degree of the numerator polynomial of Z 21 or Y 21 is less than or equal to the degree of the denominator polynomial plus one.

3. The real parts of all poles and zeros in a driving point function must be?

a) zero

b) negative

c) zero or negative

d) positive

Answer: c

Explanation: The real parts of all poles and zeros in a driving point function must be zero or negative but should not be positive and the complex or imaginary poles and zeros must occur in conjugate pairs.

4. If the real part of driving point function is zero, then the pole and zero must be?

a) complex

b) simple

c) one complex pole

d) one complex zero

Answer: b

Explanation: If the real part of driving point function is zero, then the pole and zero must be simple but should not contain any complex pole or complex zero.

5. The degree of the numerator polynomial and denominator polynomial in a driving point function may differ by?

a) 0

b) 1

c) 0 or 1

d) 2

Answer: c

Explanation: The degree of numerator polynomial and denominator polynomial in a driving point function may differ by zero or one. And the polynomials P and Q may not have any missing terms between the highest and lowest degrees unless all even or odd terms are missing.

6. The lowest degree in numerator polynomial and denominator polynomial in a driving point function may differ by at most __________

a) 0

b) 1

c) 2

d) 3

Answer: b

Explanation: The lowest degree in numerator polynomial and denominator polynomial in a driving point function may differ by at most one and the coefficients in the polynomials P and Q of network function must be real and positive.

7. The coefficients in the denominator polynomial of the transfer function must be?

a) positive

b) negative

c) positive or zero

d) negative or zero

Answer: a

Explanation: The coefficients in the denominator polynomial of the transfer function must be positive but should not be negative and the coefficients in the polynomials P and Q of transfer function must be real.

8. The coefficients in the numerator polynomial of the transfer function may be?

a) must be negative

b) must be positive

c) may be positive

d) may be negative

Answer: d

Explanation: The coefficients in the numerator polynomial of the transfer function may be negative and the complex or imaginary poles and zeros must occur in conjugate pairs.

9. The denominator polynomial in a transfer function may not have any missing terms between the highest and the lowest degree, unless?

a) all odd terms are missing

b) all even terms are missing

c) all even or odd terms are missing

d) all even and odd terms are missing

Answer: c

Explanation: The denominator polynomial in a transfer function may not have any missing terms between the highest and the lowest degree, unless all even or odd terms are missing and the polynomial P may have missing terms between the lowest and the highest degree.

10. The degree of numerator polynomial in a transfer function may be as small as _________ independent of the degree of the denominator polynomial.

a) 1

b) 2

c) 0

d) 3

Answer: c

Explanation: The degree of numerator polynomial in a transfer function may be as small as zero, independent of the degree of the denominator polynomial and for the voltage transfer ratio and the current transfer ratio, the maximum degree of P must be equal to the degree of Q.

This set of Network Theory Question Bank focuses on “Open Circuit Impedence Parameters”.

1. Two ports containing no sources in their branches are called?

a) active ports

b) passive ports

c) one port

d) three port

Answer: b

Explanation: Two ports containing no sources in their branches are called passive ports; among them are power transmission lines and transformers.

2. Two ports containing sources in their branches are called?

a) three port

b) one port

c) passive ports

d) active ports

Answer: d

Explanation: Two ports containing sources in their branches are called active ports. A voltage and current is assigned to each of the two ports.

3. In determining open circuit impedance parameters, among V 1 , V 2 , I 1 , I 2 , which of the following are dependent variables?

a) V 1 and V 2

b) I 1 and I 2

c) V 1 and I 2

d) I 1 and V 2

Answer: a

Explanation: In determining open circuit impedance parameters, among V 1 , V 2 , I 1 , I 2 ; V 1 and V 2 are dependent variables and I 1 , I 2 are independent variables i.e., dependent variables depend on independent variables.

4. In determining open circuit impedance parameters, among V 1 , V 2 , I 1 , I 2 , which of the following are independent variables?

a) I 1 and V 2

b) V 1 and I 2

c) I 1 and I 2

d) V 1 and V 2

Answer: c

Explanation: In determining open circuit impedance parameters, among V 1 , V 2 , I 1 , I 2 ; I 1 and I 2 are independent variables and V 1 , V 2 are dependent variables. Independent variables are the variables that do not depend on any other variable.

5. Which of the following expression is true in case of open circuit parameters?

a) V 1 = Z 11 V 1 + Z 12 I 2

b) V 1 = Z 11 I 1 + Z 12 V 2

c) V 1 = Z 11 I 1 + Z 12 I 2

d) V 2 = Z 11 I 1 + Z 12 I 2

Answer: c

Explanation: The expression relating the open circuit parameters Z 11 , Z 12 and currents I 1 , I 2 and voltage V 1 is V 1 = Z 11 I 1 + Z 12 I 2 .

6. Which of the following expression is true in case of open circuit parameters?

a) V 2 = Z 21 I 2 + Z 22 I 2

b) V 2 = Z 21 I 1 + Z 22 I 2

c) V 1 = Z 21 I 2 + Z 22 I 2

d) V 1 = Z 21 I 1 + Z 22 I 2

Answer: b

Explanation: The expression relating the currents I 1 , I 2 and voltage V 1 and open circuit parameters Z 21 , Z 22 is V 2 = Z 21 I 1 + Z 22 I 2 .

7. Find the Z – parameter Z 11 in the circuit shown below.

network-theory-questions-bank-q7

a) 1

b) 1.5

c) 2

d) 2.5

Answer: d

Explanation: For determining Z 11 , the current I 2 is equal to zero. Now we obtain Z eq as 1 + /=2.5Ω. So, Z 11 = 2.5Ω.

8. The value of Z 21 in the circuit shown below?

network-theory-questions-bank-q7

a) 0

b) 1

c) 2

d) 3

Answer: b

Explanation: V 2 is the voltage across the 4Ω impedance. The current through 4Ω impedance is I 1 /4. And V 2 = (I 1 /4) x 4 = I 1 . So, Z 21 = 1Ω.

9. Find the value of Z 12 in the circuit shown below.

network-theory-questions-bank-q7

a) 3

b) 2

c) 1

d) 0

Answer: c

Explanation: The current through vertical 2Ω resistor is = I 2 /2. So, V 1 = 2 x (I 2 /2). On solving and substituting we get Z 12 = 1Ω.

10. Determine the value of Z 22 in the circuit shown below.

network-theory-questions-bank-q7

a) 0

b) 1

c) 2

d) 3

Answer: c

Explanation: Open circuiting port 1, we get V 2 = I 2 ||4) => V 2 = I 2 x 2 => V 2 /I 2 = 2. Therefore the value of Z 22 is 2Ω.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Short Circuit Admittance Parameters”.

1. In determining short circuit impedance parameters, among V 1 , V 2 , I 1 , I 2 , which of the following are dependent variables?

a) V 1 and V 2

b) I 1 and I 2

c) V 1 and I 2

d) I 1 and V 2

Answer: b

Explanation: In determining short circuit impedance parameters, among V 1 , V 2 , I 1 , I 2 ; I 1 and I 2 are dependent variables and V 1 , V 2 are independent variables i.e., dependent variables depend on independent variables.

2. In determining short circuit impedance parameters, among V 1 , V 2 , I 1 , I 2 , which of the following are independent variables?

a) I 1 and V 2

b) V 1 and I 2

c) I 1 and I 2

d) V 1 and V 2

Answer: d

Explanation: In determining short circuit impedance parameters, among V 1 , V 2 , I 1 , I 2 ; V 1 and V 2 are independent variables and I 1 , I 2 are dependent variables. Independent variables are the variables that do not depend on any other variable.

3. Which of the following expression is true in case of short circuit parameters?

a) I 1 = Y 11 V 1 + Y 12 V 2

b) I 1 = Y 11 I 1 + Y 12 V 2

c) V 1 = Y 11 I 1 + Y 12 V 2

d) V 1 = Y 11 V 1 + Y 12 V 2

Answer: a

Explanation: The expression relating the short circuit parameters Y 11 , Y 12 and voltages V 1 , V 2 and current is I 1 , is I 1 = Y 11 V 1 + Y 12 V 2 .

4. Which of the following expression is true in case of short circuit parameters?

a) I 2 = Y 21 I 1 + Y 22 I 2

b) V 2 = Y 21 I 1 + Y 22 V 2

c) I 2 = Y 21 V 1 + Y 22 V 2

d) I 2 = Y 21 V 1 + Y 22 I 2

Answer: c

Explanation: The expression relating the voltages V 1 , V 2 and current is I 2 and short circuit parameters Y 11 , Y 12 is I 2 = Y 21 V 1 + Y 22 V 2 .

5. The parameters Y 11 , Y 12 , Y 21 , Y 22 are called?

a) Open circuit impedance parameters

b) Short circuit admittance parameters

c) Inverse transmission parameters

d) Transmission parameters

Answer: b

Explanation: The parameters Y 11 , Y 12 , Y 21 , Y 22 are called short circuit admittance parameters also called network functions as they are obtained by short circuiting port 1 or port 2.

6. Find the Y – parameter Y 11 in the circuit shown below.

network-theory-questions-answers-short-circuit-admittance-parameters-q6

a) 2

b) 3/2

c) 1

d) 1/2

Answer: d

Explanation: After short circuiting b-b’, the equation will be V 1 = (I 1 ) x 2. We know Y 11 = I 1 /V 1 . From the equation we get I 1 /V 1 = 2. On substituting we get Y 11 = 2 mho.

7. Find the Y – parameter Y 21 in the circuit shown below.

network-theory-questions-answers-short-circuit-admittance-parameters-q6

a) -1/4

b) 1/4

c) 1/2

d) -1/2

Answer: a

Explanation: After short circuiting b-b’, the equation will be -I 2 =I 1 × 2/4=I 1 /2 and -I 2 = V 1 /4 and on solving and substituting we get Y 21 = I 2 /V 1 =-1/4 mho.

8. Find the Y – parameter Y 22 in the circuit shown below.

network-theory-questions-answers-short-circuit-admittance-parameters-q6

a) 3/8

b) 5/8

c) 7/8

d) 9/8

Answer: b

Explanation: On short circuiting a-a’, we get Z eq = 8/5 Ω. V 2 =I 2 × 8/5. We know Y 22 = I 2 /V 2 . We got I 2 /V 1 = 5/8. ON substituting we get Y 22 = 5/8 mho.

9. Find the Y – parameter Y 12 in the circuit shown below.

network-theory-questions-answers-short-circuit-admittance-parameters-q6

a) 1/2

b) -1/2

c) -1/4

d) 1/4

Answer: c

Explanation: Short circuiting a-a’, -I 1 = 2/5 I 2 and I 2 = 5 V 2 /8. On solving -I 1 = 2/5×5/8 V 2 = V 2 /4. We know Y 12 = I 1 /V 2 . We got I 1 /V 2 = -1/4. So the value of Y 12 will be -1/4 mho.

10. Which of the following equation is true in the circuit shown below?

network-theory-questions-answers-short-circuit-admittance-parameters-q6

a) I 1 =0.5(V 1 )+0.25(V 2 )

b) I 1 =0.25(V 1 )+0.625(V 2 )

c) I 1 =-0.25(V 1 )+0.625(V 2 )

d) I 1 =0.5(V 1 )-0.25(V 2 )

Answer: d

Explanation: We got the admittance parameters as Y 11 = 0.5, Y 12 = -0.25, Y 21 = -0.25, Y 22 = 0.625. So the equations in terms of admittance parameters is I 1 =0.5(V 1 )-0.25(V 2 ) and I 2 =-0.25(V 1 )+0.625(V 2 ).

This set of Network Theory Multiple Choice Questions & Answers focuses on “Transmission Parameters”.

1.In the circuit shown below, find the transmission parameter A.

network-theory-questions-answers-transmission-parameters-q1

a) 6/5

b) 5/6

c) 3/4

d) 4/3

Answer: a

Explanation: Open circuiting b-b ‘ , V 1 = 6 I 1 , V 2 = 5I 1 . On solving V 1 /V 2 = 6/5. On substituting we get A = V 1 /V 2 =6/5.

2. In the circuit shown below, find the transmission parameter C.

network-theory-questions-answers-transmission-parameters-q1

a) 4/5

b) 3/5

c) 2/5

d) 1/5

Answer: d

Explanation: C = I 1 /V 2 |I 2 =0. By open circuiting b-b ‘ we get V 2 = 5 I 1 => I 1 /V 2 = 1/5. On substituting we get C = I 1 /V 2 =1/5 Ω.

3. In the circuit shown below, find the transmission parameter B.

network-theory-questions-answers-transmission-parameters-q1

a) 15/5

b) 17/5

c) 19/5

d) 21/5

Answer: b

Explanation: The transmission parameter B is given by B = -V 1 /I 2 |V 2 =0. Short circuiting b-b ‘ , -I 2 = 5/17 V 1 => -V 1 /I 2 = 17/5. On substituting we get B=17/5 Ω.

4. In the circuit shown below, find the transmission parameter D.

network-theory-questions-answers-transmission-parameters-q1

a) 1/5

b) 3/5

c) 7/5

d) 9/5

Answer: c

Explanation: D is a transmission parameter and is given by D = -I 1 /I 2 |V 2 =0. Short circuiting b-b ‘ , I 1 = 7/17 V 1 and-I 2 = 5/17 V 1 . So we get I 1 /I 2 = 7/5. So D=7/5.

5. The hybrid parameter h 11 is called?

a) short circuit input impedance

b) short circuit forward current gain

c) open circuit reverse voltage gain

d) open circuit output admittance

Answer: a

Explanation: h 11 =V 1 /I 1 |V 2 =0. So the hybrid parameter h 11 is called short circuit input impedance.

6. The hybrid parameter h 21 is called?

a) open circuit output admittance

b) open circuit reverse voltage gain

c) short circuit forward current gain

d) short circuit input impedance

Answer: c

Explanation: h 21 =I 2 /I 1 |V 2 =0. So the hybrid parameter h 21 is called short circuit forward current gain.

7. In the circuit shown below, find the h-parameter h 11 .

network-theory-questions-answers-transmission-parameters-q7

a) 1

b) 2

c) 3

d) 4

Answer: b

Explanation: h 11 =V 1 /I 1 |V 2 =0. So short circuiting b-b ‘ , V 1 = I 1 +1) = 2I 1 => V 1 /I 1 = 2. On substituting we get h 11 = V 1 /I 1 = 2Ω.

8. In the circuit shown below, find the h-parameter h 21 .

network-theory-questions-answers-transmission-parameters-q7

a) 1

b) -1

c) 1/2

d) -1/2

Answer: d

Explanation: Short circuiting b-b ‘ , h 21 = I 2 /I 1 when V 2 =0 and -I 2 = I 1 /2 => I 2 /I 1 = -1/2. So h 21 = -1/2.

9. In the circuit shown below, find the h-parameter h 12 .

network-theory-questions-answers-transmission-parameters-q7

a) 1/2

b) -1/2

c) 1

d) -1

Answer: a

Explanation: Open circuiting a-a ‘ we get V 1 =I y ×2 and

I y =I 2 /2 and V 2 =I x ×4 and I x =I 2 /2. On solving and substituting, we get h 12 = V 1 /V 2 =1/2.

10. In the circuit shown below, find the h-parameter h 22 .

network-theory-questions-answers-transmission-parameters-q7

a) 1

b) 2

c) 1/2

d) 3/2

Answer: c

Explanation: Open circuiting a-a ‘ we get V 1 =I y ×2 and I y =I 2 /2 and V 2 =I x ×4 and I x =I 2 /2. On solving and substituting, we get h 22 =I 2 /V 2 =1/2 Ω.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Hybrid Parameter”.

1. For the circuit given below, the value of the hybrid parameter h 11 is ___________

network-theory-questions-answers-hybrid-parameter-q1

a) 15 Ω

b) 20 Ω

c) 30 Ω

d) 25 Ω

Answer: a

Explanation: Hybrid parameter h 11 is given by, h 11 = \(\frac{V_1}{I_1}\), when V 2 =0.

Therefore short circuiting the terminal Y-Y’, we get,

V 1 = I 1 + 10)

= I 1 \Missing or unrecognized delimiter for \right+10\right)\)

= 15I 1

∴ \(\frac{V_1}{I_1}\) = 15.

Hence h 11 = 15 Ω.

2. For the circuit given below, the value of the hybrid parameter h 21 is ___________

network-theory-questions-answers-hybrid-parameter-q1

a) 0.6 Ω

b) 0.5 Ω

c) 0.3 Ω

d) 0.2 Ω

Answer: b

Explanation: Hybrid parameter h 21 is given by, h 21 = \(\frac{I_2}{I_1}\), when V 2 =0.

Therefore short circuiting the terminal Y-Y’, and applying Kirchhoff’s law, we get,

-10 I 2 – (I 2 – I 1 )10 = 0

Or, -I 2 = I 2 – I 1

Or, -2I 2 = -I 1

∴ \(\frac{I_2}{I_1} = \frac{1}{2}\)

Hence h 21 = 0.5 Ω.

3. For the circuit given below, the value of the hybrid parameter h 12 is ___________

network-theory-questions-answers-hybrid-parameter-q1

a) 6 Ω

b) 5 Ω

c) 1 Ω

d) 2 Ω

Answer: c

Explanation: Hybrid parameter h 12 is given by, h 12 = \(\frac{V_1}{V_2}\), when I 1 = 0.

Therefore short circuiting the terminal X-X’ we get,

V 1 = I A × 10

I A = \(\frac{I_2}{2}\)

V 2 = I B × 10

I B = \(\frac{I_2}{2}\)

From the above 4 equations, we get,

∴ \(\frac{V_1}{V_2} = \frac{I_2×10}{I_2×10}\) = 1

Hence h 12 = 1 Ω.

4. For the circuit given below, the value of the hybrid parameter h 22 is ___________

network-theory-questions-answers-hybrid-parameter-q1

a) 0.2 Ω

b) 0.5 Ω

c) 0.1 Ω

d) 0.3 Ω

Answer: a

Explanation: Hybrid parameter h 22 is given by, h 22 = \(\frac{I_2}{V_2}\), when I 1 = 0.

Therefore short circuiting the terminal X-X’ we get,

V 1 = I A × 10

I A = \(\frac{I_2}{2}\)

V 2 = I B × 10

I B = \(\frac{I_2}{2}\)

From the above 4 equations, we get,

∴ \(\frac{I_2}{V_2} = \frac{I_2×2}{I_2×10}\) = 0.2

Hence h 22 = 0.2 Ω.

5. In the circuit given below, the value of the hybrid parameter h 11 is _________

network-theory-questions-answers-hybrid-parameter-q5

a) 10 Ω

b) 7.5 Ω

c) 5 Ω

d) 2.5 Ω

Answer: b

Explanation: Hybrid parameter h 11 is given by, h 11 = \(\frac{V_1}{I_1}\), when V 2 = 0.

Therefore short circuiting the terminal Y-Y’, we get,

V 1 = I 1 + 5)

= I 1 \Missing or unrecognized delimiter for \right+5\right)\)

= 7.5I 1

∴ \(\frac{V_1}{I_1}\) = 7.5

Hence h 11 = 7.5 Ω.

6. In the circuit given below, the value of the hybrid parameter h 21 is _________

network-theory-questions-answers-hybrid-parameter-q5

a) 10 Ω

b) 0.5 Ω

c) 5 Ω

d) 2.5 Ω

Answer: b

Explanation: Hybrid parameter h 21 is given by, h 21 = \(\frac{I_2}{I_1}\), when V 2 = 0.

Therefore short circuiting the terminal Y-Y’, and applying Kirchhoff’s law, we get,

-5 I 2 – (I 2 – I 1 )5 = 0

Or, -I 2 = I 2 – I 1

Or, -2I 2 = -I 1

∴ \(\frac{I_2}{I_1} = \frac{1}{2}\)

Hence h 21 = 0.5 Ω.

7. For the circuit given below, the value of the hybrid parameter h 12 is ___________

network-theory-questions-answers-hybrid-parameter-q5

a) 6 Ω

b) 5 Ω

c) 1 Ω

d) 2 Ω

Answer: c

Explanation: Hybrid parameter h 12 is given by, h 12 = \(\frac{V_1}{V_2}\), when I 1 = 0.

Therefore short circuiting the terminal X-X’ we get,

V 1 = I A × 5

V 2 = I A × 5

From the above equations, we get,

∴ \(\frac{V_1}{V_2} = \frac{I_A×10}{I_A×10}\) = 1

Hence h 12 = 1 Ω.

8. For the circuit given below, the value of the hybrid parameter h 22 is ___________

network-theory-questions-answers-hybrid-parameter-q5

a) 0.2 Ω

b) 0.5 Ω

c) 0.1 Ω

d) 0.3 Ω

Answer: a

Explanation: Hybrid parameter h 22 is given by, h 22 = \(\frac{I_2}{V_2}\), when I 1 = 0.

Therefore short circuiting the terminal X-X’ we get,

V 1 = I A × 5

V 2 = I A × 5

I A = I 2

From the above equations, we get,

∴ \(\frac{I_2}{V_2} = \frac{I_2}{I_2×5}\) = 0.2

Hence h 22 = 0.2 Ω.

9. In two-port networks the parameter h 11 is called _________

a) Short circuit input impedance

b) Short circuit current gain

c) Open circuit reverse voltage gain

d) Open circuit output admittance

Answer: a

Explanation: We know that, h 11 = \(\frac{V_1}{I_1}\), when V 2 = 0.

Since the second output terminal is short circuited when the ratio of the two voltages is measured, therefore the parameter h 11 is called as Short circuit input impedance.

10. In two-port networks the parameter h 21 is called _________

a) Short circuit input impedance

b) Short circuit current gain

c) Open circuit reverse voltage gain

d) Open circuit output admittance

Answer: b

Explanation: We know that, h 21 = \(\frac{I_2}{I_1}\), when V 2 = 0.

Since the second output terminal is short circuited when the ratio of the two currents is measured, therefore the parameter h 21 is called Short circuit current gain.

11. In two-port networks the parameter h 12 is called _________

a) Short circuit input impedance

b) Short circuit current gain

c) Open circuit reverse voltage gain

d) Open circuit output admittance

Answer: c

Explanation: We know that, h 21 = \(\frac{V_1}{V_2}\), when I 1 = 0.

Since the current in the first loop is 0 when the ratio of the two voltages is measured, therefore the parameter h 12 is called as Open circuit reverse voltage gain.

12. In two-port networks the parameter h 22 is called _________

a) Short circuit input impedance

b) Short circuit current gain

c) Open circuit reverse voltage gain

d) Open circuit output admittance

Answer: d

Explanation: We know that, h 22 = \(\frac{I_2}{V_2}\), when I 1 = 0.

Since the current in the first loop is 0 when the ratio of the current and voltage in the second loop is measured, therefore the parameter h 22 is called as Open circuit output admittance.

13. For a T-network if the Short circuit admittance parameters are given as y 11 , y 21 , y 12 , y 22 , then y 11 in terms of Hybrid parameters can be expressed as ________

a) y 11 = \Missing or unrecognized delimiter for \right\)

b) y 11 = \

y 11 = –\

y 11 = \(\frac{1}{h_{11}} \)

Answer: d

Explanation: We know that the short circuit admittance parameters can be expressed in terms of voltages and currents as,

I 1 = y 11 V 1 + y 12 V 2 ………

I 2 = y 21 V 1 + y 22 V 2 ……….

And the Hybrid parameters can be expressed in terms of voltages and currents as,

V 1 = h 11 I 1 + h 12 V 2 ……….

I 2 = h 21 I 1 + h 22 V 2 ………..

Now, and can be rewritten as,

I 1 = \

And I 2 = \Missing or unrecognized delimiter for \right V_2\) ……….

∴ Comparing , and , , we get,

y 11 = \(\frac{1}{h_{11}} \)

y 12 = –\(\frac{h_{12}}{h_{11}}\)

y 21 = \(\frac{h_{21}}{h_{11}}\)

y 22 = \Missing or unrecognized delimiter for \right\).

14. For a T-network if the Short circuit admittance parameters are given as y 11 , y 21 , y 12 , y 22 , then y 12 in terms of Hybrid parameters can be expressed as ________

a) y 12 = \Missing or unrecognized delimiter for \right\)

b) y 12 = \

y 12 = –\

y 12 = \(\frac{1}{h_{11}} \)

Answer: c

Explanation: We know that the short circuit admittance parameters can be expressed in terms of voltages and currents as,

I 1 = y 11 V 1 + y 12 V 2 ………

I 2 = y 21 V 1 + y 22 V 2 ……….

And the Hybrid parameters can be expressed in terms of voltages and currents as,

V 1 = h 11 I 1 + h 12 V 2 ……….

I 2 = h 21 I 1 + h 22 V 2 ………..

Now, and can be rewritten as,

I 1 = \

And I 2 = \Missing or unrecognized delimiter for \right V_2\) ……….

∴ Comparing , and , , we get,

y 11 = \(\frac{1}{h_{11}} \)

y 12 = –\(\frac{h_{12}}{h_{11}}\)

y 21 = \(\frac{h_{21}}{h_{11}}\)

y 22 = \Missing or unrecognized delimiter for \right\).

15. For a T-network if the Short circuit admittance parameters are given as y 11 , y 21 , y 12 , y 22 , then y 22 in terms of Hybrid parameters can be expressed as ________

a) y 22 = \Missing or unrecognized delimiter for \right\)

b) y 22 = \

y 22 = –\

y 22 = \(\frac{1}{h_{11}} \)

Answer: a

Explanation: We know that the short circuit admittance parameters can be expressed in terms of voltages and currents as,

I 1 = y 11 V 1 + y 12 V 2 ………

I 2 = y 21 V 1 + y 22 V 2 ……….

And the Hybrid parameters can be expressed in terms of voltages and currents as,

V 1 = h 11 I 1 + h 12 V 2 ……….

I 2 = h 21 I 1 + h 22 V 2 ………..

Now, and can be rewritten as,

I 1 = \

And I 2 = \Missing or unrecognized delimiter for \right V_2\) ……….

∴ Comparing , and , , we get,

y 11 = \(\frac{1}{h_{11}} \)

y 12 = –\(\frac{h_{12}}{h_{11}}\)

y 21 = \(\frac{h_{21}}{h_{11}}\)

y 22 = \Missing or unrecognized delimiter for \right\).

This set of Network Theory Multiple Choice Questions & Answers focuses on “Inverse Hybrid Parameter”.

1. For the circuit given below, the value of the Inverse hybrid parameter g 11 is ___________

network-theory-questions-answers-inverse-hybrid-parameter-q1

a) 0.067 Ω

b) 0.025 Ω

c) 0.3 Ω

d) 0.25 Ω

Answer: a

Explanation: Inverse Hybrid parameter g 11 is given by, g 11 = \(\frac{I_1}{V_1}\), when I 2 = 0.

Therefore short circuiting the terminal Y-Y’, we get,

V 1 = I 1 + 10)

= I 1 \Missing or unrecognized delimiter for \right+10\right)\)

= 15I 1

∴ \(\frac{I_1}{V_1} = \frac{1}{15}\) = 0.067 Ω

Hence g 11 = 15 Ω.

2. For the circuit given below, the value of the Inverse hybrid parameter g 21 is ___________

network-theory-questions-answers-inverse-hybrid-parameter-q1

a) 0.6 Ω

b) 0.5 Ω

c) 0.3 Ω

d) 0.2 Ω

Answer: b

Explanation: Inverse Hybrid parameter g 21 is given by, g 21 = \(\frac{V_2}{V_1}\), when I 2 = 0.

Therefore short circuiting the terminal Y-Y’, and applying Kirchhoff’s law, we get,

V 1 = I 1

V 2 = I 1 10

∴ \(\frac{V_2}{V_1} = \frac{I_1 10}{I_1 20}\) = 0.5

Hence g 21 = 0.5 Ω.

3. For the 2 port network as shown below, the Z-matrix is ___________

network-theory-questions-answers-inverse-hybrid-parameter-q3

a) [Z 1 ; Z 1 + Z 2 ; Z 1 + Z 2 ; Z 3 ]

b) [Z 1 ; Z 1 ; Z 1 + Z 2 ; Z 2 ]

c) [Z 1 ; Z 2 ; Z 2 ; Z 1 + Z 2 ]

d) [Z 1 ; Z 1 ; Z 1 ; Z 1 + Z 2 ]

Answer: d

Explanation: z 11 = \(\frac{V_1}{I_1}\), when I 2 = 0

z 22 = \(\frac{V_2}{I_2}\), when I 1 = 0

z 12 = \(\frac{V_1}{I_2}\), when I 1 = 0

z 21 = \(\frac{V_2}{I_1}\), when I 2 = 0

Now, in the given circuit putting I 1 = 0, we get,

z 12 = Z 1 and z 22 = Z 1 + Z 2

And putting I 2 = 0, we get,

z 21 = Z 1 and z 11 = Z 1 .

4. Which one of the following parameters does not exist for the two-port network in the circuit given below?

network-theory-questions-answers-inverse-hybrid-parameter-q4

a) h

b) Y

c) Z

d) g

Answer: c

Explanation: Y-parameter = \(\frac{1}{Z}\)[1; -1; -1; 1]

And from the definition of the Y parameters, ∆Y = 0. Therefore the Y-parameter exists.

Since ∆Y = 0, so by property of reciprocity, ∆h = 0 and ∆g = 0.

Hence both hybrid and inverse hybrid parameters exist.

But the Z-parameters cannot exist here because if one terminal is opened the circuit will become invalid.

∴ Z- parameters do not exists.

5. In the circuit given below, the value of the Inverse hybrid parameter g 11 is _________

network-theory-questions-answers-inverse-hybrid-parameter-q5

a) 10 Ω

b) 0.133 Ω

c) 5 Ω

d) 2.5 Ω

Answer: b

Explanation: Inverse Hybrid parameter g 11 is given by, g 11 = \(\frac{I_1}{V_1}\), when I 2 = 0.

Therefore short circuiting the terminal Y-Y’, we get,

V 1 = I 1 + 5)

= I 1 \Missing or unrecognized delimiter for \right+5\right)\)

= 7.5I 1

∴ \(\frac{I_1}{V_1} = \frac{1}{7.5}\) = 0.133

Hence g 11 = 7.5 Ω.

6. In the circuit given below, the value of the Inverse hybrid parameter g 21 is _________

network-theory-questions-answers-inverse-hybrid-parameter-q5

a) 10 Ω

b) 0.5 Ω

c) 5 Ω

d) 2.5 Ω

Answer: b

Explanation: Inverse Hybrid parameter g 21 is given by, g 21 = \(\frac{V_2}{V_1}\), when I 2 = 0.

Therefore short circuiting the terminal Y-Y’, and applying Kirchhoff’s law, we get,

V 1 = I 1

V 2 = I 1 5

∴ \(\frac{V_2}{V_1} = \frac{I_1 5}{I_1 10}\) = 0.5

Hence g 21 = 0.5 Ω.

7. The short-circuit admittance matrix of a two port network is as follows.

[0; -0.5; 0.5; 0]

Then the 2 port network is ____________

a) Non-reciprocal and passive

b) Non-reciprocal and active

c) Reciprocal and passive

d) Reciprocal and active

Answer: b

Explanation: The network is non reciprocal because Y 12 ≠ Y 21 and Y 12 is also negative which means either energy storing or providing device is available. So network is active. Therefore the network is Non- reciprocal and active.

8. If a two port network is passive, then we have, with the usual notation, the relationship as _________

a) h 21 = h 12

b) h 12 = -h 21

c) h 11 = h 22

d) h 11 h 22 – h 12 h 22 = 1

Answer: d

Explanation: We know that, I 1 = y 11 V 1 + y 12 V 2 ………

I 2 = y 21 V 1 + y 22 V 2 ……….

And, V 1 = h 11 I 1 + h 12 V 2 ……….

I 2 = h 21 I 1 + h 22 V 2 ………..

Now, and can be rewritten as,

I 1 = \

And I 2 = \Missing or unrecognized delimiter for \right V_2\) ……….

Therefore using the above 6 equations in representing the hybrid parameters in terms of the Y parameters and applying ∆Y=0, we get,

h 11 h 22 – h 12 h 22 = 1 [hence proved].

9. For the circuit given below, the value of the Inverse hybrid parameter g 22 is ___________

network-theory-questions-answers-inverse-hybrid-parameter-q5

a) 7.5 Ω

b) 5 Ω

c) 6.25 Ω

d) 3 Ω

Answer: a

Explanation: Inverse Hybrid parameter g 22 is given by, g 22 = \(\frac{V_2}{I_2}\), when V 1 = 0.

Therefore short circuiting the terminal X-X’ we get,

-5 I 2 – 5 I 1 + V 2 = 0

-5 I 1 – 5(I 1 – I 2 ) = 0

Or, 2I 1 = I 2

Putting the above equation in the first equation, we get,

-7.5 I 2 = -V 2

Or, \(\frac{V_2}{I_2}\) = 7.5

Hence g 22 = 7.5 Ω.

10. In two-port networks the parameter g 11 is called _________

a) Short circuit input impedance

b) Short circuit current ratio

c) Open circuit voltage ratio

d) Open circuit input admittance

Answer: d

Explanation: We know that, g 11 = \(\frac{I_1}{V_1}\), when I 2 = 0.

Since the second voltage terminal is short circuited when the ratio of the current and voltage is measured, therefore the parameter g 11 is called as Open circuit input admittance.

11. In two-port networks the parameter g 21 is called _________

a) Short circuit input impedance

b) Short circuit current ratio

c) Open circuit voltage ratio

d) Open circuit input admittance

Answer: c

Explanation: We know that, g 21 = \(\frac{V_2}{V_1}\), when I 2 = 0.

Since the second output terminal is short circuited when the ratio of the two voltages is measured, therefore the parameter g 21 is called as Open circuit voltage ratio.

12. In two-port networks the parameter g 12 is called _________

a) Short circuit input impedance

b) Short circuit current gain

c) Open circuit reverse voltage gain

d) Open circuit output admittance

Answer: c

Explanation: We know that, g 12 = \(\frac{I_1}{I_2}\), when V 1 = 0.

Since the primary terminal is short circuited and the ratio of the two currents is measured, therefore the parameter g 12 is called as Short circuit current ratio.

13. In two-port networks the parameter g 22 is called _________

a) Short circuit input impedance

b) Short circuit current ratio

c) Open circuit voltage ratio

d) Open circuit input admittance

Answer: b

Explanation: We know that, g 22 = \(\frac{V_2}{I_2}\), when V 1 = 0.

Since the primary voltage terminal is short circuited and the ratio of the voltage and current in second loop is measured, therefore the parameter g 22 is called as Short circuit current ratio.

14. For a T-network if the Short circuit admittance parameters are given as y 11 , y 21 , y 12 , y 22 , then y 12 in terms of Inverse Hybrid parameters can be expressed as ________

a) y 12 = \Missing or unrecognized delimiter for \right\)

b) y 12 = \

y 12 = –\

y 12 = \(\frac{1}{g_{22}}\)

Answer: b

Explanation: We know that, I 1 = y 11 V 1 + y 12 V 2 ………

I 2 = y 21 V 1 + y 22 V 2 ……….

And, I 1 = g 11 V 1 + g 12 I 2 ……….

V 2 = g 21 V 1 + g 22 I 2 ………..

Now, and can be rewritten as,

I 1 = \Missing or unrecognized delimiter for \rightV_1 + \frac{g_{12}}{g_{22}} V_2\)……….

And I 2 = –\

∴ Comparing , and , , we get,

y 11 = \Missing or unrecognized delimiter for \right\)

y 12 = \(\frac{g_{12}}{g_{22}} \)

y 21 = –\(\frac{g_{21}}{g_{22}} \)

y 22 = \(\frac{1}{g_{22}}\).

15. For a T-network if the Short circuit admittance parameters are given as y 11 , y 21 , y 12 , y 22 , then y 21 in terms of Inverse Hybrid parameters can be expressed as ________

a) y 21 = \Missing or unrecognized delimiter for \right\)

b) y 21 = \

y 21 = –\

y 21 = \(\frac{1}{g_{22}}\)

Answer: c

Explanation: We know that, I 1 = y 11 V 1 + y 12 V 2 ………

I 2 = y 21 V 1 + y 22 V 2 ……….

And, I 1 = g 11 V 1 + g 12 I 2 ……….

V 2 = g 21 V 1 + g 22 I 2 ………..

Now, and can be rewritten as,

I 1 = \Missing or unrecognized delimiter for \rightV_1 + \frac{g_{12}}{g_{22}} V_2\)……….

And I 2 = –\

∴ Comparing , and , , we get,

y 11 = \Missing or unrecognized delimiter for \right\)

y 12 = \(\frac{g_{12}}{g_{22}} \)

y 21 = –\(\frac{g_{21}}{g_{22}} \)

y 22 = \(\frac{1}{g_{22}}\).

This set of Network Theory Multiple Choice Questions & Answers focuses on “Advanced Problems Involving Parameters”.

1. For the circuit given below, the value of Transmission parameter A and C are ____________

network-theory-questions-answers-advanced-problems-involving-parameters-q1

a) A = -0.7692 + j0.3461 Ω, C = 0.03461 + j0.023 Ω

b) A = 0.7692 + j0.3461 Ω, C = 0.03461 + j0.023 Ω

c) A = -0.7692 – j0.3461 Ω, C = -0.03461 + j0.023 Ω

d) A = 0.7692 – j0.3461 Ω, C = 0.023 + j0.03461 Ω

Answer: b

Explanation: V = [20 + || ] I 1

V 1 = \(\Big[20 + \frac{}{-j15}\Big]\) I 1

= [20 – j\(\frac{10}{3}\)] I 1

I 0 = \Missing or unrecognized delimiter for \right\) I 1 = \(\frac{2}{3}\)I 1

V 2 = I 0 + 20I’ 0

= –\

I_1 \)

∴ A = \(\frac{V_1}{V_2} = \frac{

I_1}{20-\frac{j40}{3}) I_1}\) = 0.7692 + j0.3461 Ω

∴ C = \(\frac{I_1}{V_2} = \frac{1}{20-j40/3}\) = 0.03461 + j0.023 Ω.

2. For the circuit given below, the value of the Transmission parameter B and D are __________

network-theory-questions-answers-advanced-problems-involving-parameters-q1

a) D = 0.5385 + j0.6923 Ω, B = -6.923 + j25.385 Ω

b) D = 0.6923 + j0.5385 Ω, B = 6.923 + j25.385 Ω

c) D = -0.6923 + j0.5385 Ω, B= 25.385 + j6.923 Ω

d) D = -0.5385 + j0.6923 Ω, B = -6.923 + j25.385 Ω

Answer: a

Explanation: Z 1 = \(\frac{}{-j15-j10-j20}\) = j10

Z 2 = \(\frac{}{-j15}\) = \(-\frac{j40}{3}\)

Z 3 = \(\frac{}{-j15}\) = j20

-I 2 = \(\frac{20-j40/3}{20-\frac{j40}{3}+j20}I_1 = \frac{3-j2}{3+j}\) I 1

∴ D = \(\frac{-I_1}{I_2} = \frac{3+j}{3-j2}\) = 0.5385 + j0.6923 Ω

V 1 = [j10 + 2] I 1

= jI 1

So, B = –\

∴ B = -6.923 + j25.385 Ω.

3. For the circuit given below, the value of the Transmission parameters A and C are _________________

network-theory-questions-answers-advanced-problems-involving-parameters-q3

a) A = 0, C = 1

b) A = 1, C = 0

c) A = Z, C = 1

d) A = 1, C = Z

Answer: b

Explanation: V 1 = V 2

Or, A = \(\frac{V_1}{I_2}\) = 1

I 1 = 0 or, C = \(\frac{I_1}{V_2}\) = 0.

4. For the circuit given below, the value of the Transmission parameters B and D are _________________

network-theory-questions-answers-advanced-problems-involving-parameters-q3

a) B = Z, D = 1

b) B = 1, D = Z

c) B = Z, D = Z

d) B = 1, D = 1

Answer: a

Explanation: V 1 = ZI 1

And I 2 = -I 1

B = \(\frac{V_1}{I_2}\)

= \(\frac{-ZI_1}{-I_1}\) = Z

D = \(\frac{-I_1}{I_2}\) = 1.

5. For the circuit given below, the value of the Transmission parameters A and C are _______________

network-theory-questions-answers-advanced-problems-involving-parameters-q5

a) A = 1, C = 0

b) A = 0, C = 1

c) A = Y, C = 1

d) A = 1, C = Y

Answer: d

Explanation: V 1 = V 2

∴ A = \(\frac{V_1}{V_2}\) = 1

And V 1 = ZI 1

∴ C = \(\frac{I_1}{V_2} = \frac{1}{Z}\) = Y.

6. For the circuit given below, the value of the Transmission parameters B and D are ________________

network-theory-questions-answers-advanced-problems-involving-parameters-q5

a) B = Y, D = 1

b) B = 1, D = 0

c) B = 0, D = 1

d) B = 0, D = Y

Answer: c

Explanation: V 1 = V 2 = 0

And I 2 = -I 1

∴ B = \(\frac{V_1}{I_2}\) = 0

∴ D = \(\frac{-I_1}{I_2}\) = 1.

7. For the circuit given below, the values of the h parameter is given as follows h = [16, 3; -2, 0.01]. The value of the ratio \(\frac{V_2}{V_1}\) is ______________

network-theory-questions-answers-advanced-problems-involving-parameters-q8

a) 0.3299

b) 0.8942

c) 1.6

d) 0.2941

Answer: d

Explanation: Replacing the given 2-port circuit by an equivalent circuit and applying nodal analysis, we get,

V 2 = (2I 1 ) = 40 I 1

Or, -10 + 20I 1 + 3V 2 = 0

Or, 10 = 20I 1 + (40I 1 ) = 140I 1

∴ I 1 = \(\frac{1}{14}\) and V 2 = \(\frac{40}{14}\)

So, V 1 = 16I 1 + 3V 2 = \(\frac{136}{14}\)

And I 2 =

(2I 1 ) = \(\frac{-8}{70}\)

∴ \(\frac{V_2}{V_1} = \frac{40}{136}\) = 0.2941.

8. For the circuit given below, the values of the h parameter is given as follows h = [16, 3; -2, 0.01]. The value of the ratio \(\frac{I_2}{I_1}\) is ______________

network-theory-questions-answers-advanced-problems-involving-parameters-q8

a) 0.3299

b) 0.8942

c) -1.6

d) 0.2941

Answer: c

Explanation: Replacing the given 2-port circuit by an equivalent circuit and applying nodal analysis, we get,

V 2 = (2I 1 ) = 40 I 1

Or, -10 + 20I 1 + 3V 2 = 0

Or, 10 = 20I 1 + (40I 1 ) = 140I 1

∴ I 1 = \(\frac{1}{14}\) and V 2 = \(\frac{40}{14}\)

So, V 1 = 16I 1 + 3V 2 = \(\frac{136}{14}\)

And I 2 =

(2I 1 ) = \(\frac{-8}{70}\)

∴ \(\frac{I_2}{I_1}\) = -1.6.

9. If for a circuit the value of the h parameter is given as h = [8, 2/3; -2/3, 4/9]. Then the value of the voltage source V is _________________

a) 2.38 V

b) 1.19 V

c) 1.6 V

d) 3.2 V

Answer: b

Explanation: 8I 1 + \(\frac{2}{3V_2}\) = 10

V 2 = \(\frac{2}{3}\)I 1

= \(\frac{2}{3}\)I 1 \

= \frac{30}{29}I_1\)

I 1 = \(\frac{29}{30}\)V 2

V 2 + \(\frac{2}{3}\)V 2 = 10

V 2 = \(\frac{300}{252}\) = 1.19 V.

10. For a 2-port network, the value of the h parameter is as h=[600, 0.04; 30, 2×10 -3 ]. Given that, Z S = 2 kΩ and Z L = 400 Ω. The value of the parameter Z in is ______________

a) 250 Ω

b) 333.33 Ω

c) 650 Ω

d) 600 Ω

Answer: b

Explanation: Z in = h ie – \(\frac{h_{re} h_{fe} R_L}{1 + h_{oe} R_L}\)

= h 11 – \(\frac{h_{12} h_{21} R_L}{1+h_{22} R_L}\)

= 600 – \(\frac{0.04×30×400}{1+2×10^{-3}×400}\) = 333.33 Ω.

11. For a 2-port network, the value of the h parameter is as h=[600, 0.04; 30, 2×10 -3 ]. Given that, Z S = 2 kΩ and Z L = 400 Ω. The value of the parameter Z out is ______________

a) 650 Ω

b) 500 Ω

c) 250 Ω

d) 600 Ω

Answer: a

Explanation: Z out = \(\frac{R_s+h_{ie}}{

h_{oe}-h_{re} h_{fe}}\)

= \(\frac{R_s+h_{11}}{

h_{22}-h_{21} h_{12}}\)

= \(\frac{2000+600}{2600×2×10^{-3}-30×0.04}\) = 650 Ω.

12. For the circuit given below, the value of the g 11 and g 21 are _________________

network-theory-questions-answers-advanced-problems-involving-parameters-q12

a) g 11 = –\(\frac{1}{R_1+R_2}\), g 21 = \

g 11 = \(\frac{1}{R_1-R_2}\), g 21 = –\

g 11 = \(\frac{1}{R_1+R_2}\), g 21 = \

g 11 = \(\frac{1}{R_1-R_2}\), g 21 = \(\frac{R_2}{R_1-R_2}\)

Answer: c

Explanation: I 1 = \(\frac{V_1}{R_1+R_2}\)

Or, g 11 = \(\frac{I_1}{V_1} = \frac{1}{R_1+R_2}\)

By voltage division, V 2 = \(\frac{R_2}{R_1+R_2}\)V 1

Or, g 21 = \(\frac{V_2}{V_1} = \frac{R_2}{R_1+R_2}\).

13. For the circuit given below, the value of the g 12 and g 22 are _______________

network-theory-questions-answers-advanced-problems-involving-parameters-q12

a) g 12 = –\(\frac{R_2}{R_1+R_2}\), g 22 = R 3 + \

g 12 = \(\frac{R_2}{R_1+R_2}\), g 22 = -R 3 + \

g 12 = –\(\frac{R_2}{R_1+R_2}\), g 22 = R 3 – \

g 12 = \(\frac{R_2}{R_1+R_2}\), g 22 = -R 3 – \(\frac{R_1 R_2}{R_1+R_2}\)

Answer: a

Explanation: I 1 = –\(\frac{R_2}{R_1+R_2}\)I 2

Or, g 12 = \(\frac{I_1}{I_2} = -\frac{R_2}{R_1+R_2}\)

Also, I 2 (R 3 + R 1 //R 2 )

= I2 \

∴ g 22 = \(\frac{V_2}{I_2} = R_3 + \frac{R_1 R_2}{R_1+R_2}\).

14. For the circuit given below, the value of g 11 and g 21 are _________________

network-theory-questions-answers-advanced-problems-involving-parameters-q14

a) g 11 = 0.0667 – j0.0333 Ω, g 21 = 0.8 + j0.4 Ω

b) g 11 = -0.0667 – j0.0333 Ω, g 21 = -0.8 – j0.4 Ω

c) g 11 = 0.0667 + j0.0333 Ω, g 21 = 0.8 + j0.4 Ω

d) g 11 = -0.0667 + j0.0333 Ω, g 21 = 0.8 – j0.4 Ω

Answer: c

Explanation: V 1 = I 1

Or, g 11 = \(\frac{I_1}{V_1} = \frac{1}{12-j6}\) = 0.0667 + j0.0333 Ω

g 21 = \(\frac{V_2}{V_1} = \frac{12I_1}{ I_1}\)

= \(\frac{2}{2-j}\) = 0.8 + j0.4 Ω.

15. For the circuit given below, the value of g 12 and g 22 are ________________

network-theory-questions-answers-advanced-problems-involving-parameters-q14

a) g 12 = 0.8 + j0.4 Ω, g 22 = 2.4 + j5.2 Ω

b) g 12 = -0.8 + j0.4 Ω, g 22 = -2.4 – j5.2 Ω

c) g 12 = 0.8 – j0.4 Ω, g 22 = 2.4 – j5.2 Ω

d) g 12 = -0.8 – j0.4 Ω, g 22 = 2.4 + j5.2 Ω

Answer: d

Explanation: I 1 = \(\frac{-12}{12-j6}\)I 2

Or, g 12 = \(\frac{I_1}{I_2}\) = -g 21 = -0.8 – j0.4 Ω

V 2 = I 2

g 22 = \(\frac{V_2}{I_2}\) = 2.4 + j5.2 Ω.

This set of Network Theory Problems focuses on “Inner Relationships of Different Parameters”.

1. For the given information Z 11 = 3, Z 12 = 1, Z 21 = 2, Z 22 = 1. Find the value of Y 11 .

a) 1

b) -1

c) 2

d) -2

Answer: a

Explanation: Y 11 = Z 22 /∆z and ∆z=3-2=1 and Z 22 = 1. So on substituting we get Y 11 = 1/1 = 1.

2. For the given information Z 11 = 3, Z 12 = 1, Z 21 = 2, Z 22 = 1. Find the value of Y 12 .

a) -2

b) 2

c) -1

d) 1

Answer: c

Explanation: Y 12 = -Z 12 /∆z and ∆z=3-2=1 and Z 12 = 1. So on substituting we get Y 12 =-1/1=-1.

3. For the given information Z 11 = 3, Z 12 = 1, Z 21 = 2, Z 22 = 1. Find the value of Y 21 .

a) 2

b) -2

c) 1

d) -1

Answer: b

Explanation: We have the relation Y 21 =-Z 21 /∆z. ∆z=3-2=1 and given Z 21 = 2. On substituting we get Y 21 =-2/1=-2.

4. For the given information Z 11 = 3, Z 12 = 1, Z 21 = 2, Z 22 = 1. Find the value of Y 22 .

a) 1

b) 2

c) 3

d) 4

Answer: c

Explanation: The relation between Y 22 and Z 11 is Y 22 = Z 11 /∆z and ∆z=3-2=1 and given Z 11 = 3. On substituting we get Y 22 = 3/1 = 3.

5. For the given information Z 11 = 3, Z 12 = 1, Z 21 = 2, Z 22 = 1. What is the value ∆y?

a) 0

b) 1

c) 2

d) 3

Answer: b

Explanation: ∆y is the determinant of y parameters. The value ∆y is (Y 11 )(Y 22 )-(Y 12 )(Y 21 ). On substituting the values we get ∆y = -=1.

6. For the given information Z 11 = 3, Z 12 = 1, Z 21 = 2, Z 22 = 1. What is the product of ∆y and ∆z is?

a) 3

b) 2

c) 1

d) 0

Answer: c

Explanation: ∆y is the determinant of y parameters and ∆z is the determinant of z parameters. And we obtained ∆y = 1 and ∆z = 1. So their product = = 1.

7. The relation between Z 11 and Y parameters is?

a) Z 11 = Y 22 /∆y

b) Z 11 = -Y 22 /∆y

c) Z 11 = Y 12 /∆y

d) Z 11 = (-Y 12 )/∆y

Answer: a

Explanation: V 1 =(Y 22 /∆y)I 1 -(Y 12 /∆y)I 2 . The relation between Z 11 and Y parameters is Z 11 = Y 22 /∆y.

8. The relation between Z 12 and Y parameters is?

a) Z 12 = Y 12 /∆y

b) Z 12 = (-Y 12 )/∆y

c) Z 12 = (-Y 22 )/∆y

d) Z 12 = Y 22 /∆y

Answer: b

Explanation: V 1 =(Y 22 /∆y)I 1 -(Y 12 /∆y)I 2 . The relation between Z 12 and Y parameters is Z 12 = (-Y 12 )/∆y.

9. The relation between Z 21 and Y parameters is?

a) Z 21 = Y 21 /∆y

b) Z 21 = Y 12 /∆y

c) Z 21 = (-Y 21 )/∆y

d) Z 21 = (-Y 12 )/∆y

Answer: c

Explanation: V 2 =(-Y 21 /∆y)I 1 +(Y 11 /∆y)I 2 . The relation between Z 21 and Y parameters is Z 21 = (-Y 21 )/∆y.

10. The relation between Z 22 and Y parameters is?

a) Z 22 = (-Y 11 )/∆y

b) Z 22 = Y 21 /∆y

c) Z 22 = (-Y 21 )/∆y

d) Z 22 = Y 11 /∆y

Answer: d

Explanation: V 2 =(-Y 21 /∆y)I 1 +(Y 11 /∆y)I 2 . The relation between Z 22 and Y parameters is Z 22 = Y 11 /∆y.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Inter Connection of Two-Port Networks”.

1. In the circuit shown below, find the Z-parameter Z 11 .

network-theory-questions-answers-inter-connection-two-port-q1

a) 1

b) 2

c) 3

d) 4

Answer: c

Explanation: The Z –parameter Z 11 is V 1 /I 1 , port 2 is open circuited. V 1 = I 1 => V 1 /I 1 = 3 and on substituting, we get Z 11 = 3Ω.

2. In the circuit shown below, find the Z-parameter Z 12 .

network-theory-questions-answers-inter-connection-two-port-q1

a) 4

b) 3

c) 2

d) 1

Answer: c

Explanation: The Z-parameter Z 12 is V 2 /I 1 |I 2 =0. On open circuiting port 2 we obtain the equation, V 1 = I 2 => V 1 /I 1 = 2. On substituting we get Z 12 = 2Ω.

3. In the circuit shown below, find the Z-parameter Z 21 .

network-theory-questions-answers-inter-connection-two-port-q1

a) 2

b) 4

c) 1

d) 3

Answer: a

Explanation: The Z-parameter Z 21 is V 2 /I 1 |I 2 =0. On open circuiting port 2, we get V 2 = I 1 => V 2 /I 1 = 2. On substituting we get Z 21 = 2Ω.

4. In the circuit shown below, find the Z-parameter Z 22 .

network-theory-questions-answers-inter-connection-two-port-q1

a) 3

b) 2

c) 4

d) 1

Answer: a

Explanation: The Z-parameter Z 21 is V 2 /I 2 |I 1 =0. This parameter is obtained by open circuiting port 1. So we get V 2 = I 2 => V 2 = 3(I 2 ) => V 2 /I 2 = 3. On substituting Z 21 = 3Ω.

5. In the circuit shown below, find the Z-parameter Z 11 .

network-theory-questions-answers-inter-connection-two-port-q5

a) 10

b) 15

c) 20

d) 25

Answer: b

Explanation: The Z –parameter Z 11 is V 1 /I 1 , port 2 is open circuited. V 1 = I 1 => V 1 /I 1 = 15 and on substituting, we get Z 11 = 2.5Ω.

6. In the circuit shown below, find the Z-parameter Z 12 .

network-theory-questions-answers-inter-connection-two-port-q5

a) 15

b) 10

c) 5

d) 1

Answer: c

Explanation: The Z-parameter Z 12 is V 2 /I 1 |I 2 =0. On open circuiting port 2 we obtain the equation, V 1 = I 2 => V 1 /I 1 = 5. On substituting we get Z 12 = 5Ω.

7. From the circuits shown below, find the combined Z-parameter Z 11 .

network-theory-questions-answers-inter-connection-two-port-q5

network-theory-questions-answers-inter-connection-two-port-q1

a) 8

b) 18

c) 28

d) 38

Answer: b

Explanation: The Z-parameter Z 11 is Z 11 = Z 11x + Z 11y and Z 11x = 3, Z 11y = 15. On substituting we get Z 11 = 3 +15 = 18Ω.

8. From the circuits shown below, find the combined Z-parameter Z 12 .

network-theory-questions-answers-inter-connection-two-port-q5

network-theory-questions-answers-inter-connection-two-port-q1

a) 4

b) 5

c) 6

d) 7

Answer: d

Explanation: The Z-parameter Z 12 is Z 12 = Z 12x + Z 12y and we have Z 12x = 2, Z 12y . On substituting we get Z 12 = 2 + 5 = 7Ω.

9. From the circuits shown below, find the combined Z-parameter Z 21 .

network-theory-questions-answers-inter-connection-two-port-q5

network-theory-questions-answers-inter-connection-two-port-q1

a) 7

b) 6

c) 5

d) 4

Answer: a

Explanation: The Z-parameter Z 21 is Z 21 = Z 21x + Z 21y and we have Z 21x = 2, Z 21y = 5. On substituting we get Z 21 = 2 + 5 = 7Ω.

10. From the circuits shown below, find the combined Z-parameter Z 22 .

network-theory-questions-answers-inter-connection-two-port-q5

network-theory-questions-answers-inter-connection-two-port-q1

a) 38

b) 28

c) 18

d) 8

Answer: b

Explanation: The Z-parameter Z 22 is Z 22 = Z 22x + Z 22y and we have Z 22x = 3, Z 22y = 25. On substituting we get Z 22 = 3 +25 = 28Ω.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Terminated Two-Port Network”.

1. Calculate the Z –parameter Z 11 in the circuit shown below.

network-theory-questions-answers-terminated-two-port-q1

a) 1.5

b) 2.5

c) 3.5

d) 4.5

Answer: b

Explanation: The Z–parameter Z 11 is V 1 /I 1 , port 2 is open circuited. V 1 = I 1 => V 1 /I 1 = 2.5 and on substituting, we get Z 11 = 2.5Ω.

2. Determine the Z-parameter Z 12 in the circuit shown below.

network-theory-questions-answers-terminated-two-port-q1

a) 1

b) 2

c) 3

d) 4

Answer: a

Explanation: The Z-parameter Z 12 is V 2 /I 1 |I 2 =0. On open circuiting port 2 we obtain the equation, V 1 = I 2 => V 1 /I 1 = 1.5. On substituting we get Z 12 = 1.5Ω.

3. Determine the Z-parameter Z 21 in the circuit shown below.

network-theory-questions-answers-terminated-two-port-q1

a) 4

b) 3

c) 2

d) 1

Answer: d

Explanation: The Z-parameter Z 21 is V 2 /I 1 |I 2 =0. On open circuiting port 2, we get V 2 = I 1 => V 2 /I 1 = 1.5. On substituting we get Z 21 = 1.5Ω.

4. Determine the Z-parameter Z 22 in the circuit shown below.

network-theory-questions-answers-terminated-two-port-q1

a) 1

b) 3

c) 2

d) 4

Answer: c

Explanation: The Z-parameter Z 21 is V 2 /I 2 |I 1 =0. This parameter is obtained by open circuiting port 1. So we get V 2 = ||4)I 2 => V 2 = 2(I 2 ) => V 2 /I 2 = 2. On substituting Z 21 = 2Ω.

5. Find the value of V 1 /I 1 in the circuit shown below.

network-theory-questions-answers-terminated-two-port-q1

a) 1.25

b) 2.25

c) 3.25

d) 4.25

Answer: b

Explanation: We have the relation V 1 /I 1 =Z 11 – Z 12 Z 21 /(Z L +Z 21 ) and Z L is the load impedance and is equal to 2Ω. On solving V 1 /I 1 =2.5-1/=2.25Ω.

6. Determine the input impedance of the network shown below.

network-theory-questions-answers-terminated-two-port-q1

a) 4.25

b) 3.25

c) 2.25

d) 1.25

Answer: b

Explanation: From the figure by inspection we can say that the source resistance is 1Ω. So Z in = (V 1 /I 1 ) + Source resistance. We had V 1 /I 1 = 2.25. On substituting Z in =1+2.25=3.25Ω.

7. Determine the value of source admittance in the circuit shown below.

network-theory-questions-answers-terminated-two-port-q7

a) 1

b) 2

c) 3

d) 4

Answer: a

Explanation: From the figure, the value of the admittance parallel to the current source is 1 mho and this is the value of source admittance. So Y s = 1 mho.

8. Find the value of I 2 /V 2 in the circuit shown below.

network-theory-questions-answers-terminated-two-port-q7

a) 7/6

b) 6/7

c) 7/12

d) 12/7

Answer: c

Explanation: The relation between I 2 /V 2 and Y-parameters is

I 2 /V 2 =/=7/12 mho.

9. The value of the Y-parameter Y 22 in the circuit shown below.

network-theory-questions-answers-terminated-two-port-q7

a) 12/7

b) 6/7

c) 7/6

d) 7/12

Answer: d

Explanation: The relation between Y 22 and I 2 /V 2 is Y 22 = I 2 /V 2 . We have the relation I 2 /V 2 = (Y 22 Y s +Y 22 Y 11 -Y 21 Y 12 )/(Y s +Y 11 ). On substituting their values in the equation we get Y 22 = 7/12 mho.

10. The value of the Z-parameter Z 22 in the circuit shown below.

network-theory-questions-answers-terminated-two-port-q7

a) 6/7

b) 7/12

c) 12/7

d) 7/6

Answer: c

Explanation: The Z-parameter Z 22 is inverse of the Y-parameter Y 22 i.e., Z 22 = 1/Y 22 . We got Y 22 = 7/12. So on substituting we get Z 22 = 12/7 mho.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Image Parameters”.

1. A network is said to be symmetrical if the relation between A and D is?

a) A = D

b) A = 2 D

c) A = 3 D

d) A = 4 D

Answer: a

Explanation: We know V 1 =AV 2 -BI 2 and I 1 =CV 2 -DI 2 . If the network is symmetrical, then the relation between A and D is A = D.

2. The relation between Z 11 and Z 12 if the network is symmetrical is?

a) Z 11 = 2 Z 12

b) Z 11 = Z 12

c) Z 11 = 3 Z 12

d) Z 11 = 4 Z 12

Answer: b

Explanation: For a network to be symmetrical A=D. So the relation between Z 11 and Z 12 for the network is symmetrical is Z 11 = Z 12 .

3. The relation between Z 12 and Z 11 and B and C parameters if the network is symmetrical is?

a) Z 11 = Z 12 = B/C

b) Z 11 = Z 12 = C/B

c) Z 11 = Z 12 = √

d) Z 11 = Z 12 = √

Answer: c

Explanation: For symmetrical network, A=D. On substituting this we get the relation between Z 12 and Z 11 and B and C parameters if the network is symmetrical is Z 11 = Z 12 = √.

4. Determine the transmission parameter A in the circuit shown below.

network-theory-questions-answers-image-parameters-q4

a) 3/4

b) 4/3

c) 5/6

d) 6/5

Answer: d

Explanation: We know V 1 =AV 2 -BI 2 and I 1 =CV 2 -DI 2 . A=(V 1 /V 2 ) |I 2 =0. On solving we get the transmission parameter A as A = 6/5.

5. Determine the transmission parameter B in the circuit shown below.

network-theory-questions-answers-image-parameters-q4

a) 17/5

b) 5/17

c) 13/5

d) 5/13

Answer: a

Explanation: The transmission parameter B is -V 1 /I 2 |V 2 =0. On short cicuiting the port 2, from the circuit we get -I 2 = V 1 => -V 1 /I 2 = 17/5. On substituting we get B = 17/5.

6. Determine the transmission parameter C in the circuit shown below.

network-theory-questions-answers-image-parameters-q4

a) 2/5

b) 1/5

c) 4/5

d) 3/5

Answer: b

Explanation: The transmission parameter C is I 1 /V 2 |I 2 =0. This parameter is obtained by open circuiting the port 2. So we get V 2 = 5I 1 => I 1 /V 2 = 1/5. On substituting, we get C = 1/5.

7. Determine the transmission parameter D in the circuit shown below.

network-theory-questions-answers-image-parameters-q4

a) 3/5

b) 4/5

c) 7/5

d) 2/5

Answer: c

Explanation: The transmission parameter D is -I 1 /I 2 |V 2 =0. This is obtained by short circuiting the port 2. We get I 1 = V 1 and -I 2 = V 1 . On solving, we get -I 1 /I 2 = 7/5. On substituting we get D = 7/5.

8. The value of Z 11 in the circuit shown below is?

network-theory-questions-answers-image-parameters-q4

a) 1.8

b) 2.8

c) 3.8

d) 4.8

Answer: c

Explanation: The relation between Z 11 and ABCD parameters is Z 11 =√. We know A = 6/5, B = 17/5, C = 1/5, D = 7/5. On substituting, Z 11 = √/) = 3.8Ω.

9. The value of Z 12 in the circuit shown below is?

network-theory-questions-answers-image-parameters-q4

a) 1.1

b) 2.2

c) 3.3

d) 4.4

Answer: d

Explanation: The relation between Z 12 and ABCD parameters is Z 12 =√. We got B = 17/5, D = 7/5, A = 6/5, C = 1/5. On substituting Z 12 = √ = √/) = 4.4Ω.

10. Determine the value of Ø in the circuit shown below.

network-theory-questions-answers-image-parameters-q4

a) 0.25

b) 0.5

c) 0.75

d) 1

Answer: c

Explanation: Ø is called image transfer constant and it is also used to describe reciprocal networks and this parameter is obtained from the voltage and current ratios. We know Ø = tanh -1 ⁡√ = tanh -1 ⁡√ = 0.75.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Series-Series Connection of Two Port Network”.

1. In the circuit given below, the value of R is ___________

network-theory-questions-answers-series-series-connection-two-port-network-q1

a) 2.5 Ω

b) 5.0 Ω

c) 7.5 Ω

d) 10.0 Ω

Answer: c

Explanation: The resultant R when viewed from voltage source = \(\frac{100}{8}\) = 12.5

∴ R = 12.5 – 10 || 10 = 12.5 – 5 = 7.55 Ω.

2. In the circuit given below, the number of chords in the graph is ________________

network-theory-questions-answers-series-series-connection-two-port-network-q2

a) 3

b) 4

c) 5

d) 6

Answer: b

Explanation: Given that, b = 6, n = 3

Number of Links is given by, b – n + 1

= 6 – 3 + 1 = 4.

3. In the circuit given below, the current through the 2 kΩ resistance is _____________

network-theory-questions-answers-series-series-connection-two-port-network-q3

a) Zero

b) 1 mA

c) 2 mA

d) 6 mA

Answer: a

Explanation: We know that when a Wheatstone bridge is balanced, no current will flow through the middle resistance.

Here, \(\frac{R_1}{R_2} = \frac{R_3}{R_4}\)

Since, R 1 = R 2 = R 3 = R 4 = 1 kΩ.

4. How many incandescent lamps connected in series would consume the same total power as a single 100 W/220 V incandescent lamp. The rating of each lamp is 200 W/220 V?

a) Not possible

b) 4

c) 3

d) 2

Answer: d

Explanation: In series power = \(\frac{1}{P}\)

Now, \(\frac{1}{P} = \frac{1}{P_1} + \frac{1}{P_2}\)

= \(\frac{1}{200} + \frac{1}{200}\)

Or, P = \(\frac{200}{2}\) = 100 W.

5. Two networks are connected in series parallel connection. Then, the forward short-circuit current gain of the network is ____________

a) Product of Z-parameter matrices

b) Sum of h-parameter matrices

c) Sum of Z-parameter matrices

d) Product of h-parameter matrices

Answer: b

Explanation: The forward short circuit current gain is given by,

h 21 = \(\frac{I_2 }{I_1 }\), when V 2 = 0

So, when the two networks are connected in series parallel combination,

[h 11 , h 12 ; h 21 , h 22 ] = [h’ 11 + h’ 11 , h’ 12 + h’ 12 ; h’ 21 + h’ 21 , h’ 22 + h’ 22 ]

So, h 21 of total network will be sum of h parameter matrices.

6. The condition for a 2port network to be reciprocal is ______________

a) Z 11 = Z 22

b) BC – AD = -1

c) Y 12 = -Y 21

d) h 12 = h 21

Answer: b

Explanation: If the network is reciprocal, then the ratio of the response transform to the excitation transform would not vary after interchanging the position of the excitation.

7. The relation AD – BC = 1, is valid for ___________

a) Both active and passive networks

b) Passive but not reciprocal networks

c) Active and reciprocal networks

d) Passive and reciprocal networks

Answer: d

Explanation: AD – BC = 1, is the condition for reciprocity for ABCD parameters, which shows that the relation is valid for reciprocal network. The ABCD parameters are obtained for the network which consists of resistance, capacitance and inductance, which indicates that it is a passive network.

8. For a 2 port network, the transmission parameters are given as 10, 9, 11 and 10 corresponds to A, B, C and D. The correct statement among the following is?

a) Network satisfies both reciprocity and symmetry

b) Network satisfies only reciprocity

c) Network satisfies only symmetry

d) Network satisfies neither reciprocity nor symmetry

Answer: a

Explanation: Here, A = 10, B = 9, C = 11, D = 10

∴ A = D

∴ Condition for symmetry is satisfied.

Also, AD – BC = –

= 100 – 99 = 1

Therefore the condition of reciprocity is satisfied.

9. In the circuit given below, the equivalent capacitance is ______________

network-theory-questions-answers-series-series-connection-two-port-network-q9

a) \

Answer: b

Explanation: The equivalent capacitance by applying the concept of series-parallel combination of the capacitance is,

Or, C EQ = \

I dt

Given, I = 2 A = constant = 2 ∫ Vdt

= 2 × Shaded area

= 2 × \

× 60 × 10

= 13.2 kJ.

10. In the circuit given below, the 60 V source absorbs power. Then the value of the current source is ____________

network-theory-questions-answers-series-series-connection-two-port-network-q10

a) 10 A

b) 13 A

c) 15 A

d) 18 A

Answer: a

Explanation: Given that, 60 V source is absorbing power, it means that current flow from positive to negative terminal in 60 V source.

Applying KVL, we get, I + I 1 = 12 A ………………

Current source must have the value of less than 12 A to satisfy equation .

11. In the circuit given below, the number of node and branches are ______________

network-theory-questions-answers-series-series-connection-two-port-network-q11

a) 4 and 5

b) 4 and 6

c) 5 and 6

d) 6 and 4

Answer: b

Explanation: In the given graph, there are 4 nodes and 6 branches.

Twig = n – 1 = 4 – 1 = 3

Link = b – n + 1 = 6 – 4 + 1 = 3.

12. A moving coil of a meter has 250 turns and a length and depth of 40 mm and 30 mm respectively. It is positioned in a uniform radial flux density of 450 mT. The coil carries a current of 160 mA. The torque on the coil is?

a) 0.0216 N-m

b) 0.0456 N-m

c) 0.1448 N-m

d) 1 N-m

Answer: a

Explanation: Given, N = 250, L = 40 × 10 -3 , d = 30 × 10 -3 m, I = 160 × 10 -3 A, B = 450 × 10 -3 T

Torque = 250 × 450 × 10 -3 × 40 × 10 -3 × 30 × 10 -3 × 160 × 10 -3

= 200 × 10 -6 N-m = 0.0216 N-m.

13. In the circuit given below, the equivalent inductance is ____________

network-theory-questions-answers-series-series-connection-two-port-network-q13

a) L 1 + L 2 – 2M

b) L 1 + L 2 + 2M

c) L 1 + L 2 – M

d) L 1 + L 2

Answer: a

Explanation: Since, in one inductor current is leaving to dot and in other inductor current is entering to dot.

So, L EQ = L 1 + L 2 – 2M.

14. In the figure given below, the pole-zero plot corresponds to _____________

network-theory-questions-answers-series-series-connection-two-port-network-q14

a) Low-pass filter

b) High-pass filter

c) Band-pass filter

d) Notch filter

Answer: d

Explanation: In pole zero plot the two transmission zeroes are located on the jω-axis, at the complex conjugate location, and then the magnitude response exhibits a zero transmission at ω – ω C .

15. In the circuit given below, the maximum power that can be transferred to the resistor R L is _____________

network-theory-questions-answers-series-series-connection-two-port-network-q15

a) 1 W

b) 10 W

c) 0.25 W

d) 0.5 W

Answer: c

Explanation: For maximum power transfer to the load resistor R L , R L must be equal to 100 Ω.

∴ Maximum power = \(\frac{V^2}{4R_L}\)

= \(\frac{10^2}{4 ×100}\) = 0.25 W.

This set of Network Theory Puzzles focuses on “Relation between Transmission Parameters with Short Circuit Admittance and Open Circuit Impedance Parameters”.

1. For a T shaped network, if the Short-circuit admittance parameters are y 11 , y 12 , y 21 , y 22 , then y 11 in terms of Transmission parameters can be expressed as ________

a) y 11 = \

y 11 = \

y 11 = – \

y 11 = \(\frac{A}{B}\)

Answer: a

Explanation: We know that, V 1 = AV 2 – BI 2 ………

I 1 = CV 2 – DI 2 ……………

And, I 1 = y 11 V 1 + y 12 V 2 ………

I 2 = y 21 V 1 + y 22 V 2 ……….

Now, and can be rewritten as, I 2 = \

And I 1 = CV 2 – D \Missing or unrecognized delimiter for \right = \frac{D}{B}V_1 + \leftMissing or unrecognized delimiter for \right V_2\) ……………

Comparing equations , and , , we get,

y 11 = \(\frac{D}{B}\)

y 12 = \(\frac{C-A}{B}\)

y 21 = – \(\frac{1}{B}\)

y 22 = \(\frac{A}{B}\).

2. For a T shaped network, if the Short-circuit admittance parameters are y 11 , y 12 , y 21 , y 22 , then y 12 in terms of Transmission parameters can be expressed as ________

a) y 12 = \

y 12 = \

y 12 = – \

y 12 = \(\frac{A}{B}\)

Answer: b

Explanation: We know that, V 1 = AV 2 – BI 2 ………

I 1 = CV 2 – DI 2 ……………

And, I 1 = y 11 V 1 + y 12 V 2 ………

I 2 = y 21 V 1 + y 22 V 2 ……….

Now, and can be rewritten as, I 2 = \

And I 1 = CV 2 – D \Missing or unrecognized delimiter for \right = \frac{D}{B}V_1 + \leftMissing or unrecognized delimiter for \right V_2\) ……………

Comparing equations , and , , we get,

y 11 = \(\frac{D}{B}\)

y 12 = \(\frac{C-A}{B}\)

y 21 = – \(\frac{1}{B}\)

y 22 = \(\frac{A}{B}\).

3. For a T shaped network, if the Short-circuit admittance parameters are y 11 , y 12 , y 21 , y 22 , then y 21 in terms of Transmission parameters can be expressed as ________

a) Y 21 = \

Y 21 = \

Y 21 = – \

Y 21 = \(\frac{A}{B}\)

Answer: c

Explanation: We know that, V 1 = AV 2 – BI 2 ………

I 1 = CV 2 – DI 2 ……………

And, I 1 = y 11 V 1 + y 12 V 2 ………

I 2 = y 21 V 1 + y 22 V 2 ……….

Now, and can be rewritten as, I 2 = \

And I 1 = CV 2 – D \Missing or unrecognized delimiter for \right = \frac{D}{B}V_1 + \leftMissing or unrecognized delimiter for \right V_2\) ……………

Comparing equations , and , , we get,

y 11 = \(\frac{D}{B}\)

y 12 = \(\frac{C-A}{B}\)

y 21 = – \(\frac{1}{B}\)

y 22 = \(\frac{A}{B}\).

4. For a T shaped network, if the Short-circuit admittance parameters are y 11 , y 12 , y 21 , y 22 , then y 22 in terms of Transmission parameters can be expressed as ________

a) y 22 = \

y 22 = \

y 22 = – \

y 22 = \(\frac{A}{B}\)

Answer: d

Explanation: We know that, V 1 = AV 2 – BI 2 ………

I 1 = CV 2 – DI 2 ……………

And, I 1 = y 11 V 1 + y 12 V 2 ………

I 2 = y 21 V 1 + y 22 V 2 ……….

Now, and can be rewritten as, I 2 = \

And I 1 = CV 2 – D \Missing or unrecognized delimiter for \right = \frac{D}{B}V_1 + \leftMissing or unrecognized delimiter for \right V_2\) ……………

Comparing equations , and , , we get,

y 11 = \(\frac{D}{B}\)

y 12 = \(\frac{C-A}{B}\)

y 21 = – \(\frac{1}{B}\)

y 22 = \(\frac{A}{B}\).

5. For a T-network if the Open circuit Impedance parameters are z 11 , z 12 , z 21 , z 22 , then z 11 in terms of Transmission parameters can be expressed as ____________

a) z 11 = \

z 11 = \

z 11 = \(\frac{D}{C}\)

Answer: a

Explanation: We know that, V 1 = z 11 I 1 + z 12 I 2 ………….

V 2 = z 21 I 1 + z 22 I 2 …………….

And V 1 = AV 2 – BI 2 ………

I 1 = CV 2 – DI 2 ……………

Rewriting and , we get,

V 2 = \

And V 1 = \Missing or unrecognized delimiter for \right – BI_2 = \frac{A}{C}I_1 + \leftMissing or unrecognized delimiter for \right I_2\) …………..

Comparing , and , , we get,

z 11 = \(\frac{A}{C}\)

z 12 = \(\frac{AD}{C – B}\)

z 21 = \(\frac{1}{C}\)

z 22 = \(\frac{D}{C}\).

6. For a T-network if the Open circuit Impedance parameters are z 11 , z 12 , z 21 , z 22 , then z 12 in terms of Transmission parameters can be expressed as ____________

a) z 12 = \

z 12 = \

z 12 = \(\frac{D}{C}\)

Answer: b

Explanation: We know that, V 1 = z 11 I 1 + z 12 I 2 ………….

V 2 = z 21 I 1 + z 22 I 2 …………….

And V 1 = AV 2 – BI 2 ………

I 1 = CV 2 – DI 2 ……………

Rewriting and , we get,

V 2 = \

And V 1 = \Missing or unrecognized delimiter for \right – BI_2 = \frac{A}{C}I_1 + \leftMissing or unrecognized delimiter for \right I_2\) …………..

Comparing , and , , we get,

z 11 = \(\frac{A}{C}\)

z 12 = \(\frac{AD}{C – B}\)

z 21 = \(\frac{1}{C}\)

z 22 = \(\frac{D}{C}\).

7. For a T-network if the Open circuit Impedance parameters are z 11 , z 12 , z 21 , z 22 , then z 21 in terms of Transmission parameters can be expressed as ____________

a) z 21 = \

z 21 = \

z 21 = \(\frac{D}{C}\)

Answer: c

Explanation: We know that, V 1 = z 11 I 1 + z 12 I 2 ………….

V 2 = z 21 I 1 + z 22 I 2 …………….

And V 1 = AV 2 – BI 2 ………

I 1 = CV 2 – DI 2 ……………

Rewriting and , we get,

V 2 = \

And V 1 = \Missing or unrecognized delimiter for \right – BI_2 = \frac{A}{C}I_1 + \leftMissing or unrecognized delimiter for \right I_2\) …………..

Comparing , and , , we get,

z 11 = \(\frac{A}{C}\)

z 12 = \(\frac{AD}{C – B}\)

z 21 = \(\frac{1}{C}\)

z 22 = \(\frac{D}{C}\).

8. For a T-network if the Open circuit Impedance parameters are z 11 , z 12 , z 21 , z 22 , then z 22 in terms of Transmission parameters can be expressed as ____________

a) z 22 = \

z 22 = \

z 22 = \(\frac{D}{C}\)

Answer: d

Explanation: We know that, V 1 = z 11 I 1 + z 12 I 2 ………….

V 2 = z 21 I 1 + z 22 I 2 …………….

And V 1 = AV 2 – BI 2 ………

I 1 = CV 2 – DI 2 ……………

Rewriting and , we get,

V 2 = \

And V 1 = \Missing or unrecognized delimiter for \right – BI_2 = \frac{A}{C}I_1 + \leftMissing or unrecognized delimiter for \right I_2\) …………..

Comparing , and , , we get,

z 11 = \(\frac{A}{C}\)

z 12 = \(\frac{AD}{C – B}\)

z 21 = \(\frac{1}{C}\)

z 22 = \(\frac{D}{C}\).

9. For a T shaped network, if the Short-circuit admittance parameters are y 11 , y 12 , y 21 , y 22 , then y 11 in terms of Inverse Transmission parameters can be expressed as ________

a) y 11 = \

y 11 = – \

y 11 = \Missing or unrecognized delimiter for \right\)

d) y 11 = \(\frac{D’}{B’}\)

Answer: a

Explanation: We know that, V 2 = A’V 1 – B’I 1 ………

I 2 = C’V 1 – D’I 1 ……………

And, I 1 = y 11 V 1 + y 12 V 2 ………

I 2 = y 21 V 1 + y 22 V 2 ……….

Now, and can be rewritten as, I 1 = – \

And I 2 = C’V 1 – D’ \Missing or unrecognized delimiter for \right = \leftMissing or unrecognized delimiter for \right V_1 + \frac{D’}{B’} V_2\) …………

Comparing equations , and , , we get,

y 11 = \(\frac{A’}{B’}\)

y 12 = – \(\frac{1}{B’}\)

y 21 = \Missing or unrecognized delimiter for \right\)

y 22 = \(\frac{D’}{B’}\).

10. For a T shaped network, if the Short-circuit admittance parameters are y 11 , y 12 , y 21 , y 22 , then y 12 in terms of Inverse Transmission parameters can be expressed as ________

a) y 12 = \

y 12 = – \

y 12 = \Missing or unrecognized delimiter for \right\)

d) y 12 = \(\frac{D’}{B’}\)

Answer: b

Explanation: We know that, V 2 = A’V 1 – B’I 1 ………

I 2 = C’V 1 – D’I 1 ……………

And, I 1 = y 11 V 1 + y 12 V 2 ………

I 2 = y 21 V 1 + y 22 V 2 ……….

Now, and can be rewritten as, I 1 = – \

And I 2 = C’V 1 – D’ \Missing or unrecognized delimiter for \right = \leftMissing or unrecognized delimiter for \right V_1 + \frac{D’}{B’} V_2\) …………

Comparing equations , and , , we get,

y 11 = \(\frac{A’}{B’}\)

y 12 = – \(\frac{1}{B’}\)

y 21 = \Missing or unrecognized delimiter for \right\)

y 22 = \(\frac{D’}{B’}\).

11. For a T shaped network, if the Short-circuit admittance parameters are y 11 , y 12 , y 21 , y 22 , then y 21 in terms of Inverse Transmission parameters can be expressed as ________

a) y 21 = \

y 21 = – \

y 21 = \Missing or unrecognized delimiter for \right\)

d) y 21 = \(\frac{D’}{B’}\)

Answer: c

Explanation: We know that, V 2 = A’V 1 – B’I 1 ………

I 2 = C’V 1 – D’I 1 ……………

And, I 1 = y 11 V 1 + y 12 V 2 ………

I 2 = y 21 V 1 + y 22 V 2 ……….

Now, and can be rewritten as, I 1 = – \

And I 2 = C’V 1 – D’ \Missing or unrecognized delimiter for \right = \leftMissing or unrecognized delimiter for \right V_1 + \frac{D’}{B’} V_2\) …………

Comparing equations , and , , we get,

y 11 = \(\frac{A’}{B’}\)

y 12 = – \(\frac{1}{B’}\)

y 21 = \Missing or unrecognized delimiter for \right\)

y 22 = \(\frac{D’}{B’}\).

12. For a T shaped network, if the Short-circuit admittance parameters are y 11 , y 12 , y 21 , y 22 , then y 22 in terms of Inverse Transmission parameters can be expressed as ________

a) y 22 = \

y 22 = – \

y 22 = \Missing or unrecognized delimiter for \right\)

d) y 22 = \(\frac{D’}{B’}\)

Answer: d

Explanation: We know that, V 2 = A’V 1 – B’I 1 ………

I 2 = C’V 1 – D’I 1 ……………

And, I 1 = y 11 V 1 + y 12 V 2 ………

I 2 = y 21 V 1 + y 22 V 2 ……….

Now, and can be rewritten as, I 1 = – \

And I 2 = C’V 1 – D’ \Missing or unrecognized delimiter for \right = \leftMissing or unrecognized delimiter for \right V_1 + \frac{D’}{B’} V_2\) …………

Comparing equations , and , , we get,

y 11 = \(\frac{A’}{B’}\)

y 12 = – \(\frac{1}{B’}\)

y 21 = \Missing or unrecognized delimiter for \right\)

y 22 = \(\frac{D’}{B’}\).

13. For a T-network if the Open circuit Impedance parameters are z 11 , z 12 , z 21 , z 22 , then z 11 in terms of Transmission parameters can be expressed as ____________

a) z 11 = \

z 11 = \

z 11 = \Missing or unrecognized delimiter for \right\)

d) z 11 = \(\frac{A’}{C’}\)

Answer: a

Explanation: We know that, V 1 = z 11 I 1 + z 12 I 2 ………….

V 2 = z 21 I 1 + z 22 I 2 …………….

And V 2 = A’V 1 – B’I 1 ………

I 2 = C’V 1 – D’I 1 ……………

Rewriting and , we get,

V 2 = A’ \Missing or unrecognized delimiter for \right – B’I_1 = \leftMissing or unrecognized delimiter for \right I_1 + \frac{A’}{C’} I_2\) …………

And V 1 = \

Comparing , and , , we get,

z 11 = \(\frac{D’}{C’}\)

z 12 = \(\frac{1}{C’}\)

z 21 = \Missing or unrecognized delimiter for \right\)

z 22 = \(\frac{A’}{C’}\).

14. For a T-network if the Open circuit Impedance parameters are z 11 , z 12 , z 21 , z 22 , then z 12 in terms of Transmission parameters can be expressed as ____________

a) z 12 = \

z 12 = \

z 12 = \Missing or unrecognized delimiter for \right\)

d) z 12 = \(\frac{A’}{C’}\)

Answer: b

Explanation: We know that, V 1 = z 11 I 1 + z 12 I 2 ………….

V 2 = z 21 I 1 + z 22 I 2 …………….

And V 2 = A’V 1 – B’I 1 ………

I 2 = C’V 1 – D’I 1 ……………

Rewriting and , we get,

V 2 = A’ \Missing or unrecognized delimiter for \right – B’I_1 = \leftMissing or unrecognized delimiter for \right I_1 + \frac{A’}{C’} I_2\) …………

And V 1 = \

Comparing , and , , we get,

z 11 = \(\frac{D’}{C’}\)

z 12 = \(\frac{1}{C’}\)

z 21 = \Missing or unrecognized delimiter for \right\)

z 22 = \(\frac{A’}{C’}\).

15. For a T-network if the Open circuit Impedance parameters are z 11 , z 12 , z 21 , z 22 , then z 22 in terms of Transmission parameters can be expressed as ____________

a) z 22 = \

z 22 = \

z 22 = \Missing or unrecognized delimiter for \right\)

d) z 22 = \(\frac{A’}{C’}\)

Answer: d

Explanation: We know that, V 1 = z 11 I 1 + z 12 I 2 ………….

V 2 = z 21 I 1 + z 22 I 2 …………….

And V 2 = A’V 1 – B’I 1 ………

I 2 = C’V 1 – D’I 1 ……………

Rewriting and , we get,

V 2 = A’ \Missing or unrecognized delimiter for \right – B’I_1 = \leftMissing or unrecognized delimiter for \right I_1 + \frac{A’}{C’} I_2\) …………

And V 1 = \

Comparing , and , , we get,

z 11 = \(\frac{D’}{C’}\)

z 12 = \(\frac{1}{C’}\)

z 21 = \Missing or unrecognized delimiter for \right\)

z 22 = \(\frac{A’}{C’}\).

This set of Network Theory written test Questions & Answers focuses on “Relation between Hybrid Parameters with Short Circuit Admittance and Open Circuit Impedance Parameters”.

1. A periodic voltage v = 1 + 4 sin ωt + 2 cos ωt is applied across a 1Ω resistance. The power dissipated is ____________

a) 1 W

b) 11 W

c) 21 W

d) 24.5 W

Answer: b

Explanation: Given that, v = 1 + 4 sin ωt + 2 cos ωt

So, Power is given by,

Power, P = \(\frac{1^2}{1} + \frac{\frac{4^2}{\sqrt{2}}}{1} + \frac{\frac{2^2}{\sqrt{2}}}{1}\)

= 11 W.

2. A constant k high pass p section has a characteristic impedance of 300 Ω at f = ∞. At f = f c , the characteristic impedance will be?

a) 0

b) ∞

c) 300 Ω

d) More than 300 Ω

Answer: b

Explanation: For constant k high pass p section is given by,

Z = \(\frac{R}{\sqrt{1 –

^2}}\)

At f = f d , denominator term is 0.

So, Z = infinite.

3. In the circuit given below, the current through R is 2 sin 8t. The value of R is ___________

network-theory-written-test-questions-answers-q3

c) –

Answer: d

Explanation: Here, Inductor is not given, hence ignoring the inductance. Let I 1 and I 2 are currents in the loop then,

I 1 = \(\frac{2 sin ⁡8t}{3} \)

= 0.66 sin 8t

Again, I 2 = \

sin 8t

So, R = .

4. For a T-network if the Short circuit admittance parameters are given as y 11 , y 21 , y 12 , y 22 , then y 21 in terms of Hybrid parameters can be expressed as ________

a) y 21 = \Missing or unrecognized delimiter for \right\)

b) y 21 = \

y 21 = –\

y 21 = \(\frac{1}{h_{11}} \)

Answer: b

Explanation: We know that, I 1 = y 11 V 1 + y 12 V 2 ………

I 2 = y 21 V 1 + y 22 V 2 ……….

And, V 1 = h 11 I 1 + h 12 V 2 ……….

I 2 = h 21 I 1 + h 22 V 2 ………..

Now, and can be rewritten as,

I 1 = \

And I 2 = \Missing or unrecognized delimiter for \right V_2\) ……….

∴ Comparing , and , , we get,

y 11 = \(\frac{1}{h_{11}} \)

y 12 = –\(\frac{h_{12}}{h_{11}} \)

y 21 = \(\frac{h_{21}}{h_{11}} \)

y 22 = \Missing or unrecognized delimiter for \right\).

5. For a T-network if the Open circuit impedance parameters are given as z 11 , z 21 , z 12 , z 22 , then z 21 in terms of Hybrid parameters can be expressed as ________

a) z 21 = \Missing or unrecognized delimiter for \right \)

b) z 21 = – \

z 21 = \

z 21 = \(\frac{1}{h_{22}} \)

Answer: b

Explanation: We know that, V 1 = z 11 I 1 + z 12 I 2 ………

V 2 = z 21 I 1 + z 22 I 2 ……….

And, V 1 = h 11 I 1 + h 12 V 2 ……….

I 2 = h 21 I 1 + h 22 V 2 ………..

Now, and can be rewritten as,

V 1 = \Missing or unrecognized delimiter for \right \frac{I_1 + h_{12}}{h_{22}}I_2\) …………

And V 2 = \

∴ Comparing , and , , we get,

z 11 = \Missing or unrecognized delimiter for \right \)

z 12 = \(\frac{h_{12}}{h_{22}} \)

z 21 = – \(\frac{h_{21}}{h_{22}} \)

z 22 = \(\frac{1}{h_{22}} \).

6. For a T-network if the Open circuit impedance parameters are given as z 11 , z 21 , z 12 , z 22 , then z 11 in terms of Hybrid parameters can be expressed as ________

a) z 11 = \Missing or unrecognized delimiter for \right \)

b) z 11 = – \

z 11 = \

z 11 = \(\frac{1}{h_{22}} \)

Answer: a

Explanation: We know that, V 1 = z 11 I 1 + z 12 I 2 ………

V 2 = z 21 I 1 + z 22 I 2 ……….

And, V 1 = h 11 I 1 + h 12 V 2 ……….

I 2 = h 21 I 1 + h 22 V 2 ………..

Now, and can be rewritten as,

V 1 = \Missing or unrecognized delimiter for \right \frac{I_1 + h_{12}}{h_{22}}I_2\) …………

And V 2 = \

∴ Comparing , and , , we get,

z 11 = \Missing or unrecognized delimiter for \right \)

z 12 = \(\frac{h_{12}}{h_{22}} \)

z 21 = – \(\frac{h_{21}}{h_{22}} \)

z 22 = \(\frac{1}{h_{22}} \).

7. For a T-network if the Open circuit impedance parameters are given as z 11 , z 21 , z 12 , z 22 , then z 12 in terms of Hybrid parameters can be expressed as ________

a) z 12 = \Missing or unrecognized delimiter for \right \)

b) z 12 = – \

z 12 = \

z 12 = \(\frac{1}{h_{22}} \)

Answer: c

Explanation: We know that, V 1 = z 11 I 1 + z 12 I 2 ………

V 2 = z 21 I 1 + z 22 I 2 ……….

And, V 1 = h 11 I 1 + h 12 V 2 ……….

I 2 = h 21 I 1 + h 22 V 2 ………..

Now, and can be rewritten as,

V 1 = \Missing or unrecognized delimiter for \right \frac{I_1 + h_{12}}{h_{22}}I_2\) …………

And V 2 = \

∴ Comparing , and , , we get,

z 11 = \Missing or unrecognized delimiter for \right \)

z 12 = \(\frac{h_{12}}{h_{22}} \)

z 21 = – \(\frac{h_{21}}{h_{22}} \)

z 22 = \(\frac{1}{h_{22}} \).

8. For a T-network if the Open circuit impedance parameters are given as z 11 , z 21 , z 12 , z 22 , then z 22 in terms of Hybrid parameters can be expressed as ________

a) z 22 = \Missing or unrecognized delimiter for \right \)

b) z 22 = – \

z 22 = \

z 22 = \(\frac{1}{h_{22}} \)

Answer: d

Explanation: We know that, V 1 = z 11 I 1 + z 12 I 2 ………

V 2 = z 21 I 1 + z 22 I 2 ……….

And, V 1 = h 11 I 1 + h 12 V 2 ……….

I 2 = h 21 I 1 + h 22 V 2 ………..

Now, and can be rewritten as,

V 1 = \Missing or unrecognized delimiter for \right \frac{I_1 + h_{12}}{h_{22}}I_2\) …………

And V 2 = \

∴ Comparing , and , , we get,

z 11 = \Missing or unrecognized delimiter for \right \)

z 12 = \(\frac{h_{12}}{h_{22}} \)

z 21 = – \(\frac{h_{21}}{h_{22}} \)

z 22 = \(\frac{1}{h_{22}} \).

9. Permeability is analogous to _____________

a) Conductivity

b) Resistivity

c) Retentivity

d) Coercivity

Answer: a

Explanation: We know that resistance and reluctance are given by,

R = \(\frac{ρL}{A}\)

And Reluctance = \(\frac{L}{μA}\)

So, Permeability is analogous to conductivity.

10. A resistance and an inductance are connected in parallel and fed from 50 Hz ac mains. Each branch takes a current of 5 A. The current supplied by source is ____________

a) 10 A

b) 7.07 A

c) 5 A

d) 0 A

Answer: b

Explanation: The current is given by,

|5 – j5| = \(\sqrt{5^2 + 5^2}\)

= \(\sqrt{50} = 5\sqrt{2}\) = 7.07 A.

11. A triangular Pulse of 50 V peak is applied to a capacitor of 0.1 F. The change of the capacitor and its waveform shape is ___________

a) 10 rectangular

b) 5 rectangular

c) 5 triangular

d) 10 triangular

Answer: c

Explanation: We know that,

Q = CV

Or, 0.1 X 50 = 5

And it is a triangular pulse.

12. A 10 μF capacitor is charged from a 5 volt source through a resistance of 10 kΩ. The charging current offer 35 m sec. If the initial voltage on C is – 3 V is ___________

a) 0.56 mA

b) 5.6 mA

c) 6 mA

d) 5 μA

Answer: a

Explanation: Initial current immediately after charging is given by,

\(\frac{V}{R} = \frac{5+3}{10000}\)

= 0.8 mA

Now, i = i 0 e -t/RC

= 0.8 mA x \(e^{\frac{t}{10k X 10 X 10^{-6}}}\)

= 0.8 X 10 -3 X \(e^{\frac{35 X 10^{-3}}{10^{-1}}}\)

= 0.56 mA.

13. For a T-network if the Open circuit impedance parameters are given as z 11 , z 21 , z 12 , z 22 , then z 11 in terms of Inverse Hybrid parameters can be expressed as ________

a) z 12 = \

z 12 = – \

z 12 = \Missing or unrecognized delimiter for \right\)

Answer: a

Explanation: We know that, V 1 = z 11 I 1 + z 12 I 2 ………

V 2 = z 21 I 1 + z 22 I 2 ……….

And, I 1 = g 11 V 1 + g 12 I 2 ……….

V 2 = g 21 V 1 + g 22 I 2 ………..

Now, and can be rewritten as,

V 1 = \

And V 2 = \Missing or unrecognized delimiter for \right I_2 – \frac{g_{21} I_1}{g_{11}}\) ………..

∴ Comparing , and , , we get,

z 11 = \(\frac{1}{g_{11}} \)

z 12 = – \(\frac{g_{12}}{g_{11}} \)

z 21 = – \(\frac{g_{21}}{g_{11}} \)

z 22 = \Missing or unrecognized delimiter for \right\).

14. For a T-network if the Open circuit impedance parameters are given as z 11 , z 21 , z 12 , z 22 , then z 12 in terms of Inverse Hybrid parameters can be expressed as ________

a) z 12 = \

z 12 = – \

z 12 = \Missing or unrecognized delimiter for \right\)

Answer: b

Explanation: We know that, V 1 = z 11 I 1 + z 12 I 2 ………

V 2 = z 21 I 1 + z 22 I 2 ……….

And, I 1 = g 11 V 1 + g 12 I 2 ……….

V 2 = g 21 V 1 + g 22 I 2 ………..

Now, and can be rewritten as,

V 1 = \

And V 2 = \Missing or unrecognized delimiter for \right I_2 – \frac{g_{21} I_1}{g_{11}}\) ………..

∴ Comparing , and , , we get,

z 11 = \(\frac{1}{g_{11}} \)

z 12 = – \(\frac{g_{12}}{g_{11}} \)

z 21 = – \(\frac{g_{21}}{g_{11}} \)

z 22 = \Missing or unrecognized delimiter for \right\).

15. For a T-network if the Open circuit impedance parameters are given as z 11 , z 21 , z 12 , z 22 , then z 22 in terms of Inverse Hybrid parameters can be expressed as ________

a) z 22 = \

z 22 = – \

z 22 = \Missing or unrecognized delimiter for \right\)

Answer: d

Explanation: We know that, V 1 = z 11 I 1 + z 12 I 2 ………

V 2 = z 21 I 1 + z 22 I 2 ……….

And, I 1 = g 11 V 1 + g 12 I 2 ……….

V 2 = g 21 V 1 + g 22 I 2 ………..

Now, and can be rewritten as,

V 1 = \

And V 2 = \Missing or unrecognized delimiter for \right I_2 – \frac{g_{21} I_1}{g_{11}}\) ………..

∴ Comparing , and , , we get,

z 11 = \(\frac{1}{g_{11}} \)

z 12 = – \(\frac{g_{12}}{g_{11}} \)

z 21 = – \(\frac{g_{21}}{g_{11}} \)

z 22 = \Missing or unrecognized delimiter for \right\).

This set of Network Theory Multiple Choice Questions & Answers focuses on “Series-Parallel Interconnection of Two Port Network”.

1. If the diameter of a wire is doubled, the current carrying capacity of the wire is ___________

a) Half

b) Twice

c) Four times

d) One-fourth

Answer: c

Explanation: Since diameter is doubled, area of cross-section becomes four times. Current carrying capacity is proportional to area of cross-section.

2. Consider an RL series circuit having resistance R = 3 Ω, inductance L = 3 H and is excited by 6V. The current after a long time after closing of switch is ____________

a) 1 A

b) 2 A

c) 0 A

d) Infinity

Answer: b

Explanation: At t = ∞ the circuit has effectively two 6Ω resistances in parallel.

So, R EQ = \(\frac{6 X 6}{6 + 6}\)

= \(\frac{36}{12}\) = 3 Ω

Given voltage = 6 V

So, current = 2 A.

3. The energy stored in a coil is 108 J. The power dissipated instantaneously across the blades of switch after it is opened in 10 ms is ____________

a) 108 W

b) 1080 W

c) 10800 W

d) 108000 W

Answer: c

Explanation: Power dissipated instantaneously across the blades of the switch is given by,

Power, P = \(\frac{Energy}{Time}\)

Given that, Energy = 108 J and time = 10 X 10 -3

So, P = \(\frac{108}{10 X 10^{-3}}\) = 10800 W.

4. A parallel RLC circuit with R 1 = 20, L 1 = \(\frac{1}{100}\) and C 1 = \(\frac{1}{200}\) is scaled giving R 2 = 10 4 , L 2 = 10 -4 and C 2 , the value of C 2 is ___________

a) 0.10 nF

b) 0.3 nF

c) 0.2 nF

d) 0.4 nF

Answer: c

Explanation: K 1 = \(\frac{R_2}{R_1}\)

= \(\frac{10^4}{20}\) = 5 X 10 2 = 500 Ω

And \(\frac{L_1}{L_2} = \frac{kω}{k_1}\)

Or, \(\frac{kω}{k_1} = \frac{L_1}{L_2} X k_1\)

Or, \(\frac{10^{-2}}{10^{-4} X 5 X 10^2}\) = 5 X 10 4

Or, C 2 = \(\frac{C_1}{kω.k_1} = \frac{0.5 X 10^{-2}}{5 X 10^4 X 500}\)

= 0.02 X 10 -8 = 0.2 nF.

5. Barletts Bisection Theorem is applicable to ___________

a) Unsymmetrical networks

b) Symmetrical networks

c) Both unsymmetrical and symmetrical networks

d) Neither to unsymmetrical nor to symmetrical networks

Answer: b

Explanation: A symmetrical network can be split into two halves. So the z parameters of the network are symmetrical as well as reciprocal of each other. Hence Barletts Bisection Theorem is applicable to Symmetrical networks.

6. The Thevenin’s equivalent of a network is a 10 V source in series with 2 Ω resistances. If a 3 Ω resistance is connected across the Thevenin’s equivalent is _____________

a) 10 V in series with 1.2 Ω resistance

b) 6 V in series with 1.2 Ω resistance

c) 10 V in series with 5 Ω resistance

d) 6 V in series with 5 Ω resistance

Answer: b

Explanation: The Thevenin equivalent voltage is given by,

V TH = \(\frac{10 X 3}{5}\)

= 6 V

And the Thevenin equivalent Resistance is given by,

R TH = \(\frac{3 X 2}{5}\) = 1.2 Ω.

7. A magnetic circuit has an iron length of 100 cm and air gap length 10 cm. If μ r = 200 then which of the following is true?

a) Mmf for iron and air gap are equal

b) Mmf for iron is much less than that for air gap

c) Mmf for iron is much more than that for air gap

d) Mmf for iron and air gap are not equal

Answer: a

Explanation: We know that, MMF for air = \(\frac{B}{4π X 10^{-7}}\) X 10

Where B is the magnetic field intensity.

Also, MMF for iron = \(\frac{B X 100}{200

}\)

= \(\frac{B X 0.5}{4π X 10^{-7}}\).

8. Two coils X and Y have self-inductances of 5 mH and 10 mH and mutual inductance of 3 mH. If the current in coils X change at a steady rate of 100 A/s, the emf induced in coil Y is ____________

a) 0.3 V

b) 0.5 V

c) 1 V

d) 1.5 V

Answer: a

Explanation: The emf is given by,

V = M\(\frac{di}{dt}\)

= \(\frac{3}{1000}\) X 100 = 0.3 V

Hence, the emf induced in coil Y is given by 0.3 V.

9. A 50 Hz current has an amplitude of 25 A. The rate of change of current at t = 0.005 after i = 0 and is increasing is ____________

a) 2221.44 A/s

b) 0

c) -2221.44 A/s

d) -3141.6 A/s

Answer: b

Explanation: The current i is given by,

i = 25 sin 314.16 t and \

= 0.

10. Consider a series RL circuit in which current 12 A is flowing through R and current 16 A is flowing through L. The current supplied by the sinusoidal current source I is ____________

a) 28 A

b) 4 A

c) 20 A

d) Cannot be determined

Answer: c

Explanation: Current I is given by,

I = \(\sqrt{16^2 + 12^2}\)

= \(\sqrt{256+144}\)

= \(\sqrt{400}\)

= 20 A.

11. Consider a circuit having resistances 16 Ω and 30 Ωis excited by a voltage V. A variable resistance R is connected across the 16 Ω resistance. The power dissipated in 30 Ω resistance will be maximum when value of R is __________

a) 30 Ω

b) 16 Ω

c) 9 Ω

d) 0

Answer: c

Explanation: We know that,

When R = 0, circuit current = \(\frac{V}{30}\) A

And Power dissipated = \(\frac{V^2}{30}\) Watts.

This is the maximum possible value which occurs for R = 0 Ω.

12. Consider a cube having resistance R on each of its sides. For this non-planar graph, the number of independent loop equations are _______________

a) 8

b) 12

c) 7

d) 5

Answer: d

Explanation: We know that the number of equations is given by,

L = B – N + 1

Where, B = Number of Branches, N = Number of Nodes

Here, B = 12 and N = 8.

So, L = 12 – 8 + 1 = 5.

13. Given two voltages, 50 ∠0 V and 75 ∠- 60° V. The sum of these voltages is ___________

a) 109 ∠- 60° V

b) 109 ∠- 25° V

c) 109 ∠- 36.6° V

d) 100 ∠- 50.1° V

Answer: c

Explanation: The voltages can be written in the form,

50 + j 0.75∠-60°

= 37.5 – j 64.95

So, sum = – j 64.95

= 87.5 – j 64.95 = 109∠-36.6°.

14. For the circuit given below, the value of z 21 parameter is ____________

network-theory-questions-answers-series-parallel-interconnection-two-port-network-q14

a) z 21 = 0.0667 Ω

b) z 21 = 2.773 Ω

c) z 21 = 1.667 Ω

d) z 21 = 0.999 Ω

Answer: a

Explanation: z 11 = \

]

z 11 = 2 + 1 ||

= 2 + \(\frac{1×\frac{11}{4}}{1+\frac{11}{4}} = 2 + \frac{11}{15}\) = 2.733

I 0 = \(\frac{1}{1+3}\) I’ 0 = \(\frac{1}{4}\) I’ 0

And I’ 0 = 1 + \(\frac{11}{4}\) I 1 = \(\frac{4}{15}\) I 1

Or, I 0 = \(\frac{1}{4} × \frac{4}{5} I_1 = \frac{1}{15} I_1\)

Or, V 2 = I 0 = \(\frac{1}{15} I_1\)

z 21 = \(\frac{V_2}{I_1} = \frac{1}{15}\) = z 12 = 0.0667

z 22 = \

= z 11 = 2.733

∴ [z] = [2.733:0.0667; 0.0667:2.733] Ω.

15. For the circuit given below, the value of z 11 parameter is ____________

network-theory-questions-answers-series-parallel-interconnection-two-port-network-q15

a) z 11 = 1.775 + j5.739 Ω

b) z 11 = 1.775 – j4.26 Ω

c) z 11 = -1.775 – j4.26 Ω

d) z 11 = 1.775 + j4.26 Ω

Answer: d

Explanation: z 1 = \(\frac{12}{12+j10-j5} = \frac{j120}{12+j5}\)

z 2 = \(\frac{j60}{12+j5}\)

z 3 = \(\frac{50}{12+j5}\)

z 12 = z 21 = z 2 = \(\frac{}{144+25}\) = -1.775 – j4.26

z 11 = z 1 + z 12 = \(\frac{}{144+25}\) + z 12 = 1.775 + j4.26

z 22 = z 3 + z 21 = \(\frac{}{144+25}\) + z 21 = 1.7758 – j5.739

∴ [z] = [1.775 + j4.26; -1.775 – j4.26; -1.775 – j4.26; 1.775 – j5.739] Ω.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Advanced Problems on Two Port Network – 1”.

1. For the circuit given below, the value of the z 12 parameter is ___________

network-theory-questions-answers-advanced-problems-two-port-network-1-q1

a) z 12 = 1 Ω

b) z 12 = 4 Ω

c) z 12 = 1.667 Ω

d) z 12 = 2.33 Ω

Answer: a

Explanation: z 11 = \

= 4Ω

I 0 = \(\frac{1}{2} I_1 \)

V 2 = 2I 0 = I 1

z 21 = \(\frac{V_2}{I_1}\) = 1Ω

z 22 = \

= 1.667Ω

So, I’ 0 = \(\frac{2}{2+10}I_2 = \frac{1}{6}I_2 \)

V 1 = 6I’ 0 = I 2

z 12 = \(\frac{V_1}{I_2}\) = 1Ω

Hence, [z] = [4:1; 1:1.667] Ω.

2. For the network of figure, z 11 is equal to ___________

network-theory-questions-answers-advanced-problems-two-port-network-1-q2

a) \

2 Ω

d) \(\frac{2}{3}\) Ω

Answer: a

Explanation: From the figure, we can infer that,

Z 11 = 1 + \(\frac{1 X 2}{3}\)

= 1 + \(\frac{2}{3}\)

= \(\frac{5}{3}\) Ω.

3. For the circuit given below, the value of z 11 parameter is ____________

network-theory-questions-answers-advanced-problems-two-port-network-1-q3

a) z 11 = 4 + j6 Ω

b) z 11 = j6 Ω

c) z 11 = -j6 Ω

d) z 11 = -j6 + 4 Ω

Answer: a

Explanation: z 12 = j6 = z 21

z 11 – z 12 = 4

Or, z 11 = z 12 + 4 = 4 + j6 Ω

And z 22 – z 12 = -j10

Or, z 22 = z 12 + -j10 = -j4 Ω

∴ [z] = [4+j6:j6; j6:-j4] Ω.

4. In a series RLC circuit excited by a voltage 3e -t u , the resistance is equal to 1 Ω and capacitance = 2 F. For the circuit, the values of I (0 + ) and I , are ____________

a) 0 and 1.5 A

b) 1.5 A and 3 A

c) 3 A and 0

d) 3 A and 1.5 A

Answer: c

Explanation: I = \

= 6 e -t – 3 e -0.5t

Putting, t = 0, we get, I = 3A

Putting t = ∞, we get, I = 0.

5. For the circuit given below, the value of z 12 parameter is ____________

network-theory-questions-answers-advanced-problems-two-port-network-1-q5

a) Z 12 = 20 Ω

b) Z 12 = 25 Ω

c) Z 12 = 30 Ω

d) z 12 = 24 Ω

Answer: a

Explanation: z 11 = \(\frac{V_1}{I_1} = \frac{I_1}{I_1}\) = 25Ω

V 0 = \(\frac{20}{25}\)V 1 = 20 I 1

-V 0 – 4I 2 + V 2 = 0

Or, V 2 = V 0 + 4I 1 = 20I 1 + 4I 1 = 24 I 1

Or, z 21 = \(\frac{V_2}{I_1}\) = 24 Ω

V 2 = I 2 = 30 I 2

Or, z 22 = \(\frac{V_2}{I_1}\) = 30 Ω

V 1 = 20I 2

Or, z 12 = \(\frac{V_1}{I_2}\) = 20 Ω

∴ [z] = [25:20; 24:30] Ω.

6. For the circuit given below, the value of the z 22 parameter is ___________

network-theory-questions-answers-advanced-problems-two-port-network-1-q1

a) z 22 = 1 Ω

b) z 22 = 4 Ω

c) z 22 = 1.667 Ω

d) z 22 = 2.33 Ω

Answer: c

Explanation: z 11 = \

= 4Ω

I 0 = \(\frac{1}{2} I_1 \)

V 2 = 2I 0 = I 1

z 21 = \(\frac{V_2}{I_1}\) = 1Ω

z 22 = \

= 1.667Ω

So, I’ 0 = \(\frac{2}{2+10}I_2 = \frac{1}{6}I_2 \)

V 1 = 6I’ 0 = I 2

z 12 = \(\frac{V_1}{I_2}\) = 1Ω

Hence, [z] = [4:1; 1:1.667] Ω.

7. For the circuit given below, the value of z 22 parameter is ____________

network-theory-questions-answers-advanced-problems-two-port-network-1-q7

a) z 22 = 0.0667 Ω

b) z 22 = 2.773 Ω

c) z 22 = 1.667 Ω

d) z 22 = 0.999 Ω

Answer: b

Explanation: z 11 = \

]

z 11 = 2 + 1 ||

= 2 + \(\frac{1×\frac{11}{4}}{1+\frac{11}{4}} = 2 + \frac{11}{15}\) = 2.733

I 0 = \(\frac{1}{1+3}\) I’ 0 = \(\frac{1}{4}\) I’ 0

And I’ 0 = 1 + \(\frac{11}{4}\) I 1 = \(\frac{4}{15}\) I 1

Or, I 0 = \(\frac{1}{4} × \frac{4}{5} I_1 = \frac{1}{15} I_1\)

Or, V 2 = I 0 = \(\frac{1}{15} I_1\)

z 21 = \(\frac{V_2}{I_1} = \frac{1}{15}\) = z 12 = 0.0667

z 22 = \

= z 11 = 2.733

∴ [z] = [2.733:0.0667; 0.0667:2.733] Ω.

8. For the circuit given below, the value of z 22 parameter is ____________

network-theory-questions-answers-advanced-problems-two-port-network-1-q3

a) z 22 = 4 + j6 Ω

b) z 22 = j6 Ω

c) z 22 = -j4 Ω

d) z 22 = -j6 + 4 Ω

Answer: c

Explanation: z 12 = j6 = z 21

z 11 – z 12 = 4

Or, z 11 = z 12 + 4 = 4 + j6 Ω

And z 22 – z 12 = -j10

Or, z 22 = z 12 + -j10 = -j4 Ω

∴ [z] = [4+j6:j6; j6:-j4] Ω.

9. For the circuit given below, the value of z 22 parameter is ____________

network-theory-questions-answers-advanced-problems-two-port-network-1-q9

a) z 22 = 1.775 + j5.739 Ω

b) z 22 = 1.775 – j4.26 Ω

c) z 22 = -1.775 – j5.739 Ω

d) z 22 = 1.775 + j4.26 Ω

Answer: c

Explanation: z 1 = \(\frac{12}{12+j10-j5} = \frac{j120}{12+j5}\)

z 2 = \(\frac{j60}{12+j5}\)

z 3 = \(\frac{50}{12+j5}\)

z 12 = z 21 = z 2 = \(\frac{}{144+25}\) = -1.775 – j4.26

z 11 = z 1 + z 12 = \(\frac{}{144+25}\) + z 12 = 1.775 + j4.26

z 22 = z 3 + z 21 = \(\frac{}{144+25}\) + z 21 = 1.7758 – j5.739

∴ [z] = [1.775 + j4.26; -1.775 – j4.26; -1.775 – j4.26; 1.775 – j5.739] Ω.

10. For the circuit given below, the value of z 22 parameter is ____________

network-theory-questions-answers-advanced-problems-two-port-network-1-q5

a) z 22 = 20 Ω

b) z 22 = 25 Ω

c) z 22 = 30 Ω

d) z 22 = 24 Ω

Answer: c

Explanation: z 11 = \(\frac{V_1}{I_1} = \frac{I_1}{I_1}\) = 25Ω

V 0 = \(\frac{20}{25}\)V 1 = 20 I 1

-V 0 – 4I 2 + V 2 = 0

Or, V 2 = V 0 + 4I 1 = 20I 1 + 4I 1 = 24 I 1

Or, z 21 = \(\frac{V_2}{I_1}\) = 24 Ω

V 2 = I 2 = 30 I 2

Or, z 22 = \(\frac{V_2}{I_1}\) = 30 Ω

V 1 = 20I 2

Or, z 12 = \(\frac{V_1}{I_2}\) = 20 Ω

∴ [z] = [25:20; 24:30] Ω.

11. A capacitor of 220 V, 50 Hz is needed for AC supply. The peak voltage rating of the capacitor is ____________

a) 220 V

b) 460 V

c) 440 V

d) 230 V

Answer: c

Explanation: We know that,

Peak voltage rating = 2

Given that the RMS voltage rating = 220 V

So, the Peak Voltage Rating = 2 X 220 V

= 440 V.

12. In the circuit given below, the value of the hybrid parameter h 21 is _________

network-theory-questions-answers-advanced-problems-two-port-network-1-q12

a) 10 Ω

b) 0.5 Ω

c) 5 Ω

d) 2.5 Ω

Answer: b

Explanation: Hybrid parameter h 21 is given by, h 21 = \(\frac{I_2}{I_1}\), when V 2 = 0.

Therefore short circuiting the terminal Y-Y’, and applying Kirchhoff’s law, we get,

-5 I 2 – (I 2 – I 1 )5 = 0

Or, -I 2 = I 2 – I 1

Or, -2I 2 = -I 1

∴ \(\frac{I_2}{I_1} = \frac{1}{2}\)

Hence h 21 = 0.5 Ω.

13. If a two port network is passive, then we have, with the usual notation, the relationship as _________

a) h 21 = h 12

b) h 12 = -h 21

c) h 11 = h 22

d) h 11 h 22 – h 12 h 22 = 1

Answer: d

Explanation: We know that, I 1 = y 11 V 1 + y 12 V 2 ………

I 2 = y 21 V 1 + y 22 V 2 ……….

And, V 1 = h 11 I 1 + h 12 V 2 ……….

I 2 = h 21 I 1 + h 22 V 2 ………..

Now, and can be rewritten as,

I 1 = \

And I 2 = \Missing or unrecognized delimiter for \right V_2\) ……….

Therefore using the above 6 equations in representing the hybrid parameters in terms of the Y parameters and applying ∆Y = 0, we get,

h 11 h 22 – h 12 h 22 = 1 [hence proved].

14. In two-port networks the parameter h 22 is called _________

a) Short circuit input impedance

b) Short circuit current gain

c) Open circuit reverse voltage gain

d) Open circuit output admittance

Answer: d

Explanation: We know that, h 22 = \(\frac{I_2}{V_2}\), when I 1 = 0.

Since the current in the first loop is 0 when the ratio of the current and voltage in second loop is measured, therefore the parameter h 12 is called as Open circuit output admittance.

15. The short-circuit admittance matrix of a two port network is as follows.

[0; -0.5; 0.5; 0] Then the 2 port network is ____________

a) Non-reciprocal and passive

b) Non-reciprocal and active

c) Reciprocal and passive

d) Reciprocal and active

Answer: b

Explanation: So, network is non reciprocal because Y 12 ≠ Y 21 and Y 12 are also negative which means either energy storing or providing device is available. So the network is active.

Therefore the network is Non- reciprocal and active.

This set of Network Theory Question Paper focuses on “Advanced Problems on Two Port Network – 2”.

1. A network contains linear resistors and ideal voltage source S. If all the resistors are made twice their initial value, then voltage across each resistor is __________

a) Halved

b) Doubled

c) Increases by 2 times

d) Remains same

Answer: d

Explanation: The voltage/resistance ratio is a constant . If K is doubled then, electric current will become half. So voltage across each resistor remains same as was initially.

a) 12 t

b) 24 t

c) 12 t 3

d) 4 t 3

Answer: d

Explanation: We know that, I = \( \frac{1}{L} \int_0^t V \,dt\)

= \(1\int_0^t 12 t^2 \,dt\)

= 4 t 3 .

3. The linear circuit element among the following is ___________

a) Capacitor

b) Inductor

c) Resistor

d) Inductor & Capacitor

Answer: c

Explanation: A linear circuit element does not change their value with voltage or current. The resistance is only one among the others does not change its value with voltage or current.

4. Consider a circuit having resistance 10 kΩ, excited by voltage 5 V and an ideal switch S. If the switch is repeatedly closed for 2 ms and opened for 2 ms, the average value of i is ____________

a) 0.25 mA

b) 0.35 mA

c) 0.125 mA

d) 1 mA

Answer: c

Explanation: Since i = \(\frac{5}{10 × 2 X 10^{-3}}\) = 0.25 × 10 -3 = 0.25 mA.

As the switch is repeatedly close, then i will be a square wave.

So average value of electric current is

= 0.125 mA.

5. In the circuit given below the value of resistance R eq is _____________

network-theory-question-papers-q5

a) 10 Ω

b) 11.86 Ω

c) 11.18 Ω

d) 25 Ω

Answer: c

Explanation: The circuit is as shown in figure below.

R eq = 5 + \(\frac{10

}{10 + 5 + R_{eq}}\)

Or, \(R_{eq}^2\) + 15R eq = 5R eq + 75 + 10R eq + 50

Or, \(R_{eq} = \sqrt{125}\) = 11.18 Ω.

6. A particular electric current is made up of two components a 10 A, a sine wave of peak value 14.14 A. The average value of electric current is __________

a) 0

b) 24.14 A

c) 10 A

d) 14.14 A

Answer: c

Explanation: Average dc electric current = 10 A.

Average ac electric current = 0 A since it is alternating in nature.

Average electric current = 10 + 0 = 10 A.

7. Given that, R 1 = 36 Ω and R 2 = 75 Ω, each having tolerance of ±5% are connected in series. The value of resultant resistance is ___________

a) 111 ± 0 Ω

b) 111 ± 2.77 Ω

c) 111 ± 5.55 Ω

d) 111 ± 7.23 Ω

Answer: c

Explanation: R 1 = 36 ± 5% = 36 ± 1.8 Ω

R 2 = 75 ± 5% = 75 ± 3.75 Ω

∴ R 1 + R 2 = 111 ± 5.55 Ω.

8. Consider a circuit having a charge of 600 C, which is delivered to 100 V source in a 1 minute. The value of Voltage source V is ___________

a) 30 V

b) 60 V

c) 120 V

d) 240 V

Answer: d

Explanation: In order for 600 C charges to be delivered to 100 V source, the electric current must be in reverse clockwise direction.

Now, I = \(\frac{dQ}{dt}\)

= \(\frac{600}{60}\) = 10 A

Applying KVL we get

V 1 + 60 – 100 = 10 × 20 ⇒ V 1 = 240 V.

9. The energy required to charge a 10 μF capacitor to 100 V is ____________

a) 0.01 J

b) 0.05 J

c) 5 X 10 -9 J

d) 10 X 10 -9 J

Answer: b

Explanation: E = \(\frac{1}{2}\) CV 2

= 5 X 10 -6 X 100 2

= 0.05 J.

10. For the circuit given below, the value of the hybrid parameter h 11 is ___________

network-theory-question-papers-q10

a) 75 Ω

b) 80 Ω

c) 90 Ω

d) 105 Ω

Answer: a

Explanation: Hybrid parameter h 11 is given by, h 11 = \(\frac{V_1}{I_1}\), when V 2 =0.

Therefore short circuiting the terminal Y-Y’, we get,

V 1 = I 1 + 50)

= I 1 \Missing or unrecognized delimiter for \right + 50\right)\)

= 75I 1

∴ \(\frac{V_1}{I_1}\) = 75.

Hence h 11 = 75 Ω.

11. For the circuit given below, the value of the hybrid parameter h 21 is ___________

network-theory-question-papers-q10

a) 0.6 Ω

b) 0.5 Ω

c) 0.3 Ω

d) 0.2 Ω

Answer: b

Explanation: Hybrid parameter h 21 is given by, h 21 = \(\frac{I_2}{I_1}\), when V 2 = 0.

Therefore short circuiting the terminal Y-Y’, and applying Kirchhoff’s law, we get,

-50 I 2 – (I 2 – I 1 )50 = 0

Or, -I 2 = I 2 – I 1

Or, -2I 2 = -I 1

∴ \(\frac{I_2}{I_1} = \frac{1}{2}\)

Hence h 21 = 0.5 Ω.

12. For the circuit given below, the value of the Inverse hybrid parameter g 11 is ___________

network-theory-question-papers-q12

a) 0.133 Ω

b) 0.025 Ω

c) 0.3 Ω

d) 0.25 Ω

Answer: a

Explanation: Inverse Hybrid parameter g 11 is given by, g 11 = \(\frac{I_1}{V_1}\), when I 2 = 0.

Therefore short circuiting the terminal Y-Y’, we get,

V 1 = I 1 + 5)

= I 1 \Missing or unrecognized delimiter for \right + 5\right)\)

= 7.5I 1

∴ \(\frac{I_1}{V_1} = \frac{1}{7.5}\) = 0.133 Ω

Hence g 11 = 0.133 Ω.

13. A resistor of 10 kΩ with the tolerance of 5% is connected in series with 5 kΩ resistors of 10% tolerance. What is the tolerance limit for a series network?

a) 9%

b) 12.04%

c) 8.67%

d) 6.67%

Answer: d

Explanation: Error in 10 kΩ resistance = 10 × \(\frac{5}{100}\) = 0.5 kΩ

Error in 5 kΩ resistance = 5 × \(\frac{10}{100}\) = 5 kΩ

Total measurement resistance = 10 + 0.5 + 5 + 0.5 = 16 kΩ

Original resistance = 10 + 5 = 15 kΩ

Error = \(\frac{16-15}{15}\) × 100 = \(\frac{1}{15}\) × 100 = 6.67%.

14. A 200 μA ammeter has an internal resistance of 200 Ω. The range is to be extended to 500μA. The shunt required is of resistance __________

a) 20.0 Ω

b) 22.22 Ω

c) 25.0 Ω

d) 50.0 Ω

Answer: c

Explanation: I sh R sh = I m R m

I sh = I – I m or, \(\frac{I}{I_m} – 1 = \frac{R_m}{R_{sh}}\)

Now, m = \(\frac{I}{I_m}\)

Or, m – 1 = \(\frac{R_m}{R_{sh}}\)

∴R sh = 25 Ω.

15. A voltmeter has a sensitivity of 1000 Ω/V reads 200 V on its 300 V scale. When connected across an unknown resistor in series with a millimeter, it reads 10 mA. The error due to the loading effect of the voltmeter is

a) 3.33%

b) 6.67%

c) 13.34%

d) 13.67%

Answer: b

Explanation: R T = \(\frac{V_T}{I_T}\)

V T = 200 V, I T = 10 A

So, R T = 20 kΩ

Resistance of voltmeter,

R V = 1000 × 300 = 300 kΩ

Voltmeter is in parallel with unknown resistor,

R X = \(\frac{R_T R_V}{R_T – R_V} = \frac{20 ×300}{280}\) = 21.43 kΩ

Percentage error = \(\frac{Actual-Apparent}{Actual}\) × 100

= \(\frac{21.43-20}{21.43}\) × 100 = 6.67%.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Filter Networks”.

1. The value of one decibel is equal to?

a) 0.115 N

b) 0.125 N

c) 0.135 N

d) 0.145 N

Answer: a

Explanation: The value of one decibel is equal to 0.115 N. One decibel = 0.115 N where N is the number of nepers and N = log e (V 1 /V 2 ).

2. A filter which passes without attenuation all frequencies up to the cut-off frequency f c and attenuates all other frequencies greater than f c is called?

a) high pass filter

b) low pass filter

c) band elimination filter

d) band pass filter

Answer: b

Explanation: A filter is called a low pass filter if it passes all frequencies up to the cut-off frequency f c without attenuation and attenuates all other frequencies greater than f c . This transmits currents of all frequencies from zero up to the cut-off frequency.

3. A filter which attenuates all frequencies below a designated cut-off frequency f c and passes all other frequencies greater than f c is called?

a) band elimination filter

b) band pass filter

c) low pass filter

d) high pass filter

Answer: d

Explanation: A filter is called high pass filter if attenuates all frequencies below a designated cut-off frequency f c and passes all other frequencies greater than f c . Thus the pass band of this filter is the frequency range above f c and the stop band is the frequency range below f c .

4. A filter that passes frequencies between two designated cut-off frequencies and attenuates all other frequencies is called?

a) high pass filter

b) band elimination filter

c) band pass filter

d) low pass filter

Answer: c

Explanation: A band pass filter passes frequencies between two designated cut-off frequencies and attenuates all other frequencies. A band pass filter has two cut-off frequencies and will have the pass band f 2 -f 1 ; f 1 is the lower cut-off frequency, f 2 is the upper cut-off frequency.

5. A filter that passes all frequencies lying outside a certain range, while it attenuates all frequencies between the two designated frequencies is called?

a) low pass filter

b) high pass filter

c) band elimination filter

d) band pass filter

Answer: c

Explanation: A band elimination filter passes all frequencies lying outside a certain range, while it attenuates all frequencies between the two designated frequencies. It is also referred to as band stop filter.

6. The expression of the characteristic impedance of a symmetrical T-section is?

a) Z OT =√(Z 1 2 /4-Z 1 Z 2 )

b) Z OT =√(Z 1 2 /4+Z 1 )

c) Z OT =√(Z 1 2 /4+Z 2 )

d) Z OT =√(Z 1 2 /4+Z 1 Z 2 )

Answer: d

Explanation: For a T-section, the value of input impedance when it is terminated in Z o is

Z in =(Z 1 /2)+(Z 2 ((Z 1 /2)+Z o ))/((Z 1 /2)+Z 2 +Z o ) and Z in =Z o . On solving, the expression of the characteristic impedance of a symmetrical T-section is Z OT =√(Z 1 2 /4+Z 1 Z 2 ).

7. The expression of the open circuit impedance Z oc is?

a) Z oc =Z 1 /2+Z 2

b) Z oc =Z 2 /2+Z 2

c) Z oc =Z 1 /2+Z 1

d) Z oc =Z 1 /2-Z 2

Answer: a

Explanation: On open circuiting the port 2 of T-section, we get the expression of the open circuit impedance Z oc as Z oc =Z 1 /2+Z 2 .

8. The expression of short circuit impedance Z sc is?

a) Z sc =(Z 1 2 -4Z 1 Z 2 )/(2Z 1 -4Z 2 )

b) Z sc =(Z 1 2 +4Z 1 Z 2 )/(2Z 1 +4Z 2 )

c) Z sc =(Z 1 2 -4Z 1 Z 2 )/(2Z 1 +4Z 2 )

d) Z sc =(Z 1 2 +4Z 1 Z 2 )/(2Z 1 -4Z 2 )

Answer: b

Explanation: On short circuiting the port 2 of T-section, we get the expression of short circuit impedance Z sc as Z sc =(Z 1 /2)+((Z 1 /2)xZ 2 )/((Z 1 /2)+Z 2 ). On solving we get Z sc =(Z 1 2 +4Z 1 Z 2 )/(2Z 1 +4Z 2 ).

9. The relation between Z OT , Z oc , Z sc is?

a) Z OT =√Z oc Z sc

b) Z oc =√(Z OT Z sc )

c) Z sc =√(Z OT Z oc )

d) Z oc =√(Z OT Z oc )

Answer: a

Explanation: Z oc =Z 1 /2+Z 2 and Z sc =(Z 1 2 +4Z 1 Z 2 )/(2Z 1 +4Z 2 ) => Z oc xZ sc =Z 1 Z 2 +Z 1 2 /4 =Z o 2 T. The relation between Z OT , Z oc , Z sc is Z OT =√Z oc Z sc .

10. The value of sinh⁡ϒ/2 in terms of Z 1 and Z 2 is?

a) sinh⁡ϒ/2=√(4Z 1 /Z 2 )

b) sinh⁡ϒ/2=√(Z 1 /Z 2 )

c) sinh⁡ϒ/2=√(Z 1 /4Z 2 )

d) sinh⁡ϒ/2=√(2Z 1 /Z 2 )

Answer: c

Explanation: sinhϒ/2=√ϒ/(1/2(1+Z 1 /2Z 2 -1))). The value of sinh⁡ϒ/2 in terms of Z 1 and Z 2 is sinh⁡ϒ/2=√(Z 1 /4Z 2 ).

This set of Network Theory Questions and Answers for Entrance exams focuses on “Classification of Pass Band and Stop Band”.

1. The relation between α, β, ϒ is?

a) α = ϒ + jβ

b) ϒ = α + jβ

c) β = ϒ + jα

d) α = β + jϒ

Answer: b

Explanation: We know that the propagation constant is a complex function and the real part of the complex propagation constant is a measure of the change in magnitude of the current or voltage in the network known as attenuation constant and imaginary part is a measure of the difference in phase between the input and output currents or voltages known as phase shift constant. ϒ = α + jβ.

2. If Z 1 , Z 2 are same type of reactance, then |Z 1 /4 Z 2 | is real, then the value of α is?

a) α = sinh -1 ⁡√(Z 1 /4 Z 2 )

b) α = sinh -1 ⁡⁡√(Z 1 /Z 2 )

c) α = sinh -1 ⁡√(4 Z 1 /Z 2 )

d) α = sinh -1 ⁡⁡√(Z 1 /2 Z 2 )

Answer: a

Explanation: Z 1 , Z 2 are same type of reactance and |Z 1 /4 Z 2 | is real. |Z 1 /4 Z 2 | > 0. The value of α is α = sinh -1 ⁡√(Z 1 /4 Z 2 ).

3. If Z 1 , Z 2 are same type of reactance, then |Z 1 /4 Z 2 | is real, then?

a) |Z 1 /4 Z 2 | = 0

b) |Z 1 /4 Z 2 | < 0

c) |Z 1 /4 Z 2 | > 0

d) |Z 1 /4 Z 2 | >= 0

Answer: c

Explanation: If Z 1 and Z 2 are same type of reactances, then √(Z 1 /4 Z 2 ) should be always positive implies that |Z 1 /4 Z 2 | > 0.

4. Which of the following expression is true if Z 1 , Z 2 are same type of reactance?

a) sinh⁡α/2 sin⁡β/2=0

b) coshα/2 sin⁡β/2=0

c) coshα/2 cos⁡β/2=0

d) sinhα/2 cos⁡β/2=0

Answer: d

Explanation: If Z 1 , Z 2 are same type of reactance, then the real part of sinhϒ/2 = sinhα/2 cos⁡β/2 + jcoshα/2 sin⁡β/2 should be zero. So sinhα/2 cos⁡β/2=0.

5. Which of the following expression is true if Z 1 , Z 2 are same type of reactance?

a) sinhα/2 cos⁡β/2=x

b) coshα/2 cos⁡β/2=0

c) coshα/2 sin⁡β/2=x

d) sinhα/2 sin⁡β/2=0

Answer: c

Explanation: If Z 1 , Z 2 are same type of reactance, then the imaginary part of sinhϒ/2 = sinhα/2 cos⁡β/2 + jcoshα/2 sin⁡β/2 should be some value. So coshα/2 sin⁡β/2=x.

6. The value of α if Z 1 , Z 2 are same type of reactance?

a) 0

b) π/2

c) π

d) 2π

Answer: a

Explanation: As sinhα/2 cos⁡β/2=0 and coshα/2 sin⁡β/2=x, the value of α if Z 1 , Z 2 are same type of reactance is α = 0.

7. The value of β if Z 1 , Z 2 are same type of reactance?

a) 2π

b) π

c) π/2

d) 0

Answer: b

Explanation: The value of β if Z 1 , Z 2 are same type of reactances, then sinhα/2 cos⁡β/2=0 and coshα/2 sin⁡β/2=x. So the value of β is β = π.

8. If Z 1 , Z 2 are same type of reactance, and if α = 0, then the value of β is?

a) β=2 sin -1 ⁡(√(Z 1 /4 Z 2 ))

b) β=2 sin -1 ⁡(√(4 Z 1 /Z 2 ))

c) β=2 sin -1 ⁡(√(4 Z 1 /Z 2 ))

d) β=2 sin -1 ⁡(√(Z 1 /Z 2 ))

Answer: a

Explanation: If α = 0, sin β/2 = x(√(Z 1 /4 Z 2 ). But sine can have a maximum value of 1. Therefore the above solution is valid only for Z 1 /4 Z 2 , and having a maximum value of unity. It indicates the condition of pass band with zero attenuation and follows the condition as -1 < Z 1 /4 Z 2 <= 0. So β=2 sin -1 ⁡(√(Z 1 /4 Z 2 )).

9. If the value of β is π, and Z 1 , Z 2 are same type of reactance, then the value of β is?

a) α=2 cosh -1 ⁡√(Z 1 /2 Z 2 )

b) α=2 cosh -1 ⁡√(Z 1 /Z 2 )

c) α=2 cosh -1 ⁡√(4 Z 1 /Z 2 )

d) α=2 cosh -1 ⁡√(Z 1 /4 Z 2 )

Answer: d

Explanation: If the value of β is π, cos β/2 = 0. So sin β/2 = ±1; cosh α/2 = x = √(Z 1 /4 Z 2 ). This solution is valid for negative Z 1 /4 Z 2 and having magnitude greater than or equal to unity. -α &lt= Z 1 /2 Z 2 <= -1. α=2 cosh -1 ⁡√(Z 1 /4 Z 2 ).

10. The relation between Z oπ , Z 1 , Z 2 , Z oT is?

a) Z oT = Z 1 Z 2 /Z oπ

b) Z oπ = Z 1 Z 2 /Z oT

c) Z oT = Z 1 Z 1 /Z oπ

d) Z oT = Z 2 Z 2 /Z oπ

Answer: b

Explanation: The characteristic impedance of a symmetrical π-section can be expressed in terms of T. Z oπ = Z 1 Z 2 /Z oT .

This set of Network Theory Multiple Choice Questions & Answers focuses on “Constant-K Low Pass Filter”.

1. A network either T or π, is said to be of the constant-k type if Z 1 and Z 2 of the network satisfies the relation?

a) Z 1 Z 2 = k

b) Z 1 Z 2 = k 2

c) Z 1 Z 2 = k 3

d) Z 1 Z 2 = k 4

Answer: b

Explanation: Z 1 ,Z 2 are inverse if their product is a constant, independent of frequency, k is real constant, that is the resistance. k is often termed as design impedance or nominal impedance of the constant k-filter.

2. In the circuit shown below, find the value of Z 1 .

network-theory-questions-answers-constant-k-low-pass-q2

a) jωL

b) 2 jωL

c) jωL/2

d) 4 jωL

Answer: a

Explanation: The constant k, T or π type filter is also known as the prototype because other more complex networks can be derived from it. From the given figure, the value of Z 1 is jωL.

3. In the circuit shown below, find the value of Z 2 .

network-theory-questions-answers-constant-k-low-pass-q2

a) jωC

b) 2 jωC

c) 1/jωC

d) 1/2 jωC

Answer: c

Explanation: From the prototype T section and prototype π section shown in figures, we get the value of Z 2 is 1/jωC.

4. The value of Z 1 Z 2 in the circuit shown below is?

network-theory-questions-answers-constant-k-low-pass-q2

a) L/C

b) C/L

c) 1/LC

d) LC

Answer: a

Explanation: Z 1 = jωL and Z 2 = 1/jωC. So the product Z 1 Z 2 is jωL x 1/jωC = L/C.

5. Determine the value of k in the circuit shown below.

network-theory-questions-answers-constant-k-low-pass-q2

a) √LC

b) √)

c) √)

d) √)

Answer: b

Explanation: We got Z 1 Z 2 = L/C. And we know Z 1 Z 2 = k 2 . So k 2 = L/C. So the value of k is √.

6. The cut-off frequency of the constant k-low pass filter is?

a) 1/√LC

b) 1/

c) √LC

d) π√LC

Answer: b

Explanation: Z 1 /4Z 2 = 0. Z 1 = jωL and Z 2 = 1/jωC. On solving the cut-off frequency of the constant k-low pass filter is f c = 1/.

7. The value of α in the pass band of constant k-low pass filter is?

a) 2 cosh -1 ⁡(f c /f)

b) cosh -1 ⁡⁡(f c /f)

c) cosh -1 ⁡(f/f c )

d) 2 cosh -1 ⁡(f/f c )

Answer: d

Explanation: The value of α in the pass band of constant k-low pass filter is α = 2 cosh -1 ⁡(f/f c ).

8. The value of β in the attenuation band of constant k-low pass filter is?

a) 0

b) π

c) π/2

d) π/4

Answer: b

Explanation: We know that in the attenuation band, Z 1 /4Z 2 < -1 i.e., f/f c < 1. So the value of β in the pass band of constant k-low pass filter is β = π.

9. The value of α in the attenuation band of constant k-low pass filter is?

a) α=2 cosh -1 ⁡(f c /f)

b) α=cosh -1 ⁡(f/f c )

c) α=2 cosh -1 ⁡(f/f c )

d) α=cosh -1 ⁡(f c /f)

Answer: c

Explanation: α = 2 cosh -1 [Z 1 /4Z 2 ] and Z 1 /4Z 2 = f/f c . On substituting we get α = 2 cosh -1 ⁡(f/f c ).

10. The value of α in the pass band of constant k-low pass filter is?

a) π

b) π/4

c) π/2

d) 0

Answer: d

Explanation: We know that in the pass band, the condition is -1 < Z 1 /4Z 2 < 0. So α = π.

This set of Network Theory Multiple Choice Questions & Answers focuses on “m-Derived T-Section”.

1.The relation between Z oT and Z oT ‘ in the circuits shown below.

network-theory-questions-answers-m-derived-t-section-q1

a) Z oT = Z oT ‘

b) Z oT = 2 Z oT ‘

c) Z oT = 3 Z oT ‘

d) Z oT = 4 Z oT ‘

Answer: a

Explanation: The relation between Z oT and Z oT ’ is Z oT = Z oT ’ where Z oT ’ is the characteristic impedance of the modified T-network.

2. The value of Z 2 ’ in terms of Z 1 , Z 2 from the circuits shown below is?

network-theory-questions-answers-m-derived-t-section-q1

a) Z 2 ‘ =Z 2 /4 m (1-m 2 )+Z 2 /m

b) Z 2 ‘ =Z 1 /4 m (1-m 2 )+Z 1 /m

c) Z 2 ‘ =Z 2 /4 m (1-m 2 )+Z 1 /m

d) Z 2 ‘ =Z 1 /4 m (1-m 2 )+Z 2 /m

Answer: d

Explanation: As Z oT = Z oT ’ , √(Z 1 2 /4+Z 1 Z 2 )=√(m 2 Z 1 2 /4+m Z 2 ‘ ). On solving, Z 2 ‘ =Z 1 /(4 m (1-m 2 ))+Z 2 /m.

3. The relation between Z oπ and Z oπ ’ in the circuits shown below is?

network-theory-questions-answers-m-derived-t-section-q3

a) Z oπ = 2 Z oπ ’

b) Z oπ = 4 Z oπ ’

c) Z oπ = Z oπ ’

d) Z oπ = 3 Z oπ ’

Answer: c

Explanation: The characteristic impedances of the prototype and its modified sections have to be equal for matching. The relation between Z oπ and Z oπ ‘ is Z oπ = Z oπ ’ .

4. The value of Z 1 ‘ in terms of Z 1 , Z 2 from the circuits shown below is?

network-theory-questions-answers-m-derived-t-section-q3

a) Z 1 ‘ =(m Z 2 (Z 2 4 m)/(1-m 2 ))/m Z 1 (Z 2 4 m/(1-m 2 ))

b) Z 1 ‘ =(m Z 1 (Z 2 4 m)/(1-m 2 ))/m Z 2 (Z 2 4 m/(1-m 2 ))

c) Z 1 ‘ =(m Z 1 (Z 2 4 m)/(1-m 2 ))/m Z 1 (Z 2 4 m/(1-m 2 ))

d) Z 1 ‘ =(m Z 1 (Z 2 4 m)/(1-m 2 ))/m Z 1 (Z 1 4 m/(1-m 2 ))

Answer: c

Explanation: As Z oπ = Z oπ ’ , √(Z 1 Z 2 /(1+Z 1 /4 Z 2 ))=√(((Z 1 ‘ Z 2 )/m)/(1+(Z 1 ‘ )/(4 Z 2 /m))). On solving, Z 1 ‘ =(m Z 1 (Z 2 4 m)/(1-m 2 ))/m Z 1 (Z 2 4 m/(1-m 2 )).

5. The value of resonant frequency in the m-derived low pass filter is?

a) f r =1/(√(LC(1+m 2 )))

b) f r =1/(√(πLC(1+m 2 )))

c) f r =1/(√(LC(1-m 2 )))

d) f r =1/(√(πLC(1-m 2 )))

Answer: d

Explanation: ω r 2 = 1/(LC(1-m 2 )). So the value of resonant frequency in the m-derived low pass filter is f r =1/√(πLC(1-m 2 )).

6. The cut-off frequency of the low pass filter is?

a) 1/√LC

b) 1/

c) 1/√L

d) 1/

Answer: b

Explanation: To determine the cut-off frequency of the low pass filter we place m = 0. So f c =1/.

7. The resonant frequency of m-derived low pass filter in terms of the cut-off frequency of low pass filter is?

a) f c /√(1-m 2 )

b) f c /√(1+m 2 )

c) f c /(π√(1-m 2 ))

d) f c /(π√(1+m 2 ))

Answer: a

Explanation: If a sharp cut-off is desired, the frequency at infinity should be near to f c . The resonant frequency of m-derived low pass filter in terms of the cut-off frequency of low pass filter is f r =f c /√(1-m 2 ).

8. The expression of m of the m-derived low pass filter is?

a) m=√(1+(f c /f r ) 2 )

b) m=√(1+(f c /f) 2 )

c) m=√(1-(f c /f r ) 2 )

d) m=√(1-(f c /f) 2 )

Answer: c

Explanation: As f r =f c /√(1-m 2 ). The expression of m of the m-derived low pass filter is m=√(1-(f c /f r ) 2 ).

9. Given a m-derived low pass filter has cut-off frequency 1 kHz, design impedance of 400Ω and the resonant frequency of 1100 Hz. Find the value of k.

a) 400

b) 1000

c) 1100

d) 2100

Answer: a

Explanation: The value of k is equal to the design impedance. Given design impedance is 400Ω. So, k = 400.

10. Given a m-derived low pass filter has cut-off frequency 1 kHz, design impedance of 400Ω and the resonant frequency of 1100 Hz. Find the value of m.

a) 0.216

b) 0.316

c) 0.416

d) 0.516

Answer: c

Explanation: m=√(1-(f c /f r ) 2 ) f c = 1000, f r = 1100. On substituting m=√ 2 )=0.416.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Attenuators”.

1. The attenuation in dB in terms of input power (P 1 ) and output power (P 2 ) is?

a) log 10 (P 1 /P 2 )

b) 10 log 10 (P 1 /P 2 )

c) log 10 (P 2 /P 1 )

d) 10 log 10 (P 2 /P 1 )

Answer: b

Explanation: The increase or decrease in power due to insertion or substitution of a new element in a network can be conveniently expressed in decibels or in nepers. The attenuation in dB in terms of input power (P 1 ) and output power (P 2 ) is Attenuation in dB = 10 log 10 (P 1 /P 2 ).

2. If V 1 is the voltage at port 1 and V 2 is the voltage at port 2, then the attenuation in dB is?

a) 20 log 10 (V 1 /V 2 )

b) 10 log 10 (V 1 /V 2 )

c) 20 log 10 (V 2 /V 1 )

d) 10 log 10 (V 2 /V 1 )

Answer: a

Explanation: If V 1 is the voltage at port 1 and V 2 is the voltage at port 2, then the attenuation in dB is Attenuation in dB = 20 log 10 (V 1 /V 2 ) where V 1 is the voltage at port 1 and V 2 is the voltage at port 2.

3. What is the attenuation in dB assuming I 1 is the input current and I 2 is the output current leaving the port?

a) 10 log 10 (I 1 /I 2 )

b) 10 log 10 (I 2 /I 1 )

c) 20 log 10 (I 2 /I 1 )

d) 20 log 10 (I 1 /I 2 )

Answer: d

Explanation: Assuming I 1 is the input current and I 2 is the output current leaving the port, the attenuation in dB is Attenuation in dB = 20 log 10 (I 1 /I 2 ) where I 1 is the input current and I 2 is the output current leaving the port.

4. The value of one decibel is equal to?

a) log 10

b) 10 log 10

c) 20 log 10

d) 40 log 10

Answer: c

Explanation: The value of one decibel is equal to 20 log 10 . One decibel = 20 log 10 where N is the attenuation.

5. The value of N in dB is?

a) N = anti log

b) N = anti log

c) N = anti log

d) N = anti log

Answer: c

Explanation: The value of N in dB can be expressed as N = anti log.

6. In the circuit shown below, find the value of I 1 /I 2 .

network-theory-questions-answers-attenuators-q6

a) (R 1 -R 2 +R 0 )/R 2

b) (R 1 +R 2 +R 0 )/R 2

c) (R 1 -R 2 -R 0 )/R 2

d) (R 1 +R 2 -R 0 )/R 2

Answer: b

Explanation: R 2 (I 1 -I 2 )=I 2 (R 1 +R 0 )

=> I 2 (R 2 +R 1 +R 0 )I 1 R 2 . On solving, I 1 /I 2 =(R 1 +R 2 +R 1 )/R 2 .

7. Determine the value of N in the circuit shown below.

network-theory-questions-answers-attenuators-q6

a) (R 1 +R 2 -R 0 )/R 2

b) (R 1 -R 2 -R 0 )/R 2

c) (R 1 +R 2 +R 0 )/R 2

d) (R 1 -R 2 +R 0 )/R 2

Answer: c

Explanation: N = I 1 /I 2 . We got I 1 /I 2 = (R 1 +R 2 +R 1 )/R 2 . So on substituting we get N = (R 1 +R 2 +R 0 )/R 2 .

8. The value of the characteristic impedance R 0 in terms of R 1 and R 2 and R 0 in the circuit shown below is?

network-theory-questions-answers-attenuators-q6

a) R 1 +R 2 (R 1 +R 0 )/(R 1 +R 0 +R 2 )

b) R 1 + R 2 (R 1 +R 0 )/(R 1 +R 0 +R 2 )

c) R 2 + R 2 (R 1 +R 0 )/(R 1 +R 0 +R 2 )

d) R 0 +R 2 (R 1 +R 2 )/(R 1 +R 0 +R 2 )

Answer: b

Explanation: The value of the characteristic impedance R 0 in terms of R 1 and R 2 and R 0 when it is terminated in a load of R 0 is R 0 =R 1 + R 2 (R 1 +R 0 )/(R 1 +R 0 +R 2 ).

9. Determine the value of R 1 in terms of R 0 and N in the circuit shown below is?

network-theory-questions-answers-attenuators-q6

a) R 1 = R 0 /

b) R 1 = R 0 /

c) R 1 = R 0 /

d) R 1 = R 0 /

Answer: a

Explanation: R 0 = R 1 +(R 1 +R 0 )/N. On solving, the value of R 1 in terms of R 0 and N is R 1 = R 0 /.

10. Determine the value of R 2 in terms of R 0 and N in the circuit shown below is?

network-theory-questions-answers-attenuators-q6

a) R 2 = NR 0 /(N 2 -1)

b) R 2 = 2 NR 0 /(N 2 -1)

c) R 2 = 3 NR 0 /(N 2 -1)

d) R 2 = 4 NR 0 /(N 2 -1)

Answer: b

Explanation: NR 2 = R 1 +R 0 +R 2 . On substituting the value of R 1 , we get the value of R 2 in terms of R 0 and N as R 2 = 2 NR 0 /(N 2 -1).

This set of Network Theory Multiple Choice Questions & Answers focuses on “Inverse Network”.

1. The impedances Z 1 and Z 2 are said to be inverse if?

a) Z 1 Z 2 = R 0

b) Z 1 + Z 2 = R 0

c) 1/Z 1 + 1/Z 2 = R 0

d) Z 1 Z 2 = R 0 2

Answer: d

Explanation: The impedances Z 1 and Z 2 are said to be inverse if the geometric mean of the two impedances is a real number.

2. An inverse network may be obtained by?

a) Converting each series branch into another series branch

b) Converting each series branch into another parallel branch

c) Converting each parallel branch into another series branch

d) None of the mentioned

Answer: c

Explanation: An inverse network may be obtained by converting each parallel branch into another series branch and vice-versa and not by converting each series branch into another series branch and not by converting each series branch into another parallel branch.

3. An inverse network may be obtained by converting each resistance element R into a corresponding resistive element of value?

a) R 0 2 /R

b) R/R 0 2

c) R 0 /R

d) R/R 0

Answer: a

Explanation: To obtain the inverse network we have to convert each resistance element R into a corresponding resistive element of value R 0 2 /R.

4. An inverse network may be obtained by converting each inductance L into a capacitance of value?

a) L/R 0

b) L/R 0 2

c) R 0 /L

d) R 0 2 /L

Answer: b

Explanation: An inverse network may be obtained by converting each inductance L should be converted into a capacitance of value L/R 0 2 to obtain the inverse network.

5. An inverse network may be obtained by converting each capacitance C into an inductance of value?

a) CR 0 2

b) CR 0

c) R 0 2 /C

d) C/R 0 2

Answer: a

Explanation: An inverse network is obtained by converting each capacitance C into an inductance of value CR 0 2 where R 0 is resistance.

6. Consider the network shown below. Find the value of capacitance C 1 ‘ after converting the inductance L 1 into a capacitance.

network-theory-questions-answers-inverse-network-q6

a) R 0 2 /L 1

b) R 0 /L 1

c) L 1 /R 0 2

d) L 1 /R 0

Answer: c

Explanation: An inverse network may be obtained by converting Each inductance L should be converted into a capacitance of value L/R 0 2 to obtain the inverse network. The value of capacitance C 1 ‘ after converting the inductance into a capacitance is L 1 /R 0 2 . C 1 ’ = L 1 /R 0 2 .

7. Consider the network shown below, find the value of inductance L 1 ‘ after converting the capacitance into an inductance.

network-theory-questions-answers-inverse-network-q6

a) C 1 /R 0 2

b) R 0 2 /C 1

c) C 1 R 0

d) C 1 R 0 2

Answer: d

Explanation: An inverse network is obtained by converting each capacitance C into an inductance of value CR 0 2 where R 0 is resistance. The value of inductance L 1 ‘ after converting the capacitance into an inductance is L 1 ‘ = C 1 R 0 2 .

8. Consider the network shown below, find the value of resistance R 1 ‘ after converting the resistance R 1 .

network-theory-questions-answers-inverse-network-q6

a) R 1 /R 0

b) R 0 /R 1

c) R 1 /R 0 2

d) R 0 2 /R 1

Answer: d

Explanation: To obtain the inverse network we have to convert each resistance element R into a corresponding resistive element of value R 0 2 /R. The value of resistance R 1 ‘ after converting R 1 is R 1 ‘ = R 0 2 /R 1 .

9. The value of the capacitance C 2 ‘ after converting the inductor into the C 2 ‘ in the network shown below.

network-theory-questions-answers-inverse-network-q6

a) L 2 /R 0 2

b) L 2 /R 0

c) R 0 2 /L 2

d) R 0 /L 2

Answer: a

Explanation: An inverse network may be obtained by converting Each inductance L should be converted into a capacitance of value L/R 0 2 to obtain the inverse network. The value of the capacitance C 2 ‘ after converting the inductor into the capacitance is

C 2 ‘ = L 2 /R 0 2 .

10. The value of the inductor L 2 ‘ after converting the capacitor into the L 2 ‘ in the network shown below.

network-theory-questions-answers-inverse-network-q6

a) R 0 2 /C 2

b) C 2 R 0 2

c) C 2 R 0

d) R 0 2 /C 2

Answer: b

Explanation: An inverse network is obtained by converting each capacitance C into an inductance of value CR 0 2 where R 0 is resistance. The value of the inductor L 2 ‘ after converting the capacitor into the inductance is L 2 ‘ = C 2 R 0 2 .

This set of Network Theory Multiple Choice Questions & Answers focuses on “Series Equalizer”.

1. The value of attenuation D is equal to?

a) log 10

b) 10 log 10

c) 20 log 10

d) 40 log 10

Answer: a

Explanation: The value of attenuation D is equal to log 10 . Attenuation D = log 10 where N is input to output power ratio of the load.

2. The value of N in terms of attenuation D is?

a) antilog

b) antilog

c) antilog

d) antilog

Answer: b

Explanation: The value of N in terms of attenuation D is antilog. N = antilog where D is attenuation in decibels.

3. The input to output power ratio of the load is the ratio of the________ to the __________

a) Maximum power delivered to the load when the equalizer is not present, power delivered to the load when equalizer is present

b) Power delivered to the load when equalizer is present, maximum power delivered to the load when the equalizer is not present

c) Maximum power delivered to the load when the equalizer is present, power delivered to the load when equalizer is not present

d) Power delivered to the load when equalizer is not present, maximum power delivered to the load when the equalizer is present

Answer: a

Explanation: The input to output power ratio of the load is the ratio of the maximum power delivered to the load when the equalizer is not present to the power delivered to the load when equalizer is present.

4. The N is defined as?

a) output power/ input power

b) input power/ output power

c) output power at inductor/ input power

d) output power at capacitor/ input power

Answer: b

Explanation: The N is defined as the ratio of input power to the output power. N = P i /P l where P i is input power and P l is output power.

5. The expression of input power of a series equalizer is?

a) V max 2 /R o

b) V max 2 /2R o

c) V max 2 /3R o

d) V max 2 /4R o

Answer: d

Explanation: The expression of input power of a series equalizer is P i =(V max /2R o ) 2 R o =V max 2 /4R o .

6. The expression of current flowing in a series equalizer is?

a) V max /√((R o ) 2 +(X 1 ) 2 )

b) V max /√((2R o ) 2 +(X 1 ) 2 )

c) V max /√((2R o ) 2 +(2X 1 ) 2 )

d) V max /√((R o ) 2 +(2X 1 ) 2 )

Answer: c

Explanation: When the equalizer is connected, the expression of current flowing in a series equalizer is I 1 = V max /√((2R o ) 2 +(2X 1 ) 2 ) where V max is voltage applied to the network and R o is resistance of the load as well as source and 2X 1 is the reactance of the equalizer.

7. What is the power at the load of a series equalizer?

a) [V max 2 /(R o 2 +X 1 2 )]R o

b) [V max 2 /(2(R o 2 +X 1 2 ))]R o

c) [V max 2 /(3(R o 2 +X 1 2 ))]R o

d) [V max 2 /(4(R o 2 +X 1 2 ))]R o

Answer: d

Explanation: The power at the load of a series equalizer is P=(V max /√((2R o ) 2 +(2X 1 ) 2 )) 2 R o = [V max 2 /(4(R o 2 +X 1 2 ))]R o .

8. Determine the value of N in the series equalizer.

a) 1+ X 1 2 /R o 2

b) X 1 2 /R o 2

c) 1+ R o 2 /X 1 2

d) R o 2 /X 1 2

Answer: a

Explanation: The N is defined as the ratio of input power to the output power.

N=P i /P l = (V max 2 /4R o )/[V max 2 /(4(R o 2 +X 1 2 ))]R o =1+X 1 2 /R o 2 .

9. The expression of N in a full series equalizer considering Z 1 as inductor and Z 2 as capacitor is?

a) R o 2 /(ωL 1 ) 2

b) 1+ R o 2 /(ωL 1 ) 2

c) (ω 2 L 1 2 )/R o 2

d) 1+ (ω 2 L 1 2 )/R o 2

Answer: d

Explanation: The expression of N in a full series equalizer considering Z 1 as inductor and Z 2 as capacitor is N = 1 + X 1 2 /R o 2 = 1+ (ω 2 L 1 2 )/R o 2 .

10. The expression of N in a full series equalizer considering Z 1 as capacitor and Z 2 as inductor is?

a) 1+ (ω 2 L 1 2 )/R o 2

b) (ω 2 L 1 2 )/R o 2

c) 1+ R o 2 /(ωL 1 ) 2

d) R o 2 /(ωL 1 ) 2

Answer: c

Explanation: The expression of N in a full series equalizer considering Z 1 as capacitor and Z 2 as inductor is N = 1+ R o 2 /X 2 2 = 1+R o 2 /(ωL 1 ) 2 .

This set of Network Theory Multiple Choice Questions & Answers focuses on “Shunt Equalizer”.

1. In the shunt equalizer, the current flowing from the source is?

a) V max (2R o +jX 1 )/2R o (R o +jX 1 )

b) V max (R o +jX 1 )/R o (R o +jX 1 )

c) V max (R o +jX 1 )/2R o (R o +jX 1 )

d) V max (2R o +jX 1 )/R o (R o +jX 1 )

Answer: a

Explanation: In the shunt equalizer, the current flowing from the source is I s = V max /(R o +(R o ||jX 1 /2)). On solving, I s = V max (2R o +jX 1 )/2R o (R o +jX 1 ).

2. What is the load current in terms of source current in the shunt equalizer?

a) I s jX 1 /(R o +jX 1 )

b) I s jX 1 /(R o +2jX 1 )

c) I s jX 1 /(2R o +2jX 1 )

d) I s jX 1 /(2R o +jX 1 )

Answer: d

Explanation: The load current in terms of source current in the shunt equalizer is I l = I s (jX 1 /2)/(R o +jX 1 /2)). On solving, I l = I s jX 1 /(2R o +jX 1 ).

3. What is the load current in terms of V max in the shunt equalizer?

a) (V max jX 1 )/(R o (2R o +jX 1 ))

b) (V max jX 1 )/(2R o (2R o +jX 1 ))

c) (V max jX 1 )/(2R o (R o +jX 1 ))

d) (V max jX 1 )/(R o (R o +jX 1 ))

Answer: c

Explanation: On substituting I s in the load current equation we get the load current in terms of V max in the shunt equalizer as I l = (V max jX 1 )/(2R o (R o +jX 1 )).

4. The input power in shunt equalizer is?

a) V max 2 /R o

b) V max 2 /2R o

c) V max 2 /3R o

d) V max 2 /4R o

Answer: d

Explanation: The expression of input power of a shunt equalizer is P i =(V max /2R o ) 2 R o =V max 2 /4R o .

5. What is the power at the load of a shunt equalizer?

a) [(V max 2 X 1 2 )/(R o (R o 2 +X 1 2 ))]

b) [(V max 2 X 1 2 )/(2R o (R o 2 +X 1 2 ))]

c) [(V max 2 X 1 2 )/(3R o (R o 2 +X 1 2 ))]

d) [(V max 2 X 1 2 )/(4R o (R o 2 +X 1 2 ))]

Answer: d

Explanation: The power at the load of a series equalizer is P=((V max jX 1 )/(2R o (R o +jX 1 ))) 2 R o =[(V max 2 X 1 2 )/(4R o (R o 2 +X 1 2 ))].

6. The value of N in shunt equalizer is?

a) 1+ X 1 2 /R o 2

b) X 1 2 /R o 2

c) 1+ R o 2 /X 1 2

d) R o 2 /X 1 2

Answer: c

Explanation: The N is defined as the ratio of input power to the output power. N=P i /P l =(V max 2 /4R o )/((V max 2 X 1 2 )/4R o (R o 2 +X 1 2 ))=1+ R o 2 /X 1 2 .

7. The propagation constant of a symmetrical T-section and π-section are the same.

a) True

b) False

Answer: a

Explanation: The propagation constant of a symmetrical T-section and π-section are the same.

8. The attenuation is not sharp in the stop band for an m-derived filter.

a) True

b) False

Answer: b

Explanation: The attenuation is sharp in the stop band for an m-derived filter. So the given statement is not true.

9. The bridged-T phase equalizer consists of?

a) Only pure inductors

b) Only pure capacitors

c) Only pure resistors

d) Only pure reactance

Answer: d

Explanation: The bridged-T phase equalizer consists of only pure reactances. So the bridged T-circuit consists of only inductive or capacitive elements not resistive elements.

10. A lattice phase equalizer is a constant equalizer which satisfies the equation?

a) Z 1 Z 2 = R o

b) Z 1 + Z 2 = R o

c) 1/Z 1 +1/Z 2 = R o

d) Z 1 Z 2 = R o 2

Answer: d

Explanation: The lattice phase equalizer consists of only reactive components. So a lattice phase equalizer is a constant equalizer if the following equation is satisfied. Z 1 Z 2 = R o 2 .

This set of Network Theory Multiple Choice Questions & Answers focuses on “Advanced Problems on Filters – 1”.

1. In the circuit given below, C= C 1 = C 2 . The gain of the multiple-feedback band-pass filter is ___________

network-theory-questions-answers-advanced-problems-filters-1-q1

a) A 0 = \

A 0 = \

A 0 = \(\frac{R_1}{2R_2}\)

Answer: c

Explanation: The total output C = C INPUT + C OUTPUT that is the gain capacitor.

∴ The total Resistance is equal to the Resistance input and Resistance output.

Again, the total resistance gain = \(\frac{R_1 R_2}{R_1 + R_2}\)

Hence, the gain = A 0 = \(\frac{R_2}{2R_1}\).

2. Two network functions are given below.

H 1 = \(\frac{1}{s^2 + \sqrt{2}s+1}\), H 2 = \Missing open brace for subscript Low-pass

b) Band-pass

c) High-pass

d) Band reject

Answer: a

Explanation: | H 1 | 2 = \(\frac{1}{[^2 + \sqrt{2} jω+1] [^2 + \sqrt{2} jω+1]}\)

Similarly, | H 2 | 2 = \(\frac{1}{1 + ω^6}\)

Therefore at ω = 0, 1 and ∞, we have | H | 2 = 1, \(\frac{1}{2}\) and 0 respectively.

Hence, the filter is a Low-pass filter.

3. A particular band-pass function has a network function as H = \Missing open brace for subscript \

\(\frac{\sqrt{3}}{4}\)

Answer: d

Explanation: H = \(\frac{Ks}{s^2+as+b}\)

Then, quality factor is given as \(\frac{\sqrt{b}}{a}\)

Here, b = 3, a = 4

∴ Q = \(\frac{\sqrt{3}}{4}\).

4. In which of the filter circuits given below, will the bandwidth be equal to the critical frequency?

a) Low-pass

b) High-pass

c) Band-pass

d) Band-stop

Answer: a

Explanation: Bandwidth can be calculated by considering,

Largest positive value – Smallest Positive Value

Here, in case of the Low-pass filter only, the largest positive value will of course be the critical frequency, beyond which frequencies have to be blocked. Hence, the bandwidth in a Low-pass filter equals the critical frequency.

5. In the circuit given below, the Roll-off of the filter is _______________

network-theory-questions-answers-advanced-problems-filters-1-q5

a) 20 dB/decade

b) 40 dB/decade

c) 60 dB/decade

d) 80 dB/decade

Answer: a

Explanation: The given filter is a first order Band Pass Filter.

Also, the Roll-off of the filter is depends upon the order of the filter.

For a first order it is 20dB/decade, for second order it is 40dB/decade, and so on.

Therefore, the Roll-off of the filter = 20 dB/decade.

6. The circuit given below, represents which filter?

network-theory-questions-answers-advanced-problems-filters-1-q5

a) Low-pass

b) High-pass

c) Band-pass

d) Band-stop

Answer: b

Explanation: From the given circuit, we can infer that Roll off of the Filter circuit is 80dB/decade. This Roll-off value is obtained as second order high pass filter followed by another 2nd order HPF results in an HPF.

Therefore the circuit represents a High-pass Filter.

7. A Low-pass filter circuit has a cut-off frequency of 1.23 kHz. The bandwidth of the filter is ______________

a) 2.46 kHz

b) 1.23 kHz

c) 0.615 kHz

d) 3.69 kHz

Answer: b

Explanation: The bandwidth is defined as the highest cut-off frequency to the lowest cut-off frequency. Here the lowest cut-off frequency is Zero.

For a Low-pass filter therefore, Cut-off Frequency = Bandwidth of the filter

∴ Bandwidth = 1.23 kHz.

8. For the circuit given below, the Roll-off value of the filter is _____________

network-theory-questions-answers-advanced-problems-filters-1-q5

a) 20 dB/decade

b) 40 dB/decade

c) 60 dB/decade

d) 80 dB/decade

Answer: d

Explanation: The given filter is a first order Band Pass Filter. Also, the Roll-off of the filter is depends upon the order of the filter. For a first order it is 20dB/decade, for second order it is 40dB/decade, and so on.

Therefore, the Roll-off of the filter = 20 dB/decade.

Roll of first order low pass Butterworth filter is 20dB/decade.

Now here two stages of second order Low-pass Butterworth filter are cascaded

∴ Roll-off = 20*4 = 80 dB/decade.

9. For providing a Roll-off greater than 20dB/decade/pole, filters with which characteristics are useful?

a) Butterworth

b) Chebyshev

c) Bessel

d) Butterworth & Bessel

Answer: b

Explanation: Roll off is a term commonly refers to the steepness of the transmission function wrt to the frequency.

For a Chebyshev filter, the Roll-off value greater than 20. This characteristic feature is useful when a rapid roll-off is required because it provides a Roll-off rate of more than 20.

On the other hand, both Butterworth and Bessel have the Roll-off rate less than or equal to 20 dB/decade/pole.

10. In a moving iron meter, the deflection torque is proportional to?

a) Square of the current through the coil

b) Current through the coil

c) Sine of measurand

d) The Square root of the measurand

Answer: a

Explanation: We know that T d = \(\frac{1}{2} I^2 \frac{dl}{dθ}\)

Hence, the deflection torque is proportional to the square of the current through the coil.

11. The filter which passes all frequencies above fc by attenuating significantly, all frequencies below f c is _______________

a) Low-pass

b) High-pass

c) Band-pass

d) Band-stop

Answer: b

Explanation: A high-pass filter is one which passes all frequencies above f c by attenuating significantly, all frequencies below f c .

12. The circuit given below represents which type of filter circuit?

network-theory-questions-answers-advanced-problems-filters-1-q12

a) Low-pass Filter

b) High-pass Filter

c) Band-pass Filter

d) Band-stop Filter

Answer: c

Explanation: The given circuit is a first order Band Pass Filter. Also, the Roll-off of the filter depends upon the order of the filter. For a first order it is 20dB/decade, for second order it is 40dB/decade, and so on.

13. For the circuit given below, the cut-off frequency of the filter is ________________

network-theory-questions-answers-advanced-problems-filters-1-q12

a) 3225.8 Hz

b) 7226 Hz

c) 3225.8 kHz

d) 7226 kHz

Answer: c

Explanation: We know that, F = \(\frac{1}{2π×

}\)

Where, R 1 = 4700 Ω, R 2 = 4700 Ω, C 1 = 0.047 × 10 -6 F, C 2 = 0.047×10 -6 F

∴ F = \(\frac{1}{2π××

}\)

= \(\frac{1}{2π×48796.81×10^{-12}}\)

= \(\frac{10^6}{2π×0.04879681}\)

= \(\frac{10^6}{0.30654156042}\) = 3225.8 kHz.

14. The circuit given below represents which type of filter circuit?

network-theory-questions-answers-advanced-problems-filters-1-q14

a) Low-pass Filter

b) High-pass Filter

c) Band-pass Filter

d) Band-stop Filter

Answer: b

Explanation: We know that the position of Resistance and Capacitance determines whether it is Low-pass Filter or High-Pass Filter. If R is connected directly to source and the capacitor connected in parallel to it, then it is a Low-pass Filter and if the position of R and C are inter change then high pass filter is formed.

Here since R and C are in series and also R is not connected directly to the power source, hence the filter is a High-pass Filter.

15. For the circuit given below, the cut-off frequency of the filter is ________________

network-theory-questions-answers-advanced-problems-filters-1-q14

a) 5283 kHz

b) 5283 Hz

c) 2653.1 kHz

d) 2653.1 Hz

Answer: d

Explanation: We know that, F = \(\frac{1}{2π×R×C}\)

Where, R = 1200 Ω, C 1 = 0.05×10 -6 F

∴ F = \(\frac{1}{2π×1200×C}\)

= \(\frac{1}{2π×60×10^{-6}}\)

= \(\frac{10^6}{2π×60}\) = 2653.1 Hz.

This set of Network Theory Technical Interview Questions & Answers focuses on “Advanced Problems on Filters – 2”.

1. In a parallel RL circuit, 10 A current enters into the resistor R and 15 A current enters into the Inductor L. The total current I is _____________

a) 15.5 A

b) 25.05 A

c) 18.03 A

d) 30.15 A

Answer: c

Explanation: Currents in resistance and inductance are out of phase by 90°.

Hence, I = \(I_1^2 + I_2^2\)

Or, I = [10 2 + 15 2 ] 0.5

Or, I = \(\sqrt{100+225} = \sqrt{325}\)

= 18.03 A.

2. In the circuit given below, a dc circuit fed by a current source. With respect to terminals AB, Thevenin’s voltage and Thevenin’s resistance are ____________

network-theory-technical-interview-questions-answers-q2

a) V TH = 5 V, R TH = 0.75 Ω

b) V TH = 7.14 V, R TH = 2.14 Ω

c) V TH = 7.5 V, R TH = 2.5 Ω

d) V TH = 5 V, R TH = 1 Ω

Answer: b

Explanation: V TH = \(\frac{5 X 10}{35}\) X 5 = 7.14 V

Also, R TH = \(\frac{5 X 15}{35}\) = 2.14 Ω.

3. In the circuit given below, the value of R is ____________

network-theory-technical-interview-questions-answers-q3

a) 12 Ω

b) 6 Ω

c) 3 Ω

d) 1.5 Ω

Answer: b

Explanation: The resistance of parallel combination is given by,

R eq = \(\frac{40}{3}\) – 10 = 3.33 Ω

Or, \(\frac{1}{3.33} = \frac{1}{12} + \frac{1}{15} + \frac{1}{15}\)

Or, R = 6 Ω.

4. In the circuit given below, M = 1. The resonant frequency is _______________

network-theory-technical-interview-questions-answers-q4

a) 145.3 Hz

b) 0.1453 Hz

c) 1.453 Hz

d) 14.53 Hz

Answer: d

Explanation: I EQ = L 1 + L 2 + 2M

L EQ = 5 + 5 + 2 × 1

= 10 + 2 = 12 H

∴ F O = \(\frac{1}{2π\sqrt{LC}}\)

= \(\frac{1}{2π\sqrt{12 × 0.1}}\)

= 0.1453 Hz.

5. A thermistor is used for ____________

a) Over voltage protection

b) Temperature alarm circuit

c) Automatic light control

d) Automatic sensor

Answer: b

Explanation: The resistance of a thermistor decreases with increases in temperature. It is used to monitor the hot spot temperature of electric machines.

6. Two network functions are given below.

H 1 = \(\frac{1}{s^2 + 2s + 1}\), H 2 = \Missing open brace for subscript Low-pass

b) Band-pass

c) High-pass

d) Band reject

Answer: a

Explanation: | H 1 | 2 = \(\frac{1}{[^2 + 2jω + 1] [^2 + 2 jω + 1]}\)

Similarly, | H 2 | 2 = \(\frac{1}{1 + ω^6}\)

Therefore at ω = 0, 1 and ∞, we have | H | 2 = 1, \(\frac{1}{2}\) and 0 respectively.

Hence, the filter is a Low-pass filter.

7. A particular band-pass function has a network function as H = \Missing open brace for subscript \

\(\frac{1}{4}\)

Answer: d

Explanation: H = \(\frac{Ks}{s^2+as+b}\)

Then, quality factor is given as \(\frac{\sqrt{b}}{a}\)

Here, b = 1, a = 2, K = 1

∴ Q = \(\frac{1}{2}\) = 0.5.

8. The filter which passes all frequencies above fc by attenuating significantly, all frequencies above f c is _______________

a) Low-pass

b) High-pass

c) Band-pass

d) Band-stop

Answer: b

Explanation: A high-pass filter is one which passes all frequencies above f c by attenuating significantly, all frequencies below f c .

9. For a Band Pass Filter, the slope of the filter is given as 80dB/decade. The order of the Band Pass Filter is __________

a) 10

b) 8

c) 4

d) 6

Answer: b

Explanation: The Bode plot is a logarithmic plot which helps in fitting a large scale of values into a small scale by the application of logarithm. Plotting the slope 80dB/decade on the Bode plot, we get n=8. Hence the order of the Band Pass Filter is 8.

10. In which of the filter circuits given below, will the bandwidth be equal to the critical frequency?

a) Low-pass

b) High-pass

c) Band-pass

d) Band-stop

Answer: a

Explanation: Bandwidth can be calculated by considering,

Largest positive value – Smallest Positive Value

11. For providing a Roll-off greater than 60dB/decade/pole, filters with which characteristics are useful?

a) Butterworth

b) Chebyshev

c) Bessel

d) Butterworth & Bessel

Answer: b

Explanation: Roll off is a term commonly refers to the steepness of the transmission function with respect to the frequency. For a Chebyshev filter, the Roll-off value greater than 60. This characteristic feature is useful when a rapid roll-off is required because it provides a Roll-off rate more than 60. On the other hand, both Butterworth and Bessel have the Roll-off rate less than or equal to 60 dB/decade/pole.

12. A Low-pass filter circuit has a cut-off frequency of 50 kHz. The bandwidth of the filter is ______________

a) 24.6 kHz

b) 50 kHz

c) 61.5 kHz

d) 36.9 kHz

Answer: b

Explanation: The bandwidth is defined as the highest cut-off frequency to the lowest cut-off frequency. Here the lowest cut-off frequency is Zero.

For a Low-pass filter therefore, Cut-off Frequency = Bandwidth of the filter

∴ Bandwidth = 50 kHz.

13. Given a system function H = \Missing open brace for subscript 1

b) 2

c) 0

d) 5

Answer: c

Explanation: H = \

= \

J = L = \

= \(\frac{1}{s^2+100} . \frac{10}{s+10}\)

V SS = lim s→0 ⁡ sV

= 0.

14. Current I = – 1 + \) A is passed through three meters. The respective readings will be?

a) \ 1 A, \ – 1 A, \ -1 A, 1 A and 1 A

Answer: c

Explanation: We know that a PMMC instrument reads only DC value and since it is a centre zero type, so it will give – 1.

So, rms = \(\sqrt{1^2 +

^2} = \sqrt{2}\) A

Moving iron also reads rms value, so its reading will also be \(\sqrt{2}\) A.

15. For a Band Pass Filter, the slope of the filter is given as 40dB/decade. The order of the Band Pass Filter is __________

a) 2

b) 3

c) 4

d) 6

Answer: c

Explanation: The Bode plot is a logarithmic plot which helps in fitting a large scale of values into a small scale by the application of logarithm. Plotting the slope 40dB/decade on the Bode plot, we get n=4. Hence order of the Band Pass Filter is 4.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Hurwitz Polynomials”.

1. The denominator polynomial in a transfer function may not have any missing terms between the highest and the lowest degree, unless?

a) all odd terms are missing

b) all even terms are missing

c) all even or odd terms are missing

d) all even and odd terms are missing

Answer: c

Explanation: All the quotients in the polynomial P are positive. The denominator polynomial in a transfer function may not have any missing terms between the highest and the lowest degree, unless all even or odd terms are missing.For example P = s 3 +3s is Hurwitz because all quotient terms are positive and all even terms are missing.

2. The roots of the odd and even parts of a Hurwitz polynomial P lie on ____________

a) right half of s plane

b) left half of s-plane

c) on jω axis

d) on σ axis

Answer: c

Explanation: The roots of the odd and even parts of a Hurwitz polynomial P lie on jω axis not on right half of s plane or on left half of s-plane.

3. If the polynomial P is either even or odd, then the roots of P lie on __________

a) on σ axis

b) on jω axis

c) left half of s-plane

d) right half of s plane

Answer: b

Explanation: If the polynomial P is either even or odd, then the roots of P lie on jω axis not on right half of s plane or on left half of s-plane.

4. If the ratio of the polynomial P and its derivative gives a continued fraction expansion with ________ coefficients, then the polynomial P is Hurwitz.

a) all negative

b) all positive

c) positive or negative

d) positive and negative

Answer: b

Explanation: If the ratio of the polynomial P and its derivative P ‘ gives a continued fraction expansion with all positive coefficients, then the polynomial P is Hurwitz. If all the quotients in the continued fraction expansion are positive, the polynomial P is positive.

5. Consider the polynomial P=s 4 +3s 2 +2. The given polynomial P is Hurwitz.

a) True

b) False

Answer: a

Explanation: P=s 4 +3s 2 +2 => P ‘ =4s 3 +6s

After doing the continued fraction expansion, we get all the quotients as positive. So, the polynomial P is Hurwitz.

6. When s is real, the driving point impedance function is _________ function and the driving point admittance function is _________ function.

a) real, complex

b) real, real

c) complex, real

d) complex, complex

Answer: b

Explanation: When s is real, the driving point impedance function is real function and the driving point admittance function is real function because the quotients of the polynomials P and Q are real. When Z is determined from the impedances of the individual branches, the quotients are obtained by adding together, multiplying or dividing the branch parameters which are real.

7. The poles and zeros of driving point impedance function and driving point admittance function lie on?

a) left half of s-plane only

b) right half of s-plane only

c) left half of s-plane or on imaginary axis

d) right half of s-plane or on imaginary axis

Answer: c

Explanation: The poles and zeros of driving point impedance function and driving point admittance function lie on left half of s-plane or on imaginary axis of the s-plane.

8. For real roots of s k , all the quotients of s in s 2 +ω k 2 of the polynomial P are __________

a) negative

b) non-negative

c) positive

d) non-positive

Answer: b

Explanation: For real roots of s k , all the quotients of s in s 2 +ω k 2 of the polynomial P are non-negative. So by multiplying all factors in P we find that all quotients are positive.

9. The real parts of the driving point function Z and Y are?

a) positive and zero

b) positive

c) zero

d) positive or zero

Answer: d

Explanation: The real parts of the driving point impedance function Z and driving point admittance function Y are positive or zero.

10. For the complex zeros to appear in conjugate pairs the poles of the network function are ____ and zeros of the network function are ____________

a) complex, complex

b) complex, real

c) real, real

d) real, complex

Answer: c

Explanation: P and Q are real when s is real. So the poles of the network function are real and zeros of the network function are real, the complex zeros to appear in conjugate pairs.

This set of Network Theory Questions and Answers for Campus interviews focuses on “Frequency Response of Reactive One-Ports”.

1. Based on the location of zeros and poles, a reactive one-port can have ____________ types of frequency response.

a) 1

b) 2

c) 3

d) 4

Answer: d

Explanation: A reactive one-port can have four types of frequency response based on the location of zeros and poles.

frequency response with two external poles

frequency response with two external zeros

frequency response with an external zero ω = 0 and an external poles at ω = ∞

frequency response with an external zero ω = ∞ and an external poles at ω = 0.

2. A driving point impedance with poles at ω = 0, ω = ∞ must have ___________ term in the denominator polynomial.

a) s

b) s+1

c) s+2

d) s+3

Answer: a

Explanation: As there is a pole at ω = 0, =s. Poles are written in the denominator of Z. So there will be s term in the denominator polynomial in a driving point impedance function Z.

3. A driving point impedance with poles at ω = 0, ω = ∞ must have excess ___________ term in the numerator polynomial.

a) s 1 +ω n 1

b) s 1 +ω n 2

c) s 2 +ω n 2

d) s 2 +ω n 1

Answer: c

Explanation: The driving point impedance of the one-port is infinite, and it will not pass either direct current or alternating current of an infinitely high frequency.

4. A driving point impedance with zeros at ω = 0, ω = ∞ must have ___________ term in the numerator polynomial.

a) s+3

b) s+2

c) s+1

d) s

Answer: d

Explanation: As there is a zero at ω = 0, =s. Zeros are written in the numerator of Z. So there will be s term in the numerator polynomial in a driving point impedance function Z.

5. A driving point impedance with zeros at ω = 0, ω = ∞ must have an excess ___________ term in the denominator polynomial.

a) s 2 +ω n 1

b) s 2 +ω n 2

c) s 1 +ω n 2

d) s 1 +ω n 1

Answer: b

Explanation: The driving point impedance of the one-port is zero, and it will pass both direct current and an alternating current of an infinitely high frequency.

6. A driving point impedance with zero at ω = 0 and pole at ω = ∞ must have ___________ term in the numerator polynomial.

a) s+1

b) s

c) s+3

d) s+2

Answer: b

Explanation: As ω = 0, =s. The numerator of Z contains poles and denominator contains zeros. So there will be s term in the numerator polynomial.

7. A driving point impedance with zero at ω = 0 and pole at ω = ∞ must have ___________ term in the numerator polynomial.

a) s 1 +ω n 1

b) s 2 +ω n 1

c) s 1 +ω n 2

d) s 2 +ω n 2

Answer: d

Explanation: If a pole is at ω = ∞, there will be an equal number of s 2 +ω n 2 type terms in the numerator polynomial and the denominator polynomial.

8. A driving point impedance with zero at ω = 0 and pole at ω = ∞ must have ___________ term in the denominator polynomial.

a) s 2 +ω n 2

b) s 1 +ω n 1

c) s 2 +ω n 1

d) s 1 +ω n 2

Answer: a

Explanation: If there is a zero at ω = 0 and pole at ω = ∞, the one-port will pass direct current and block the alternating current of an infinitely high frequency.

9. A driving point impedance with pole at ω = 0 and zero at ω = ∞ must have ___________ term in the denominator polynomial.

a) s

b) s+3

c) s+1

d) s+2

Answer: a

Explanation: s-jω = ) = s. As pole is at ω = 0, there will be s term in the denominator polynomial.

10. A driving point impedance with pole at ω = 0 and zero at ω = ∞ must have ____________ term in the numerator and denominator.

a) s 1 +ω n 2

b) s 2 +ω n 2

c) s 1 +ω n 1

d) s 2 +ω n 1

Answer: b

Explanation: If a pole at ω = 0 and zero at ω = ∞, the one-port will block the direct current and pass the alternating current of an infinitely high frequency.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Synthesis of Reactive One-Ports by Foster’s Method”.

1. The driving point impedance of a one-port reactive network is given by Z=5(s 2 +4)(s 2 +25)/s(s 2 +16). After taking the partial fractions, find the coefficient of 1/s.

a) 25/4

b) 50/4

c) 100/4

d) 125/4

Answer: d

Explanation: Since there is an extra term in the numerator compared to the denominator and also an s term in the denominator, the two poles exists at 0 and infinity. Therefore the network consists of first element and last element.

By taking the partial fraction expansion of Z, we obtain A=5(s 2 +4)(s 2 +25)/((s 2 +16))|s=0

=/16=125/4.

2. The driving point impedance of a one-port reactive network is given by Z=5(s 2 +4)(s 2 +25)/s(s 2 +16). After taking the partial fractions, find the coefficient of .

a) 135/4

b) 145/4

c) 155/4

d) 165/4

Answer: a

Explanation: On taking the partial fraction expansion,

B=5 (s 2 +4)(s 2 +25)/s|s=-j4

= 135/8.

3. The driving point impedance of a one-port reactive network is given by Z=5(s 2 +4)(s 2 +25)/s(s 2 +16). After taking the partial fractions, what is the value of H from Z ?

a) 3

b) 4

c) 5

d) 6

Answer: c

Explanation: By inspection, the value of H is 5.

So H = 5.

4. The driving point impedance of a one-port reactive network is given by Z=5(s 2 +4)(s 2 +25)/s(s 2 +16). After taking the partial fractions, what is the value of C 0 ?

a) 1/125

b) 4/125

c) 2/125

d) 3/125

Answer: b

Explanation: The coefficient of 1/s is 125/4.

And C 0 =1/P 0

= 1/=4/125 Farad.

5. The driving point impedance of a one-port reactive network is given by Z=5(s 2 +4)(s 2 +25)/s(s 2 +16). After taking the partial fractions, what is the value of L ∞ ?

a) 2

b) 3

c) 4

d) 5

Answer: d

Explanation: We know L ∞ = H

= 5 H.

6. The driving point impedance of a one-port reactive network is given by Z=5(s 2 +4)(s 2 +25)/s(s 2 +16). After taking the partial fractions, what is the value of C 2 ?

a) 4/270

b) 8/270

c) 12/270

d) 16/270

Answer: b

Explanation: We know C 2 = 1/2P 2

= 8/=8/270 F.

7. The driving point impedance of a one-port reactive network is given by Z=5(s 2 +4)(s 2 +25)/s(s 2 +16). After taking the partial fractions, what is the value of L 2 ?

a) 135/60

b) 135/62

c) 135/64

d) 135/66

Answer: c

Explanation: We know L 2 = 2P 2 /ω n 2

= /=135/64 H.

8. For performing second Foster form, after splitting the Z in The driving point impedance of a one-port reactive network is given by Z=5(s 2 +4)(s 2 +25)/s(s 2 +16), what is the coefficient of s/((s 2 +4)) is?

a) 1/35

b) 2/35

c) 3/35

d) 4/35

Answer: b

Explanation: The value of A is (s(s 2 +16)))/(s 2 +25) at s=-j2

On solving we get the value of A as 2/35.

So the coefficient of s/((s 2 +4)) is 2/35.

9. For performing second Foster form, after splitting the Z in The driving point impedance of a one-port reactive network is given by Z=5(s 2 +4)(s 2 +25)/s(s 2 +16), what is the coefficient of s/((s 2 +4)) is?

a) 4/35

b) 3/35

c) 2/35

d) 1/35

Answer: c

Explanation: The coefficient of s/((s 2 +4)) is B=((s(s 2 +16)))/(s 2 +4) at s=-j5. On solving we get B = 2/35.

10. Determine the value of L 1 by performing second foster form for Z=5(s 2 +4)(s 2 +25)/s(s 2 +16).

a) 35/4

b) 35/3

c) 35/2

d) 35

Answer: a

Explanation: P 1 = 2/35.

We know L 1 = 1/2P 1

= 35/4 H.

This set of Network Theory Questions and Answers for Aptitude test focuses on “Synthesis of Reactive One-Ports by Cauer Method”.

1. The driving point impedance of an LC network is given by Z=(2s 5 +12s 3 +16s)/(s 4 +4s 2 +3). By taking the continued fraction expansion using first Cauer form, find the value of L 1 .

a) s

b) 2s

c) 3s

d) 4s

Answer: b

Explanation: The first Cauer form of the network is obtained by taking the continued fraction expansion of given Z. And we get he first quotient as 2s.

So, L 1 = 2s.

2. Find the first reminder obtained by taking the continued fraction expansion in the driving point impedance of an LC network is given by Z=(2s 5 +12s 3 +16s)/(s 4 +4s 2 +3). By taking the continued fraction expansion using first Cauer form.

a) 4s 3 +10s

b) 12s 3 +10s

c) 4s 3 +16s

d) 12s 3 +16s

Answer: a

Explanation: On taking the continued fraction expansion, the first reminder obtained is 4s 3 +10s.

3. The driving point impedance of an LC network is given by Z=(2s 5 +12s 3 +16s)/(s 4 +4s 2 +3). By taking the continued fraction expansion using first Cauer form, find the value of C 2 .

a) 1

b) 1/2

c) 1/3

d) 1/4

Answer: d

Explanation: The second quotient obtained on taking the continued fraction expansion is s/4 and this is the value of sC 2 . So the value of C 2 = 1/4.

4. The driving point impedance of an LC network is given by Z=(2s 5 +12s 3 +16s)/(s 4 +4s 2 +3). By taking the continued fraction expansion using first Cauer form, find the value of L 3 .

a) 8

b) 8/3

c) 8/5

d) 8/7

Answer: b

Explanation: By taking the continued fraction expansion, the third quotient is 8s/3.

sL 3 = 8s/3.

So L 3 = 8/3H.

5. The driving point impedance of an LC network is given by Z=(2s 5 +12s 3 +16s)/(s 4 +4s 2 +3). By taking the continued fraction expansion using first Cauer form, find the value of C 4 .

a) 1/2

b) 1/4

c) 3/4

d) 1

Answer: c

Explanation: We get the fourth quotient as 3s/4.

So sC 4 = 3s/4.

C 4 = 3/4F.

6. The driving point impedance of an LC network is given by Z=(2s 5 +12s 3 +16s)/(s 4 +4s 2 +3). By taking the continued fraction expansion using first Cauer form, find the value of L 5 .

a) 2

b) 2/5

c) 2/7

d) 2/3

Answer: d

Explanation: On taking the continued fraction expansion fifth quotient is 2s/3.

sL 5 = 2s/3

So L 5 = 2/3H.

7. The driving point impedance of an LC network is given by Z=(s 4 +4s 2 +3)/(s 3 +2s). By taking the continued fraction expansion using second Cauer form, find the value of C 1 .

a) 2/3

b) 2/2

c) 1/2

d) 4/2

Answer: a

Explanation: To obtain the second Cauer form, we have to arrange the numerator and the denominator of given Z in ascending powers of s before starting the continued fraction expansion.

By taking the continued fraction expansion we get the first quotient as 3/2s.

So 1/sC 1 = 3/2s

C 1 = 2/3F.

8. The driving point impedance of an LC network is given by Z=(s 4 +4s 2 +3)/(s 3 +2s). By taking the continued fraction expansion using second Cauer form, find the value of L 2 .

a) 1/5

b) 2/5

c) 3/5

d) 5/4

Answer: d

Explanation: On taking the continued fraction expansion the second quotient is 4/5s.

1/sL 2 = 4/5s

So L 2 = 5/4H.

9. The driving point impedance of an LC network is given by Z=(s 4 +4s 2 +3)/(s 3 +2s). By taking the continued fraction expansion using second Cauer form, find the value of C 3 .

a) 25/s

b) 2/25s

c) 25/3s

d) 25/4s

Answer: b

Explanation: The third quotient is 25/2s.

1/sC 3 = 25/2s.

C 3 = 2/25F.

10. The driving point impedance of an LC network is given by Z=(s 4 +4s 2 +3)/(s 3 +2s). By taking the continued fraction expansion using second Cauer form, find the value of L 4 .

a) 5

b) 2/5

c) 3/5

d) 4/5

Answer: a

Explanation: We obtain the fourth quotient as 1/5s.

1/sL 4 = 1/5s

L 4 = 5H.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Synthesis of R-L Network by the Foster Method”.

1. Consider a function Z=5/. Find the value of R 1 after performing the first form of Foster method.

a) 1/3

b) 2/3

c) 3/3

d) 4/3

Answer: d

Explanation: After splitting the given function into partial according to the properties of first form of Foster method, we get

Z=4/3+)+2s/.

So, R 1 = 4/3Ω.

2. Consider a function Z=5/. Find the value of R 1 .

a) 4/3

b) 5/3

c) 3/5

d) 3/4

Answer: b

Explanation: On taking the partial fractions, we get P 1 as 5/3 and we know R 1 =P 1 . So the value of R 1 is 5/3Ω.

R 1 = 5/3Ω.

3. Consider a function Z=5/. Find the value of L 1 after performing the first form of Foster method.

a) 5/9

b) 9/5

c) 4/9

d) 9/4

Answer: a

Explanation: We know R 1 =(L 1 ) and as R 1 is 5/3Ω. So the value of L 1 is 5/9 H.

L 1 = 5/9 H.

4. Consider a function Z=5/. Find the value of R 2 after performing the first form of Foster method.

a) 1

b) 2

c) 3

d) 4

Answer: b

Explanation: We obtain P 2 as 2.

R 2 =P 2 .

So the value of R 2 is 2Ω.

R 2 = 2Ω.

5. Consider a function Z=5/. Find the value of L 2 after performing the first form of Foster method.

a) 4/5

b) 3/5

c) 2/5

d) 1/5

Answer: c

Explanation: As R 2 =(L 2 )

So the value of L 2 is 2/5 H.

L 2 = 2/5 H.

6. Consider the admittance function, Y=((2s 2 +16s+30))/(s 2 +6s+8). Determine the value of L 1 after performing the second form of Foster method.

a) 1/3

b) 2/3

c) 3/3

d) 4/3

Answer: a

Explanation: After splitting the given function into partial according to the properties of first form of Foster method, we get Y=2+3/+1/.

We know L 1 =1/P 1 and as P 1 = 3, the value of L 1 is 1/3H.

L 1 = 1/3H.

7. Consider the admittance function, Y=((2s 2 +16s+30))/(s 2 +6s+8). Determine the value of R 1 after performing the second form of Foster method.

a) 4/3

b) 3/3

c) 2/3

d) 1/3

Answer: c

Explanation: R 1 = 2/P 1 and as P 1 is 3, the value of R 1 is 2/3Ω.

R 1 = 2/3Ω.

8. Consider the admittance function, Y=((2s 2 +16s+30))/(s 2 +6s+8). Determine the value of R 2 after performing the second form of Foster method.

a) 1

b) 2

c) 3

d) 4

Answer: d

Explanation: On taking the partial fractions we get P 2 as 1. And we know R 2 = 4/P 2 . So the value of R 2 is 4Ω.

R 2 = 4Ω.

9. Consider the admittance function, Y=((2s 2 +16s+30))/(s 2 +6s+8). Determine the value of L 2 after performing the second form of Foster method.

a) 4

b) 1

c) 2

d) 3

Answer: b

Explanation: We got P 2 = 1

And L 2 =1/P 2

So the value of L 2 is 1H.

L 2 = 1H.

10. Consider the admittance function, Y=((2s 2 +16s+30))/(s 2 +6s+8). Determine the value of R ∞ after performing the second form of Foster method.

a) 3

b) 1

c) 2

d) 4

Answer: c

Explanation: On performing partial fractions we get the value of R ∞ is 2Ω.

R ∞ = 2Ω.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Synthesis of R-L Network by Cauer Method”.

1. Consider the impedance function, Z=)/). Find the value of R 1 after converting into first Cauer form.

a) 1

b) 2

c) 3

d) 4

Answer: a

Explanation: To find out the first Cauer form, we have to take the continued fraction expansion of Z .

On solving, we get the first quotient as 1. So the value of R 1 as 1Ω.

2. Consider the impedance function, Z=)/). Find the value of L 2 after converting into first Cauer form.

a) 1

b) 1/2

c) 1/4

d) 1/8

Answer: c

Explanation: On taking the continued fraction expansion of Z , the second quotient is s/4. So the value of L 2 is 1/4 H.

L 2 = 1/4 H.

3. Consider the impedance function, Z=)/). Find the value of R 2 after converting into first Cauer form.

a) 1/4

b) 2/4

c) 3/4

d) 4/4

Answer: c

Explanation: We get the third quotient on performing continued fraction expansion of Z as 4/3 and is the value of 1/R 2 . So the value of R 2 is 3/4Ω .

R 2 = 3/4Ω .

4. Consider the impedance function, Z=)/). Find the value of L 3 after converting into first Cauer form.

a) 4/3

b) 3/4

c) 4/5

d) 5/4

Answer: b

Explanation: The fourth quotient obtained is 3s/4. And this is the value of sL 3 . So the value of L 3 is 3/4 H.

L 3 = 3/4 H.

5. Consider the impedance function, Z=)/). Find the value of R 3 after converting into first Cauer form.

a) 4

b) 3

c) 2

d) 1

Answer: b

Explanation: The value of 1/R 3 obtained by continued fraction expansion 1/3. So the value of R 3 is 3Ω.

6. Consider the impedance function, Z=(2s 2 +8s+6)/(s 2 +8s+12). Find the value of R 1 after converting into second Cauer form.

a) 1

b) 3/4

c) 1/2

d) 1/4

Answer: c

Explanation: To find out the second Cauer form, we have to write the impedance function in ascending powers and by taking the continued fraction expansion of Z .

On solving, the first quotient obtained is 1/2. So the value of R 1 as 1/2 Ω.

7. Consider the impedance function, Z=(2s 2 +8s+6)/(s 2 +8s+12). Find the value of L 1 after converting into second Cauer form.

a) 1/3

b) 2/3

c) 3/3

d) 4/3

Answer: a

Explanation: The second quotient obtained is 3/s and this is the value of 1/sL 1 . So the value of L 1 is 1/3 H.

L 1 = 1/3 H.

8. Consider the impedance function, Z=(2s 2 +8s+6)/(s 2 +8s+12). Find the value of R 2 after converting into second Cauer form.

a) 6/7

b) 7/6

c) 7/8

d) 8/7

Answer: d

Explanation: On taking the continued fraction expansion, we get third quotient as 8/7. So the value of R 2 is 8/7Ω.

R 2 = 8/7Ω.

9. Consider the impedance function, Z=(2s 2 +8s+6)/(s 2 +8s+12). Find the value of L 2 after converting into second Cauer form.

a) 5/50

b) 10

c) 5/49

d) 49/5

Answer: c

Explanation: We obtain the fourth quotient i.e., 1/sL 2 as 49/5s. So the value of L 2 is 5/49 H.

L 2 = 5/49H.

10. Consider the impedance function, Z=(2s 2 +8s+6)/(s 2 +8s+12). Find the value of R 3 after converting into second Cauer form.

a) 1/5

b) 14/5

c) 5/14

d) 5

Answer: b

Explanation: The value of R 3 is 14/5Ω as the fifth quotient obtained is 5/14.

R 3 = 5/14Ω.

This set of Network Theory Assessment Questions and Answers focuses on “Synthesis of R-C Network by Foster Method”.

1. Consider the impedance function Z=3/. Find the value of R 1 after realizing by first Foster method.

a) 9/2

b) 2/9

c) 9

d) 1/9

Answer: a

Explanation: The first Foster form can be realized by taking the partial fraction of Z .

On solving, we get

Z =3+)+).

So the value of R 1 is 9/2Ω.

2. Consider the impedance function Z=3/. Find the value of C 1 after realizing by first Foster method.

a) 1/9

b) 9

c) 2/9

d) 9/2

Answer: c

Explanation: After taking the partial fractions, as P 1 = 1/C 1 , we get the 1/C 1 value as 9/2. So the value of C 1 is 2/9 F.

C 1 = 2/9 F.

3. Consider the impedance function Z=3/. Find the value of C 2 after realizing by first Foster method.

a) 1/3

b) 3

c) 3/2

d) 2/3

Answer: d

Explanation: We know P 2 = 1/C 2 . The value of P 2 is 3/2. So the value of C 2 is 2/3 F.

C 2 = 2/3 F.

4. Consider the impedance function Z=3/. Find the value of R 2 after realizing by first Foster method.

a) 1

b) 1/2

c) 1/4

d) 1/8

Answer: b

Explanation: We got 1/R 2 C 2 = 3. And the value of C 2 is 2/3 F. So the value of R 2 is 1/2 Ω.

R 2 = 1/2 Ω.

5. Consider the impedance function Z=3/. Find the value of R ∞ after realizing by first Foster method.

a) 1

b) 2

c) 3

d) 4

Answer: c

Explanation: On taking partial fraction of Z, we get the value of R ∞ is 3Ω.

R ∞ = 3Ω.

6. Consider the impedance function Y=(s 2 +4s+3)/(3s 2 +18s+24). Find the value of R 0 after realizing by second Foster method.

a) 4

b) 8

c) 12

d) 16

Answer: b

Explanation: The second Foster Form can be realized by taking the reciprocal of the impedance function and by taking partial fractions. So we get 1/R 0 as 1/8. So the value of R 0 is 8Ω.

R 0 = 8Ω.

7. Consider the impedance function Y=(s 2 +4s+3)/(3s 2 +18s+24). Find the value of R 1 after realizing by second Foster method.

a) 16

b) 12

c) 8

d) 4

Answer: b

Explanation: On taking the partial fractions we get P 1 as 1/12. And we know that R 1 =1/P 1 . So the value of R 1 is 12Ω.

R 1 = 12Ω.

8. Consider the impedance function Y=(s 2 +4s+3)/(3s 2 +18s+24). Find the value of C 1 after realizing by second Foster method.

a) 1/24

b) 1/12

c) 1/6

d) 1/3

Answer: a

Explanation: The value of C 1 is 1/(R 1 )). As R 1 is 12Ω, the value of C 1 is 1/24 F.

C 1 = 1/24 F.

9. Consider the impedance function Y=(s 2 +4s+3)/(3s 2 +18s+24). Find the value of R 2 after realizing by second Foster method.

a) 5

b) 6

c) 7

d) 8

Answer: d

Explanation: We get the value of P 2 on taking the partial fractions as 1/8. And we know R 2 = 1/P 2 So the value of R 2 is 8Ω.

So, R 2 = 8Ω.

10. Consider the impedance function Y=(s 2 +4s+3)/(3s 2 +18s+24). Find the value of C 2 after realizing by second Foster method.

a) 1/16

b) 1/8

c) 1/32

d) 1/64

Answer: c

Explanation: We obtained R 2 as 8Ω and we know C 2 = 1/(R 2 )). So the value of C 2 is 1/32 F.

C 2 = 1/32 F.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Synthesis of R-C Network by Cauer Method”.

1. Consider the impedance functionZ=(s 2 +6s+8)/(s 2 +3s). Find the value of R 1 after performing the first Cauer form.

a) 1

b) 2

c) 3

d) 4

Answer: a

Explanation: To find the first Cauer form, we take the continued fraction expansion by the divide, invert, divide procedure.

On performing this we get the first quotient is 1Ω.

So, R 1 = 1Ω.

2. Consider the impedance functionZ=(s 2 +6s+8)/(s 2 +3s). Find the first reminder obtained by taking the continued fraction expansion after performing the first Cauer form.

a) s + 8

b) 2s + 8

c) 3s + 8

d) 4s + 8

Answer: c

Explanation: The continued fraction expansion is done by the divide, invert, divide procedure. So the first reminder obtained is 3s + 8.

3. Consider the impedance functionZ=(s 2 +6s+8)/(s 2 +3s). Find the value of R 2 after performing the first Cauer form.

a) 4

b) 3

c) 6

d) 9

Answer: d

Explanation: On performing the continued fraction expansion we get the third quotient as 9.

So the value of R 2 is 9Ω.

R 2 = 9Ω.

4. Consider the impedance functionZ=(s 2 +6s+8)/(s 2 +3s). Find the value of C 1 after performing the first Cauer form.

a) 1/4

b) 1/3

c) 1/2

d) 1

Answer: b

Explanation: The second quotient of the continued fraction expansion is s/3.

So the value of C 1 is 1/3 F.

C 1 = 1/3 F.

5. Consider the impedance functionZ=(s 2 +6s+8)/(s 2 +3s). Find the value of C 2 after performing the first Cauer form.

a) 1/6

b) 1/12

c) 1/24

d) 1/48

Answer: c

Explanation: We obtain the fourth quotient on performing continued fraction expansion as s/24.

So the value of C 2 is 1/24 F.

C 2 = 1/24 F.

6. Consider the impedance function Z=(s 2 +6s+8)/(s 2 +3s). Find the value of C 1 after performing the second Cauer form.

a) 1/2

b) 3/8

c) 1/4

d) 1/8

Answer: b

Explanation: The second Cauer network can be obtained by arranging the numerator and denominator polynomials of Z in ascending powers of s. After performing the continued fraction expansion, we get

C 1 = 3/8 F.

7. Consider the impedance function Z=(s 2 +6s+8)/(s 2 +3s). Find the first reminder obtained by taking the continued fraction expansion after performing the second Cauer form.

a) 10s/3+s 2

b) s/3+s 2

c) 10s/3+3s 2

d) s/3+3s 2

Answer: a

Explanation: On performing the continued fraction expansion after arranging the numerator and denominator polynomials of Z in ascending powers of s, the first reminder is 10s/3+s 2 .

8. Consider the impedance function Z=(s 2 +6s+8)/(s 2 +3s). Find the value of R 1 after performing the second Cauer form.

a) 9/10

b) 10/9

c) 8/9

d) 9/8

Answer: b

Explanation: The second quotient on performing continued fraction expansion is 9/10. This is the value of 1/R 1 . So the value of R 1 is 10/9Ω.

R 1 = 10/9Ω.

9. Consider the impedance function Z=(s 2 +6s+8)/(s 2 +3s). Find the value of C 2 after performing the second Cauer form.

a) 3

b) 3/10

c) 3/100

d) 3/1000

Answer: c

Explanation: On performing continued fraction expansion, the third quotient is 100/3s. So the value of C 2 is 3/100 F.

C 2 = 3/100 F.

10. Consider the impedance function Z=(s 2 +6s+8)/(s 2 +3s). Find the value of R 1 after performing the second Cauer form.

a) 10

b) 1

c) 100

d) 1000

Answer: a

Explanation: The final quotient is 1/10. So the value of R 1 is 1/1/10.So the value of R 1 is 10Ω.

R 1 = 10Ω.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Advanced Problems on Network Theory – 1”.

1. Branch current and loop current relation is expressed in matrix form shown below, where Ij represents branch current and Ik represents loop current.

[I 1 ; I 2 ; I 3 ; I 4 ; I 5 ; I 6 ; I 7 ; I 8 ] = [0 0 1 0; -1 -1 -1 0; 0 1 0 0; 1 0 0 0; 0 0 -1 -1; 1 1 0 -1; 1 0 0 0; 0 0 0 1] [I 1 ; I 2 ; I 3 ; I 4 ] The rank of the incidence matrix is?

a) 4

b) 5

c) 6

d) 8

Answer: a

Explanation: Number of branches b = 8

Number of links l = 4

Number of twigs t = b – l = 4

Rank of matrix = n – 1 = t = 4.

2. A capacitor, used for power factor correction in a single phase circuit decreases which of the following?

a) Power factor

b) Line current

c) Both Line current and Power factor

d) Neither Line current nor Power factor

Answer: b

Explanation: We know that a capacitor is used to increase the Power factor. However, with decrease in line current the power factor is increased. Hence line current decreases.

3. D is the distance between the plates of a parallel plate capacitor. The dielectric constants are ∈ 1 and ∈ 2 respectively. The total capacitance is proportional to ____________

a) \

∈ 1 – ∈ 2

c) \

∈ 1 ∈ 2

Answer: a

Explanation: The combination is equal to two capacitors in series.

So, C = \(\frac{\Big[∈_0 ∈_1 \leftMissing or unrecognized delimiter for \right\Big]\Big[∈_0 ∈_2 \leftMissing or unrecognized delimiter for \right\Big]}{∈_0 ∈_1 \frac{A}{0.5d} + ∈_0 ∈_1 \frac{A}{0.5d}}\)

Hence, C is proportional to \(\frac{∈_1 ∈_2}{∈_1+∈_2}\).

4. A two branch circuit has a coil of resistance R 1 , inductance L 1 in one branch and capacitance C 2 in the second branch. If R is increased, the dynamic resistance is going to ___________

a) Increase

b) Decrease

c) Remains constant

d) May increase or decrease

Answer: b

Explanation: We know that,

Dynamic resistance = \(\frac{L_1}{R_1 C_2}\)

So, if R 1 is increased, keeping Inductance and Capacitance same, so The Dynamic resistance will decrease, as the denomination is increasing.

5. A 1 μF capacitor is connected to 12 V batteries. The energy stored in the capacitor is _____________

a) 12 x 10 -6 J

b) 24 x 10 -6 J

c) 60 x 10 -6 J

d) 72 x 10 -6 J

Answer: d

Explanation: We know that,

Energy, E = 0.5 CV 2

= 0.5 X 1 X 10 -6 X 144

= 72 x 10 -6 J.

6. For the two circuits shown below, the relation between I A and I B is ________

network-theory-questions-answers-advanced-problems-network-theory-1-q6

a) I B = I A + 6

b) I B = I A + 2

c) I B = 1.5I A

d) I B = I A

Answer: c

Explanation: In the circuit of figure (I B ), transforming 3A source into 18 V source, all sources are 1.5 times of that in circuit (I A ). Hence, I B = 1.5I A .

network-theory-questions-answers-advanced-problems-network-theory-1-q6a

7. For the circuit given below, the current I in the circuit is ________

network-theory-questions-answers-advanced-problems-network-theory-1-q7

a) –j1 A

b) J1 A

c) Zero

d) 20 A

Answer: a

Explanation: X EQ = sL + \(\frac{R×1/sC}{R+1/sC} = sL + \frac{R}{1+sRC}\)

I O = \(\frac{V}{X_{EQ}}\)

∴ I = \(\frac{X_C}{X_C+R}\) I O

= \(\frac{1/sC}{\frac{1}{sC}+R} × \frac{V}{\frac{sL+R}{}}\)

= \(\frac{1}{1+sRC} × \frac{V}{\frac{sL+R}{}}\)

= \(\frac{V}{sL+R}\)

= \(\frac{V}{j×10^3×20×10^{-3}

}\)

= \(\frac{V}{20j

+1}\)

= \(\frac{V}{20j-1+1} = \frac{20}{20j}\) = -j1 A.

8. An AC source of RMS voltage 20 V with internal impedance Z S = Ω feeds a load of impedance Z L = Ω in the circuit given below. The reactive power is _________

network-theory-questions-answers-advanced-problems-network-theory-1-q8

a) 8 VAR

b) 16 VAR

c) 28 VAR

d) 32 VAR

Answer: b

Explanation: Current I = \

= 2∠-arc tan⁡

Power consumed by load = |I| 2 Z L

= 4

= 28 + j16

∴ The reactive power = 16 VAR.

9. In the circuit given below, R I = 1 MΩ, R O = 10 Ω, A = 10 6 and V I = 1μV. Then the output voltage, input impedance and output impedance respectively are _________

network-theory-questions-answers-advanced-problems-network-theory-1-q9

a) 1 V, ∞ and 10 Ω

b) 1 V, 0 and 10 Ω

c) 1V, 0 and ∞

d) 10 V, ∞ and 10 Ω

Answer: a

Explanation: V O = AV I = 10 6 × 10 -6 = 1 V

V 1 = Z 11 I 1 + Z 12 I 2

V 2 = Z 21 I 1 + Z 22 I 2

Here, I 1 = 0

Z 11 = \(\frac{V_1}{I_1} = \frac{V_O}{0}\) = ∞

Z 22 = \(\frac{V_2}{I_2} = \frac{AV_I}{I_2}\)

Or, Z 22 = \(\frac{1}{I_2}\) = R O = 10 Ω.

10. If operator ‘a’ = 1 ∠120°. Then is equal to ____________

a) \ \ \ \(\sqrt{3}\)∠60°

Answer: b

Explanation: Given that, ‘a’ = 1 ∠120°

So, 1 – a = 1 – 1∠120°

= 1 + 0.5 – j 0.866

= 1.5 – j 0.866

= 3∠-30°.

11. For making a capacitor, the dielectric should have __________

a) High relative permittivity

b) Low relative permittivity

c) Relative permittivity = 1

d) Relative permittivity neither too high nor too low

Answer: a

Explanation: Relative permittivity is for ideal dielectric which is air. Achieving such a precise dielectric is very difficult.

Low relative permittivity will lead to low value of capacitance.

High relative permittivity will lead to a higher value of capacitance.

12. In the circuit shown below, the voltage V will be __________

network-theory-questions-answers-advanced-problems-network-theory-1-q12

a) – 3V

b) Zero

c) 3 V

d) 5 V

Answer: a

Explanation: By applying KVL, I = 1 A

V AB – 2 × 1 + 5 = 0

Or, V AB = -3 V.

13. If A = 3 + j1, then A 4 is equal to __________

a) 3.16 ∠18.4°

b) 100 ∠73.72°

c) 100 ∠18.4°

d) 3.16 ∠73.22°

Answer: b

Explanation: Given A = 3 + j1

So, 3 + j1 = 10∠18.43°

Or, 3 + j1 = 4 ∠4 X 18.43°

= 100∠73.72°.

14. In the figures given below, Value of R A , R B and R C are 20 Ω, 10 Ω and 10 Ω respectively. The resistances R 1 , R 2 and R 3 in Ω are ________

network-theory-questions-answers-advanced-problems-network-theory-1-q14

a) 2.5, 5 and 5

b) 5, 2.5 and 5

c) 5, 5 and 2.5

d) 2.5, 5 and 2.5

Answer: a

Explanation: R 1 = \(\frac{R_B R_C}{R_A+R_B+R_C} = \frac{100}{40}\) = 2.5 Ω

R 2 = \(\frac{R_A R_C}{R_A+R_B+R_C} = \frac{200}{40}\) = 5 Ω

R 3 = \(\frac{R_B R_A}{R_A+R_B+R_C} = \frac{200}{40}\) = 5 Ω.

15. The resistance of a thermistor decreases with increases in __________

a) temperature

b) circuit

c) light control

d) sensors

Answer: b

Explanation: The resistance of a thermistor decreases with increases in temperature. Hence, it is used to monitor hot spot temperature of electric machines.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Advanced Problems on Network Theory – 2”.

1. In the circuit given below, the value of R is _________

network-theory-questions-answers-advanced-problems-network-theory-2-q1

a) 10 Ω

b) 18 Ω

c) 24 Ω

d) 12 Ω

Answer: d

Explanation: By KCL,

∴ \(\frac{V_P-40}{1} + \frac{V_P-100}{14} + \frac{V_P}{2}\) = 0

Or, 22 V P = 660

∴ V P = 30 V

Potential difference between node x and y = 60 V

∴ -I – 5 + \(\frac{40-30}{I}\) = 0

Or, I = 5 A

∴ R = \(\frac{60}{5}\)= 12 Ω.

2. In the circuit given below, the resistance between terminals A and B is 7Ω, between terminals B and C is 12Ω and between terminals C and A is 10Ω. The remaining one terminal in each case is assumed to be open. Then the value of R A and R B are _________

network-theory-questions-answers-advanced-problems-network-theory-2-q2

a) R A = 9 Ω and R B = 7 Ω

b) R A = 2.5 Ω and R B = 4.5 Ω

c) R A = 3 Ω and R B = 3 Ω

d) R A = 5 Ω and R B = 1 Ω

Answer: b

Explanation: Given R A + R B = 7 with C open

R B + R C = 12 with A as open

R A + R C = 10 with B open

Then, R A + R B + R C = \(\frac{29}{2}\) = 14.5

Hence, R A = 2.5 Ω, R B = 4.5 Ω and R C = 7.5 Ω.

3. Currents I 1 , I 2 and I 3 meet at a junction in a circuit. All currents are marked as entering the node. If I 1 = -6sin mA and I 2 = 8 cos mA, the I 3 is ________

a) 10 cos mA

b) 14 cos mA

c) -14 sin mA

d) -10 cos mA

Answer: d

Explanation: Applying KCL, we get, I 1 + I 2 + I 3 = 0

∴ -6 sin + 8 cos + I 3 = 0

∴ I 3 = 6 sin – 8 cos

= 10[sin .sin – cos cos ]

=-10[cos cos – sin sin ]

= -10 cos

[As, cos = cosA.cosB – sinA.sinB].

4. Viewed from the terminals A and B the circuit given below can be reduced to an equivalent circuit with a single voltage source in series with a resistor with ________

network-theory-questions-answers-advanced-problems-network-theory-2-q4

a) 5 V source in series with 10 Ω resistor

b) 1 V source in series with 2.4 Ω resistor

c) 15 V source in series with 2.4 Ω resistor

d) 1 V source in series with 10 Ω resistor

Answer: b

Explanation: Applying Thevenin’s Theorem R EQ = 6 || 4

= \(\frac{6 × 4}{6 + 4}\)

= \(\frac{24}{10}\) = 2.4 Ω

V AB = 10 – 6 ×

= 1 V.

5. In the circuit given below, the voltage across the 2Ω resistor is ________

network-theory-questions-answers-advanced-problems-network-theory-2-q5

a) 3.41 V

b) -3.41 V

c) 3.8 V

d) -3.8 V

Answer: b

Explanation: Applying KCL to node A, \(\frac{V_A-10}{10} + \frac{V_A}{20} + \frac{V_A}{7}\) = 0

Or, V A = 1

Or, V A = 3.41 V

The voltage across the 2 Ω resistor due to 10 V source is V 2 = \(\frac{V_A}{7} × 2\) = 0.97 V

V 2Ω due to 20 V source, \(\frac{V_A}{10} + \frac{V_A}{20} + \frac{V_A-20}{7}\) = 0

Or, 0.1 V A + 0.05V A + 0.143V A = 2.86

∴ V A = \(\frac{2.86}{0.293}\) = 9.76 V

V 2Ω = \(\frac{V_A-20}{7}\) × 2 = -2.92 V

The current in 2 Ω resistor = 2 × \(\frac{5}{5+8.67}\)

= \(\frac{10}{13.67}\) = 0.73 A

The voltage across the 2 Ω resistor = 0.73 × 2 = 1.46 V

V 2Ω = 0.97 – 2.92 -1.46 = -3.41 V.

6. In the circuit given below, the value of I X using nodal analysis is _______

network-theory-questions-answers-advanced-problems-network-theory-2-q6

a) -2.5 A

b) 2.5 A

c) 5 A

d) -5 A

Answer: b

Explanation: Applying KCL, we get, I 1 + 5 = I 2 + I 3

∴ \(\frac{10 – V_1}{1} + 5 = \frac{V_1}{2} + \frac{V_2}{6}\)

∴ 30 – 3V 1 + 15 = 3V 1 + V 2

∴ 6 V 1 + V 2 = 45

From voltage source, V 2 – V 1 = 10

Now, 7 V 1 = 35, V 1 = 5 V

And V 2 = 15 V

∴ I X = \(\frac{V_2}{6} = \frac{15}{6}\) = 2.5 A.

7. In the circuit given below, the values of V 1 and V 2 respectively are _________

network-theory-questions-answers-advanced-problems-network-theory-2-q7

a) 0 and 5V

b) 5 and 0 V

c) 5 and 5 V

d) 2.5 and 2.5 V

Answer: d

Explanation: I 1 = \(\frac{V_1-V_2}{2}\)

Applying KCL at node 1, 5 = \(\frac{V_1}{1} + \frac{V_1-V_2}{2} + V_1 \)

10 = 2V 1 + V 1 – V 2 + 2V 1

Or, 10 = 5V 1 – V 2

KCL at node 2, \(\frac{V_1-V_2}{2}\) + V 1 + 2I 1 = V 2

∴ 1.5 V 1 – V 2 = 0

∴ V 1 = V 2 = 2.5 V.

8. In the circuit given below, the voltage across the 18 Ω resistor is 90 V. The voltage across the combined circuit is _________

network-theory-questions-answers-advanced-problems-network-theory-2-q8

a) 125 V

b) 16 V

c) 24 V

d) 40 V

Answer: a

Explanation: Current through the 18 Ω resistance = \(\frac{90}{18}\) = 5 A

Equivalent resistance of 3 Ω and 7 Ω banks = \(\frac{3 × 6}{3 + 6}\) = 2 Ω

Since, this 2 Ω resistance is in series with 18 Ω resistance, therefore total resistance = 18 + 2 = 20 Ω

This 20 Ω resistance is in parallel with 5 Ω resistance = \(\frac{5 × 20}{5 + 20}\) = 4 Ω

Hence, total resistance of the circuit = 1 + 4 = 5 Ω

Current through this branch = 5 A

∴ Voltage across dc= 5 × 20 = 100 V

Hence current through 5 Ω resistance = \(\frac{100}{5}\) = 20 A

∴ Total current = 20 + 5 = 25 A

Since, total resistance of the circuit is 5 Ω therefore, voltage E = 25 × 5 = 125 V.

9. In the circuit given below, the value of R in the circuit, when the current is zero in the branch CD is _________

network-theory-questions-answers-advanced-problems-network-theory-2-q9

a) 10 Ω

b) 20 Ω

c) 30 Ω

d) 40 Ω

Answer: d

Explanation: The current in the branch CD is zero if the potential difference in the branch CD is zero.

That is, V C = V D

Or, V 10 = V C = V D = V A × \(\frac{10}{15}\)

V R = V A × \(\frac{R}{20+R} \)

And V 10 = V R

∴ V A × \(\frac{10}{15}\) = V A × \(\frac{R}{R+20} \) ∴ R = 40 Ω.

10. Two capacitors of 0.5 μF and 1.5 μF capacitance are connected in parallel across a 110 V dc battery. The charges across the two capacitors after getting charged is ___________

a) 55 μC each

b) 275 μC each

c) 55 μC and 275 μC respectively

d) 275 μC and 55 μC respectively

Answer: c

Explanation: Q 1 = 0.5 x 10 -6 x 110

= 55 μC.

Also, Q 2 = 2.5 x 10 -6 x 110

= 275 μC.

11. Consider a circuit having resistances 2 Ω and 2 Ω in series with an inductor of inductance 2 H. The circuit is excited by a voltage of 12 V. A switch S is placed across the first resistance. Battery has remained switched on for a long time. The current i after switch is closed at t=0 is _____________

a) 6

b) 6 – 3e -t

c) 6 + 3e -t

d) 3 – 6e -t

Answer: b

Explanation: From the figure, we can infer that,

I = \Missing or unrecognized delimiter for \right\)

= 6 – 3e -t .

12. A resistance R is connected to a voltage source V having internal resistance R I . A voltmeter of resistance R 2 is used to measure the voltage across R. The reading of the voltmeter is _____________

a) \

\(\frac{V_S RR_2}{R_1 R_2 + R_1 R + RR_2}\)

Answer: d

Explanation: Effective resistance of R and R m is

R eq = \(\frac{RR_m}{R + R_m}\)

Therefore the reading is \(\frac{V}{R_1 + \frac{R}{R + R_2}}\Big[\frac{RR_2}{R+R_2}\Big]\)

= \(\frac{V_S RR_2}{R_1 R_2 + R_1 R + RR_2}\).

13. When a lead acid battery is being charged, the specific gravity of the electrolyte will ___________

a) Decrease

b) Increase

c) Either Increase or Decrease

d) Neither Increase nor Decrease

Answer: b

Explanation: We know that the specific gravity of electrolyte is highest when battery is fully charged and is lowest when discharged. So, the specific gravity of the electrolyte will increase when a lead acid battery is being charged.

14. A series RLC circuit has a resonant frequency of 550 Hz. The maximum voltage across C is likely to occur at a frequency of ___________

a) 1000 Hz

b) 2000 Hz

c) 1025 Hz

d) 500 Hz

Answer: d

Explanation: We know that, maximum voltage across capacitance occurs at a frequency slightly less than resonant frequency.

Here, given that resonant frequency = 550 Hz.

So, out of the given options 500 is the lowest nearest integer to 550.

Hence, 975 Hz.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Advanced Problems Involving Complex Circuit Diagram – 1”.

1. The current wave shape is in the form of a square terminating at t = 4sec. The voltage across the element increases linearly till t = 4 sec and then becomes constant. The element is ____________

a) Resistance

b) Inductance

c) Capacitance

d) Semi-conductor

Answer: c

Explanation: We know that, when a current pulse is applied to a capacitor, the voltage will have a waveform which rises linearly and then becomes constant towards the end of pulse. Hence, the element is a capacitor.

2. An infinite ladder is constructed with 1 Ω and 2 Ω resistor shown below. The current I flowing through the circuit is ___________

network-theory-questions-answers-advanced-problems-involving-complex-circuit-diagram-1-q2

a) 8.18 A

b) 0 A

c) 9 A

d) 10 A

Answer: c

Explanation: Equivalent Resistance, R EQ = R = 1 +

Or, R EQ = 2

So, I = \(\frac{18}{2}\) A = 9 A.

3. In the circuit given below, the phase angle of the current I with respect to the voltage V 1 is __________

network-theory-questions-answers-advanced-problems-involving-complex-circuit-diagram-1-q3

a) 0°

b) +45°

c) -45°

d) -90°

Answer: d

Explanation: Net voltage applied to the circuit is 200∠0° V

I 1 = \(\frac{200∠0°}{10.0}\)

= 20∠0° = 20

I 2 = \(\frac{200∠0°}{10∠90°}\)

= 20∠-90° = -j20

I = I 1 + I 2 = 20 = 20\(\sqrt{2}\)∠45°

Voltage V 1 = 100

= 100\(\sqrt{2}\)∠45°

∴ Required phase angle = -45° – 45° = -90°.

4. Consider a circuit having 3 identical Ammeters A 1 , A 2 , A 3 parallel to one another. The 1 st Ammeter is in series with a resistance, the 2 nd Ammeter is in series with a capacitor and the circuit is excited by a voltage V. If A 1 and A 3 read 5 and 13 A respectively, reading of A 2 will be?

a) 8 A

b) 13 A

c) 18 A

d) 12 A

Answer: d

Explanation: We can infer from the circuit,

A 2 = \(\sqrt{13^2 – 5^2}\)

Or, A 2 = \(\sqrt{169 – 25}\)

Or, A 2 = \(\sqrt{144}\)

Or, A 2 = 12 A.

5. In the circuit given below, the value of V 1 is __________

network-theory-questions-answers-advanced-problems-involving-complex-circuit-diagram-1-q5

a) 32.2 V

b) -25.23 V

c) 29.25 V

d) -29.25 V

Answer: b

Explanation: \(\frac{V_A}{30} + \frac{V_A-40}{12} + \frac{V_A-V_B}{8}\) = 0

Or, 29 V A – 15 V B = 400

Also, \(\frac{V_B-V_A}{8} + \frac{V_B-120}{8}\) + 6 = 0

Or, V A = 65.23 V, V B = 99.44 V

V 1 = 40-65.23 = -25.23 V.

6. For the three coupled coils shown in figure, KVL equation is ____________

network-theory-questions-answers-advanced-problems-involving-complex-circuit-diagram-1-q6

a) V = (L 1 + L 2 + L 3 ) \

V = (L 1 – L 2 – L 3 – M 13 ) \

V = (L 1 + L 2 + L 3 + 2M 12 – 2M 23 – 2M 13 ) \

V = (L 1 + L 2 + L 3 – 2M 12 + 2M 23 + 2M 13 ) \(\frac{di}{dt}\)

Answer: c

Explanation: M 12 is positive while M 23 and M 13 are negative because of dots shown in figure.

So, the KVL equation is given by,

V = (L 1 + L 2 + L 3 + 2M 12 – 2M 23 – 2M 13 ) \(\frac{di}{dt}\).

7. A circuit excited by voltage V has a resistance R which is in series with an inductor and capacitor, which are connected in parallel. The voltage across the resistor at the resonant frequency is ___________

a) 0

b) \

Answer: a

Explanation: Dynamic resistance of the tank circuit, Z DY = \(\frac{L}{RLC}\)

But given that R L = 0

So, Z DY = \(\frac{L}{0XC}\) = ∞

Therefore current through circuit, I = \(\frac{V}{∞}\) = 0

∴ V D = 0.

8. In the circuit given below, the value of resistance R is _________

network-theory-questions-answers-advanced-problems-involving-complex-circuit-diagram-1-q8

a) 10 Ω

b) 18 Ω

c) 24 Ω

d) 12 Ω

Answer: d

Explanation: Using KVL in loop 1, we get, 100 – 14I – 30 = 0

Or, I = \(\frac{70}{14}\) = 5 A

Then, V P – V Q = 14I – .1

= 70 – 10 = 60 V

∴ R = \(\frac{60}{10-I} = \frac{60}{5}\) = 12 Ω.

9. The current flowing through the resistance R in the circuit in the figure has the form 2 cos 4t, where R is ____________

network-theory-questions-answers-advanced-problems-involving-complex-circuit-diagram-1-q9

c) –

Answer: d

Explanation: Inductor is not given, hence ignoring the inductance. Let I 1 and I 2 are currents in the loop then,

I 1 = \(\frac{2 cos⁡4t}{3}\)

= 0.66 cos 4t

Again, I 2 = \

cos 4t

10. In the circuit given below, the voltage V AB is _________

network-theory-questions-answers-advanced-problems-involving-complex-circuit-diagram-1-q10

a) 6 V

b) 25 V

c) 10 V

d) 40 V

Answer: a

Explanation: For finding the Thevenin Equivalent circuit across A-B we remove the 5 Ω resistor.

Then, I = \(\frac{10+50}{15}\) = 4 A

V OC = 50 – =10 V

And R EQ = \(\frac{10×5}{10+5} = \frac{10}{3}\) Ω

Current I 1 = \(\frac{10}{10/3+5} = \frac{6}{5}\)

Hence, V AB = \(\frac{6}{5 × 5}\) = 6 V.

11. In the circuit given below, the magnitudes of V L and V C are twice that of V K . Calculate the inductance of the coil, given that f = 50.50 Hz.

network-theory-questions-answers-advanced-problems-involving-complex-circuit-diagram-1-q11

a) 6.41 mH

b) 5.30 mH

c) 3.18 mH

d) 2.31 mH

Answer: c

Explanation: V L = V C = 2 VR

∴ Q = \(\frac{V_L}{V_R}\) = 2

But we know, Q = \(\frac{ωL}{R} = \frac{1}{ωCR}\)

∴ 2 = \(\frac{2πf × L}{5}\)

Or, L = 3.18 mH.

12. In the circuit given below, the current source is 1 A, voltage source is 5 V, R 1 = R 2 = R 3 = 1 Ω, L 1 = L 2 = L 3 = 1 H, C 1 = C 2 = 1 F. The current through R 3 is _________

network-theory-questions-answers-advanced-problems-involving-complex-circuit-diagram-1-q12

a) 1 A

b) 5 A

c) 6 A

d) 8 A

Answer: b

Explanation: At steady state, the circuit becomes,

network-theory-questions-answers-advanced-problems-involving-complex-circuit-diagram-1-q12a

∴ The current through R 3 = \(\frac{5}{1}\) = 5 A.

13. In the circuit given below, the capacitor is initially having a charge of 10 C. 1 second after the switch is closed, the current in the circuit is ________

network-theory-questions-answers-advanced-problems-involving-complex-circuit-diagram-1-q13

a) 14.7 A

b) 18.5 A

c) 40.0 A

d) 50.0 A

Answer: a

Explanation: Using KVL, 100 = \(R\frac{dq}{dt} + \frac{q}{C}\)

100 C = \(RC\frac{dq}{dt}\) + q

Or, \(\int_{q_o}^q \frac{dq}{100C-q} = \frac{1}{RC} ∫_0^t \,dt\)

100C – q = (100C – q o )e -t/RC

I = \(\frac{dq}{dt} = \frac{

}{RC} e^{-1/1}\)

∴ e -t/RC = 40e -1 = 14.7 A.

14. A circuit is given in the figure below. The Norton equivalent as viewed from terminals x and x’ is ___________

network-theory-questions-answers-advanced-problems-involving-complex-circuit-diagram-1-q14

a) 6 Ω and 1.333 A

b) 6 Ω and 0.833 A

c) 32 Ω and 0.156 A

d) 32 Ω and 0.25 A

Answer: b

Explanation: We, draw the Norton equivalent of the left side of xx’ and source transformed right side of yy’.

network-theory-questions-answers-advanced-problems-involving-complex-circuit-diagram-1-q14a

V xx’ = V N = \(\frac{\frac{4}{8} + \frac{8}{24}}{\frac{1}{8} + \frac{1}{24}}\) = 5V

∴ R N = 8 ||

= \(\frac{8×24}{8+24}\) = 6 Ω

∴ I N = \(\frac{V_N}{R_N} = \frac{5}{6}\) = 0.833 A.

15. In the circuit given below, the current source is 1 A, voltage source is 5 V, R 1 = R 2 = R 3 = 1 Ω, L 1 = L 2 = L 3 = 1 H, C 1 = C 2 = 1 F. The current through the voltage source V is _________

network-theory-questions-answers-advanced-problems-involving-complex-circuit-diagram-1-q12

a) 1 A

b) 3 A

c) 2 A

d) 4 A

Answer: d

Explanation: At steady state, the circuit becomes,

network-theory-questions-answers-advanced-problems-involving-complex-circuit-diagram-1-q12a

∴ The current through the voltage source V = 5 – 1 = 4 A.

This set of Network Theory Multiple Choice Questions & Answers focuses on “Advanced Problems Involving Complex Circuit Diagram – 2”.

1. In the circuit given below, the KVL for first loop is ___________

network-theory-questions-answers-advanced-problems-involving-complex-circuit-diagram-2-q1

a) V = R 1 i 1 + L 1 \

V = R 1 i 1 – L 1 \

V = R 1 i 1 + L 1 \

V = R 1 i 1 – L 1 \(\frac{di_1}{dt}\) + M \(\frac{di_2}{dt}\)

Answer: c

Explanation: We know that, in general, the KVL is of the form V = R 1 i 1 + L 1 \(\frac{di_1}{dt}\) + M \(\frac{di_2}{dt}\)

But here, M term is negative because i 1 , is entering the dotted terminal and i 2 , is leaving the dotted terminal.

So, V = R 1 i 1 + L 1 \(\frac{di_1}{dt}\) – M \(\frac{di_2}{dt}\).

2. In a parallel RL circuit, 12 A current enters into the resistor R and 16 A current enters into the Inductor L. The total current I the sinusoidal source is ___________

a) 25 A

b) 4 A

c) 20 A

d) Cannot be determined

Answer: c

Explanation: Currents in resistance and inductance are out of phase by 90°.

Hence, I = \(I_1^2 + I_2^2\)

Or, I = [12 2 + 16 2 ] 0.5

Or, I = \(\sqrt{144+256} = \sqrt{400}\)

= 20 A.

3. Consider a series RLC circuit having resistance = 1Ω, capacitance = 1 F, considering that the capacitor gets charged to 10 V. At t = 0 the switch is closed so that i = e -2t . When i = 0.37 A, the voltage across capacitor is _____________

a) 1 V

b) 6.7 V

c) 0.37 V

d) 0.185 V

Answer: b

Explanation: We know that, during discharge of capacitor,

V C = V R

Now, V R = 0.67 X 10 = 6.7 V

So, V C = 6.7 V.

4. A waveform is of the form of a trapezium, which increases linearly with the linear slope till θ = \Missing open brace for subscript 2 V

b) 0 V

c) 4 V

d) 3 V

Answer: d

Explanation: The average value of the waveform = \(\frac{2 X Area \,of \,1st \,triangle\, + \,Area \,of\, 2nd \,triangle}{π}\)

= \(\frac{2 X \frac{π}{3} X \frac{1}{2} X 6 + 6

}{π}\)

= \(\frac{2π + π}{π}\) = 3 Volt.

5. For a series RLC circuit excited by a unit step voltage, V c is __________

a) 1 – e -t/RC

b) e -t/RC

c) e t/RC

d) 1

Answer: a

Explanation: At t = 0, V c = 0 and at t = ∞, V c = 1.

This condition can be satisfied only by (1 – e -t/RC ).

6. In the circuit given below, a dc circuit fed by a current source. With respect to terminals AB, Thevenin’s voltage and Thevenin’s resistance are ____________

network-theory-questions-answers-advanced-problems-involving-complex-circuit-diagram-2-q6

a) V TH = 5 V, R TH = 0.75 Ω

b) V TH = 0.5 V, R TH = 0.75 Ω

c) V TH = 2.5 V, R TH = 1 Ω

d) V TH = 5 V, R TH = 1 Ω

Answer: b

Explanation: V TH = \(\frac{1 X 20}{40}\) X 1 = 0.5 V

Also, R TH = \(\frac{1 X 30}{40}\) = 0.75 Ω.

7. In the circuit given below, the value of R is ____________

network-theory-questions-answers-advanced-problems-involving-complex-circuit-diagram-2-q7

a) 12 Ω

b) 6 Ω

c) 3 Ω

d) 1.5 Ω

Answer: b

Explanation: The resistance of parallel combination is given by,

R eq = \(\frac{40}{3} \) – 10 = 3.33 Ω

Or, \(\frac{1}{3.33} = \frac{1}{12} + \frac{1}{15} + \frac{1}{15}\)

Or, R = 6 Ω.

8. A circuit consists of an excitation voltage V S , a resistor network and a resistor R. For different values of R, the values of V and I are as given, R = ∞, V = 5 volt; R = 0, I = 2.5 A; when R = 3 Ω, the value of V is __________

a) 1 V

b) 2 V

c) 3 V

d) 5 V

Answer: c

Explanation: When R = ∞, V = 5v,

Then, V oc = 5V and the circuit is open

When R = 0, I = 2.5A

Then, I sc = 2.5 and the circuit is short circuited.

So, R eq = \(\frac{V_{OC}}{I_{SC}}\)

= \(\frac{5}{2.5}\) = 2 Ω

Hence the voltage across 3 Ω is 3 volt.

9. Three inductors each 30 mH are connected in delta. The value of inductance or each arm of equivalent star is _____________

a) 10 mH

b) 15 mH

c) 30 mH

d) 90 mH

Answer: a

Explanation: We know that if an inductor L is connected in delta, then the equivalent star of each arm = \(\frac{L X L}{L+L+L}\)

Given that, L = 30 mH

= \(\frac{30 X 30}{30+30+30}\)

= \(\frac{900}{90}\) = 10 mH.

10. In a series RLC circuit having resistance R = 2 Ω, and excited by voltage V = 1 V, the average power is 250 mW. The phase angle between voltage and current is ___________

a) 75°

b) 60°

c) 15°

d) 45°

Answer: d

Explanation: VI cos θ = 0.25 or I cos θ = 0.25

Or, Z cosθ = 2

Or, \(\frac{V}{I}\) cos⁡θ = 2

Or, cos θ = \(\frac{1}{\sqrt{2}}\)

So, from the above equations, cos θ = 0.707 and θ = 45°.

11. In the circuit given below, the equivalent capacitance is _________________

network-theory-questions-answers-advanced-problems-involving-complex-circuit-diagram-2-q11

a) 1.6 F

b) 3.1 F

c) 0.5 F

d) 4.6 F

Answer: b

Explanation: C CB = \Missing or unrecognized delimiter for \right\) + C 5 = 7.5 F

Now, C AB = \Missing or unrecognized delimiter for \right\) + C 6 = 8 F

C XY = \(\frac{C_{AB} × C_4}{C_{AB} + C_4}\) = 3.1 F.

12. In the circuit given below, the equivalent capacitance is ______________

network-theory-questions-answers-advanced-problems-involving-complex-circuit-diagram-2-q12

a) 5.43 μF

b) 4.23 μF

c) 3.65 μF

d) 5.50 μF

Answer: a

Explanation: The 2 μF capacitor is in parallel with 1 μF capacitor and this combination is in series with 0.5 μF.

Hence, C 1 = \(\frac{0.5}{0.5+2+1}\)

= \(\frac{1.5}{3.5}\) = 0.43

Now, C 1 is in parallel with the 5 μF capacitor.

∴ C EQ = 0.43 + 5 = 5.43 μF.

13. In the circuit given below, the voltage across AB is _______________

network-theory-questions-answers-advanced-problems-involving-complex-circuit-diagram-2-q13

a) 250 V

b) 150 V

c) 325 V

d) 100 V

Answer: c

Explanation: Loop current I 1 = \(\frac{50}{20}\) = 2.5 A

I 2 = \(\frac{100}{20}\) = 5 A

V AB = + 100 +

= 125 + 100 + 100

= 325 V.

14. The number of non-planar graph of independent loop equations is ______________

network-theory-questions-answers-advanced-problems-involving-complex-circuit-diagram-2-q14

a) 8

b) 12

c) 3

d) 5

Answer: c

Explanation: The total number of independent loop equations are given by L = B – N + 1 where,

L = number of loop equations

B = number of branches = 10

N = number of nodes = 8

∴ L = 10 – 8 + 1 = 3.

15. In the circuit given below, M = 20. The resonant frequency is _______________

network-theory-questions-answers-advanced-problems-involving-complex-circuit-diagram-2-q15

a) 4.1 Hz

b) 41 Hz

c) 0.41 Hz

d) 0.041 Hz

Answer: d

Explanation: I EQ = L 1 + L 2 + 2M

L EQ = 10 + 20 + 2 × \(\frac{1}{20}\) = 30.1 H

∴ F O = \(\frac{1}{2π\sqrt{LC}}\)

= \(\frac{1}{2π\sqrt{30.1 × 0.5}}\)

= 0.041 Hz.