Electric Circuits Pune University MCQs
Electric Circuits Pune University MCQs
This set of Electric Circuits Interview Questions and Answers focuses on “The International System of Units, Voltage and Current, Power and Energy”.
1. Which of the following is a defined quantity?
a) Pressure
b) Polarity
c) Money
d) Length
Answer: d
Explanation: The SI units are based on 7 defined quantities namely length, mass, time, electric current, thermodynamic temperature, amount of substance and also on the luminous intensity.
2. The basic unit for luminous intensity is
a) Ampere
b) Candela
c) Coulomb
d) Radian
Answer: b
Explanation: The unit for luminous intensity is candela and its symbol is cd.
3. Which of the following is a correct relation?
a) Giga>Mega>Tera
b) Mega>Tera>Giga
c) Tera>Mega>Giga
d) Tera>Giga>Mega
Answer: d
Explanation: Tera – 10 12
Giga – 10 9
Mega – 10 6 .
4. Charge is
a) Unipolar
b) Bipolar
c) Tripolar
d) Non – Polar in nature
Answer: b
Explanation: Charge is bipolar since it can be expressed in terms of positive and negative.
5. Separation of charge creates
a) Current
b) Voltage
c) Resistance
d) Friction
Answer: b
Explanation: An electric force called voltage is created by separation of charge where as an electric fluid called current is created by motion of charge.
6. The energy per unit charge is
a) Voltage
b) Power
c) Current
d) Work
Answer: a
Explanation: Voltage v=dw/dq and its SI unit is volt.
7. If charge q = 3t 2 + 2 then current is given by
a) 6t + 2
b) 3t 2
c) 6t
d) 3t 2 + 2
Answer: c
Explanation: I = dq/dt = d/dt(3t 2 + 2) = 6t.
8. If energy w = 200t 2 + 99 et + 2 then calculate at 0.1 sec
a) 148.52
b) 149.41
c) 149.95
d) 148.39
Answer: b
Explanation: p=dw/dt
=400t +99e t
=400 + 99e
=149.41
9. For the following circuit power is given by
electric-circuits-interview-questions-answers-q9
a) –Vi
b) Vi
c) 0
d) 1
Answer: a
Explanation: P = vi. The algebraic sign of power depends on movement of charge through the drop and rise of voltage.
10. ‘Positive Power’ meaning ___________
a) Power is being delivered to circuit
b) Power is being extracted from circuit
c) No power supply
d) Input and output powers are equal
Answer: a
Explanation: P > 0 means +ve Power
Being delivered
P < 0 means –ve Power
Being extracted.
This set of Electric Circuits Multiple Choice Questions & Answers focuses on “Voltage and Current Sources, Electrical Resistance”.
1) The symbol used for representing Independent sources
a) Diamond
b) Square
c) Circle
d) Triangle
Answer: c
Explanation: Independent sources are represented by circle
Dependent sources are represented by Diamond.
2. Controlled sources are also known as
a) Independent sources
b) Dependent sources
c) Ideal sources
d) Voltage sources
Answer: a
Explanation: Voltage V = dw/dq and its SI unit is Volt.
3.
electric-circuits-questions-answers-voltage-current-sources-resistance-q3
I 3 = α v x .This is
a) Voltage control voltage source
b) Current control voltage source
c) Voltage control current source
d) Current control current source
Answer: b
Explanation: i 3 =αv x means ix value depends on v x
Controlled voltage is v x .
4. Inductor is _______________ element.
a) Active
b) Passive
c) Linear
d) Polar
Answer: b
Explanation: Passive element means it could not generate electricity.
5.
electric-circuits-questions-answers-voltage-current-sources-resistance-q5
Which of the above is valid?
a) 1
b) 2
c) Both
d) Neither 1 nor 2
Answer: a
Explanation: Same Voltage
6.
electric-circuits-questions-answers-voltage-current-sources-resistance-q6
The above circuit is valid.
a) False
b) True
Answer: b
Explanation: Independent current source supplied current through terminals a and b. Dependent source supplies voltage across the same pair of terminals and an ideal current source supplies same current regardless of voltage, similarly an ideal voltage source supplies same voltage irrespective of current, so this is an allowable connection.
7. The opposing capacity of materials against the current flow is
a) Conductance
b) Inductance
c) Susceptance
d) Resistance
Answer: d
Explanation: The opposing capacity of materials against the current flow is resistance.
8. The conductance of a 923Ω resistance is
a) 1.08 * 10 -3 mho
b) 1.08 * 10 -4 mho
c) 1.02 * 10 -3 mho
d) 1.02 * 10 -4 mho
Answer: a
Explanation: c=1/R =1/923 = 1.08 * 10 -3 mho
9. The current passing through a circuit is 7.2A and the power at the terminals is 27 watts.
Existence is ___________ ohms.
a) 0.5402
b) 0.5208
c) 0.5972
d) 0.5792
Answer: b
Explanation: p = vi
= i
= i 2 R
R = P/i 2
=27/ 2
= 0.5208Ω.
10. Relation between power, voltage and conductance
a) V = P 2 .G
b) V = P 2 /G
c) P = v 2 /G
d) P = V 2 G
Answer: d
Explanation: P = vi
= v
= v 2 /R
= v 2 G.
This set of Electric Circuits Multiple Choice Questions & Answers focuses on “Kirchhoff’s Laws”.
1.KCL is based on the fact that
a) There is a possibility for a node to store energy.
b) There cannot be an accumulation of charge at a node.
c) Charge accumulation is possible at node
d) Charge accumulation may or may not be possible.
Answer: b
Explanation: Since the node is not a circuit element, any charge which enters node must leave immediately.
2. Relation between currents according to KCL is
electric-circuits-questions-answers-kirchhoffs-laws-q2
a) i 1 =i 2 =i 3 =i 4 =i 5
b) i 1 +i 4 +i 3 =i 5 +i 2
c) i 1 -i 5 =i 2 -i 3 -i 4
d) i 1 +i 5 =i 2 +i 3 +i 4
Answer: d
Explanation: According to KCL, entering currents=leaving currents.
3. The algebraic sum of voltages around any closed path in a network is equal to ____________
a) Infinity
b) 1
c) 0
d) Negative polarity
Answer: c
Explanation: According to KVL, the sum of voltages around the closed path in a network is zero.
4.
electric-circuits-questions-answers-kirchhoffs-laws-q4
Calculate potential difference between x and y
a) 4.275v
b) -4.275v
c) 4.527v
d) -4.527v
Answer: b
Explanation:
electric-circuits-questions-answers-kirchhoffs-laws-q4a
I 1 = 3/3+5 = 3/8 = 0.375Ω
I 2 = 4/5 = 0.8Ω
V xy = v x – v y
V x + 5I 1 + 4 – 2I 2 – v y = 0
V x – v y = 2I2 – 4 – 5I 1 = -4.275Ω
5.
electric-circuits-questions-answers-kirchhoffs-laws-q5
Find R
a) 17.5 Ω
b) 17.2 Ω
c) 17.4 Ω
d) 17.8 Ω
Answer: a
Explanation:
electric-circuits-questions-answers-kirchhoffs-laws-q5a
KVL: 70 – 5I – 7 = 0
I = 7A
KVL to 2nd loop: 7 – 2R = 0
R=17.5Ω
6. Determine currents I 1 , I 2 and I 3 .
electric-circuits-questions-answers-kirchhoffs-laws-q6
a) -3.3A, -8.5A, 2.4A
b) 3A, -8A, 2A
c) 3.3A, 8.5A, -2.4A
d) 3.2A, 8.6A, 2.3A
Answer: c
Explanation:
electric-circuits-questions-answers-kirchhoffs-laws-q6a
I 1 = I 1 – I 2 + 8 + I 3 + 3
I 2 – I 3 = 11 -> 1
And -11 I1 – 7(I 1 – I 2 ) = 0
-18 I 1 + 7 I 2 = 0 -> 2
And -11 I 1 – 15 I 3 =0 -> 3
Solving I 1 = 3.32A I 2 = 8.5A I 3 = -2.4A.
7. All _____________ are loops but _______________ are not meshes
a) Loops, Meshes
b) Meshes, loops
c) Branches, loops
d) Nodes, Branches
Answer: b
Explanation: A mesh cannot be divided further in loops.
8.
electric-circuits-questions-answers-kirchhoffs-laws-q8
Solve for I.
a) -0.5A
b) 0.5A
c) -0.2A
d) 0.2A
Answer: a
Explanation: V eq = 10 + 5 -20 = -5u
R eq = 5 + 2 + 3 = 10Ω
I = V/R = -5/10 = -0.5A.
9. The basic laws for analyzing an electric circuit are :-
a) Einstein’s theory
b) Newtons laws
c) Kirchhoff’s laws
d) Faradays laws
Answer: c
Explanation: Kirchhoff’s laws are used for analyzing an electric circuit.
10. A junction whell two more than two network elements meet is known as a ______________
a) Node
b) Branch
c) Loop
d) Mesh
Answer: a
Explanation: Node is a junction where two or more than two network elements meet.
This set of Electric Circuits Multiple Choice Questions & Answers focuses on “Analysis of a Circuit Containing Dependent Sources”.
1.Dependent sources are _____________ types.
a) 3
b) 2
c) 4
d) 1
Answer: c
Explanation: Dependent sources are 4 types. Voltage controlled voltage/current source and current controlled current/voltage source.
2. In case of a dependent voltage/current source, the value of this voltage/current source depends on _________
a) Voltage/current sources of an external circuit
b) Voltage/current source present somewhere in the circuit
c) Only on voltage sources
d) Only on current sources
Answer: b
Explanation: The name dependent itself tells us that they are dependent on some other source. A dependent voltage/current source depends on the value of the voltage/current source present somewhere in the circuit itself.
3.
electric-circuits-questions-answers-circuit-containing-dependent-sources-q3
Find i 0 and v 0 in the above circuit..
a) 26A, 260v
b) 28A, 280v
c) 27A, 275v
d) 29A, 285v
Answer: b
Explanation: Applying KVL in loop1: 300= 3i ∆ + 10i 0 ———-
and i 0 = i ∆ +3i ∆ =4i ∆
→ 300= 43i ∆ ,on solving i ∆ =6.976A, i 0 =27.90A, v 0 =279V.
4.The value of the voltage controlled current source i a =βv a given β=0.3 and v a =9.5mV.
a) 2.5 mA
b) 2.85 mA
c) 1.75 mA
d) 1.2 mA
Answer: b
Explanation: i a = 0.3*9.5*10 -3 =2.85mA.
5.Find I 0 in the following circuit, R 1 = 1.1 kilo ohms, R 2 =3.25 kilo ohms, V= 3.7 v.
electric-circuits-questions-answers-circuit-containing-dependent-sources-q5
a) 1.5 mA
b) 2 mA
c) 0.5 mA
d) 1.2 mA
Answer: d
Explanation: By using the fact that the current is same in series connection resistors and voltage is same parallel, the above problem can be solved. I0 is divided into αIx and Ix. So, calculation these two gives the required current value.
6. The value of the current controlled voltage source ,given β=0.8 and i a =9.5mA, is ___________
a) 8mV
b) 7.6mV
c) 0.0011mV
d) 0.0051mV
Answer: b
Explanation: v a =βi a
7. In a VCIS which is the controlled source and which one is the dependent source?
a) V-contorller, I-dependent
b) V-dependent, I-controller
c) Both V and I are controllers
d) Both V and I are dependent
Answer: a
Explanation: VCIS: Voltage-controlled current source. i a =βv a , current value depends on the voltage value so voltage source is the controller and current is the dependent source.
8. In an ICVS which is the controlled source and which one is the dependent source?
a) V-controller, I-dependent
b) V-dependent, I-controller
c) Both V and I are controllers
d) Both V and I are dependent
Answer: b
Explanation: ICVS :Current-controlled voltage source. v a =βi a , voltage value depends on the current value so current source is the controller and voltage is the dependent source.
9. What is the other name for Dependent sources?
a) Uncontrolled sources
b) Time response elements
c) Steady state elements
d) Controlled sources
Answer: d
Explanation: Dependant sources are also known as Controlled sources as there are controlled by other elements present in the circuit.
10. The analysis of a circuit containing dependent sources can be done using nodal and mesh analysis.
a) True
b) False
Answer: a
Explanation: The analysis of a circuit containing dependent sources can be completed using nodal and mesh analysis with the help of Kirchhoff’s laws and also by using various circuit theorems.
This set of Electric Circuits Questions and Answers for Freshers focuses on “The Voltage Divider and Current Divider Circuits”.
1. Where voltage division problem arises
a) Series connected resistors
b) Parallel connected resistors
c) When resistors are equal
d) Both series and parallel resistors.
Answer: a
Explanation: In series, voltage is the difference and current same.
2. Where current division problem arises
a) Series connected resistors
b) Parallel connected resistors
c) When resistors are equal
d) Both series and parallel resistors.
Answer: b
Explanation: In parallel voltage is same and current is the difference.
3. If there are 3 Resistors R 1 , R 2 and R 3 in series and V is total voltage and I is total current then Voltage across R 2 is
a) V R 3 / R 1 + R 2 + R 3
b) V R 2 / R 1 + R 2 + R 3
c) V R 1 /R 1 + R 2 + R 3
d) V
Answer: b
Explanation: V 2 =I R 2
= V R 2 / R 1 + R 2 + R 3 .
4.
electric-circuits-questions-answers-freshers-q4
Calculate Voltage across 2Ω Resistor where supply v= 10volts.
a) 2V
b) 3V
c) 10V
d) 4V
Answer: d
Explanation: I = 10/5 = 2A
V 2 = 10
V 2 = I.R 2
= 2
4V.
5.
electric-circuits-questions-answers-freshers-q5
Calculate i =?
a) -1A
b) +2A
c) 8A
d) -5A
Answer: b
Explanation: i = 1/1+3
= 2A.
6. For a parallel connected resistor R 1 , R 2 and a voltage of V volts. Current across the first resistor is given by
a) I R 1
b) I R 2
c) I R 1 / R 1 + R 2
d) I R 2 / R 1 + R 2
Answer: d
Explanation: I 1 = V / R 1
R = R 1 . R 2 / R 1 + R 2
= I . R 1 . R 2 / R 1 . R 1 + R 2
I 1 = I R 2 / R 1 + R 2 .
7. R 1 = 1Ω, R 2 = 3Ω, R 3 = 5Ω and R 4 = 7Ω connected in series. Total voltage = 20V, Current I, V2 =?
a) I = 1.23, V 2 = 3.75
b) I = 1.25, V 2 = 3.75
c) I = 1.15, V 2 = 3.73
d) I = 1.16, V 2 = 3.72
Answer: b
Explanation: I = 20/ 1 + 3 + 5 + 7 = 1.25A
V 2 = V. R 2 / R 1 + R 2 + R 3 + R 4
= 20/16
= 3.75V.
8. R 1 = 1Ω, R 2 = 3Ω, R 3 = 5Ω and R 4 = 7Ω connected in parallel. Total Current = 23A. Then V, I 1 , I 2 =?
a) 12.26v, 1.725, 2.875
b) 12.23v, 2.875, 1.725
c) 11.26v, 1.95, 1.74
d) 11.23v, 1.74, 1.95
Answer: a
Explanation: V = I/R
V = I (R 1 + R 2 ) R 1 R 2 = 12.26v
I1 = IR 2 / R 1 + R 2 = 1.725A
I 2 = IR 1 / R 1 + R 2
= 2.875A.
9. Voltage division is necessary for parallel resistance networks
a) True
b) False
Answer: b
Explanation: In parallel, connection voltage is same so no division is required.
10. Why is current division necessary?
a) In series current is the same
b) In parallel current differs
c) Because the voltage is also different
d) Because of Kirchhoff’s laws.
Answer: b
Explanation: In parallel current differs.
This set of Electric Circuits Multiple Choice Questions & Answers focuses on “Measuring Voltage and Current”.
1. ____________ helps in current measurement by placing it in ____________ with the circuit element.
a) Voltameter, Parallel
b) Ammeter, series
c) Voltmeter, series
d) Ammeter, parallel
Answer: b
Explanation: In series, current is same. So Ammeter is placed in series and is used to measure current.
2. An ideal voltmeter has ___________ equivalent resistance and ideal ammeter has ___________ equivalent resistance.
a) Unity, Unity
b) Zero, infinite
c) Infinite, Zero
d) Zero, Zero
Answer: c
Explanation: An ideal voltmeter has Infinite equivalent resistance and ideal ammeter has zero equivalent resistance.
3. Continuous voltages current signals are measured using
a) Tachometers
b) Sonometers
c) Analog meters
d) Digital meters
Answer: d
Explanation: Digital meters are used to measure current voltage signals at discrete points in time known as sampling times.
4. Digital meters are preferable than analog meters.
a) True
b) False
Answer: a
Explanation: Features like easy connection, Introduction of less resistance into the circuit to which they all connected and also due to read out mechanism digital meter are preferred.
5. A 20mv, 1mA d’Arsonval movement is used in an ammeter whose full-scale reading is 10 mA. Determine R A .
a) 2.222Ω
b) 6.667Ω
c) 5.92Ω
d) 3.333Ω
Answer: a
Explanation:
electric-circuits-questions-answers-measuring-voltage-current-q5
1 mA flowing through coil implies that 9mA must be diverted through RA.
V = ir
20 * 10 -3 = 9 * 10 -3 R A
R A = 2.222Ω.
6. A 25mv, 2mA d’Arsonval movement is to be used in voltmeter whose full scale reading is 100v. The resistance inserted by 100v meter into circuit is ___________
a) 1 * 10 5 Ω
b) 1 * 10 6 Ω
c) 1 * 10 4 Ω
d) 1 * 10 3 Ω
Answer: a
Explanation: v = iR
R = v/i
= 100/1mA
= 100,000Ω.
7. An ideal voltmeter functions as __________ circuit
a) A short
b) An open
c) A power
d) An infinite
Answer: b
Explanation: An ideal voltmeter offers an infinite equivalent resistance. So acts as an open circuit.
8. An ideal ammeter functions as __________ circuit
a) A short
b) An open
c) A power
d) An infinite
Answer: a
Explanation: An ideal ammeter offers a zero equivalent resistance. So acts a short circuit.
9. A 100mv, 5mA d’Arsonval movement is to be used in an ammeter whose full-scale reading is 1A. Calculate RA.
a) 0.7 ohms
b) 0.5 ohms
c) 0.1 ohms
d) 0.2 ohms
Answer: c
Explanation: 5mA is flowing through the coil which implies 995mA are diverted through R A .
V = iR
= 100 * 10 -3
= 995 * 10 -3 R A
RA = 0.100Ω.
10. A 122mv, 12mA d’Arsonval movement is to be used in voltmeter whose full scale reading is 120v. The resistance inserted by 120v _____________
a) 1200Ω
b) 12000Ω
c) 1000Ω
d) 10,000Ω
Answer: d
Explanation: R =120/12 * 10 -3
= 10,000Ω.
This set of Electric Circuits Interview Questions and Answers for freshers focuses on “Measuring Resistance the Wheatstone Bridge and Delta-to-Wye Equivalent Circuits”.
1. The Wheatstone Bridge is mainly used to measure ______________
a) Currents
b) Voltages
c) Node potentials
d) Resistances
Answer: d
Explanation: Resistances can be measured by various methods. Wheatstone bridge is one such method. In this method resistances in the range of 1Ω to 1 MΩ can be measured.
2. The relation between the resistances in the given Wheatstone bridge circuit is _____________
electric-circuits-interview-questions-answers-freshers-q2
a) P/S = R/Q
b) PR = QS
c) P/Q = R/S
d) PQ = RS
Answer: c
Explanation: The relation is P/Q=R/S or PS=QR.
3. Find the unknown resistance value in given circuit.
electric-circuits-interview-questions-answers-freshers-q3
a) 10.2Ω
b) 11.7Ω
c) 10.5Ω
d) 11.5Ω
Answer: a
Explanation: A/B=C/D. Using this D= 10.2Ω.
4. Lower resistances are difficult to measure using Wheatstone bridge circuit because of ____________
a) Leakage currents
b) I 2 R effects
c) Power dissipation
d) Thermal breakdown
Answer: b
Explanation: A standard Wheatstone bridge couldn’t measure lower resistances because of thermoelectric voltages which are generated at the junctions of the dissimilar metals and also because of thermal heating effects- that is, i 2 R effects.
5. If P/Q=1, unknown resistance S=1000Ω and R could be varied from 0 to 100Ω then the bridge could be ___________
a) A balanced circuit
b) A rectified circuit
c) An unbalanced circuit
d) An identical circuit
Answer: c
Explanation: P/Q=R/S. If P/Q=1 then according to given range of R and S, the bridge circuit could never be a balanced one.
6. The other name for Delta connection is ___________
a) Star connection
b) Pi connection
c) T connection
d) Y connection
Answer: b
Explanation: Delta connection is also known as Pi connection because the ∆ can be shaped into π without disturbing the electrical equivalence of both the structures.
7. Star connection can also be called as Y T connection.
a) True
b) False
Answer: a
Explanation: Star connection can also be called as Y T connection because the star can be shaped into Y or T without disturbing the electrical equivalence of both the structures.
8. If R 2 = R C R A / (R A +R B +R C ) then R 3 equals?
a) R A R B / (R A +R B +R C )
b) R C R A / (R A +R B +R C )
c) R B R C / (R A +R B +R C )
d) R X R A / (R A +R B +R C )
Answer: a
Explanation: R 3 = R A R B / (R A +R B +R C ).
9. Convert the given Delta circuit to star circuit and give the R a , R b and R c values.
electric-circuits-interview-questions-answers-freshers-q9
a) R a =5Ω, R b = 4.5Ω, R c =4.67Ω
b) R a =4Ω, R b =4.30Ω, R c =4.66Ω
c) R a =3Ω, R b =4Ω, R c =5Ω
d) R a =5.2Ω, R b =4.2Ω, R c =4.89Ω
Answer: b
Explanation: By using the standard formulae the delta circuit can be converted into star circuit.
10. Find V AB if i AB = 5A.
electric-circuits-interview-questions-answers-freshers-q10
a) 32.76V
b) 35.56V
c) 36.12V
d) 34.21V
Answer: d
Explanation: By converting the star circuits into the delta and then measuring the equivalent resistance, voltage value can be calculated using this resistance and the given current value.
11. Convert the given star network into Pi network and calculate the sum of all the resistances in the obtained Pi network.
electric-circuits-interview-questions-answers-freshers-q11
a) 125.5Ω
b) 122.5Ω
c) 127.8Ω
d) 129.8Ω
Answer: b
Explanation: Conversion of given network into delta gives the resistances.
After that sum of the resistances equals 122.5Ω.
12. The star and delta networks would be electrically equal if resistances measured between any pair of terminals __________
a) Is different
b) Greater in star
c) Greater in delta
d) Is equal
Answer: d
Explanation: The star and delta networks would be electrically equal if a resistance measured between any pair of terminals is same.
13. A Wheatstone bridge is balanced when the galvanometer shows __________ reading.
a) 0A
b) 1A
c) Infinity
d) -1A
Answer: a
Explanation: A Wheatstone bridge is balanced when the galvanometer shows 0A reading when resistors obey P/Q=R/S.
14. __________ are difficult to measure using Wheatstone bridge.
a) Higher resistances
b) Currents
c) Lower resistances
d) Voltages
Answer: c
Explanation: Specifically Kelvin Bridge is used for measuring lower resistances.
15. What will be the resistance between B and C when the network given below is converted into delta?
electric-circuits-interview-questions-answers-freshers-q15
a) 13Ω
b) 8.66Ω
c) 6.5Ω
d) 7.33Ω
Answer: b
Explanation: Resistance between B and C = 2+4+ /3).
This set of Electric Circuits Multiple Choice Questions & Answers focuses on “The Node-Voltage Method and Dependent Sources and Some Special Cases”.
1. Nodal analysis is mainly based on __________
a) KCL
b) KVL
c) Wheatstone bridge principle
d) Faraday’s electric laws
Answer: a
Explanation: Nodal analysis or Node-Voltage method is done by identifying the currents at the node and thereby forming equations.
2. If there are n nodes, then how many node-voltage equations are required?
a) n
b) n+1
c) n-1
d) 1
Answer: c
Explanation: If there are n nodes then n-1 nodal equations are required to describe the circuit.
3. Find VA and VB using Node-Voltage method in the given circuit.
electric-circuits-questions-answers-node-voltage-dependent-sources-q3
a) 2.5V, 3.6V
b) 2.87V, 3.25V
c) 2.65V, 3.47V
d) 3.15V, 2.76V
Answer: b
Explanation: Node A: V A /2 + (V A -1)/2 + (V A -V B )/1 =2
Node B: (V B -2)/2 + (VB-V A )/1 =1
By solving the above equations required voltages are obtained.
4. A supernode is between _____________
a) Essential node and reference node
b) Two reference nodes
c) Two essential nodes
d) Essential node and neutral path
Answer: c
Explanation: A supernode is between two essential nodes.
5. Find V3 in the circuit given below.
electric-circuits-questions-answers-node-voltage-dependent-sources-q5
a) 4.833V
b) 2.616V
c) -4.833V
d) -2.616V
Answer: a
Explanation: supernode: V 3 -V 2 = 5V
Node1: 166V 1 -100V 2 -66V 3 =132
Node3: -166V 1 +265V 2 +99V 3 =0
On solving the required voltage is obtained.
6. Find all the node voltages in the given circuit containing dependent sources.
electric-circuits-questions-answers-node-voltage-dependent-sources-q6
a) 10V, 20V, 30V, 40V
b) 15V, 25V, 32V, 45V
c) 10V, -20V, 30V, -40V
d) -15V, 25V, -35V, 45V
Answer: a
Explanation: At supernode: V C -V B =5i X
And ix = (V B -V A )/5. On solving remaining nodes and forming equations, the required voltage values at nodes are obtained.
7. What is the voltage at 2nd terminal in the given circuit?
electric-circuits-questions-answers-node-voltage-dependent-sources-q7
a) 132.57V
b) 137.25V
c) 173.25V
d) 123.57V
Answer: b
Explanation: Given voltage source 135V is in between essential node and reference node. So that implies V 1 =135V. Using this, V 2 can be calculated.
8. If there are 5 nodes then the no of nodal equations are ___________
a) 5
b) 0
c) 1
d) 4
Answer: d
Explanation: If there are n nodes then n-1 nodal equations are required to describe the circuit.
9. If there are Node-Voltage equations then the number of nodes in the circuit are __________
a) N+2
b) N+1
c) N
d) N-1
Answer: a
Explanation: If there are n nodes then n-1 nodal equations are required to describe the circuit. So, given N-2+3 i.e. N+1 nodal equations and it implies N+2 nodes.
10. The reference node is also known as __________
a) Essential node
b) Principle node
c) Datum node
d) Neutral node
Answer: c
Explanation: The node taken for reference in the network is known as reference node or datum node.
11. If there are 9 nodes, then how many node-voltage equations are required?
a) 9
b) 10
c) 8
d) 1
Answer: c
Explanation: If there are n nodes then n-1 nodal equations are required to describe the circuit.
12. There are 13 branches in a complicated network and nearly 8 nodes. How many equations are required to solve the circuit in node-voltage method?
a) 7
b) 13
c) 5
d) 6
Answer: a
Explanation: Branches number is not required in this method. Only nodes number is required.
This set of Electric Circuits Questions and Answers for Experienced people focuses on “The Mesh-Current Method and Dependent Sources and Some Special Cases”.
1. The loop which does not contain any other inner loop is known as _____________
a) A node
b) A mesh
c) A branch
d) A super mesh
Answer: b
Explanation: A mesh is defined as a loop which does not contain any other loop within it.
2. If there are 6 branches and 4 essential nodes, how many equations are required to describe a circuit in mesh-current method?
a) 3
b) 6
c) 4
d) 2
Answer: a
Explanation: In Mesh-Current method, b- equations are required to describe the circuit. b= the number of branches and n= the number of essential nodes.
3. Find the current flowing through 5Ω resistor in the given circuit.
electric-circuits-questions-answers-experienced -q3
a) 0.57A
b) 0.64A
c) 0.78A
d) 0.89A
Answer: c
Explanation: There are 3 meshes in the given circuit. Assuming currents I 1 , I 2 , I 3 in the 3 meshes and by applying KVL, equations will be obtained which on solving gives the respective currents flowing in the circuits.
4. A Super Mesh analysis could be done when there is a common _____________ between any two loops.
a) Voltage source
b) Current source
c) Resistor
d) Both voltage and current source
Answer: b
Explanation: A Super Mesh analysis could be done when there is a common current source between any two loops.
5. Calculate the current flowing through 10Ω resistor in the circuit shown below.
electric-circuits-questions-answers-experienced -q5
a) ±0.435A
b) ±0.985A
c) 1.217A
d) 2.782A
Answer: a
Explanation: Loop2 and loop3 forms a supermesh.
Supermesh: I 3 -I 2 =4
Loop1: 11I 1 -10I 2 =2
KVL at Supermesh: -2I 1 +3I 2 +3I 3 =0
Solving these gives the currents flowing in the circuit and current through 10Ω resistor is either I 1 -I 2 or I 2 -I 1 .
6. Find the power delivered by the voltage source in the network given below.
electric-circuits-questions-answers-experienced -q6
a) 65Watts
b) 72Watts
c) 63Watts
d) 76Watts
Answer: c
Explanation: 3 loops and a supermesh between loop1 and loop3. Using KVL currents are found out. I 1 =9A, I 2 =2.5A, I 3 =2A. As voltage source is in 1st loop, Power delivered by voltage source=V*I 1 .
7. The Mesh-Current method is applicable only for ___________
a) Non-linear networks
b) Equivalent networks
c) Non-planar networks
d) Planar networks
Answer: d
Explanation: The Mesh-Current method is applicable only for Planar networks. A network is said to be planar if there are no crossovers in it and it can be drawn freely on a plane surface.
8. Find the value of V X in the circuit given below.
electric-circuits-questions-answers-experienced -q8
a) –0.8A
b) +0.8A
c) -4.8A
d) +4.8A
Answer: a
Explanation: Applying KVL, currents could be found out. I 1 =0.4A, I 2 =2.4A. V X =-I 1 R 1 .
9. A Supermesh is formed between two loops which share a common voltage source.
a) True
b) False
Answer: b
Explanation: Meshes that share a current source with other meshes, none of which contains a current source in the outer loop, forms a supermesh.
10. If 4 equations are required to describe a circuit by Mesh-Current method and there are n nodes. How many branches are there in the network?
a) n+5
b) n+3
c) n
d) n-1
Answer: b
Explanation: Standard formulae: b-
Given b- =4 -> b=4+ =n+3.
11. If there are 16 branches and 5 essential nodes, how many equations are required to describe a circuit in mesh-current method?
a) 12
b) 16
c) 21
d) 9
Answer: a
Explanation: In Mesh-Current method, b- equations are required to describe the circuit. b=the number of branches and n= the number of essential nodes.
12. Determine the current through 3Ω resistor in the network given below.
electric-circuits-questions-answers-experienced -q8
a) 2A
b) 3A
c) 4A
d) -2A
Answer: c
Explanation: Mesh1 and Mesh2 form a super mesh. Assuming currents I 1 and I 2 and applying KVL, the current through required resistor is found out.
13. Mesh analysis is best suitable for _____________
a) Current sources
b) Voltage sources
c) Complex elements
d) Unilateral elements
Answer: a
Explanation: Mesh analysis is best suitable for Current sources.
This set of Electric Circuits Multiple Choice Questions & Answers focuses on “TThe Node-Voltage Method Versus the Mesh-Current Method”.
1. Which method is best for voltage sources?
a) Mesh analysis
b) Nodal analysis
c) Superposition principle
d) Differentiation method
Answer: b
Explanation: Every voltage source connected to the reference node reduces the equations to be solved. Thus, the node-voltage method is best for voltage sources.
2. When there is a current source between two loops which method is preferred?
a) Mesh-voltage analysis
b) Node-current analysis
c) Supermesh
d) Supernode
Answer: c
Explanation: Supermesh is taken into consideration when there is a current source n between two loops and is considered as one single loop.
3. Determine the current through 5Ω resistor in the network given below.
electric-circuits-questions-answers-node-voltage-mesh-current-q3
a) 3.38A
b) 6.01A
c) 3.27A
d) 1.27A
Answer: a
Explanation: This problem can be solved quickly by using the mesh-current method. 3loops=3 KVL equations. Solving them gives respective currents.
4. Find the power supplied by the dependent voltage source in the circuit given below.
electric-circuits-questions-answers-node-voltage-mesh-current-q4
a) 400W
b) 383W
c) 412W
d) 148W
Answer: b
Explanation: 3loops=3KVL equations. Solving them gives currents flowing in the circuit. I 1 =5A, I 2 =-1.47A, I 3 =0.56A .Power supplied by dependent voltage source =0.4V 1 (I 1 -I 2 ).
5. Determine the voltage V 2 of the network given.
electric-circuits-questions-answers-node-voltage-mesh-current-q5
a) 0V
b) 1V
c) 4/7V
d) -4/7V
Answer: a
Explanation: As there are voltage sources and nodes, node-voltage method is best suitable. Solving gives V 1 =4/7V and V 2 =0 .
6. If there are b branches and n nodes, then how many node-voltage equations are required?
a) n
b) b+1
c) n-1
d) b
Answer: c
Explanation: If there are n nodes then n-1 nodal equations are required to describe the circuit. Branches do nothing in this case.
7. There are 6 branches and 2 essential nodes then 3 equations are required to describe a circuit in the mesh-current method.
a) True
b) False
Answer: b
Explanation: In Mesh-Current method, b- equations are required to describe the circuit. b= the number of branches and n= the number of essential nodes. So, 6- =5.
8. Determine V 1 of the given network.
electric-circuits-questions-answers-node-voltage-mesh-current-q8
a) -0.17V
b) 4.83V
c) 5V
d) 2.62V
Answer: d
Explanation: Supernode: V 3 -V 2 =5. Applying KCL at node1 and at supernode gives the equations which on solving, required voltages are obtained.
9. When there is a voltage source between two nodes which method is preferred?
a) Mesh-voltage analysis
b) Node-current analysis
c) Supermesh
d) Supernode
Answer: d
Explanation: Supernode is taken into consideration when there is a voltage source n between two nodes.
10. Which is the best-preferred method to calculate currents flowing in the circuit?
a) Mesh-voltage analysis
b) Node-current analysis
c) Superposition principle
d) Duality principle
Answer: a
Explanation: By KVL, currents can be easily found out in mesh-voltage method.
This set of Electric Circuits Multiple Choice Questions & Answers focuses on “Source Transformations”.
1. By using source transformation voltage source in series resistor is replaced by __________
a) Voltage source in series with a resistor
b) Current source in parallel with a resistor
c) Voltage source in parallel with a resistor
d) Current source in series with a resistor
Answer: b
Explanation: In Source transformation, a voltage source in series with a resistor is replaced by a current source in parallel with the same resistor and vice versa.
2. Source Transformation is _____________
a) Unilateral
b) Unique
c) Bilateral
d) Complicated
Answer: c
Explanation: In Source transformation, a voltage source in series with a resistor is replaced by a current source in parallel with the same resistor and vice versa. So, it is bilateral.
3. If there are two resistors in parallel and in series with a voltage source then ___________
a) Parallel resistor has no effect
b) Series resistor has no effect
c) Both has their respective effects
d) Both has no effect on the voltage source
Answer: a
Explanation: In source transformation, the voltage source in series with a resistor to be replaced by a current source in parallel with the same resistor and vice versa. So other resistors are redundant and have no effect.
4. Using source transformation, calculate the voltage.
electric-circuits-questions-answers-source-transformations-q4
a) 4.33V
b) 39V
c) 0.230V
d) 36V
Answer: b
Explanation: V=IR=13*3=39V.
5. Which element has no effect in the given circuit?
electric-circuits-questions-answers-source-transformations-q5
a) 7Ω
b) 10Ω
c) Both 7Ω and 10Ω
d) Voltage source.
Answer: b
Explanation: Voltage in series with a resistor in replaced by a current source but here 10Ω is in parallel. So, it is redundant and has no effect.
6. The value of current source is __________ after replacing the given network with a single current source and a resistor.
electric-circuits-questions-answers-source-transformations-q6
a) 70V
b) 60V
c) 90V
d) 80V
Answer: d
Explanation: In the given circuit 9Ω resistor has no effect.
10*6=60V, 60V+20V=80V.
7. If there is a 12A current source in series with 2Ω and in parallel with a 4Ω resistor, then voltage V=?
a) 24V
b) 48V
c) 3V
d) 6V
Answer: b
Explanation: 2Ω resistor is redundant. 12*4=48V.
8. Find the current flowing through 4Ω resistor shown in network below.
electric-circuits-questions-answers-source-transformations-q8
a) 1.33A
b) 2.35A
c) 1.66A
d) 2.66A
Answer: c
Explanation: By using source transformation the above network is reduced and then by current division rule I 4Ω = 5* =1.66A.
9. Calculate the power delivered by the 50V source.
electric-circuits-questions-answers-source-transformations-q9
a) 274W
b) 276W
c) 285W
d) 291W
Answer: a
Explanation: By using source transformation the above network is reduced and current in the circuit is found out and later power delivered by 50V source= 50*current in the circuit= 50*5.48A= 274W.
10. Source transformation can be used for dependent sources.
a) True
b) False
Answer: a
Explanation: Source transformation can be used for dependent sources. However, the controlling variable must not be tampered with any way since the controlled source operation depends on it.
11. Using source transformation, calculate vm.
electric-circuits-questions-answers-source-transformations-q11
a) 2v
b) -2v
c) 1v
d) -1v
Answer: b
Explanation: Using source transformation, the network is reduced and at last voltage is obtained.
12. Find the voltage value Vm in the circuit given below.
electric-circuits-questions-answers-source-transformations-q12
a) -3V
b) 3V
c) 2.1V
d) -2.1V
Answer: a
Explanation: Using source transformation, the voltage source in series with a resistor to be replaced by a current source in parallel with the same resistor and vice versa.
13. Source transformation technique is mainly based on __________ law.
a) Newton’s
b) Kirchhoff’s
c) Ohm’s
d) Einstein’s
Answer: c
Explanation: Ohm’s law: V=iR. By using this, the voltage/ current sources are reduced.
14. In source transformation,
a) Voltage sources remain same
b) Current sources remain same
c) Both voltage and current sources undergo change
d) Resistances/Impedances remain same
Answer: d
Explanation: In source transformation, only the particular voltage/current sources change whereas the resistances remain same.
15. If there are five 20V voltage sources in parallel, then in source transformation __________
a) All are considered
b) Only one is considered
c) All are ignored
d) Only 2 are considered
Answer: b
Explanation: In parallel, voltages are same. So, only is considered and rest are ignored.
This set of Electric Circuits Interview Questions and Answers for Experienced people focuses on “Thevenin and Norton Equivalents, More on Deriving a Thevenin Equivalent”.
1. Find the voltage across 24Ω resistor by using Thevenin’s theorem.
electric-circuits-interview-questions-answers-experienced-q1a
a) 8V
b) 9V
c) 1V
d) 6V
Answer: a
Explanation: 1. Remove 24Ω resistor and calculate the voltage across the open circuit.
2. Calculate the thevenin’s resistance and by using it, the thevenin’s current.
3. V 24Ω =I*R .
2. Calculate Thevenin’s voltage for the network shown below where the voltage source is 4V.
electric-circuits-interview-questions-answers-experienced-q2
a) 6V
b) 4.71V
c) 5V
d) 1V
Answer: c
Explanation: In the circuit given, thevenin’s voltage is nothing but the open circuit voltage which is V x . Applying KVL, it is obtained.
3. Find the Thevenin’s resistance for the network given.
electric-circuits-interview-questions-answers-experienced-q3
a) 6.75Ω
b) 5.85Ω
c) 4.79Ω
d) 1.675Ω
Answer: a
Explanation: Remove all the voltage/current sources and calculate the equivalent resistance.
4. Find the current through Ω resistor.
electric-circuits-interview-questions-answers-experienced-q4
a) 0.9-j0.2 A
b) 0.78-j0.1 A
c) 2.7-j0.5 A
d) 1A
Answer: a
Explanation: 1. Remove the 5+j4 Ω branch and calculate thevenin’s voltage.
2. Calculate Z th .
3. Current= (V th / (Z th +Z).
5. The voltage across 6Ω resistor is __________
electric-circuits-interview-questions-answers-experienced-q5
a) 7.5V
b) 6.78V
c) 20V
d) 8.5V
Answer: d
Explanation: Remove the resistor across which voltage is to be calculated and short circuit it. By using short circuit current and resistance calculate the current across 6Ω resistor and thereby voltage. .
6. Find the Norton’s current for the circuit given below.
electric-circuits-interview-questions-answers-experienced-q6
a) 5A
b) 3.33A
c) 4A
d) 1.66A
Answer: c
Explanation: I N = + .
7. Calculate IN for the given network.
electric-circuits-interview-questions-answers-experienced-q7
a) 0A
b) 1A
c) 4.37A
d) 0.37A
Answer: a
Explanation: Using nodal analysis V x is calculated. I N =V x /4.
8. Calculate R Th for the network given.
electric-circuits-interview-questions-answers-experienced-q8
a) 8Ω
b) 7Ω
c) 2Ω
d) 1Ω
Answer: b
Explanation: 5//20 and then in series with 3Ω resistor.
9. Thevenin’s equivalent circuit consists of a ____________
a) Voltage source in series with a resistor
b) Current source in parallel with a resistor
c) Voltage source in parallel with a resistor
d) Current source in series with a resistor
Answer: a
Explanation: Thevenin’s equivalent circuit contains a Voltage source in series with a resistor.
10. Norton’s equivalent circuit consists of a _____________
a) Voltage source in series with a resistor
b) Current source in parallel with a resistor
c) Both voltage and current sources
d) Current source in series with a resistor
Answer: b
Explanation: Norton’s equivalent circuit consists of a Current source in parallel with a resistor.
11. Thevenin’s voltage is equal to ____________
a) Short circuit voltage
b) Open circuit current
c) Open circuit voltage
d) Short circuit current
Answer: c
Explanation: Thevenin’s voltage is equal to open circuit voltage.
12. Norton’s current is equal to ____________
a) Short circuit voltage
b) Open circuit current
c) Open circuit voltage
d) Short circuit current
Answer: d
Explanation: Norton’s current is equal to Short circuit current.
13. Thevenin’s resistance R Th = ___________
a) V Th /I SC
b) VSC/I Th
c) V Th /I Th
d) V SC /I SC
Answer: a
Explanation: Thevenin’s resistance is defined as the ratio of open circuit voltage to the short circuit current across the terminals of the original circuit.
14. What is the expression forthe thevenin’s current if there is an external resistance in series with the R Th ?
a) V Th /I Th
b) V Th / (R Th -R)
c) V Th / (R Th +R)
d) V Th /R Th
Answer: c
Explanation: I Th = V Th / (R Th +R).
15. One can find the thevenin’s resistance simply by removing all voltage/current sources and calculating equivalent resistance.
a) False
b) True
Answer: b
Explanation: Yes. One can find the thevenin’s resistance simply by removing all voltage/current sources and calculating equivalent resistance.
This set of Electric Circuits Multiple Choice Questions & Answers focuses on “Maximum Power Transfer and Superposition”.
1. Which of the following is the example to describe the efficiency of power transfer?
a) Communication systems
b) Power utility systems
c) Instrumentation systems
d) Telecom systems
Answer: b
Explanation: Power utility systems are good examples for this case as they are concerned with the generation, transmission and distribution of power in large quantities.
2. In a network maximum power transfer occurs when __________
a) R Th = -R L
b) R Th /R L =0
c) R Th = R L
d) R Th +R L = 1
Answer: c
Explanation: Maximum power transfer occurs when load resistance equals the thevenin’s resistance.
3. Maximum power in terms of the thevenin’s voltage and load resistance __________
a) (V Th ) 2 /4R L
b) (V Th ) 2 *4R L
c) (V Th ) 2 +4R L
d) 4R L / (V Th ) 2
Answer: a
Explanation: P max = v*i= i*R*i= i2R L = (V Th /(R Th +R L ))2*R L . Max power occurs when R L =R Th .
4. Calculate the maximum power delivered across R L of the circuit given.
electric-circuits-questions-answers-maximum-power-transfer-superposition-q4
a) 900W
b) 1025W
c) 2025W
d) 1500W
Answer: c
Explanation: Pmax= (V Th /(R Th +R L )) 2 *R L (R Th =R L )
V Th = *540= 450V
R Th =/180= 25Ω .
5. Determine the maximum power delivered to the load in the network given.
electric-circuits-questions-answers-maximum-power-transfer-superposition-q5
a) 4.68W
b) 5.75W
c) 3.16W
d) 6.84W
Answer: a
Explanation: Load is given across node V 2 and reference path. It implies the thevenin’s voltage is V 2 . By using nodal analysis this voltage is found out.
R Th =RL= 10Ω//5Ω and in series with 2Ω and then parallel with 3Ω=1.92Ω
Max power = (V Th ) 2 /4R L = 4.688W.
6. The Superposition principle is obeyed by ____________
a) Linear networks
b) Non-linear networks
c) Lateral networks
d) Trilateral networks
Answer: a
Explanation: A linear system obeys Superposition Principle. In a linear network parameters are constant i/e/ won’t change with voltage and current.
7. According to Superposition principle response in one element is the algebraic sum of responses by individual sources acting alone.
a) False
b) True
Answer: b
Explanation: According to Superposition principle response in one element is the algebraic sum of responses by individual sources acting alone while other sources are non-operative.
8. Find the current in the 3Ω resistor of the given network using Superposition principle.
electric-circuits-questions-answers-maximum-power-transfer-superposition-q8
a) 2.5A
b) 3.125A
c) 6.525A
d) 5.625A
Answer: d
Explanation: 1.When 20v source acting alone: current source is replaced by open circuit. R eq = 5+3=8Ω and Current I= 20/8=2.5A
2. When 5A source acting alone: 20v source is replaced by a short circuit. By current division rule, I 3 = 25/8=3.125A
Total current through I 3 = 2.5+3.125=5.625A.
9. Find the current in 5Ω resistor near 12V source using superposition principle in network given
electric-circuits-questions-answers-maximum-power-transfer-superposition-q9
a) 2.9A
b) -2.9A
c) 1A
d) -1A
Answer: b
Explanation: Only 3 sources are considered , other is dependent .
12v source: I=0.6A
10A source: I=-2.5A through 5Ω resistor and I=7.5A
20v source: I=-1A
Total current = 0.6+ + = -2.9A.
10. If there are 5 sources in a network out of which 3 are dependent and 2 are independent. For superposition principle ___________ sources are considered.
a) 5
b) 3
c) 2
d) 0
Answer: c
Explanation: Only independent sources are considered while using Superposition principle. Dependent sources are never deactivated while using this principle.
11. Find the current in the 1Ω resistor of the given circuit.
electric-circuits-questions-answers-maximum-power-transfer-superposition-q11
a) 4A
b) 1.33A
c) 2A
d) 0.66A
Answer: a
Explanation: when all the sources are acting alone the corresponding currents are found out using current division rule and finally all are summated to get the required current through the1Ω resistor. .
12. Find the value of RL in given circuit.
electric-circuits-questions-answers-maximum-power-transfer-superposition-q12
a) 4Ω
b) 5Ω
c) 3Ω
d) 1.66Ω
Answer: c
Explanation: R Th =R L = + 2=3Ω.
13. Calculate the maximum power delivered to load in the network given.
electric-circuits-questions-answers-maximum-power-transfer-superposition-q13
a) 1.56W
b) 1.66W
c) 2.33W
d) 2.79W
Answer: a
Explanation: 1.Calculation of V Th .
2. Calculation of Norton’s current.
3. Calculation of R Th . (R Th =V Th /I N )
4. R Th =R L and P Max = V Th /4R L .
14. In AC networks, maximum power is delivered when __________
a) Z L *Z S *=0
b) Z L +Z S *=1
c) Z L =-Z S *
d) Z L =Z S *
Answer: d
Explanation: Max power is delivered when load impedance equals complex conjugate of the source impedance.
15. Superposition principle states that at a time __________ source acts.
a) All the given sources
b) Only voltage sources
c) Only one source
d) Only current sources
Answer: c
Explanation: Only one source acts at a time. Remaining sources are non-operative during this period.
This set of Electric Circuits test focuses on “Operational Amplifier Terminals,Terminal Voltages and Currents”.
1. Op-amp was introduced by __________
a) Fairchild
b) Maxwell
c) Rutherford
d) Sahani
Answer: a
Explanation: Op-amp was introduced by Fairchild semiconductor in 1968.
2. The number of terminals in an Op-amp ______________
a) 6
b) 2
c) 5
d) 3
Answer: c
Explanation: Inverting input, the Non-inverting input, Output, Positive power supply, Negative power supply.
3. The Op-amp is a type of ___________
a) Differential amplifier
b) Integrated amplifier
c) Isolation amplifier
d) Feedback amplifier
Answer: a
Explanation: The Op-amp is a type of differential amplifier.
4. In the circuit of Op-amp given V- stands for _________
electric-circuits-questions-answers-test-q4
a) Non-inverting input
b) Non-inverting output
c) Inverting input
d) Inverting output
Answer: c
Explanation: V+: Non-inverting input and V- : Inverting input.
5. When the input voltage difference is small in magnitude, the Op-amp behaves as ____________
a) Non-linear device
b) Linear device
c) Complex device
d) Bipolar device
Answer: b
Explanation: When │vp-vn│is small then Op-amp acts as a linear device as the output voltage is a linear function of input voltages.
6. If the output voltage is not a linear function of input voltage then ____________
a) Op-amp acts a linear device
b) Op-amp acts as a non-linear device
c) Op-amp acts a polar device
d) Op-amp acts as an inverter
Answer: b
Explanation: If output voltage is not a linear function of input voltage then Op-amp acts as a non-linear device.
7. The negative feedback causes the input voltage difference to ____________
a) 1
b) Increase
c) Decrease
d) 0
Answer: c
Explanation: Negative feedback means a signal is fed back from output terminals to the non-inverting input terminals and this results in a decrease in input voltage difference.
8. Find the gain for the following circuit.
electric-circuits-questions-answers-test-q8
a) -2
b) 2
c) -1
d) 1
Answer: a
Explanation: In this circuit, the only node is at the negative terminal of the Op-amp and by ideal rules of Op-amp, Vn= Vp =0. Gain= V out /V in = -R 2 /R 1 .
9. Calculate the gain for the Op-amp given.
electric-circuits-questions-answers-test-q9
a) 0.719
b) 2.572
c) 1.390
d) 1.237
Answer: c
Explanation: Gain= V out /V in = -R 2 /R 1 = -5.98*10 -3 /4.3*10 -3 .
10. Given Op-amp is ideal. Calculate v o if v a =1v and v b =0v.
electric-circuits-questions-answers-test-q10
a) -4v
b) -2.5v
c) 4v
d) 2.5v
Answer: b
Explanation: In the given circuit, a negative feedback exists between Op-amp’s output and its inverting input (voltage here is 0, as v p =v b =0 and v n =v p ). Node-voltage equation is i 50 =i 125 =i 0 .
i 50 = (v a -v n )/50 =1/50 mA.
I 125 = (v 0 -v n )/125 = v 0 /125 mA.
1/50 + v 0 /125 = 0.
v 0 is -2.5volts.
This set of Electric Circuits Multiple Choice Questions & Answers focuses on “The Inverting and Non-Inverting Amplifier Circuit”.
1. The opamp in the Inverting circuit is in __________
a) Linear region
b) Saturation
c) Cut-off region
d) Non-linear region
Answer: a
Explanation: We assume that the opamp is in linear region.
2. In an Inverting Amplifier circuit, the output voltage v o is expressed as a function of ____________
a) Input current
b) Output current
c) Source voltage
d) Source current
Answer: c
Explanation: The goal of an inverting circuit is to express output voltage v o as a function of source voltage vs.
3. The other name for Gain is ____________
a) Scaling factor
b) Output
c) Amplifying factor
d) Scaling level
Answer: a
Explanation: The gain is also known as scaling factor and it is the ratio of R f /R s in case of an Inverting amplifying circuit.
4. If V CC = 12V and vs=1mV, then R f /R s is _____________
a) >12000
b) <12000
c) 12000
d) 1
Answer: b
Explanation: R f /R s ≤ │V CC /vs│.
5. In the expression v o = -Av n , A is called ______________
a) Closed loop gain
b) Closed loop fault
c) Open loop fault
d) Open loop gain
Answer: d
Explanation: A is called open loop gain.
6. The circuits of an inverting and Non-Inverting amplifying comprises of __________ and _______ number of resistors.
a) 3, 2
b) 2, 3
c) 2, 2
d) 3, 3
Answer: b
Explanation: Inverting amplifying circuit- R s , R f .
Non-Inverting amplifying circuit – R s , R f , R g .
7. The condition for a Non-inverting amplifying circuit to operate in linear region operation _____________
a) (R s +R f )/R s < │V CC /v g │
b) (R s +R f )/R s ≠ │V CC /v g │
c) (R s +R f )/R s > │V CC /v g │
d) (R s +R f )/R s = │V CC /v g │
Answer: a
Explanation: Assume that opamp is ideal. The condition for the linear region operation in a Non-inverting amplifying circuit is (R s +R f )/R s <│VCC/vg│.
8. If R s = 3Ω, Rf= 6Ω then the relation between v o and v g in case of a Non-Inverting amplifying circuit.
a) v o = 9v g
b) v o = 6v g
c) v o = 3v g
d) v o = v g
Answer: c
Explanation: v o = ((R s +R f )/R s ) *v g .
9. If R s = 5Ω, R f = 25Ω and -2.5V ≤ v g ≤ 2.5V. What are the smallest power supply voltages that could be applied and still have opamp in linear region?
a) ±9V
b) ±2.5V
c) ±6V
d) ±15V
Answer: d
Explanation: vo= ((R s +R f )/R s ) *v g . By substituting the values, we have v o =6v g .
v o =6 = -15
v o =6 =15.
10. If an inverting amplifying circuit has a gain of 10 and ±15V power supplies are used. The values of input for which opamp would be in the linear region?
a) ±1.25
b) ±1.5V
c) ±2.25
d) ±0.5
Answer: b
Explanation: Gain= R f /R s = 10 and v o = (-R f /R s )*v s .
→ v o = -10v s and given -12V≤ v o ≤ 12V.
→ -15= -10v s . So, v s = 1.5V
→ 15=-10v s . So, v s =-1.5V.
11. If the gain of an inverting amplifying circuit is 13 and ±22V power supplies are used. What range of input values allows the opamp to be in linear region?
a) ±1.69
b) ±1.35V
c) ±2.28
d) ±0.5
Answer: a
Explanation: Gain= Rf/Rs= 13 and v o = (-R f /R s )*v s .
→ v o = -13v s and given -22V≤ v o ≤ 22V.
→ -22= -13v s . So, v s =1.692 V
→ 22=-13v s . So, v s =-1.692V.
12. The input applied to an Inverting amplifier is ______________
a) Equal to output
b) Equal to Inverted output
c) Not equal to output
d) Output is equal to input
Answer: b
Explanation: The name itself indicates it is an Inverting amplifier. So, the input applied is inverted and is given as output. Suppose the input applied is sinusoidal then, the output is
electric-circuits-questions-answers-inverting-non-inverting-amplifier-q12
13. In R 1 =10kΩ, R f =100kΩ, v 1 =1V. A load of 25kΩ is connected to the output terminal. Calculate i 1 and v o .
electric-circuits-questions-answers-inverting-non-inverting-amplifier-q13
a) 0.5mA, 10V
b) 0.1mA, 10V
c) 0.1mA, -10V
d) 0.5mA, -10V
Answer: c
Explanation: i 1 = v 1 /R 1 = 1V/10kΩ = 0.1mA
V 0 = -(R f /R 1 )*v 1 = -*1V = -10V.
This set of Electric Circuits Multiple Choice Questions & Answers focuses on “Inductor and Capacitor”.
1. The symbol used for inductance is __________
a)
electric-circuits-questions-answers-inductor-capacitor-q1
b)
electric-circuits-questions-answers-inductor-capacitor-q1b
c)
electric-circuits-questions-answers-inductor-capacitor-q1c
d)
electric-circuits-questions-answers-inductor-capacitor-q1d
Answer: c
Explanation: electric-circuits-questions-answers-inductor-capacitor-q1c is the symbol used to represent inductance.
2. The symbol used for capacitance is _____________
a)
electric-circuits-questions-answers-inductor-capacitor-q1d
b)
electric-circuits-questions-answers-inductor-capacitor-q2
c)
electric-circuits-questions-answers-inductor-capacitor-q1c
d)
electric-circuits-questions-answers-inductor-capacitor-q1b
Answer: b
Explanation: electric-circuits-questions-answers-inductor-capacitor-q2 is the symbol used to represent capacitance.
3. The formula used to find the capacitance C is __________
a) Q/v
b) Qv
c) Q-v
d) Q + v
Answer: a
Explanation: Q=cv. Q-charge, V-voltage, c-capacitance.
4. The capacitor doesn’t allow sudden changes in ___________
a) Voltage
b) Current
c) Resistance
d) Capacitance
Answer: a
Explanation: Any small change in voltage occurs within zero time across the gives an infinite current which is practically impossible. So, in a fixed capacitor, the voltage cannot change abruptly.
5. The Inductor doesn’t allow sudden changes in ___________
a) Voltage
b) Current
c) Resistance
d) Inductance
Answer: b
Explanation: Any small change in current occurs within zero time across the gives an infinite voltage which is practically impossible. So, in a fixed inductor, the voltage cannot change abruptly.
6. The expression for energy of an inductor ____________
a) ½ LI
b) L/2I
c) ½ L 2 I
d) ½ LI 2
Answer: d
Explanation: E=∫p dt
=∫ LI*.dt
= L∫I dI
= ½ LI 2 .
7. The units for inductance is _________ and capacitance is ___________
a) Faraday, Henry
b) Coulomb, Faraday
c) Henry, Faraday
d) Henry, Coulomb
Answer: c
Explanation: The unit for inductance is ‘Henry’ and capacitance is ‘Faraday’.
8. The voltage applied to a pure capacitor of 50*10 -6 F is as shown in figure. Calculate the current for 0-1msec.
electric-circuits-questions-answers-inductor-capacitor-q8
a) 5A
b) 1A
c) -5A
d) -1A
Answer: a
Explanation: For 0≤t≤1msec,
V =m*t
→100= 1*10 -3 *m →m= 1*10 5
→ V = 1*10 5 t
Current I = c. d )/dt = 50*10 -6 * (d (1*10 5 t)/dt) = 5A.
9. If a capacitor of capacitance 9.2F has a voltage of 22.5V across it. Calculate the energy of the capacitor.
a) 5062.5W
b) 506.25W
c) 50.625W
d) 50625W
Answer: b
Explanation: E= ½ cv 2 .
10. The voltage applied to the 212mH inductor is given by v= 15e -5t v. Calculate the current.
a) 16.782e -10t
b) 15.75e -5t
c) 11.27e -10t
d) 14.15e -5t
Answer: d
Explanation: Current I= 1/L 0∫t v*dt.
11. A voltage across a capacitor of 0.5F is defined by
V = [0, t<0
2t, 0<t<2s
4e - , t>2s] Find i .
a) -2e - A
b) -4e - A
c) -20e - A
d) -12e - A
Answer: a
Explanation: i= C*.
12. If the voltage across a capacitor is constant, then current passing through it is ________
a) 1
b) 0
c) -1
d) Infinity
Answer: b
Explanation: I= c*.
13. An Inductor works as a ___________ circuit for DC supply.
a) Open
b) Short
c) Polar
d) Non-polar
Answer: b
Explanation: Induced voltage across an inductor is zero if the current flowing through it is constant. I.e. Inductor works as a short circuit for DC supply.
14. The insulating medium between the two plates of capacitor is known as __________
a) Electrode
b) Capacitive medium
c) Conducting medium
d) Dielectric
Answer: d
Explanation: The conducting surfaces are called electrodes and the insulating medium is called Dielectric.
15. If the current flowing through an inductor of inductance 0.3Henry is 5.3t 2 +4.7t. Calculate the power.
a) 0
b) 1
c) 16.854t 3 +22.41t 2 +6.62t
d) 15.3t 3 +27.8t 2 +19
Answer: c
Explanation: P= L*i*.
