Basic Electrical Engineering Pune University MCQs

 This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Series Circuits”.

1. Find the current in the circuit.

basic-electrical-engineering-questions-answers-series-circuits-q1

a) 1 A

b) 2 A

c) 3 A

d) 4 A

Answer: b

Explanation: I=V/R. Total resistance R = 20+40=60ohm. V=120V. I=120/60=2A.

2. In a series circuit, which of the parameters remain constant across all circuit elements such as resistor, capacitor and inductor etcetera?

a) Voltage

b) Current

c) Both voltage and current

d) Neither voltage nor current

Answer: b

Explanation: In a series circuit, the current across all elements remain the same and the total voltage of the circuit is the sum of the voltages across all the elements.

3. Voltage across the 60ohm resistor is______

basic-electrical-engineering-questions-answers-series-circuits-q3

a) 72V

b) 0V

c) 48V

d) 120V

Answer: b

Explanation: The 60ohm resistance is shorted since current always choses the low resistance path. Voltage across short circuit is equal to zero, hence voltage across the resistor is 0.

4. Find the voltage across the 6 ohm resistor.

basic-electrical-engineering-questions-answers-series-circuits-q4

a) 150V

b) 181.6V

c) 27.27V

d) 54.48V

Answer: c

Explanation: Total current I=150/=V.

V across 6 ohm = 6*I = 6*V = 27.27V.

5. If there are two bulbs connected in series and one blows out, what happens to the other bulb?

a) The other bulb continues to glow with the same brightness

b) The other bulb stops glowing

c) The other bulb glows with increased brightness

d) The other bulb also burns out

Answer: b

Explanation: Since the two bulbs are connected in series, if the first bulb burns out there is a break in the circuit and hence the second bulb does not glow.

6. What is the value of x if the current in the circuit is 5A?

basic-electrical-engineering-questions-answers-series-circuits-q6

a) 15 ohm

b) 25 ohm

c) 55 ohm

d) 75 ohm

Answer: a

Explanation: Total voltage=sum of voltages across each resistor. =>150=10*5+5*5+5*x. Solving the equation, we get x=15 ohm.

7. A voltage across a series resistor circuit is proportional to?

a) The amount of time the circuit was on for

b) The value of the resistance itself

c) The value of the other resistances in the circuit

d) The power in the circuit

Answer: b

Explanation: V=IR hence the voltage across a series resistor circuit is proportional to the value of the resistance.

8. Many resistors connected in series will?

a) Divide the voltage proportionally among all the resistors

b) Divide the current proportionally

c) Increase the source voltage in proportion to the values of the resistors

d) Reduce the power to zero

Answer: a

Explanation: In a series circuit, the current remains the same across all resistors hence the voltage divides proportionally among all resistors.

9. What is the voltage measured across a series short?

a) Infinite

b) Zero

c) The value of the source voltage

d) Null

Answer: b

Explanation: A short is just a wire. The potential difference between two points of a wire is zero hence the voltage measured is equal to zero.

10. What happens to the current in the series circuit if the resistance is doubled?

a) It becomes half its original value

b) It becomes double its original value

c) It becomes zero

d) It becomes infinity

Answer: a

Explanation: I=V/R. If R becomes 2R then I becomes I/2 i.e. half of its original value.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Parallel Networks”.

1. If two bulbs are connected in parallel and one bulb blows out, what happens to the other bulb?

a) The other bulb blows out as well

b) The other bulb continues to glow with the same brightness

c) The other bulb glows with increased brightness

d) The other bulb stops glowing

Answer: b

Explanation: If one bulb blows out, it acts as an open circuit. Current does not flow in that branch but it continues to flow in the other branch of the parallel circuit. Hence the other bulb continues to glow. Also the voltage across other bulb remains the same due to which power delivered to it remains the same so it continues to glow with the same brightness.

2. Calculate the current across the 20 ohm resistor.

basic-electrical-engineering-questions-answers-parallel-networks-q2

a) 10A

b) 20A

c) 6.67A

d) 36.67A

Answer: a

Explanation: I=V/R. Since in parallel circuit, voltage is same across all resistors. Hence across the 20 ohm resistor, V=200V so I=200/20=10A.

3. In a parallel circuit, with a number of resistors, the voltage across each resistor is ________

a) The same for all resistors

b) Is divided equally among all resistors

c) Is divided proportionally across all resistors

d) Is zero for all resistors

Answer: a

Explanation: In parallel circuits, the current across the circuits vary whereas the voltage remains the same.

4. The current in each branch of a parallel circuit is proportional to _________

a) The amount of time the circuit is on for

b) Proportional to the value of the resistors

c) Equal in all branches

d) Proportional to the power in the circuit

Answer: b

Explanation: I=V/R. In a parallel circuit, the voltage across each resistor is equal, hence the value of the current is proportional  to the value of the resistance.

5. Calculate the total current in the circuit.

basic-electrical-engineering-questions-answers-parallel-networks-q5

a) 20 A

b) 10 A

c) 11.43 A

d) 15 A

Answer: c

Explanation: The 1 ohm and 2 ohm resistor are in series which is in parallel to the 3 ohm resistor. The equivalent of these resistances  is in series with the 4 ohm and 5 ohm resistor. Total R = 21/2 ohm. I=V/R=120/=240/21=11.43 A.

6. The voltage across the open circuit is?

basic-electrical-engineering-questions-answers-parallel-networks-q6

a) 100V

b) Infinity

c) 90V

d) 0V

Answer: a

Explanation: The voltage across all branches in a parallel circuit is the same as that of the source voltage. Hence the voltage across the 10 ohm resistor and the open circuit is the same=100V.

7. The voltage across the short is?

basic-electrical-engineering-questions-answers-parallel-networks-q7

a) 135V

b) Infinity

c) Zero

d) 11.25V

Answer: c

Explanation: The voltage across a short is always equal to zero whether it is connected in series or parallel.

8. If the current through x ohm resistance in the circuit is 5A, find the value of x.

basic-electrical-engineering-questions-answers-parallel-networks-q8

a) 27 ohm

b) 5 ohm

c) 12 ohm

d) 135 ohm

Answer: a

Explanation: R=V/I. In this circuit I=5A and V=135V. Therefore, R=135/5=27 ohm.

9. The currents in the three branches of a parallel circuit are 3A, 4A and 5A. What is the current leaving it?

a) 0A

b) Insufficient data provided

c) The largest one among the three values

d) 12A

Answer: d

Explanation: The total current leaving a node is the same as the current that enters it. Total I=I1+I2+I3=3+4+5=12A.

10. The total resistance between A and B are?

basic-electrical-engineering-questions-answers-parallel-networks-q10

a) 20 ohm

b) 5 ohm

c) 80 ohm

d) 0 ohm

Answer: b

Explanation: The resistors are connected in parallel, hence the equivalent resistance = 1/=5A.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Series Circuits and Parallel Networks”.

1. It is preferable to connect bulbs in series or in parallel?

a) Series

b) Parallel

c) Both series and parallel

d) Neither series nor parallel

Answer: b

Explanation: Bulbs are connected in parallel so that even if one of the bulbs blow out, the others continue to get a current supply.

2. Calculate the total resistance between the points A and B.

basic-electrical-engineering-questions-answers-series-circuits-parallel-networks-q2

a) 7 ohm

b) 0 ohm

c) 7.67 ohm

d) 0.48 ohm

Answer: c

Explanation: 1 ohm in parallel with 2 ohm give 2/3 ohm equivalent which is in series with 4 ohm and 3 ohm so total resistance between A and B = 4 + 2/3 + 3 = 23/3 = 7.67 ohm.

3. Calculate the equivalent resistance between A and B.

basic-electrical-engineering-questions-answers-series-circuits-parallel-networks-q3

a) 60 ohm

b) 15 ohm

c) 12 ohm

d) 48 ohm

Answer: c

Explanation: 5 ohm and 15 ohm are connected in series to give 20 ohm.10ohm and 20 ohm are connected in series to give 30 ohm. Now both equivalent resistances  are in parallel to give equivalent resistance 20*30/ = 12 ohm.

4. Calculate the resistance between A and B.

basic-electrical-engineering-questions-answers-series-circuits-parallel-networks-q4

a) 3.56 ohm

b) 7 ohm

c) 14.26 ohm

d) 29.69 ohm

Answer: a

Explanation: The 1 ohm, 2 ohm and 3 ohm resistors are connected in parallel. Its equivalent resistance is in series with the 4 ohm resistor and the parallel connection of the 5 ohm and 6 ohm resistor. The equivalent resistance of this combination is 80/11 ohm. This is in parallel with 7 ohm to give equivalent resistance between A and B is 3.56 ohm.

5. Batteries are generally connected in______

a) Series

b) Parallel

c) Either series or parallel

d) Neither series nor parallel

Answer: a

Explanation: Batteries are generally connected in series so that we can obtain the desired voltage since voltages add up once they are connected in series.

6. In a _________ circuit, the total resistance is greater than the largest resistance in the circuit.

a) Series

b) Parallel

c) Either series or parallel

d) Neither series nor parallel

Answer: a

Explanation: In series circuits, the total resistance is the sum of all the resistance in the circuit, hence the total is greater than the largest resistance.

7. In a ____________ circuit, the total resistance is smaller than the smallest resistance in the circuit.

a) Series

b) Parallel

c) Either series or parallel

d) Neither series nor parallel

Answer: b

Explanation: in a parallel circuit, the equivalent resistance=1/sum of the reciprocals of all the resistances in the circuit. Hence it is smaller than the smallest resistance in the circuit.

8. Which is the most cost efficient connection?

a) Series

b) Parallel

c) Either series or parallel

d) Neither series nor parallel

Answer: a

Explanation: The advantage of series-connections is that they share the supply voltage, hence cheap low voltage appliances may be used.

9. Calculate the equivalent resistance between A and B.

basic-electrical-engineering-questions-answers-series-circuits-parallel-networks-q9

a) 2 ohm

b) 4 ohm

c) 6 ohm

d) 8 ohm

Answer: b

Explanation: R=||5)+1.5)||4. The 2 and the 3 ohm resistor are in series. The equivalent of these two resistors is in parallel with the 5 ohm resistor. The equivalent of these three resistances is in series with the 1.5 ohm resistor. Finally, the equivalent of these resistances is in parallel with the 4 ohm resistor.

10. Calculate the equivalent resistance between A and B.

basic-electrical-engineering-questions-answers-series-circuits-parallel-networks-q10

a) 6.67 ohm

b) 46.67 ohm

c) 26.67 ohm

d) 10.67 ohm

Answer: a

Explanation: R=20||20||20=6.67 ohm. The three 20 ohm resistors are in parallel and re-sistance is measured across this terminal.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Kirchhoff’s Current Law”.

1. Find the value of v if v1=20V and value of current source is 6A.

basic-electrical-engineering-questions-answers-kirchhoffs-current-law-q1

a) 10V

b) 12V

c) 14V

d) 16V

Answer: b

Explanation: The current through the 10 ohm resistor=v1/10=2A.Applying KCL at node 1: i5=i10+i2. i2=6-2=4A.

Thus the drop in the 2 ohm resistor = 4×2 = 8V.

v1=20V; hence v2=20-v across 2 ohm resistor=20-8=12V

v2=v since they are connected in parallel.

v=12V.

2. Calculate the current A.

basic-electrical-engineering-questions-answers-kirchhoffs-current-law-q2

a) 5A

b) 10A

c) 15A

d) 20A

Answer: c

Explanation: KCl states that the total current leaving the junction is equal to the current entering it. In this case, the current entering the junction is 5A+10A=15A.

3. Calculate the current across the 20 ohm resistor.

basic-electrical-engineering-questions-answers-kirchhoffs-current-law-q3

a) 20A

b) 1A

c) 0.67A

d) 0.33A

Answer: d

Explanation: Assume lower terminal of 20 ohm at 0V and upper terminal at V volt and applying KCL, we get V/10 +V/20 =1. V=20/3V So current through 20 ohm = V/20 = /20 =1/3=0.33V.

4. Calculate the value of I3, if I1= 2A and I2=3A.

basic-electrical-engineering-questions-answers-kirchhoffs-current-law-q4

a) -5A

b) 5A

c) 1A

d) -1A

Answer: a

Explanation: According to KCl, I1+I2+I3=0. Hence I3=-=-5A.

5. Find the value of i2, i4 and i5 if i1=3A, i3=1A and i6=1A.

basic-electrical-engineering-questions-answers-kirchhoffs-current-law-q5a

a) 2,-1,2

b) 4,-2,4

c) 2,1,2

d) 4,2,4

Answer: a

Explanation: At junction a: i1-i3-i2=0. i2=2A.

At junction b: i4+i2-i6=0. i4=-1A.

At junction c: i3-i5-i4=0. i5=2A.

6. What is the value of current if a 50C charge flows in a conductor over a period of 5 seconds?

a) 5A

b) 10A

c) 15A

d) 20A

Answer: b

Explanation: Current=Charge/Time. Here charge = 50c and time = 5seconds, so current = 50/5 = 10A.

7. KCL deals with the conservation of?

a) Momentum

b) Mass

c) Potential Energy

d) Charge

Answer: d

Explanation: KCL states that the amount of charge entering a junction is equal to the amount of charge leaving it, hence it is the conservation of charge.

8. KCL is applied at _________

a) Loop

b) Node

c) Both loop and node

d) Neither loop nor node

Answer: b

Explanation: KCL states that the amount of charge leaving a node is equal to the amount of charge entering it, hence it is applied at nodes.

9. KCL can be applied for __________

a) Planar networks

b) Non-planar networks

c) Both planar and non-planar

d) Neither planar nor non-planar

Answer: c

Explanation: KCL is applied for different nodes of a network whether it is planar or non-planar.

10. What is the value of the current I?

basic-electrical-engineering-questions-answers-kirchhoffs-current-law-q10

a) 8A

b) 7A

c) 6A

d) 5A

Answer: a

Explanation: At the junction, I-2+3-4-5=0. Hence I=8A.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Kirchhoff’s Voltage Law”.

1. Calculate the value of V1 and V2.

basic-electrical-engineering-questions-answers-kirchhoffs-voltage-law-q1

a) 4V, 6V

b) 5V, 6V

c) 6V, 7V

d) 7V, 8V

Answer: a

Explanation: Using KVL, 12-V1-8=0. V1= 4V.

8-V2-2=0. V2=6V.

2. KVL deals with the conservation of?

a) Mass

b) Momentum

c) Charge

d) Energy

Answer: d

Explanation: KVL states that the sum of the potential energy and taken with the right sign is equal to zero, hence it is the conservation of energy since energy doesn’t enter or leave the system.

3. Calculate the voltage across the 10 ohm resistor.

basic-electrical-engineering-questions-answers-kirchhoffs-voltage-law-q3

a) 12V

b) 4V

c) 10V

d) 0V

Answer: b

Explanation: Total resistance = 5+10+15 = 30 ohm. Current in the circuit is 12/30 A.

Voltage across 10 ohm resistor is 10* = 4V.

4. Find the value of the currents I1 and I2.

basic-electrical-engineering-questions-answers-kirchhoffs-voltage-law-q4

a) 0.3, 0.1

b) -0.1, -0.3

c) -0.3, -0.1

d) 0.1, 0.2

Answer: d

Explanation: Using KVL in loop 1, 10-100 i1=0. i1=0.1A

Using KVL in outer loop, -100i2+20=0 i2=0.2A.

5. The sum of the voltages over any closed loop is equal to __________

a) 0V

b) Infinity

c) 1V

d) 2V

Answer: a

Explanation: According to KVL, the sum of the voltage over any closed loop is equal to 0.

6. What is the basic law that has to be followed in order to analyze the circuit?

a) Newton’s laws

b) Faraday’s laws

c) Ampere’s laws

d) Kirchhoff’s law

Answer: d

Explanation: Kirchhoff’s laws, namely Kirchhoff’s Current Law and Kirchhoff’s Voltage law are the basic laws in order to analyze a circuit.

7. Every____________ is a ____________ but every __________ is not a __________

a) Mesh, loop, loop, mesh

b) Loop, mesh, mesh, loop

c) Loop, mesh, loop, mesh

d) Mesh, loop, mesh, loop

Answer: a

Explanation: According to Kirchhoff’s Voltage Law, Every mesh is a loop but every loop is not a mesh. Mesh is a special case of loop which is planar.

8. What is the voltage across the 5 ohm resistor if current source has current of 17/3 A?

basic-electrical-engineering-questions-answers-kirchhoffs-voltage-law-q8

a) 2.32V

b) 5.21V

c) 6.67V

d) 8.96V

Answer: b

Explanation: Assuming i1 and i2 be the currents in loop 1 and 2 respectively. In loop 1, 4+2i1+3+4+5=0

In loop 2, i2-4i1-5=0 =>-4i1+10i2=5.

Solving these equations simultaneously i2=1.041A and i1=1.352A

V=i2*5= 5.21V.

9. Calculate VAB.

basic-electrical-engineering-questions-answers-kirchhoffs-voltage-law-q9

a) 3.5V

b) 12V

c) 9.5V

d) 6.5V

Answer: a

Explanation: For branch A: VAC=15*20/=7.5V

For branch B: VBC= 10*20/=4V

Applying KVL to loop ABC:

VAB+VBC+VCA=0

VAB=3.5V.

10. KVL is applied in ____________

a) Mesh analysis

b) Nodal analysis

c) Both mesh and nodal

d) Neither mesh nor nodal

Answer: a

Explanation: Mesh analysis helps us to utilize the different voltages in the circuit as well as the IR products in the circuit which is nothing but KVL.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Power”.

1. Which of the following is not an expression power?

a) P=VI

b) P=I 2 R

c) P=V 2 /R

d) P=I/R

Answer d

Explanation: Power is the product of voltage and current. Writing I in terms of V, we get P=V 2 /R and writing V in terms of I, we get P=I 2 r.

2. Which of the following statements are true?

a) Power is proportional to voltage only

b) Power is proportional to current only

c) Power is neither proportional to voltage nor to the current

d) Power is proportional to both the voltage and current

Answer: d

Explanation: Power is proportional to both voltage and current.

3. A 250V bulb passes a current of 0.3A. Calculate the power in the lamp.

a) 75W

b) 50W

c) 25W

d) 90W

Answer: a

Explanation: Here, V = 250v and I = 0.3A. P=VI. Which implies that, P=250*0.3=75W.

4. Kilowatt-hour is a unit of?

a) Current

b) Power

c) Energy

d) Resistance

Answer: c

Explanation: Power is the energy per unit time. That is, P=E/t. If the unit of power in kW and the unit of time is an hour, then the unit of energy=unit of power*unit of time=kWh.

5. Calculate the power in the 20 ohm resistance.

basic-electrical-engineering-questions-answers-power-q5

a) 2000kW

b) 2kW

c) 200kW

d) 2W

Answer: b

Explanation: Here V = 200v and Resistance = 20ohm. P=V 2 /R= 200 2 /20=2000W=2kW.

6. A current of 5A flows in a resistor of 2 ohms. Calculate the energy dissipated in 300 seconds in the resistor.

a) 15kJ

b) 15000kJ

c) 1500J

d) 15J

Answer: a

Explanation: P=I 2 R =5 2 *2=50W.

E= Pt=50*300=15000J=15kJ.

7. Calculate the power across each 20 ohm resistance.

basic-electrical-engineering-questions-answers-power-q7

a) 1000W, 1000W

b) 500W, 500W

c) 1000kW, 1000kW

d) 500kW, 500kW

Answer: b

Explanation: This is a series connected circuit hence the current across each resistance is the same. To find current: I=V/R=200/20=5A.

To find power: P=I 2 R=5 2 *20=500W. Since both the resistors have a resistance of 20 ohms, the power across both is the same.

8. Calculate the power across each 10 ohm resistance.

basic-electrical-engineering-questions-answers-power-q8

a) 1000kW, 1000kW

b) 1kW, 1kW

c) 100W, 100W

d) 100kW, 100kW

Answer: b

Explanation: This is parallel connected circuit, hence the voltage across each of the resistors is the same. P =(V 2 )/R=(100 2 )/10 = 1000W=1kW. Since both the resistors receive the same amount of voltage, the power in both is the same.

9. Calculate the work done in a resistor of 20 ohm carrying 5A of current in 3 hours.

a) 1.5J

b) 15J

c) 1.5kWh

d) 15kWh

Answer: c

Explanation: To find power: P=I 2 R=5 2 *20=500W=0.5kW.

To find Work done: W=Pt=0.5*3=1.5kWh.

10. The SI unit of power is?

a) kW

b) J/s

c) Ws

d) J/h(joules per hour

Answer: b

Explanation: Power = energy/time

SI unit of power = SI unit of energy/SI unit of time = joule/second.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Energy”.

1. Which among the following is a unit for electrical energy?

a) V

b) kWh

c) Ohm

d) C

Answer: b

Explanation: Kilowatt is a unit of power and hour is a unit of time. Energy is the product of power and time, hence the unit for Energy is kWh.

2. A bulb has a power of 200W. What is the energy dissipated by it in 5 minutes?

a) 60J

b) 1000J

c) 60kJ

d) 1kJ

Answer: c

Explanation: Here, Power = 200w and time = 5min. E=Pt => E= 200*5= 1000Wmin=60000Ws= 60000J= 60kJ.

3. Out of the following, which one is not a source of electrical energy?

a) Solar cell

b) Battery

c) Potentiometer

d) Generator

Answer: c

Explanation: Solar cell converts light energy to electrical energy. Battery converts chemical energy to electrical energy. Generator generates electrical energy using electromagnetic induction. A potentiometer is an instrument used for measuring voltage and consumes electrical energy instead of generating it.

4. Calculate the energy dissipated by the circuit in 50 seconds.

basic-electrical-engineering-questions-answers-energy-q4

a) 50kJ

b) 50J

c) 100j

d) 100kJ

Answer: a

Explanation: Here V = 100 and R = 10. Power in the circuit= V 2 /R = 100 2 /10 = 1000W.

Energy= Pt= 1000*50 = 50000J = 50kJ.

5. Which among the following is an expression for energy?

a) V 2 It

b) V 2 Rt

c) V 2 t/R

d) V 2 t 2 /R

Answer: c

Explanation: Expression for power = VI, substituting I from ohm’s law we can write, P=V 2 /R. Energy is the product of power and time, hence E=Pt = V 2 t/R.

6. Calculate the energy in the 10 ohm resistance in 10 seconds.

basic-electrical-engineering-questions-answers-energy-q6

a) 400J

b) 40kJ

c) 4000J

d) 4kJ

Answer: b

Explanation: Since the resistors are connected in parallel, the voltage across both the resistors are the same, hence we can use the expression P=V 2 /R. P=200 2 /10= 4000W. E=Pt = 4000*10=40000Ws = 40000J = 40kJ.

7. A battery converts___________

a) Electrical energy to chemical energy

b) Chemical energy to electrical energy

c) Mechanical energy to electrical energy

d) Chemical energy to mechanical energy

Answer: b

Explanation: A battery is a device in which the chemical elements within the battery react with each other to produce electrical energy.

8. A current of 2A flows in a wire offering a resistance of 10ohm. Calculate the energy dissipated by the wire in 0.5 hours.

a) 72Wh

b) 72kJ

c) 7200J

d) 72kJh

Answer b

Explanation: Here I  = 2A and Resistance = 10ohm. Power = I 2 R = 2 2 *10=40. Energy = Pt = 40*0.5*60*60 = 72000J=72kJ.

9. Calculate the energy in the 5 ohm resistor in 20 seconds.

basic-electrical-engineering-questions-answers-energy-q9

a) 21.5kJ

b) 2.15kJ

c) 2.15J

d) 21.5kJ

Answer: a

Explanation: The current in the circuit is equal to the current in the 5 ohm resistor since it a series connected circuit, hence I=220/=14.67A. P=I 2 R = 14.67 2 *5=1075.8W. E=Pt = 1075.8*20 = 21516J=21.5kJ.

10. Practically, if 10kJ of energy is supplied to a device, how much energy will the device give back?

a) Equal to10kJ

b) Less than 10kJ

c) More than 10kJ

d) Zero

Answer: b

Explanation: Practically, if 10kJ of energy is supplied to a system, it returns less than the supplied energy because, some of the energy is lost as heat energy, sound energy etc.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Resistivity”.

1. Materials which easily allow the passage of electric current are known as ______

a) Insulators

b) Conductors

c) Dielectrics

d) Semi-conductors

Answer: b

Explanation: Conductors are materials, which freely allow the passage of electrons through it. If electrons can flow freely through a material, it implies that even current can flow freely through that material since current is the rate of flow of electrons.

2. A wire of length 2m and another wire of length 5m are made up of the same material and have the same area of cross section, which wire has higher resistance?

a) Both have equal resistance

b) The 2m wire has higher resistance

c) The 5m wire has higher resistance

d) The value of resistance cannot be determined from the given data

Answer: c

Explanation: Resistance is directly proportional to the length of the wire, hence as the length of the wire increases, resistance increases.

3. A wire having an area of cross section = 10sqm and another wire having an area of cross section= 15sqm, have the same length and are made up of the same material. Which wire has more resistance?

a) Both have equal resistance

b) The 10sqm wire has higher resistance

c) The 15sqm wire has higher resistance

d) The value of resistance cannot be determined from the given data

Answer: b

Explanation: Resistance is inversely proportional to the area of cross-section. As an area of cross-section increases, resistance decreases. Hence the 10sqm wire has a higher resistance than the 15sqm wire.

4. Which of the following statements are true with regard to resistance?

a) Resistance is directly proportional to a length of the wire

b) Resistance is directly proportional to an area of cross section of the wire

c) Resistance is inversely proportional to the length of the wire

d) Resistance is inversely proportional to the resistivity of the wire

Answer: a

Explanation: The expression for resistance is: Resistance=Resistivity*length of wire/ area of cross section of the wire. Hence resistance is directly proportional to length.

5. A wire has the same resistance as the one given in the figure. Calculate its resistivity if the length of the wire is 10m and its area of cross section is 2m.

basic-electrical-engineering-questions-answers-resistivity-q5

a) 16 ohm-metre

b) 8 ohm-metre

c) 16 kiloohm-metre

d) 8 kiloohm-metre

Answer: b

Explanation: From the given circuit, R=V/I = 200/5 = 40ohm.

Resistivity= Resistance*Area of cross section/ Length of the wire.

Resistivity= 40*2/10= 8 ohm-metre.

6. Which, among the following is a unit for resistivity?

a) ohm/metre

b) ohm/metre 2

c) ohm-metre

d) ohm-metre 2

Answer: c

Explanation: Resistivity = Resistance* Length/area of cross section.

Unit of resistivity = ohm*(m 2 )/m = ohm-m.

7. What is the resistivity of Copper?

a) 1.59*10 -8 ohm-m

b) 2.7*10 -8 ohm-m

c) 7.3*10 -8 ohm-m

d) 5.35*10 -8 ohm-m

Answer: a

Explanation: Resistivity is a material property. Different materials have different resistivity. Resistivity of copper is 1.72*10 -8 ohm-m.

8. Calculate the ratio of the resistivity of 2 wires having the same length and same resistance with area of cross section 2m 2 and 5m 2 respectively.

a) 5:7

b) 2:7

c) 2:5

d) 7:5

Answer: c

Explanation: Resistivity = R*A/L

Since resistance and length of the two wires are same so resistivity is directly proportional to area of cross section. Ratio of area is 2:5 so the ratio of resistivity is also 2:5.

9. Which of the following statements are true with regard to resistivity?

a) Resistance depends on the temperature

b) Resistance does not depend on the temperature

c) Resistivity depend on the length

d) Resistivity depend on area of cross section

Answer: a

Explanation: Resistivity is material property. It depends only on temperature.

For the same material with different length and area, resistivity remains the same until temperature remains constant.

10. The reciprocal of resistivity is________

a) Conductance

b) Resistance

c) Conductivity

d) Impedance

Answer: c

Explanation: The expression for resistivity is = RA/l. The expression for conductivity = Cl/A; C=1/R => Conductivity = l/ = 1/resistivity. Hence, conductivity is the reciprocal of resistivity.

This set of Basic Electrical Engineering Interview Questions and Answers focuses on “Temperature Coefficient of Resistance”.

1. The resistance of pure metals ___________

a) Increases with an increase in temperature

b) Decreases with an increase in temperature

c) Remains the same with an increase in temperature

d) Becomes zero with an increase in temperature

Answer: a

Explanation: In a conductor, the valence band and conduction band overlap each other, there is an excess of electrons in the conduction band. When the temperature increases, there is an overcrowding of electrons in the conduction band hence reducing the mobility and hence resistance increases.

2. The resistance of insulators __________

a) Increases with an increase in temperature

b) Decreases with an increase in temperature

c) Remains the same with an increase in temperature

d) Becomes zero with an increase in temperature

Answer: b

Explanation: In the case of an insulator, the energy gap between the conduction band and the valence band is very large. When the temperature is increased, the electrons move from the conduction band to the valence band and hence it starts conducting. When conductance increases, resistance decreases, since C=1/R. Thus, when the temperature increases, resistance decreases in insulators.

3. Which of the following statements are true about metals?

a) Metals have a positive temperature coefficient

b) Metals have a negative temperature coefficient

c) Metals have zero temperature coefficient

d) Metals have infinite temperature coefficient

Answer: a

Explanation: The resistance of metals increases with an increase in temperature thus, it has a positive temperature coefficient.

4. Which of the following statements are true about insulators?

a) Insulators have a positive temperature coefficient

b) Insulators have a negative temperature coefficient

c) Insulators have zero temperature coefficient

d) Insulators have infinite temperature coefficient

Answer: b

Explanation: Insulators have a negative temperature coefficient because as temperature increases, the resistance of insulators decreases.

5. What is the unit of temperature coefficient?

a) ohm/centigrade

b) ohm-centigrade

c) centigrade -1

d) centigrade

Answer: c

Explanation: R=Reff[1+temp. coeff].

From the given expression: / = temp. coeff. Hence, the unit is the reciprocal of that of temperature = centigrade -1 .

6. A copper coil has a resistance of 200 ohms when its mean temperature is 0 degree centigrade. Calculate the resistance of the coil when its mean temperature is 80 degree centigrade. Temperature coefficient of copper is 0.004041 centigrade -1

a) 264.65 ohm

b) 264.65 kilo-ohm

c) 286.65 ohm

d) 286.65 kilo-ohm

Answer: a

Explanation: R=R 0  = 200 = 264.65 ohm.

7. The temperature of a coil cannot be measured by which of the following methods?

a) Thermometer

b) Increase in resistance of the coil

c) Thermo-junctions embedded in the coil

d) Calorimeter

Answer: d

Explanation: Calorimeter measures the amount of heat and not the temperature of the coil. The temperature of a coil is mainly measured by thermometer. Resistance of coil increase with an increase in temperature of coil so we can measure temperature using this method. Another method is the formation of thermocouple inside coil due to high temperature at one end and low temperature at other ends.

8. The rise or fall in resistance with the rise in temperature depends on ________

a) The property of the conductor material

b) The current in the metal

c) Property of material as well current in that material

d) Does not depend on any factor

Answer: a

Explanation: The rise or fall in resistance with a rise in temperature depends upon the property of the material. Hence it rises with temperature in metals and falls with temperature in insulators and semi-conductors.

9. If the temperature is increased in semi-conductors such that the resistance incessantly falls, it is termed as _______

a) Avalanche breakdown

b) Zener breakdown

c) Thermal runway

d) Avalanche runway

Answer c

Explanation: When the temperature keeps increasing, the resistance keeps falling continuously and hence the current to increase. This causes the heat in the semi-conductor to rise. This causes the temperature to increase further and the resistance to further decrease. This process continues and until there is sufficient heat to destroy the structure of the semi-conductor completely. This is known as a thermal runway.

10. Materials having resistance almost equal to zero is _______

a) Semi-conductor

b) Conductor

c) Superconductors

d) Insulators

Answer: c

Explanation: When the temperature of the material falls to absolute zero, the resistance falls to zero and hence there are no I 2 R losses. Since resistance is zero, conductance is almost infinity and hence these materials are known as superconductors.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Kirchhoff’s Laws and Network Solutions”.

1. Find the value of I1, I2 and I3.

basic-electrical-engineering-questions-answers-kirchhoffs-laws-network-solutions-q1

a) -0.566A, 1.29A, -1.91A

b) -1.29A, -0.566A, 1.91A

c) 1.29A, -0.566A, -1.91A

d) 1.91A, 0.566A, 1.29A

Answer: c

Explanation: Using the matrix method:

Matrix =

 =

 =

Solving this matrix equation, we get I1 = 1.29A, I2 = -0.566A and I3 = -1.91A.

2. Find the value of V, if the value of I3= 0A.

basic-electrical-engineering-questions-answers-kirchhoffs-laws-network-solutions-q2

a) 1.739 V

b) 6.5 V

c) 4.5V

d)2.739V

Answer: a

Explanation: 5-3I1+2I2=0, 9I2-2I1=0, -4I2+V=0

On solving,V=1.739V.

3. Find the value of R if the power in the circuit is 1000W.

basic-electrical-engineering-questions-answers-kirchhoffs-laws-network-solutions-q3

a) 10 ohm

b) 9 ohm

c) 8 ohm

d) 7 ohm

Answer: c

Explanation: To find the value of I:

VI=P =>100I=1000 => I=10A.

Voltage across the 2 ohm resistor = 20V.

Voltage across the R resistor = 100-20= 80V.

R=V/I => R=80/10 = 8A.

4. Find the current in the 4 ohm resistor.

basic-electrical-engineering-questions-answers-kirchhoffs-laws-network-solutions-q4

a) 5A

b) 0A

c) 2.2A

d) 20A

Answer: b

Explanation: The 4 ohm resistor gets shorted since current always prefers the low resistance path. All the current flows to the branch which is connected in parallel to the 4 ohm branch, hence no current flows in the 4 ohm resistance.

5. Nodal analysis is generally used to determine______

a) Voltage

b) Current

c) Resistance

d) Power

Answer: a

Explanation: Nodal analysis uses Kirchhoff’s Current Law to find all the node voltages. Hence it is a method used to determine the voltage.

6. Mesh analysis is generally used to determine_________

a) Voltage

b) Current

c) Resistance

d) Power

Answer: b

Explanation: Mesh analysis uses Kirchhoff’s Voltage Law to find all the mesh currents. Hence it is a method used to determine current.

7. What is the current in the circuit?

basic-electrical-engineering-questions-answers-kirchhoffs-laws-network-solutions-q7

a) 0A

b) 15A

c) 5A

d) 10A

Answer: a

Explanation: If we move in the clockwise direction, we get the total voltage to be equal to: -10-20+30 = 0V. Since I=V/R = 0/4=0, I=0A.

8. Does the 15A source have any effect on the circuit?

basic-electrical-engineering-questions-answers-kirchhoffs-laws-network-solutions-q8

a) Yes

b) No

c) Cannot be determined

d) Yes, only when the 10V source is removed

Answer: b

Explanation: The 15A current source has a lower resistance path associated with it and hence it keeps moving in that particular loop. It does not leave that loop and enter the circuit, hence the circuit is not affected by it.

9. KVL is associated with____________

a) Mesh analysis

b) Nodal analysis

c) Both mesh and nodal

d) Neither mesh nor nodal

Answer: a

Explanation: KVL employs mesh analysis to find the different mesh currents by finding the IR products in each mesh.

10. KCL is associated with_________

a) Mesh analysis

b) Nodal analysis

c) Both mesh and nodal

d) Neither mesh nor nodal

Answer: b

Explanation: KCL employs nodal analysis to find the different node voltages by finding the value if a current in each branch.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Mesh Analysis”.

1. Find the value of the currents I1, I2 and I3 flowing clockwise in the first, second and third mesh respectively.

basic-electrical-engineering-questions-answers-mesh-analysis-q1

a) 1.54A, -0.189A, -1.195A

b) 2.34A, -3.53A, -2.23A

c) 4.33A, 0.55A, 6.02A

d) -1.18A, -1.17A, -1.16A

Answer: a

Explanation: The three mesh equations are:

-3I1+2I2-5=0

2I1-9I2+4I3=0

4I2-9I3-10=0

Solving the equations, we get I1= 1.54A, I2=-0.189 and I3= -1.195A.

2. Find the value of the currents I1 and I2 flowing clockwise in the first and second mesh respectively.

basic-electrical-engineering-questions-answers-mesh-analysis-q2

a) 0.96A, 1.73A

b) 0.96A, -1.73A

c) -0.96A, 1.73A

d) -0.96A, -1.73A

Answer: b

Explanation: The two mesh equations are:

5I1-3I2=10

-3I1+7I2=-15

Solving the equations simultaneously, we get I1=0.96A and I2=-1.73A.

3. Find the value of V if the current in the 3 ohm resistor=0.

basic-electrical-engineering-questions-answers-mesh-analysis-q3

a) 3.5V

b) 6.5V

c) 7.5V

d) 8.5V

Answer: c

Explanation: Taking the mesh currents in the three meshes as I1, I2 and I3, the mesh equations are:

3I1+0I2+0V=5

-2I1-4I2+0V=0

0I1+9I2+V=0

Solving these equations simultaneously and taking the value of I2=0, we get V=7.5V.

4. Find the value of V1 if the current through the 1 ohm resistor=0A.

basic-electrical-engineering-questions-answers-mesh-analysis-q4

a) 83.33V

b) 78.89V

c) 87.87V

d) 33.33V

Answer: a

Explanation: Taking I1, I2 and I3 as the currents in the three meshes and taking I3=0 since it is the current across the 1 ohm resistor, the three mesh equations are:

15I1-5I2=V1

-5I1+10I2=0

3I2=10

Solving these equations simultaneously we get V1= 83.33V.

5. Calculate the mesh currents I1 and I2 flowing in the first and second meshes respectively.

basic-electrical-engineering-questions-answers-mesh-analysis-q5

a) 1.75A, 1.25A

b) 0.5A, 2.5A

c) 2.3A, 0.3A

d) 3.2A, 6.5A

Answer: a

Explanation: In this circuit, we have a super mesh present.

Let I1 and I2 be the currents in loops in clockwise direction. The two mesh equations are:

I 2 -I 1 =3

-5I 1 -3I 2 =5

Solving these equations simultaneously, we get I 1 = -1.75A and I 2 = 1.25A.

Since no specific direction given so currents in loop 1 and loop 2 are 1.75A and 1.25A respectively.

6. I1 is the current flowing in the first mesh. I2 is the current flowing in the second mesh and I3 is the current flowing in the top mesh. If all three currents are flowing in the clockwise direction, find the value of I1, I2 and I3.

basic-electrical-engineering-questions-answers-mesh-analysis-q6

a) 7.67A, 10.67A, 2A

b) 10.67A, 7.67A, 2A

c) 7.67A, 8.67A, 2A

d) 3.67A, 6.67A, 2A

Answer: a

Explanation: The two meshes which contain the 3A current is a super mesh. The three mesh equations therefore are:

I 3 =2A

I 2 -I 1 =3

-2I 1 -I 2 =-26

Solving these equations simultaneously we get:

I 1 =7.67A, I 2 =10.67A and I 3 =2A.

7. Calculate the mesh currents.

basic-electrical-engineering-questions-answers-mesh-analysis-q7

a) 7A, 6A, 6.22A

b) 2A, 1A, 0.57A

c) 3A, 4A, 5.88A

d) 6A, 7A, 8.99A

Answer: b

Explanation: The two meshes which contain the 3A source, act as a supper mesh. The mesh equations are:

I 1 -I 3 =-3

4I 1 -14I 2 +11I 3 =10

4I 1 -28I 2 +10I 3 =0

Solving these equations simultaneously, we get the three currents as

I 1 =-1A, I 2 =0.57A, I 3 =2A

So currents are 2A, 1A, 0.57A.

8. Mesh analysis employs the method of ___________

a) KVL

b) KCL

c) Both KVL and KCL

d) Neither KVL nor KCL

Answer: a

Explanation: KVL employs mesh analysis to find the different mesh currents by finding the IR products in each mesh.

9. Mesh analysis is generally used to determine _________

a) Voltage

b) Current

c) Resistance

d) Power

Answer: b

Explanation: Mesh analysis uses Kirchhoff’s Voltage Law to find all the mesh currents. Hence it is a method used to determine current.

10. Mesh analysis can be used for __________

a) Planar circuits

b) Non-planar circuits

c) Both planar and non-planar circuits

d) Neither planar nor non-planar circuits

Answer: a

Explanation: If the circuit is not planar, the meshes are not clearly defined. In planar circuits, it is easy to draw the meshes hence the meshes are clearly defined.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Nodal Analysis”.

1. Find the value of the node voltage V.

basic-electrical-engineering-questions-answers-nodal-analysis-q1

a) -60V

b) 60V

c) 40V

d) -40V

Answer: a

Explanation: The node equation is:

-2+8+V/10=0 => 6 + v/10 = 0 => v = -10*6 = -60V

Solving this equation, we get V = -60V.

2. Calculate the node voltages V1 and V2.

basic-electrical-engineering-questions-answers-nodal-analysis-q2

a) 12V, 13V

b) 26.67V, 11.33V

c) 11.33V, 26.67V

d) 13V, 12V

Answer: c

Explanation: The nodal equations are:

2V1-V2=-4

-4V1+5V2=88

Solving these equations simultaneously, we get V1=11.33V and V2=26.67V.

3. Find the node voltage V.

basic-electrical-engineering-questions-answers-nodal-analysis-q3

a) 1V

b) 2V

c) 3V

d) 4V

Answer: d

Explanation: The nodal equation is:

/2+/3+V/1=0

Solving for V, we get V=4V.

4. Calculate the node voltages.

basic-electrical-engineering-questions-answers-nodal-analysis-q4a

a) 30.77V, 7.52V, 18.82V

b) 32.34V, 7.87V, 8.78V

c) 34.34V, 8.99V, 8.67V

d) 45.44V, 6.67V, 7.77V

Answer: a

Explanation: The nodal equations, considering V1, V2 and V3 as the first, second and third node respectively, are:

-8+/3-3+/4=0

3+V2+/7+/3=0

-2.5+/7+/4+V3/5=0

Solving the equations simultaneously, we get V1=30.77V, V2=7.52V and V3=18.82V.

5. Find the value of V1 and V2.

basic-electrical-engineering-questions-answers-nodal-analysis-q5

a) 87.23V, 29.23V

b) 23.32V, 46.45V

c) 64.28V, 16.42V

d) 56.32V, 78, 87V

Answer: c

Explanation: The nodal equations are:

0.3V1-0.2V2=16

-V1+3V2=-15

Solving these equations simultaneously, we get V1=64.28V and V2=16.42V.

6. Nodal analysis is generally used to determine_______

a) Voltage

b) Current

c) Resistance

d) Power

Answer: a

Explanation: Nodal analysis uses Kirchhoff’s Current Law to find all the node voltages. Hence it is a method used to determine the voltage.

7. If there are 10 nodes in a circuit, how many equations do we get?

a) 10

b) 9

c) 8

d) 7

Answer: b

Explanation: One node is taken as reference node so, the number of equations we get is always one less than the number of nodes in the circuit, hence for 10 nodes we get 9 equations.

8. Nodal analysis can be applied for________

a) Planar networks

b) Non-planar networks

c) Both planar and non-planar networks

d) Neither planar nor non-planar networks

Answer: c

Explanation: Nodal analysis can be applied for both planar and non-planar networks since each node, whether it is planar or non-planar, can be assigned a voltage.

9. How many nodes are taken as reference nodes in a nodal analysis?

a) 1

b) 2

c) 3

d) 4

Answer: a

Explanation: In the nodal analysis, one node is treated as the reference node and the voltage at that point is taken as 0.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Superposition Theorem”.

1. In superposition theorem, when we consider the effect of one voltage source, all the other voltage sources are ____________

a) Shorted

b) Opened

c) Removed

d) Undisturbed

Answer: a

Explanation: In superposition theorem when we consider the effect of one voltage source, all the other voltage sources are shorted and current sources are opened.

2. In superposition theorem, when we consider the effect of one current source, all the other voltage sources are ____________

a) Shorted

b) Opened

c) Removed

d) Undisturbed

Answer: a

Explanation: In superposition theorem, whether we consider the effect of a voltage or current source, voltage sources are always shorted and current sources are always opened.

3. In superposition theorem, when we consider the effect of one voltage source, all the other current sources are ____________

a) Shorted

b) Opened

c) Removed

d) Undisturbed

Answer: b

Explanation: In superposition theorem when we consider the effect of one voltage source, all the other current sources are opened and voltage sources are shorted.

4. In superposition theorem, when we consider the effect of one current source, all the other current sources are ____________

a) Shorted

b) Opened

c) Removed

d) Undisturbed

Answer: b

Explanation: In superposition theorem, whether we consider the effect of a voltage or current source, current sources are always opened and voltage sources are always shorted.

5. Find the value of Vx due to the 16V source.

basic-electrical-engineering-questions-answers-superposition-theorem-q5

a) 4.2V

b) 3.2V

c) 2.3V

d) 6.3V

Answer: b

Explanation: When we consider the 16V source, we short the 10V source and open the 15A and 3A source. From the resulting series circuit we can use voltage divider to find Vx.

Vx = 16*20/=3.2A.

6. Find Vx due to the 3A source.

basic-electrical-engineering-questions-answers-superposition-theorem-q5

a) 56V

b) 78V

c) 38V

d) 48V

Answer: d

Explanation: Due to the 3A source, we short the 16V and 10V source and open the 15A source. From the resulting circuit, we can use current divider to find the current in the 20 ohm branch and then multiply it with the resistance to find the voltage.

I20 = 3*80/=2.4A

Vx=20*2.4=48V.

7. Find the value of Vx due to the 10V source.

basic-electrical-engineering-questions-answers-superposition-theorem-q5

a) 1V

b) 2V

c) 3V

d) 4V

Answer: b

Explanation: Due to the effect of the 10V source, we short the 16V source and open the 3A and 15A source. From the resulting series circuit, we can use voltage divider to find the value of Vx.

Vx=10*20/=2V.

8. Find the voltage due to the 15A source.

basic-electrical-engineering-questions-answers-superposition-theorem-q5

a) 0V

b) 2V

c) 4V

d) 6V

Answer: a

Explanation: Due to 15 A current source, 10V and 16V sources get shorted and the 3A source acts as an open circuit. Since the 10V source is shorted, it acts as a low resistance path and current flows only within that loop and do not flow to the 20 ohm resistor. Hence the voltage is 0V.

9. Superposition theorem is valid for _________

a) Linear systems

b) Non-linear systems

c) Both linear and non-linear systems

d) Neither linear nor non-linear systems

Answer: a

Explanation: Superposition theorem is valid only for linear systems because the effect of a single source cannot be individually calculated in a non-linear system.

10. Superposition theorem does not work for ________

a) Current

b) Voltage

c) Power

d) Works for all: current, voltage and power

Answer: c

Explanation: Power across an element is not equal to the power across it due to all the other sources in the system. The power in an element is the product of the total voltage and the total current in that element.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Thevenin’s Theorem”.

1. Calculate the Thevenin resistance across the terminal AB for the following circuit.

basic-electrical-engineering-questions-answers-thevenins-theorem-q1

a) 4.34 ohm

b) 3.67 ohm

c) 3.43 ohm

d) 2.32 ohm

Answer: b

Explanation: Thevenin resistance is found by opening the circuit between the specified terminal and shorting all voltage sources.

When the 10V source is shorted, we get:

Rth=+3=3.67 ohm.

2. Calculate V th for the given circuit.

basic-electrical-engineering-questions-answers-thevenins-theorem-q1

a) 5.54V

b) 3.33V

c) 6.67V

d) 3.67V

Answer: c

Explanation: 4 ohm is removed and then v across 2 ohm is calculated by voltage divider 2*10/ = 6.67V. Voltage between A and B i.e. V th is equal to voltage across 4 ohm resistance since no current flow through 3 ohm resistance. So, V th = 6.67V.

3. Calculate the current across the 4 ohm resistor.

basic-electrical-engineering-questions-answers-thevenins-theorem-q1

a) 0.86A

b) 1.23A

c) 2.22A

d) 0.67A

Answer: a

Explanation: Thevenin resistance is found by opening the circuit between the specified terminal and shorting all voltage sources.

When the 10V source is shorted, we get:

Rth=+3=3.67 ohm.

Vth is calculated by opening the specified terminal.

Using voltage divider, Vth= 2*10/=6.67V.

On drawing the Thevenin equivalent circuit, we get Rth, 4 ohm and Vth in series.

Applying Ohm’s law, I=Vth/ = 0.86A.

4. The Thevenin voltage is the__________

a) Open circuit voltage

b) Short circuit voltage

c) Open circuit and short circuit voltage

d) Neither open circuit nor short circuit voltage

Answer: a

Explanation: Thevenin voltage is obtained by opening the specified terminals so it is open circuit voltage. It is not the short circuit voltage because if specified terminals are shorted voltage is equal to zero.

5. Thevenin resistance is found by ________

a) Shorting all voltage sources

b) Opening all current sources

c) Shorting all voltage sources and opening all current sources

d) Opening all voltage sources and shorting all current sources

Answer: c

Explanation: Ideal current sources have infinite internal resistance hence behave like an open circuit whereas ideal voltage sources have zero internal resistance hence behave as a short circuit.

6. Thevenin’s theorem is true for __________

a) Linear networks

b) Non-Linear networks

c) Both linear networks and nonlinear networks

d) Neither linear networks nor non-linear networks

Answer: a

Explanation: Thevenin’s theorem works for only linear circuit elements and not non-linear ones such as BJT, semiconductors etc.

7. In Thevenin’s theorem Vth is __________

a) Sum of two voltage sources

b) A single voltage source

c) Infinite voltage sources

d) 0

Answer: b

Explanation: Thevenin’s theorem states that a combination of voltage sources, current sources and resistors is equivalent to a single voltage source V and a single series resistor R.

8. Vth is found across the ____________ terminals of the network.

a) Input

b) Output

c) Neither input nor output

d) Either input or output

Answer: b

Explanation: According to Thevenin’s theorem, Vth is found across the output terminals of a network and not the input terminals.

9. Which of the following is also known as the dual of Thevenin’s theorem?

a) Norton’s theorem

b) Superposition theorem

c) Maximum power transfer theorem

d) Millman’s theorem

Answer: a

Explanation: Norton’s theorem is also known as the dual of Thevenin’s theorem because in Norton’s theorem we find short circuit current which is the dual of open circuit voltage-what we find in Thevenin’s theorem.

10. Can we use Thevenin’s theorem on a circuit containing a BJT?

a) Yes

b) No

c) Depends on the BJT

d) Insufficient data provided

Answer: b

Explanation: We can use Thevenin’s theorem only for linear networks. BJT is a non-linear network hence we cannot apply Thevenin’s theorem for it.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Norton’s Theorem”.

1. The Norton current is the_______

a) Short circuit current

b) Open circuit current

c) Open circuit and short circuit current

d) Neither open circuit nor short circuit current

Answer: a

Explanation: Norton current is obtained by shorting the specified terminals. So, it is the short circuit current. It is not the open circuit current because if specified terminals get open circuited then current is equal to zero.

2. Norton resistance is found by?

a) Shorting all voltage sources

b) Opening all current sources

c) Shorting all voltage sources and opening all current sources

d) Opening all voltage sources and shorting all current sources

Answer: c

Explanation: Ideal current sources have infinite internal resistance hence behave like an open circuit whereas ideal voltage sources have zero internal resistances hence behave as a short circuit. So, to obtain Norton resistance, all voltage sources are shorted and all current sources are opened.

3. Norton’s theorem is true for __________

a) Linear networks

b) Non-Linear networks

c) Both linear networks and nonlinear networks

d) Neither linear networks nor non-linear networks

Answer: a

Explanation: Norton’s theorem works for only linear circuit elements and not non-linear ones such as BJT, semiconductors etc.

4. In Norton’s theorem Isc is__________

a) Sum of two current sources

b) A single current source

c) Infinite current sources

d) 0

Answer: b

Explanation: Norton’s theorem states that a combination of voltage sources, current sources and resistors is equivalent to a single current source I N and a single parallel resistor R N .

5. Isc is found across the ____________ terminals of the network.

a) Input

b) Output

c) Neither input nor output

d) Either input or output

Answer: b

Explanation: According to Norton’s theorem, Isc is found through the output terminals of a network and not the input terminals.

6. Can we use Norton’s theorem on a circuit containing a BJT?

a) Yes

b) No

c) Depends on the BJT

d) Insufficient data provided

Answer: b

Explanation: We can use Norton’s theorem only for linear networks. BJT is a non-linear network hence we cannot apply Norton’s theorem for it.

7. Calculate the Norton resistance for the following circuit if 5 ohm is the load resistance.

basic-electrical-engineering-questions-answers-nortons-theorem-q7

a) 10 ohm

b) 11 ohm

c) 12 ohm

d) 13 ohm

Answer: c

Explanation: Shorting all voltage sources and opening all current sources we have:

RN=+10 = 12 ohm.

8. Calculate the short circuit current is the 5 ohm resistor is the load resistance.

basic-electrical-engineering-questions-answers-nortons-theorem-q7

a) 0.72A

b) 0.32A

c) 0.83A

d) 0.67A

Answer: a

Explanation: Since the 5 ohm is the load resistance, we short it and find the resistance through the short.

If we apply source transformation between the 6 ohm resistor and the 1A source, we get a 6V source in series with a 6 ohm resistor. Now we have two meshes. Let us consider I1 flowing in the first mesh and I2 flowing in the second mesh.

The mesh equations are:

9I1-6I2=4

-6I1+16I2=6

On solving these equations simultaneously, we get I2=0.72A, which is the short circuit current.

9. Find the current in the 5 ohm resistance using Norton’s theorem.

basic-electrical-engineering-questions-answers-nortons-theorem-q7

a) 1A

b) 1.5A

c) 0.25A

d) 0.5A

Answer: d

Explanation: Shorting all voltage sources and opening all current sources we have:

R N =+10 = 12 ohm.

Since the 5 ohm is the load resistance, we short it and find the resistance through the short.

If we apply source transformation between the 6 ohm resistor and the 1A source, we get a 6V source in series with a 6 ohm resistor. Now we have two meshes. Let us consider I1 flowing in the first mesh and I2 flowing in the second mesh.

The mesh equations are:

9I1-6I2=4

-6I1+16I2=6

On solving these equations simultaneously, we get I2=0.72A, which is the short circuit current.

Connecting the current source in parallel to R N which is in turn connected in parallel to the load resistance=5ohm, we get Norton’s equivalent circuit.

Using current divider: I = 0.72*12/ = 0.5 A.

10. Which of the following is also known as the dual of Norton’s theorem?

a) Thevenin’s theorem

b) Superposition theorem

c) Maximum power transfer theorem

d) Millman’s theorem

Answer: a

Explanation: Thevenin’s theorem is also known as the dual of Norton’s theorem because in Norton’s theorem we find short circuit current which is the dual of open circuit voltage-what we find in Thevenin’s theorem.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Source Transformations”.

1. A voltage source connected in series with a resistor can be converted to a?

a) Current source in series with a resistor

b) Current source in parallel with a resistor

c) Voltage source in parallel with a resistor

d) Cannot be modified

Answer: b

Explanation: A voltage source connected in series can be converted to a current source connected in parallel using the relation obtained from Ohm’s law, that is V=IR. This equation shows that a voltage source connected in series has the same impact as a current source connected in parallel.

2. Calculate the total current in the circuit.

basic-electrical-engineering-questions-answers-source-tansformations-q2

a) 2.3mA

b) 4.3mA

c) 3.3mA

d) 1.3mA

Answer: c

Explanation: The 9mA source connected in parallel to the 5 kohm resistor can be converted to a 45V source in series with a 5 kohm resistor. Applying mesh analysis, we get:

I=/ = 3.3mA.

3. Find the value of voltage once source transformation is applied to the circuit.

basic-electrical-engineering-questions-answers-source-tansformations-q3

a) 10V

b) 30V

c) 50V

d) 70V

Answer: c

Explanation: Using ohm’s law, we can use the relation: V=IR.

Thus V=10*5 = 50V.

4. Once the circuit is transformed to a voltage source where will the resistance be connected?

basic-electrical-engineering-questions-answers-source-tansformations-q3

a) In series with the voltage source

b) In parallel with the voltage source

c) The resistance is removed from the circuit

d) Resistance is multiplied by 10 and connected in series with the source

Answer: a

Explanation: The resistance is connected in series with the voltage source because we are transforming a current source connected in parallel to a resistor to a voltage source connected in series with it.

5. What will the value of the current be once source transformation is applied to the circuit?

basic-electrical-engineering-questions-answers-source-tansformations-q5

a) 10A

b) 20A

c) 30A

d) 40A

Answer: a

Explanation: Using ohm’s law, we can use the relation: V=IR.

Thus I=V/R.

I=220/22=10A.

6. Once the circuit is transformed into a current source where will the resistance be connected?

basic-electrical-engineering-questions-answers-source-tansformations-q5

a) In series with the current source

b) In parallel with the current source

c) The resistance is removed from the circuit

d) Resistance is multiplied by 10 and connected in parallel with the source

Answer: b

Explanation: When we perform source transformation on a circuit, we transform a voltage source connected in series with a resistor to a current source connected in parallel to it. This is due to the relation we get by Ohm’s law, that is V=IR.

7. A current source connected in parallel with a resistor can be converted to a?

a) Current source in series with a resistor

b) Voltage source in series with a resistor

c) Voltage source in parallel with a resistor

d) Cannot be modified

Answer: b

Explanation: A current source connected in parallel can be converted to a voltage source connected in series using the relation obtained from Ohm’s law, that is V=IR. This equation shows that a current source connected in parallel has the same impact as a voltage source connected in series.

8. A source transformation is_________

a) Unilateral

b) Bilateral

c) Unique

d) Cannot be determined

Answer: b

Explanation: A source transformation is bilateral because a voltage source can be converted to a current source and vice-versa.

9. In source transformation________

a) Voltage source remains the same

b) Current sources remain the same

c) Both voltage and current source remain the same

d) Resistances remain the same

Answer: d

Explanation: In source transformation, the value of the voltage and current sources change when changed from voltage to current source and current to voltage source but the value of the resistance remains the same.

10. If there are 3 10V sources connected in parallel then on source transformation__________

a) The effect of all the sources is considered

b) The effect of only one source is considered

c) The effect of none of the sources is considered

d) The effect of only 2 sources is considered.

Answer: b

Explanation: When voltages are connected in parallel, the effect of only one source is considered because the effect of the voltage remains the same when connected in parallel.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Delta Star Transformation”.

1. The value of the 3 resistances when connected in star connection is_________

basic-electrical-engineering-questions-answers-delta-star-transformation-q1

a) 2.32ohm,1.22ohm, 4.54ohm

b) 3.55ohm, 4.33ohm, 5.67ohm

c) 2.78ohm, 1.67ohm, 0.83ohm

d) 4.53ohm, 6.66ohm, 1.23ohm

Answer: c

Explanation: Following the delta to star conversion:

R1=10*5/ = 2.78 ohm

R2=10*3/ = 1.67 ohm

R3=5*3/ = 0.83 ohm.

2. Which, among the following is the right expression for converting from delta to star?

a) R1=Ra*Rb/, R2=Rb*Rc/, R3=Rc*Ra/

b) R1=Ra/, R2=Rb/, Rc=/

c) R1=Ra*Rb*Rc/, R2=Ra*Rb/, R3=Ra/

d) R1=Ra*Rb*Rc/, R2=Ra*Rb*Rc/, R3=Ra*Rb*Rc/

Answer: a

Explanation: After converting to star, each star connected resistance is equal to the ratio of product of the resistances it is connected to and the total sum of the resistances. Hence R1=Ra*Rb/, R2=Rb*Rc/, R3=Rc*Ra/.

3. Find the equivalent star network.

basic-electrical-engineering-questions-answers-delta-star-transformation-q3

a) 2.3ohm, 2.3ohm, 2.3ohm

b) 1.2ohm, 1.2ohm, 1.2ohm

c) 3.3ohm, 3.3ohm, 3.3ohm

d) 4.5ohm, 4.5ohm, 4.5ohm

Answer: b

Explanation: The 6 ohm and 9 ohm resistances are connected in parallel. Their equivalent resistances are: 6*9/=3.6 ohm.

The 3 3.6 ohm resistors are connected in delta. Converting to star:

R1=R2=R3= 3.6*3.6/=1.2 ohm.

4. Star connection is also known as__________

a) Y-connection

b) Mesh connection

c) Either Y-connection or mesh connection

d) Neither Y-connection nor mesh connection

Answer: a

Explanation: The star connection is also known as the Y-connection because its formation is like the letter Y.

5. Rab is the resistance between the terminals A and B, Rbc between B and C and Rca between C and A. These 3 resistors are connected in delta connection. After transforming to star, the resistance at A will be?

a) Rab*Rac/

b) Rab/

c) Rbc*Rac/

d) Rac/

Answer: a

Explanation: When converting from delta to star, the resistances in star connection is equal to the product of the resistances it is connected to, divided by the total sum of the resistance.

Hence Rab*Rac/.

6. Rab is the resistance between the terminals A and B, Rbc between B and C and Rca between C and A. These 3 resistors are connected in delta connection. After transforming to star, the resistance at B will be?

a) Rac/

b) Rab/

c) Rbc*Rab/

d) Rab/

Answer: c

Explanation: When converting from delta to star, the resistances in star connection is equal to the product of the resistances it is connected to, divided by the total sum of the resistance.

Hence Rab*Rbc/.

7. Rab is the resistance between the terminals A and B, Rbc between B and C and Rca between C and A. These 3 resistors are connected in delta connection. After transforming to star, the resistance at C will be?

a) Rac/

b) Rab/

c) Rbc*Rac/

d) Rab/

Answer: c

Explanation: When converting from delta to star, the resistances in star connection is equal to the product of the resistances it is connected to, divided by the total sum of the resistance.

Hence Rac*Rbc/.

8. Find the current in the circuit.

basic-electrical-engineering-questions-answers-delta-star-transformation-q8

a) 0.54A

b) 0.65A

c) 0.67A

d) 0.87A

Answer: a

Explanation: The 3 5 ohm resistors are connected in delta. Changing it to star:

R1=R2=R3 = 1.67 ohm.

One of the 1.67 ohm resistors are connected in series with the 2 ohm resistor and another 1.67 ohm resistor is connected in series to the 3 ohm resistor.

The resulting network has a 1.67 ohm resistor connected in series with the parallel connection of the 3.67 and 4.67 resistors.

The equivalent resistance is 3.725A.

I=2/3.725 = 0.54A.

9. If a 6 ohm, 2ohm and 4ohm resistor is connected in delta, find the equivalent star connection.

a) 1ohm, 2ohm, 3ohm

b) 2ohm, 4ohm, 7ohm

c) 5ohm, 4ohm, 2ohm

d) 1ohm, 2ohm, 2/3ohm

Answer: d

Explanation: Using the delta to star conversion formula:

R1=2*6/

R2=2*4/

R3=4*6/.

10. If a 4ohm, 3ohm and 2ohm resistor is connected in delta, find the equivalent star connection.

a) 8/9ohm, 4/3ohm, 2/3ohm

b) 8/9ohm, 4/3ohm, 7/3ohm

c) 7/9ohm, 4/3ohm, 2/3ohm

d) 8/9ohm, 5/3ohm, 2/3ohm

Answer: a

Explanation: Using the delta-star conversion formula:

R1=4*3/

R2=2*3/

R3=2*4/.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Star Delta Transformation”.

1. Find the equivalent delta circuit.

basic-electrical-engineering-questions-answers-star-delta-transformation-q1

a) 9.69 ohm, 35.71 ohm, 6.59 ohm

b) 10.69 ohm, 35.71 ohm, 6.59 ohm

c) 9.69 ohm, 34.71 ohm, 6.59 ohm

d) 10.69 ohm, 35.71 ohm, 7.59 ohm

Answer: a

Explanation: Using the star to delta conversion:

R 1 = 4.53+6.66+4.53*6.66/1.23 = 35.71 ohm

R 2 = 4.53+1.23+4.53*1.23/6.66 = 6.59 ohm

R 3 = 1.23+6.66+1.23*6.66/4.53 = 9.69 ohm.

2. Which, among the following is the correct expression for star-delta conversion?

a) R1=Ra*Rb/, R2=Rb*Rc/, R3=Rc*Ra/b)

b) R1=Ra/, R2=Rb/, Rc=/

c) R1=Ra+Rb+Ra*Rb/Rc, R2=Rc+Rb+Rc*Rb/Ra, R3=Ra+Rc+Ra*Rc/Rb

d) R1=Ra*Rb/Rc, R2=Rc*Rb/Ra, R3=Ra*Rc/Rb

Answer: c

Explanation: After converting to delta, each delta connected resistance is equal to the sum of the two resistance it is connected to+product of the two resistances divided by the remaining resistance. Hence R1=Ra+Rb+Ra*Rb/Rc, R2=Rc+Rb+Rc*Rb/Ra, R3=Ra+Rc+Ra*Rc/Rb.

3. Find the equivalent resistance between X and Y.

basic-electrical-engineering-questions-answers-star-delta-transformation-q3

a) 3.33 ohm

b) 4.34 ohm

c) 5.65 ohm

d) 2.38 ohm

Answer: d

Explanation: The 3 2ohm resistors are connected in star, changing them to delta, we have R1=R2=R3= 2+2+2*2/2=6 ohm.

The 3 6ohm resistors are connected in parallel to the 10 ohm 5 ohm and 10ohm resistors respectively.

This network can be further reduced to a network consisting of a 3.75ohm and 2.73ohm resistor connected in series whose resultant is intern connected in parallel to the 3.75 ohm resistor.

4. Delta connection is also known as____________

a) Y-connection

b) Mesh connection

c) Either Y-connection or mesh connection

d) Neither Y-connection nor mesh connection

Answer: b

Explanation: Delta connection is also known as mesh connection because its structure is like a mesh, that is, a closed loop which is planar.

5. Ra is resistance at A, Rb is resistance at B, Rc is resistance at C in star connection. After transforming to delta, what is resistance between B and C?

a) Rc+Rb+Rc*Rb/Ra

b) Rc+Rb+Ra*Rb/Rc

c) Ra+Rb+Ra*Rc/Rb

d) Rc+Rb+Rc*Ra/Rb

Answer: a

Explanation: After converting to the delta, each delta connected resistance is equal to the sum of the two resistances it is connected to+product of the two resistances divided by the remaining resistance. Hence, resistance between B and C = Rc+Rb+Rc*Rb/Ra.

6. Ra is resistance at A, Rb is resistance at B, Rc is resistance at C in star connection. After transforming to delta, what is resistance between A and C?

a) Ra+Rb+Ra*Rb/Rc

b) Ra+Rc+Ra*Rc/Rb

c) Ra+Rb+Ra*Rc/Ra

d) Ra+Rc+Ra*Rb/Rc

Answer: b

Explanation: After converting to the delta, each delta connected resistance is equal to the sum of the two resistances it is connected to+product of the two resistances divided by the remaining resistance. Hence, resistance between A and C = Ra+Rc+Ra*Rc/Rb.

7. Ra is resistance at A, Rb is resistance at B, Rc is resistance at C in star connection. After transforming to delta, what is resistance between A and B?

a) Rc+Rb+Ra*Rb/Rc

b) Ra+Rb+Ra*Rc/Rb

c) Ra+Rb+Ra*Rb/Rc

d) Ra+Rc+Ra*Rc/Rb

Answer: c

Explanation: After converting to the delta, each delta connected resistance is equal to the sum of the two resistances it is connected to+product of the two resistances divided by the remaining resistance. Hence, resistance between A and B = Ra+Rb+Ra*Rb/Rc.

8. If a 1ohm 2ohm and 32/3ohm resistor is connected in star, find the equivalent delta connection.

a) 34 ohm, 18.67 ohm, 3.19 ohm

b) 33 ohm, 18.67 ohm, 3.19 ohm

c) 33 ohm, 19.67 ohm, 3.19 ohm

d) 34 ohm, 19.67 ohm, 3.19 ohm

Answer: a

Explanation: Using the formula for delta to star conversion:

Using the formula for delta to star conversion:

R1=1+2+1*2/

R2=1+32/3+1*/2

R3=2+32/3+2*/1.

9. If an 8/9ohm, 4/3ohm and 2/3ohm resistor is connected in star, find its delta equivalent.

a) 4ohm, 3ohm, 2ohm

b) 1ohm, 3ohm, 2ohm

c) 4ohm, 1ohm, 2ohm

d) 4ohm, 3ohm, 1ohm

Answer: a

Explanation: Using the formula for the star to delta conversion:

R1=8/9+4/3+*/

R2=8/9+2/3+*/

R3=2/3+4/3+*/.

10. Find the equivalent resistance between A and B.

basic-electrical-engineering-questions-answers-star-delta-transformation-q10

a) 32ohm

b) 31ohm

c) 30ohm

d) 29ohm

Answer: d

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

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Maximum Power Transfer”.

1. The maximum power drawn from source depends on __________

a) Value of source resistance

b) Value of load resistance

c) Both source and load resistance

d) Neither source or load resistance

Answer: b

Explanation: The maximum power transferred is equal to E 2 /4*R L . So, we can say maximum power depends on load resistance.

2. The maximum power is delivered to a circuit when source resistance is __________ load resistance.

a) Greater than

b) Equal to

c) Less than

d) Greater than or equal to

Answer: b

Explanation: The circuit can draw maximum power only when source resistance is equal to the load resistance.

3. If source impedance is a complex number Z, then load impedance is equal to _________

a) Z’

b) -Z

c) -Z’

d) Z

Answer: a

Explanation: When Source impedance is equal to Z, its load impedance is the complex conjugate of Z which is Z’. Only under this condition, maximum power can be drawn from the circuit.

4. If ZL=Zs’, then RL=?

a) -RL

b) Rs

c) -Rs

d) 0

Answer: b

Explanation: Rs is the real part of the complex number ZL. Hence when we find the complex conjugate the real part remains the same whereas the complex part acquires a negative sign.

5. Calculate the value of RL across A and B.

basic-electrical-engineering-questions-answers-maximum-power-transfer-q5

a) 3.45ohm

b) 2.91ohm

c) 6.34ohm

d) 1.54ohm

Answer: b

Explanation: On shorting the voltage sources:

RL=3||2+4||3 = 1.20+1.71 = 2.91 ohm.

6. Calculate Eth.

basic-electrical-engineering-questions-answers-maximum-power-transfer-q5

a) 3.43V

b) 4.57V

c) 3.23V

d) 5.34V

Answer: b

Explanation: The two nodal equations are:

/3+VA/2=0

/4+VB/3=0

On solving the two equations, we get VA=4V, VB=8.571V.

VAB = VA-VB = 4V – 8.571V = -4.57V.

E th = 4.57V.

7. Calculate the maximum power transferred.

basic-electrical-engineering-questions-answers-maximum-power-transfer-q5

a) 1.79W

b) 4.55W

c) 5.67W

d) 3.78W

Answer: a

Explanation: On shorting the voltage sources:

RL=3||2+4||3 =1.20+1.71=2.91 ohm.

The two nodal equations are:

/3+VA/2=0

/4+VB/3=0

On solving the two equations, we get VA=4V, VB=8.571V.

VAB=VA-VB = 4V – 8.571V = -4.57V.

E th =4.57V

The maximum power transferred = Eth 2 /4RL. Substituting the given values in the formula, we get Pmax = 1.79W.

8. Does maximum power transfer imply maximum efficiency?

a) Yes

b) No

c) Sometimes

d) Cannot be determined

Answer: b

Explanation: Maximum power transfer does not imply maximum efficiency. If the load resistance is smaller than source resistance, the power dissipated at the load is reduced while most of the power is dissipated at the source then the efficiency becomes lower.

9. Under the condition of maximum power efficiency is?

a) 100%

b) 0%

c) 30%

d) 50%

Answer: d

Explanation: Efficiency=*100.

Power Output=I 2 R L , Power Input=I 2 (R L +R S )

Under maximum power transfer conditions, R L =R S

Power Output=I 2 RL; Power Input=2*I 2 RL

Thus efficiency=50%.

10. Name some devices where maximum power has to be transferred to the load rather than maximum efficiency.

a) Amplifiers

b) Communication circuits

c) Both amplifiers and communication circuits

d) Neither amplifiers nor communication circuits

Answer: c

Explanation: Maximum power transfer to the load is preferred over maximum efficiency in both amplifiers and communication circuits since in both these cases the output voltage is more than the input.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Capacitors”.

1. What is the relation between current and voltage in a capacitor?

a) I=1/C*integral

b) I=CdV/dt

c) I=1/CdV/dt

d) I=Ct

Answer: b

Explanation: Current=rate of change of charge

I=dQ/dt. Q=CV. C is constant for a given capacitor so I=CdV/dt.

2. If 2V is supplied to a 3F capacitor, calculate the charge stored in the capacitor.

a) 1.5C

b) 6C

c) 2C

d) 3C

Answer: b

Explanation: Q is directly proportional to V. The constant of proportionality in this case is C, that is, the capacitance. Hence Q=CV.

Q=3*2=6C.

3. Calculate the current in the capacitor having 2V supply voltage and 3F capacitance in 2seconds.

a) 2A

b) 5A

c) 6A

d) 3A

Answer: d

Explanation: Q is directly proportional to V. The constant of proportionality in this case is C, that is, the capacitance. Hence Q=CV.

Q=3*2=6C.

I=Q/t = 6/2 = 3A.

4. A 4microF capacitor is charged to 120V, the charge in the capacitor would be?

a) 480C

b) 480microC

c) 30C

d) 30microC

Answer: b

Explanation: Q is directly proportional to V. The constant of proportionality in this case is C, that is, the capacitance. Hence Q=CV.

Q=4*120=480microC.

5. For high frequencies, capacitor acts as _________

a) Open circuit

b) Short circuit

c) Amplifier

d) Rectifier

Answer: b

Explanation: Capacitive impedance is inversely proportional to frequency. Hence at very high frequencies, the impedance is almost equal to zero, hence it acts as a short circuit and there is no voltage across it.

6. For very low frequencies, capacitor acts as ________

a) Open circuit

b) Short circuit

c) Amplifier

d) Rectifier

Answer: a

Explanation: Capacitive impedance is inversely proportional to frequency. Hence at very low frequencies the impedance is almost infinity and hence acts as an open circuit and no current flows through it.

7. A capacitor consists of_________

a) Two conductors

b) Two semiconductors

c) Two dielectrics

d) Two insulators

Answer: a

Explanation: A capacitor consists of two conductors connected in parallel to each other so that it can store charge in between the plates.

8. Capacitor preferred when there is high frequency in the circuits is __________

a) Electrolyte capacitor

b) Mica capacitor

c) Air capacitor

d) Glass capacitor

Answer: b

Explanation: Mica capacitors are preferred for high frequency circuits because they have low ohmic losses and less reactance.

9. Capacitance increases with ________

a) Increase in plate area

b) Decrease in plate area

c) Increase in distance between the plates

d) Increase in density of the material

Answer: a

Explanation: Capacitance is directly proportional to the plate area. Hence as the plate area increases, the capacitance also increases.

10. Capacitance increases with __________

a) Increase in distance between the plates

b) Decrease in plate area

c) Decrease in distance between the plates

d) Increase in density of the material

Answer: c

Explanation: Capacitance is inversely proportional to the distance between the two parallel plates. Hence, as the distance between the plate decreases, the capacitance increases.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Charge and Voltage”.

1. Which among the following expressions relate charge, voltage and capacitance of a capacitor?

a) Q=C/V

b) Q=V/C

c) Q=CV

d) C=Q 2 V

Answer: c

Explanation: Q is directly proportional to V. The constant of proportionality in this case is C, that is, the capacitance. Hence Q=CV.

2. If a 2F capacitor has 1C charge, calculate the voltage across its terminals.

a) 0.5V

b) 2V

c) 1.5V

d) 1V

Answer: a

Explanation: Q is directly proportional to V. The constant of proportionality in this case is C, that is, the capacitance. Hence Q=CV. V=Q/C=1/2 V=0.5V.

3. What is the voltage across a capacitor at the time of switching, that is, when t=0?

a) Infinity

b) 0V

c) Cannot be determined

d) 1V

Answer: b

Explanation: At the time of switching, when t=0, the capacitor acts as a short circuit. The voltage across a short is always equal to zero hence the voltage across the capacitor is equal to zero.

4. What is the voltage across the capacitor if the switch is closed and steady state is reached?

basic-electrical-engineering-questions-answers-charge-voltage-q4

a) 8V

b) 0V

c) 10V

d) Infinity

Answer: c

Explanation: When steady state is reached, the capacitor acts as a open circuit and the 10V is connected in parallel to it. Hence Vc=10V.

5. If one plate of a parallel plate capacitor is charged to positive charge the other plate is charged to?

a) Positive

b) Negative

c) Positive or negative

d) Not charged

Answer: b

Explanation: If one plate is charged to positive, the other plate is automatically charged to negative so that it can store electrical charge.

6. When the voltage across a capacitor increases, what happens to the charge stored in it?

a) Increases

b) Decreases

c) Becomes zero

d) Cannot be determined

Answer: a

Explanation: When the voltage across a capacitor increases, the charge stored in it also increases because a charge is directly proportional to voltage, capacitance being the constant of proportionality.

7. When will capacitor fully charged?

a) When the voltage across its plates is half the voltage from ground to one of its plates

b) When the current through the capacitor is a 1/root2 time its value

c) When the supply voltage is equal to the capacitor voltage

d) Never

Answer: c

Explanation: When the capacitor voltage is equal to the supply voltage the current stops flowing through the circuit and the charging phase is over.

8. What happens to the current flow in a fully charged capacitor?

a) Current flow stops

b) Current flow doubles

c) Current flow becomes half its original value

d) Current flow becomes one-fourth its original value

Answer: a

Explanation: When a capacitor is fully charged, it does not store any more charge. There is no change in charge with time. Current is the rate of change of charge, hence it becomes zero, or stops.

9. Calculate the capacitance of a capacitor that stores 40microC of charge and has a voltage of 2V.

a) 20F

b) 20microF

c) 10F

d) 10microF

Answer: b

Explanation: Q is directly proportional to V. The constant of proportionality in this case is C, that is, the capacitance. Hence C=Q/V.

C=40microC/2V=20microF.

10. What happens to the capacitance when the voltage across the capacitor increases?

a) Decreases

b) Increases

c) Becomes 0

d) No effect

Answer: d

Explanation: Q is directly proportional to V. The constant of proportionality in this case is C, that is, the capacitance. Capacitance is a constant so it will not change on changing voltage.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Capacitance”.

1. A power factor of a circuit can be improved by placing which, among the following, in a circuit?

a) Inductor

b) Capacitor

c) Resistor

d) Switch

Answer: b

Explanation: Power factor = Real power/Apparent power = kW/kVA

By adding a capacitor in a circuit, an additional kW load can be added to the system without altering the kVA. Hence, the power factor is improved.

2. When the supply frequency increases, what happens to the capacitive reactance in the circuit?

a) Increases

b) Decreases

c) Remains the same

d) Becomes zero

Answer: b

Explanation: The expression for capacitive reactance is: Xc=1/. This relation shows that frequency is inversely related to capacitive reactance. Hence, as supply frequency increases, the capacitive reactance decreases.

3. Calculate the time constant of a series RC circuit consisting of a 100microF capacitor in series with a 100ohm resistor.

a) 0.1 sec

b) 0.1 msec

c) 0.01 sec

d) 0.01 msec

Answer: c

Explanation: The time constant of a RC circuit= R*C= 100*10 -6 *100=0.01 sec.

4. Capacitors charge and discharge in __________ manner.

a) Linear

b) Constant

c) Square

d) Exponential

Answer: d

Explanation: Capacitors charge and discharge in an exponential manner because of the relation: X C =1/ and Q=CV ∴ Q=V/(2πf X C )

X C is complex which can be written in the form of exponent through euler formula.

5. Air has a dielectric constant of ___________

a) Unity

b) Zero

c) Infinity

d) Hundred

Answer: a

Explanation: Dielectric constant of air is the same as that of a vacuum which is equal to unity. Dielctric constant of air is taken as the reference to measure the dielectric constant of all other materials.

6. What is the value of capacitance of a capacitor which has a voltage of 4V and ha 8C of charge?

a) 2F

b) 4F

c) 6F

d) 8F

Answer: a

Explanation: Q is directly proportional to V. The constant of proportionality in this case is C, that is, the capacitance. Hence Q=CV. From the relation, C=Q/V= 8/4=2F.

7. Unit of capacitance is___________

a) Volts

b) Farad

c) Henry

d) Newton

Answer: b

Explanation: Volts is the unit of voltage, Henry for inductance and Newton for a force. Hence the unit for capacitance is Farad.

8. What will happen to the capacitor just after the source is removed?

a) It will not remain in its charged state

b) It will remain in its charged state

c) It will start discharging

d) It will become zero

Answer: b

Explanation: As soon as the source is removed, the capacitor does not start discharging it remains in the same charged state.

9. Which among the following equations is incorrect?

a) Q=CV

b) Q=C/V

c) V=Q/C

d) C=Q/V

Answer: b

Explanation: Q is directly proportional to V. The constant of proportionality in this case is C, that is, the capacitance. Hence Q=CV. From the given relation we can derive all the equations except for Q=C/V.

10. Capacitance is directly proportional to__________

a) Area of cross section between the plates

b) Distance of separation between the plates

c) Both area and distance

d) Neither area nor distance

Answer: a

Explanation: The relation between capacitance, area and distance between the plates is: C=epsilon*A/D. According to this relation, the capacitance is directly proportional to the area.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Capacitors in Parallel”.

1. What is the total capacitance when three capacitors, C1, C2 and C3 are connected in parallel?

a) C1/

b) C1+C2+C3

c) C2/

d) 1/C1+1/C2+1/C3

Answer: b

Explanation: When capacitors are connected in parallel, the total capacitance is equal to the sum of the capacitance of each of the capacitors. Hence Ctotal=C1+C2+C3.

2. Calculate the total capacitance.

basic-electrical-engineering-questions-answers-capacitors-parallel-q2

a) 10F

b) 15F

c) 13F

d) 20F

Answer: c

Explanation: The equivalent capacitance when capacitors are connected in parallel is the sum of all the capacitors = 1+2+10 = 13F.

3. Calculate the voltage across AB if the total charge stored in the combination is 13C.

basic-electrical-engineering-questions-answers-capacitors-parallel-q2

a) 1V

b) 2V

c) 3V

d) 4V

Answer: a

Explanation: The equivalent capacitance when capacitors are connected in parallel is the sum of all the capacitors = 1+2+10 = 13F. V = Q/C = 13/13 = 1V.

4. Calculate the charge in the 2F capacitor.

basic-electrical-engineering-questions-answers-capacitors-parallel-q4

a) 200C

b) 100C

c) 300C

d) 400C

Answer: a

Explanation: Since the capacitors are connected in parallel, the voltage across each is the same, it does not get divided. Q = CV = 2*100 = 200C.

5. Calculate the charge in the 1F capacitor.

basic-electrical-engineering-questions-answers-capacitors-parallel-q4

a) 200C

b) 100C

c) 300C

d) 400C

Answer: b

Explanation: Since the capacitors are connected in parallel, the voltage across each is the same, it does not get divided. Q = CV = 1*100 = 100C.

6. Calculate the total charge of the system.

basic-electrical-engineering-questions-answers-capacitors-parallel-q4

a) 200C

b) 100C

c) 300C

d) 400C

Answer: c

Explanation: The equivalent capacitance when capacitors are connected in parallel is the sum of all the capacitors=1+2=3F. Q = CV = 3*100 = 300V.

7. When capacitors are connected in parallel, the total capacitance is always __________ the individual capacitance values.

a) Greater than

b) Less than

c) Equal to

d) Cannot be determined

Answer: a

Explanation: When capacitors are connected in parallel, the total capacitance is equal to the sum of the capacitance of each of the capacitors. Hence Ctotal=C1+C2+C3. Since it is the sum of all the capacitance values, the total capacitance is greater the individual capacitance values.

8. When capacitors are connected in parallel, what happens to the effective plate area?

a) Increases

b) Decreases

c) Remains the same

d) Becomes zero

Answer: a

Explanation: When capacitors are connected in parallel, the top plates of each of the capacitors are connected together while the bottom plates are connected to each other. This effectively increases the top plate area and the bottom plate area.

9. Three capacitors having a capacitance equal to 2F, 4F and 6F are connected in parallel. Calculate the effective parallel.

a) 10F

b) 11F

c) 12F

d) 13F

Answer: c

Explanation: When capacitors are connected in parallel, the total capacitance is equal to the sum of the capacitance of each of the capacitors. Hence Ctotal = C1+C2+C3 = 2+4+6 = 12F.

10. Two capacitors having capacitance value 4F, three capacitors having capacitance value 2F and 5 capacitors having capacitance value 1F are connected in parallel, calculate the equivalent capacitance.

a) 20F

b) 19F

c) 18F

d) 17F

Answer: b

Explanation: When capacitors are connected in parallel, the total capacitance is equal to the sum of the capacitance of each of the capacitors. Hence Ctotal=4+4+2+2+2+1+1+1+1+1=19F.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Capacitors in Series”.

1. What is the total capacitance when two capacitors C1 and C2 are connected in series?

a) /C1C2

b) 1/C1+1/C2

c) C1C2/

d) C1+C2

Answer: c

Explanation: When capacitors are connected in series, the equivalent capacitance is:

1/Ctotal=1/C1+1/C2, therefore Ctotal = C1C2/.

2. N capacitors having capacitance C are connected in series, calculate the equivalent capacitance.

a) C/N

b) C

c) CN

d) N/C

Answer: d

Explanation: When capacitors are connected in series, the equivalent capacitance is:

1/Ctotal = 1/C+1/C+1/C+……..N times.

1/Ctotal=N/C.

Ctotal=C/N.

3. When capacitors are connected in series, the equivalent capacitance is ___________ each individual capacitance.

a) Greater than

b) Less then

c) Equal to

d) Insufficient data provided

Answer: b

Explanation: When capacitors are connected in series, the equivalent capacitance is:

1/Ctotal=1/C1+1/C2. Since we find the reciprocals of the sum of the reciprocals, the equivalent capacitance is less than the individual capacitance values.

4. What is the equivalent capacitance?

basic-electrical-engineering-questions-answers-capacitors-series-q4

a) 1.5F

b) 0.667F

c) 2.45F

d) 2.75F

Answer: b

Explanation: When capacitors are connected in series,

1/C total = 1/C1+1/C2 = 1/2+1 = 3/2

C total = 2/3 = 0.667F.

5. When capacitors are connected in series ___________ remains the same.

a) Voltage across each capacitor

b) Charge

c) Capacitance

d) Resistance

Answer: b

Explanation: When capacitors are connected in series, the charge remains the same because the same amount of current flow exists in each capacitor.

6. When capacitors are connected in series _______________ Varies

a) Voltage across each capacitor

b) Charge

c) Capacitance

d) Resistance

Answer: a

Explanation: When capacitors are connected in series, the voltage varies because the voltage drop across each capacitor is different.

7. Four 10F capacitors are connected in series, calculate the equivalent capacitance.

a) 1.5F

b) 2.5F

c) 3.5F

d) 0.5F

Answer: b

Explanation: When capacitors are connected in series,

1/C total =1/C1+1/C2+1/C3+1/C4=1/10+1/10+1/10+1/10=4/10F.

C total =10/4=2.5F.

8. Calculate the charge in the circuit.

basic-electrical-engineering-questions-answers-capacitors-series-q8

a) 66.67C

b) 20.34C

c) 25.45C

d) 30.45C

Answer: a

Explanation: When capacitors are connected in series, the equivalent capacitance is:

1/C total =1/C1+1/C2 = 1/2+1=3/2

C total =2/3 F

Q=CV=*100 = 200/3 C=66.67C.

9. Calculate the voltage across the 1F capacitor.

basic-electrical-engineering-questions-answers-capacitors-series-q8

a) 33.33V

b) 66.67V

c) 56.56V

d) 23.43V

Answer: b

Explanation: When capacitors are connected in series,

1/C total =1/C1+1/C2 = 1/2+1=3/2

Q = CV = *100 = 66.67C.

V across the 1F capacitor = 66.67/1 = 66.67V.

10. Calculate the voltage across the 2F capacitor.

basic-electrical-engineering-questions-answers-capacitors-series-q8

a) 33.33V

b) 66.67V

c) 56.56V

d) 23.43V

Answer: a

Explanation: When capacitors are connected in series,

1/C total =1/C1+1/C2 = 1/2+1 = 3/2

Q = CV = *100 = 66.67C.

V across the 2F capacitor = 66.67/2 = 33.33V.

This set of Basic Electrical Engineering Questions and Answers for Freshers focuses on “Distribution of Voltage Across Capacitors in Series”.

1. The total voltage drop across a series of capacitors is __________

a) The voltage drop across any one of the capacitors

b) The sum of the voltage drop across each of the capacitors

c) The product of the voltage drop across each of the capacitors

d) Zero

Answer: b

Explanation: The total voltage drop across a series of capacitors is equal to the sum of the voltage drop across each of the capacitors because when capacitors are connected in series, the voltage drops across each capacitor.

2. Capacitors C1, C2 and C3 have voltage drops 2V, 3V and 5V respectively. Calculate the total voltage in the circuit.

a) 10V

b) 2V

c) 5V

d) 0V

Answer: a

Explanation: When capacitors are connected in series:

Vtotal=V12+V2+V3 = 2+3+5 = 10V.

3. What is the voltage across the 2F capacitor?

basic-electrical-engineering-questions-answers-freshers-q3

a) 240V

b) 200V

c) 220V

d) 120V

Answer: d

Explanation: Capacitors are in series.

1/C=1/2+1/4+1/6, therefore, C=F.

Q = C*V = 220* = 240C.

V across 2F capacitor = Q/C = 240/2 = 120V.

4. What is the voltage across the 4F capacitor?

basic-electrical-engineering-questions-answers-freshers-q3

a) 120V

b) 60V

c) 100V

d) 220V

Answer: b

Explanation: Capacitors are in series.

1/C=1/2+1/4+1/6, therefore, C=F.

Q = C*V = 220* = 240C.

V across 4F capacitor = Q/C = 240/4 = 60V.

5. Calculate the voltage across the 6F capacitor.

basic-electrical-engineering-questions-answers-freshers-q3

a) 120V

b) 60V

c) 40V

d) 220V

Answer: c

Explanation: Capacitors are in series.

1/C=1/2+1/4+1/6, therefore, C=F.

Q = C*V = 220* = 240C.

V across 6F capacitor = Q/C = 240/6 = 40V.

6. When capacitors are connected in series, which of the following rules are applied?

a) Voltage divider

b) Current divider

c) Both voltage divider and current divider

d) Neither voltage divider nor current divider

Answer: a

Explanation: Voltage divider is the rule applied when capacitors are connected in series because when capacitors are connected in series, the voltage is different across each capacitor.

7. A capacitor does not allow sudden changes in _________

a) Current

b) Voltage

c) Resistance

d) Inductance

Answer: b

Explanation: Capacitor does not allow sudden changes in voltage because these changes occur in zero time which results in the current being infinity, which is not possible.

8. Which of the following expressions is correct with respect to the voltage across capacitors in series?

a) V1/V2=C2/C1

b) V2/V1=C2/C1

c) V1*V2=C1*C2

d) V1/C1=V2/C2

Answer: a

Explanation: When capacitors are connected in series, the charge across each capacitor remains the same whereas the voltage across each varies. When two capacitors are connected in series:

Q=V1C1; Q=V2C2. Thus: V1/V2=C2/C1.

9. Two 4F capacitors are connected in series, calculate the voltage across each if the total voltage is 20V.

a) 10V

b) 5V

c) 20V

d) 0V

Answer: a

Explanation: The two capacitors have the same capacitance, hence the voltage gets divides equally. V across each=Total voltage/2 = 20/2 = 10V.

10. Two capacitors having voltage 2F and 4F are connected in series. This combination is connected to a 100V supply, calculate the voltage across the 2F capacitor.

a) 66.67V

b) 33.33V

c) 100V

d) 0V

Answer: a

Explanation: C total = 2*4/ = 4/3F

Q = CV = *100 = 400/3 C.

V across 2F capacitor = Q/C = /2 = 200/3 = 66.67F.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Capacitance and the Capacitor”.

1. Capacitor is a device used to__________

a) store electrical energy

b) vary the resistance

c) store magnetic energy

d) dissipate energy

Answer: a

Explanation: Capacitor is used to store the charge. It stores electrical energy between the plates.

2. Capacitor stores which type of energy?

a) kinetic energy

b) vibrational energy

c) potential energy

d) heat energy

Answer: c

Explanation: Capacitor store charge in between the plates. This charge is stationary so we can say capacitor store potential energy.

3. Capacitor blocks__________ after long time.

a) alternating current

b) direct current

c) both alternating and direct current

d) neither alternating nor direct current

Answer: b

Explanation: Capacitor blocks direct current at steady state and pass alternating current.

4. Why does capacitor block dc signal at steady state?

a) due to high frequency of dc signal

b) due to zero frequency of dc signal

c) capacitor doesnot pass any current at steady state

d) due to zero frequency of dc signal

Answer: d

Explanation: Frequency of dc signal is zero. So, Capacitive reactance X C =1/2πfc becomes infinite and capacitor behaves as open circuit for dc signal. Hence, capacitor block dc signal.

5. If a parallel plate capacitor of plate area 2m 2 and plate separation 1m store the charge of 1.77*10 -11 C. What is the voltage across the capacitor?

a) 1V

b) 2V

c) 3V

d) 4V

Answer: a

Explanation: C=€ 0 A/d

On substituting values of d, A, we get C=2€ 0 .

Q=CV

V=1 V.

6. Which of the following is a passive device?

a) Transistor

b) Rectifier

c) Capacitor

d) Vaccuum Tubes

Answer: c

Explanation: Capacitor is a passive device as it consumes power rest all generate power so, they are active devices.

7. What is the value of capacitance of a capacitor which has a voltage of 4V and has 16C of charge?

a) 2F

b) 4F

c) 6F

d) 8F

Answer: b

Explanation: Q is directly proportional to V. The constant of proportionality in this case is C, that is, the capacitance. Hence Q=CV. From the relation, C=Q/V= 16/4=4F.

8. For which medium capacitance is high?

a) Air

b) Mica

c) Water

d) Metal

Answer: d

Explanation: Metals are assumed to have a high value of dielectric constant so they have high capacitance.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Electric Fields”.

1. The conventional direction of electric field is ________

a) Positive to negative

b) Negative to positive

c) No specific direction

d) Direction cannot be determined

Answer: a

Explanation: The conventional direction of field lines is from positive to negative. The field lines originate at the positive charge and terminate at the negative charge.

2. Electric field originates at __________

a) Positive charge

b) Negative charge

c) Neither positive nor negative

d) Both positive and negative

Answer: a

Explanation: Electric field originates at the positive charge and terminates at the negative charge. The conventional direction of the field is from positive to negative.

3. Electric field terminates at ________

a) Positive charge

b) Negative charge

c) Neither positive nor negative

d) Both positive and negative

Answer: b

Explanation: Electric field originates at the positive charge and terminates at the negative charge. The conventional direction of the field is from positive to negative.

4. Which among the following statements is true with regard to electric field lines?

a) Electric field lines always intersect

b) Electric field lines may or may not intersect

c) Electric field lines can be seen

d) Electric field lines never intersect

Answer: d

Explanation: Electric field lines can never intersect because tangent at any point on electric field lines represent the direction of electric field and if they intersect at a point it means that at that point there are two different directions for electric field which is not possible.

5. Which, among the following, is the field where electric charge experiences a force?

a) Electric field

b) Magnetic field

c) Gravitational field

d) Electric, magnetic and gravitational field

Answer: a

Explanation: Charges experience a force in an electric field because charges come under the influence of a field which already has charges- electric field.

6. A field that spreads outwards in all directions is __________

a) Linear

b) Radial

c) Weak

d) Strong

Answer: b

Explanation: A radial field is one which spreads in all directions. This field is known as the radial field because it spreads out radially from a source.

7. In uniform fields, all points have ________ field strength.

a) Zero

b) Same

c) Infinity

d) Different

Answer: b

Explanation: A uniform field is one as the word suggests in which the field is spread over an area and at every point in the field strength of the field is the same.

8. Which, among the following is the correct expression for an electric field?

a) E=F/C

b) E=F*C

c) E=F/Q

d) E=F*Q

Answer: c

Explanation: Electric field is the force per unit charge hence, the correct expression among the following is: E=F/Q.

9. What happens when one material is rubbed against another?

a) The material becomes electrically neutral

b) The material becomes electrically charged

c) The material becomes negatively charged

d) The material becomes positively charged

Answer: b

Explanation: When one material is rubbed against another, there is a transfer of charges from one material to another hence the material becomes electrically charged.

10. The insulant between the two plates of a capacitor is called _______

a) Conductor

b) Semi-conductor

c) Dielectric

d) Superconductor

Answer: c

Explanation: The material between the two plates of a capacitor is an insulator, more specifically known as a dielectric.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Electric Field Strength and Electric Flux Density”.

1. Gauss law is applicable for_________

a) Point charge

b) Sheet charge

c) Line charge

d) Point, sheet and line charge

Answer: d

Explanation: Gauss law states that the total electric flux through any closed surface is equal to the charge enclosed by that surface. Hence it is applicable for all point, surface and volume.

2. “Total electric flux through any closed surface is equal to the charge enclosed by that surface divided by permittivity”. This is the statement for?

a) Gauss law

b) Lenz law

c) Coloumb’s law

d) Faraday’s law

Answer: a

Explanation: Total electric flux through any closed surface is equal to the charge enclosed by that surface divided by permittivity is the statement for Gauss law because among the four laws, Gauss law deals with electric flux.

3. Electric flux density is a function of_______

a) Volume

b) Charge

c) Current

d) Voltage

Answer: b

Explanation: Electric flux density is the charge per unit area. Hence it is a function of charge and not any of the other values.

4. As charge increases, what happens to flux density?

a) Increases

b) Decreases

c) Remains constant

d) Becomes zero

Answer: a

Explanation: Electric flux density is the charge per unit area. The expression for flux density is:

D=Q/A. Electric flux is directly proportional to charge, hence as charge increases, electric flux also increases.

5. As area increases, what happens to electric flux density?

a) Increases

b) Decreases

c) Remains constant

d) Becomes zero

Answer: b

Explanation: Electric flux density is the charge per unit area. The expression for flux density is:

D=Q/A. Electric flux is inversely proportional to area, hence an area increases, electric flux decreases.

6. Which, among the following, is the correct expression for electric flux density?

a) D=epsilon*E

b) D=epsilon/E

c) D 2 =epsilon*E

d) D=epsilon*E 2

Answer: a

Explanation: Electric flux density is directly proportional to the electric field, epsilon being the constant of proportionality. Hence D=epsilon*E.

7. Strength of the electric field is ___________

a) Directly proportional to the force applied

b) Inversely proportional to the force applied

c) Directly proportional to the square of the force applied

d) Inversely proportional to the square of the force applied

Answer: a

Explanation: Electric field intensity is the force per unit charge, hence it is directly proportional to the force applied.

8. The force applied to a conductor is 10N if the charge in the conductor is 5C, what is the electric field intensity?

a) 10V/m

b) 2V/m

c) 3V/m

d) 15V/m

Answer: b

Explanation: Electric field intensity is the force per unit charge. The formula is:

E = F/Q = 10/5 = 2V/m.

9. What is the electric flux density in free space if the electric field intensity is 1V/m?

a) 7.76*10 -12 C/m 2

b) 8.85*10 -12 C /m 2

c) 1.23*10 -12 C /m 2

d) 3.43*10 -12 C /m 2

Answer: b

Explanation: The formula for electric filed density is:

D=epsilon*E = 1*8.85*10 -12 = 8.85*10 -12 C /m 2 .

10. If the charge in a conductor is 16C and the area of cross section is 4m 2 . Calculate the electric flux density.

a) 64C/m 2

b) 16C/m 2

c) 4C/m 2

d) 2C/m 2

Answer: c

Explanation: Electric Flux density is the charge per unit area. The formula is:

D = Q/A = 16/4 = 4C/m 2 .

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Relative Permittivity”.

1. In order to obtain a high value for capacitance, the permittivity of the dielectric medium should be?

a) Low

b) High

c) Zero

d) Unity

Answer: b

Explanation: Form the expression:

C=epsilon*A/d.

From this expression, it is seen that capacitance is directly proportional to the permittivity, hence for capacitance value to be high, permittivity value should be high.

2. Find the capacitance of a capacitor whose area of cross section of the plates is 4m 2 and distance of separation between the plates is 2m. The capacitor is placed in vacuum.

a) 1.77*10 -11 F

b) 1.34*10 -11 F

c) 2.33*10 -11 F

d) 5.65*10 -11 F

Answer: a

Explanation: The expression for finding the value of capacitance is:

C=epsilon*A/d.

The medium is free space hence, epsilon = 8.85*10 -12 .

Therefore, C=8.85*10 -12 *4/2 = 1.77*10 -12 F = 1.77*10 -11 F.

3. What is relative permittivity?

a) Equal to the absolute permittivity

b) Ratio of actual permittivity to absolute permittivity

c) Ratio of absolute permittivity to actual permittivity

d) Equal to the actual permittivity

Answer: b

Explanation: Relative permittivity is the ratio of actual permittivity to the absolute permittivity. As the actual permittivity increases, the relative permittivity also increases.

4. What happens to relative permittivity when actual permittivity decreases?

a) Increases

b) Decreases

c) Remains the same

d) Becomes zero

Answer: b

Explanation: Relative permittivity is the ratio of actual permittivity to the absolute permittivity. Relative permittivity is directly proportional to actual permittivity. Hence, as actual permittivity decreases, relative permittivity also decreases.

5. What is the relative permittivity when the actual permittivity is 4F/m?

a) 4.57*10 -11

b) 4.57*10 12

c) 4.57*10 11

d) 4.57*10 -12

Answer: c

Explanation: Relative permittivity= Actual permittivity/ Absolute permittivity.

Relative permittivity = 4/(8.85*10 -12 ) = 4.57*10 11 .

6. What happens to absolute permittivity when relative permittivity increases?

a) Increases

b) Decreases

c) Remains the same

d) Becomes zero

Answer: c

Explanation: Absolute permittivity does not depend on the value of relative permittivity. Absolute permittivity is the permittivity of free space and it is a constant value = 8.85*10 -12 F/m.

7. Calculate the actual permittivity of a medium whose relative permittivity is 5.

a) 4.43*10 -11 F/m

b) 4.43*10 -12 F/m

c) 4.43*10 11 F/m

d) 4.43*10 12 F/m

Answer: a

Explanation: Actual permittivity = Relative permittivity*absolute permittivity.

Actual permittivity = 5*8.85*10 -12 = 4.43*10 -11 F/m.

8. What is the unit for relative permittivity?

a) F/m

b) Fm

c) F/m 2

d) No unit

Answer: d

Explanation: Relative permittivity is the ratio of actual permittivity to the relative permittivity of the medium. Since it is a ratio, and we know that a ratio does not have any unit, relative permittivity does not have any unit.

9. Which, among the following, will be unity in free space?

a) Absolute permittivity

b) Relative permittivity

c) Actual permittivity

d) Both absolute and relative permittivity

Answer: b

Explanation: Relative permittivity is constant for a particular medium. For air or free space, it is unity. Absolute permittivity does not depend on the medium, its value is always constant=8.85*10 -12 F/m. Actual permittivity is the product of relative permittivity and absolute permittivity.

10. Which, among the following, do not have any unit?

a) Absolute permittivity

b) Relative permittivity

c) Actual permittivity

d) Both absolute and relative permittivity

Answer: b

Explanation: Relative permittivity is the ratio of actual permittivity to the relative permittivity of the medium. Since it is a ratio, and we know that a ratio does not have any unit, relative permittivity does not have any unit.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Capacitance of a Multi Plate Capacitor”.

1. Which is the correct expression for capacitance of a multi plate capacitor?

a) C=absolute permittivity*A/d

b) C=Actual permittivity**A/d

c) C=Actual permittivity**A/d

d) Actual permittivity**A/d

Answer: b

Explanation: The correct expression is: C=Actual permittivity**A/d.

Where, n=number of plates, A=area of cross section of the plates, d=distance of separation between the plates.

2. What happens to the capacitance of a multi plate capacitor when the area of cross section of the plate decreases?

a) Increases

b) Decreases

c) Remains constant

d) Becomes zero

Answer: b

Explanation: When the area of cross section decreases, the capacitance also decreases since it is related by the formula C=Actual permittivity**A/d. Here, we can see that the capacitance is directly proportional to the area of cross section.

3. What happens to the capacitance of a multi plate capacitor when the distance of separation between the plate increases?

a) Increases

b) Decreases

c) Remains constant

d) Becomes zero

Answer: b

Explanation: When the distance of separation between the plates increases, the capacitance decreases since it is related by the formula C=Actual permittivity**A/d. Here, we can see that the capacitance is inversely proportional to the distance of separation.

4. What happens to the capacitance of a multi plate capacitor when the number of plates increases?

a) Increases

b) Decreases

c) Remains constant

d) Becomes zero

Answer: a

Explanation: When the number of capacitors increases, the capacitance also increases since it is related by the formula C=Actual permittivity**A/d. Here, we can see that the capacitance is directly proportional to the number of capacitors.

5. Find the capacitance of a multi plate capacitor whose actual permittivity= 5F/m, n=3, A=4m 2 and d=2m.

a) 10F

b) 20F

c) 30F

d) 40F

Answer: b

Explanation: The formula for capacitance of a multi plate capacitor: C=Actual permittivity**A/d.

Thus, C=5**4/2 = 20F.

6. Find the capacitance of a multi plate capacitor whose relative permittivity=5, n=3, A=4m 2 and d=2m.

a) 1.77*10 -10 F

b) 1.77*10 10 F

c) 1.77*10 -11 F

d) 1.77*10 11 F

Answer: a

Explanation: The formula for capacitance of a multi plate capacitor: C=Relative permittivity*absolute permittivity**A/d.

C = 5*8.85*10 -12 **4/2=1.77*10 -10 .

7. Find the number of plates in the multi plate capacitor having C=20F absolute permittivity=5F/m, A=4m 2 and d=2m.

a) 1

b) 2

c) 3

d) 4

Answer: c

Explanation: The formula for capacitance of a multi plate capacitor: C=Actual permittivity**A/d.

Substituting the given values in the equation, we get n=3.

8. Calculate the distance between the plates of the capacitor having C=20F, actual permittivity = 5F/m n=3 and A=4m 2 .

a) 1m

b) 2m

c) 3m

d) 4m

Answer: b

Explanation: The formula for capacitance of a multi plate capacitor: C=Actual permittivity**A/d.

Substituting the given values in the equation, we get d=2m.

9. Calculate the area of cross section of the multi plate capacitor having C=20F, actual permittivity=5F/m n=3 and d=2m.

a) 1m 2

b) 2m 2

c) 3m 2

d) 4m 2

Answer: d

Explanation: The formula for capacitance of a multi plate capacitor: C=Actual permittivity**A/d.

Substituting the given values in the equation, we get A=4m 2 .

10. Calculate the number of plates in the multi plate capacitor having C=1.77*10 -10 F relative permittivity=5, A=4m 2 and d=2m.

a) 1

b) 2

c) 3

d) 4

Answer: c

Explanation: The formula for capacitance of a multi plate capacitor: C=Relative permittivity*absolute permittivity**A/d.

Substituting the given values in the equation, we get n=3.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Composite Dielectric Capacitor”.

1. Potential drop in a dielectric is equal to _______

a) Electric field strength*thickness

b) Electric field strength*area of a cross section

c) Electric field strength

d) Zero

Answer: a

Explanation: When a dielectric is introduced between the two plates of a parallel plate capacitor, the potential difference decreases by the value of the product of electric field strength*thickness which is the potential difference of the dielectric.

2. The electric field strength is 10N/C and the thickness of the dielectric is 3m. Calculate the potential drop in the dielectric.

a) 10V

b) 20V

c) 30V

d) 40V

Answer: c

Explanation: The potential drop in a dielectric= electric field strength*area of cross section = 10*3 = 30V.

3. The electric fields of dielectrics having the same cross sectional area in series are related to their relative permittivities in which way?

a) Directly proportional

b) Inversely proportional

c) Equal

d) Not related

Answer: b

Explanation: Let us consider two plates having fields E1 and E2 and relative permittivities e1 and e2. Then, E1=Q/ and E2=Q/, where e0=absolute permittivity and A=area of cross section. From the given expression, we see that E1/E2=e2/e1, hence the electric field is inversely proportional to the relative permittivities.

4. What happens to the capacitance when a dielectric is introduced between its plates?

a) Increases

b) Decreases

c) Remains the same

d) Becomes zero

Answer: a

Explanation: The capacitance of a capacitance increases when a dielectric is introduced between its plates because the capacitance is related to the dielectric constant k by the equation:

C=k∈ 0 A/d.

5. Calculate the relative permittivity of the second dielectric if the relative permittivity of the first is 4. The electric field strength of the first dielectric is 8V/m and that of the second is 2V/m.

a) 32

b) 4

c) 16

d) 8

Answer: c

Explanation: The relation between the two electric fields and the relative permittivities is:

E1/E1=e2/e1. Substituting the given values, we get e2=16.

6. What happens to the potential drop between the two plates of a capacitor when a dielectric is introduced between the plates?

a) Increases

b) Decreases

c) Remains the same

d) Becomes zero

Answer: b

Explanation: When a dielectric is introduced between the two plates of a parallel plate capacitor, the potential difference decreases because the potential difference of the dielectric is subtracted from it.

7. If the potential difference across the plates of a capacitor is 10V and a dielectric having thickness 2m is introduced between the plates, calculate the potential difference after introducing the dielectric. The electric field strength is 2V/m.

a) 4V

b) 6V

c) 8V

d) 10V

Answer: b

Explanation: When a dielectric is introduced between the plates of a capacitor, its potential difference decreases.

New potential difference= potential difference without dielectric-potential difference of dielectric = 10-2*2 = 6V.

8. Calculate the capacitance if the dielectric constant=4, area of cross section= 10m 2 and the distance of separation between the plates is 5m.

a) 7.08*10 -11 F

b) 7.08*10 11 F

c) 7.08*10 -12 F

d) 7.08*10 -10 F

Answer: a

Explanation: The expression to find capacitance when a dielectric is introduced between the plates is:

C=ke0A/d. Substituting the given values in the equation, we get C = 7.08*10 -11 F.

9. A dielectric is basically a ________

a) Capacitor

b) Conductor

c) Insulator

d) Semiconductor

Answer: c

Explanation: A dielectric is basically an insulator because it has all the properties of an insulator.

10. What happens to the potential difference between the plates of a capacitor as the thickness of the dielectric slab increases?

a) Increases

b) Decreases

c) Remains the same

d) Becomes zero

Answer: b

Explanation: When a dielectric is introduced between the plates of a capacitor, its potential difference decreases.

New potential difference= potential difference without dielectric-potential difference of dielectric. Hence as the thickness of the dielectric slab increases, a larger value is subtracted from the original potential difference.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Charging and Discharging Currents”.

1. Which of the following depends on charging and discharging rate of a capacitor?

a) Time constant

b) Current

c) Power

d) Voltage

Answer: a

Explanation: The time constant in a circuit consisting of a capacitor is the product of the resistance and the capacitance. Smaller the time constant, faster is the charging and discharging rate and vice versa.

2. What is the initial current while charging a capacitor?

a) High

b) Low

c) 0

d) Cannot be determined

Answer: a

Explanation: The initial current of a capacitor is very high because the voltage source will transport charges from one plate of the capacitor to the other plate.

3. What is the final current while charging a capacitor?

a) High

b) Zero

c) Infinity

d) Low

Answer: b

Explanation: The final current is almost equal to zero while charging a capacitor because the capacitor is charged up to the source voltage.

4. A capacitor is charged to a voltage of 400V and has a resistance of 20ohm. Calculate the initial value of charging current.

a) 10A

b) 0A

c) Infinity

d) 20A

Answer: d

Explanation: When the capacitor is charging the initial value if the current is V/R=400/20 = 20A.

5. A capacitor is charged to a voltage of 400V and has a resistance of 20ohm. Calculate the initial value of the discharge current.

a) 10A

b) 0A

c) Infinity

d) 20A

Answer: b

Explanation: When the capacitor is discharging the value of the initial current is zero.

6. A capacitor is charged to a voltage of 400V and has a resistance of 20ohm. Calculate the final value of the discharge current.

a) 10A

b) 0A

c) Infinity

d) 20A

Answer: d

Explanation: In a discharging circuit, the final voltage is equal to zero for capacitor. For a resistor, final voltage is 400V.So,final current = V/R = 400/20 = 20A.

7. When will be capacitors fully charged?

a) When voltage is zero

b) When the supply voltage is equal to the capacitor voltage

c) When voltage is infinity

d) When capacitor voltage is equal to half the supply voltage

Answer: b

Explanation: When the capacitor voltage is equal to the source voltage, it means that all the charges have moved from one plate of the capacitor to the other.

8. What happens to the capacitor when the capacitor voltage is equal to the source voltage?

a) The charging phase of the capacitor is over

b) The discharging phase of the capacitor is over

c) The capacitor is switched off

d) The capacitor is switched on

Answer: c

Explanation: When the capacitor voltage is equal to the source voltage, it means that all the charges have moved from one plate of the capacitor to the other. Hence the capacitor is fully charged and we say it gets switched off.

9. A capacitor is charged to a voltage of 400V and has a resistance of 20ohm. Calculate the final value of charging current.

a) 10A

b) 0A

c) Infinity

d) 20A

Answer: b

Explanation: When the capacitor is charging, the final voltage of the capacitor becomes equal to the voltage of source. Hence, the current becomes equal to zero.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Growth and Decay”.

1. The charging time constant of a circuit consisting of a capacitor is the time taken for the charge in the capacitor to become __________ % of the initial charge.

a) 33

b) 63

c) 37

d) 36

Answer: b

Explanation: We know that: Q=Q 0 (1-e -t/RC ).

When RC=t, we have: Q=Q 0 (1-e -1 ) = 0.63*Q 0 .

Hence the time constant is the time taken for the charge in a capacitive circuit to become 0.63 times its initial charge.

2. The discharging time constant of a circuit consisting of a capacitor is the time taken for the charge in the capacitor to become __________ % of the initial charge.

a) 33

b) 63

c) 37

d) 36

Answer: c

Explanation: We know that: Q=Q 0 (1-e -t/RC ).

When RC=t, we have: Q=Q 0 (1-e -1 ) = 0.37*Q 0 .

Hence the time constant is the time taken for the charge in a capacitive circuit to become 0.37 times its initial charge.

3. A circuit has a resistance of 2 ohm connected in series with a capacitance of 6F. Calculate the charging time constant.

a) 3

b) 1

c) 12

d) 8

Answer: c

Explanation: The charging time constant in a circuit consisting of a capacitor and resistor in series is the product of the resistance and capacitance = 2*6 = 12.

4. A circuit has a resistance of 5 ohm connected in series with a capacitance of 10F. Calculate the discharging time constant.

a) 15

b) 50

c) 5

d) 10

Answer: b

Explanation: The discharging time constant in a circuit consisting of a capacitor and resistor in series is the product of the resistance and capacitance = 5*10 = 50.

5. What is the value of current in a discharging capacitive circuit if the initial current is 2A at time t=RC.

a) 0.74A

b) 1.26A

c) 3.67A

d) 2.89A

Answer: b

Explanation: At time t=RC, that is the time constant, we know that the value of current at that time interval is equal to 63% of the initial charge in the discharging circuit. Hence, I = 2*0.63 = 1.26A.

6. What is the value of current in a charging capacitive circuit if the initial current is 2A at time t=RC.

a) 0.74A

b) 1.26A

c) 3.67A

d) 2.89A

Answer: a

Explanation: At time t=RC, that is the time constant, we know that the value of current at that time interval is equal to 37% of the initial charge in the discharging circuit. Hence, I = 2*0.37 = 0.74A.

7. While discharging, what happens to the current in the capacitive circuit?

a) Decreases linearly

b) Increases linearly

c) Decreases exponentially

d) Increases exponentially

Answer: d

Explanation: The equation for the value of current in a discharging capacitive circuit is:

I=I 0 (1-e -t/RC ). From this equation, we can see that the current is exponentially increasing.

8. While discharging, what happens to the voltage in the capacitive circuit?

a) Decreases linearly

b) Increases linearly

c) Decreases exponentially

d) Increases exponentially

Answer: c

Explanation: The equation for the value of voltage in a discharging capacitive circuit is:

V=V 0 (e -t/RC ). From this equation, we can see that the voltage is exponentially decreasing.

9. While charging, what happens to the current in the capacitive circuit?

a) Decreases linearly

b) Increases linearly

c) Decreases exponentially

d) Increases exponentially

Answer: c

Explanation: The equation for the value of current in a charging capacitive circuit is:

I=I0(e -t/RC ). From this equation, we can see that the current is exponentially decreasing.

10. While charging, what happens to the voltage in the capacitive circuit?

a) Decreases linearly

b) Increases linearly

c) Decreases exponentially

d) Increases exponentially

Answer: d

Explanation: The equation for the value of voltage in a charging capacitive circuit is:

V=V0(1-e -t/RC ). From this equation, we can see that the voltage is exponentially increasing.

This set of Basic Electrical Engineering Interview Questions and Answers for freshers focuses on “Discharge of a Capacitor Through a Resistor”.

1. An 8microF capacitor is connected in series with a 0.5 megaohm resistor. The DC voltage supply is 200V. Calculate the time constant.

a) 1s

b) 2s

c) 3s

d) 4s

Answer: d

Explanation: The time constant is the product of the resistance and capacitance in a series RC circuit.

Therefore, time constant = 8*10 -6 *4*10 6 =4s.

2. An 8microF capacitor is connected in series with a 0.5 megaohm resistor. The DC voltage supply is 200V. Calculate the initial charging current.

a) 100 microA

b) 500 microA

c) 400 microA

d) 1000microA

Answer: c

Explanation: In a series RC circuit, the initial charging current is:

I=V/R = 200/(0.5*10 6s ) = 400*10 -6 A = 400 microA.

3. A 8 microF capacitor is connected in series with a 0.5 megaohm resistor. The DC voltage supply is 200V. Calculate the time taken for the potential difference across the capacitor to grow to 160V.

a) 6.93s

b) 7.77s

c) 2.33s

d) 3.22s

Answer: a

Explanation: From the previous explanations, we know that the initial current is 400mA and the time constant is 4s. Substituting the values of capacitor voltage, initial voltage, initial current and time constant in the equation: V=V 0 (1-e -t/RC )

Substituting V=160V, V0=200V, RC=4s we get,

t=6.93s.

4. An 8microF capacitor is connected in series with a 0.5 megaohm resistor. The DC voltage supply is 200V. Calculate the voltage in the capacitor 4s after the power is supplied.

a) 123.4V

b) 126.4V

c) 124.5V

d) 132.5V

Answer: b

Explanation: We can get the value of the potential difference across the capacitor in 4s, from the following equation:

Vc=V(1-e -t /RC). Substituting the values in the given equation, we get Vc = 126.4V.

5. An 8microF capacitor is connected in series with a 0.5 megaohm resistor. The DC voltage supply is 200V. Calculate the current in the capacitor 4s after the power is supplied.

a) 79 microA

b) 68 microA

c) 48 microA

d) 74 microA

Answer: d

Explanation: In the given question, the time constant is equal to the time taken= 4s. Hence the value of current will be 37% of its initial value = I=0.37*200 = 74 microA.

6. A circuit has a resistance of 2 ohms connected in series with a capacitance of 6F. Calculate the discharging time constant.

a) 3

b) 1

c) 12

d) 8

Answer: c

Explanation: The discharging time constant in a circuit consisting of a capacitor and resistor in series is the product of the resistance and capacitance = 2*6 = 12.

7. What is the energy in a capacitor if the voltage is 5V and the charge is10C?

a) 25J

b) 35J

c) 54J

d) 55J

Answer: a

Explanation: We know that Q/V=C. Hence the value of capacitance is 2F.

U=/2 = /2 = 25 J.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Transients in CR Networks”.

1. A CR network is one which consists of _________

a) A capacitor and resistor connected in parallel

b) A capacitor and resistor connected in series

c) A network consisting of a capacitor only

d) A network consisting of a resistor only

Answer: b

Explanation: A CR network is one which consists of a capacitor connected in series with a resistor. The capacitor discharges or charges through the resistor.

2. At DC, capacitor acts as _________

a) Open circuit

b) Short circuit

c) Resistor

d) Inductor

Answer: a

Explanation: Capacitive Reactance X C = 1/

For DC, f=0 so, X C becomes infinite. Hence for dc, the capacitor acts as an open circuit.

3. In an RC series circuit, when the switch is closed and the circuit is complete, what is the response?

a) Response does not vary with time

b) Decays with time

c) Increases with time

d) First increases, then decrease

Answer: b

Explanation: In an RC series circuit, the response decays with time because according to the equation, there is an exponential decrease in the response.

4. If the switch is closed at t=0, what is the current in the circuit?

basic-electrical-engineering-questions-answers-transients-cr-networks-q4

a) 0A

b) 10A

c) 20A

d) Infinity

Answer: b

Explanation: As soon as the switch is closed at t=0, the capacitor acts as a short circuit. The current in the circuit is:

I=V/R = 100/10 = 10A.

5. Calculate the voltage across the capacitor at t=0.

basic-electrical-engineering-questions-answers-transients-cr-networks-q4

a) 0V

b) 10V

c) 20V

d) Infinity

Answer: a

Explanation: When the switch is closed at t=0, the capacitor has no voltage across it since it has not been charged. The capacitor acts as a short circuit and the voltage across it is zero.

6. Calculate di/dt if the switch is closed at t=0.

basic-electrical-engineering-questions-answers-transients-cr-networks-q4

a) -9.9A/s

b) -10A/s

c) 0A/s

d) -0.1A/s

Answer: d

Explanation: Applying KVL to the given circuit, we get:

i=i 0 e -t/RC = e -t/100

i=10 e -t/100

di/dt = - e -t/100

di/dt=-0.1A/s.

7. Calculate d 2 i/dt 2 from the given circuit.

basic-electrical-engineering-questions-answers-transients-cr-networks-q4

a) 10 -6 A/s 2

b) 10 -3 A/s 2

c) 10 6 A/s 2

d) 10 3 A/s 2

Answer: b

Explanation: Applying KVL to the given circuit, we get:

100+10i+1/10*integraldt)=0

Differentiating once, we get:

10di/dt+1/10*i.

Differentiating once again, we get:

10d 2 i/dt 2 +10di/dt=0.

Substituting the values of di/dt from the previous explanation, we get d 2 i/dt 2 =10 -3 A/s 2 .

8. The current equation for the given circuit is?

basic-electrical-engineering-questions-answers-transients-cr-networks-q4

a) i=10e  t A

b) i=10e  t A

c) i=10e  t A

d) i=100e  t A

Answer: a

Explanation: The KVL equation is:

100+10i+1/10*integraldt)=0

On applying Laplace transform to this equation, we get:

100/s=I/10s+10I

Solving the equation, we get:

i=10e  t A.

9. The expression for the current in an RC circuit is?

a) i=e t/RC

b) i=e -t/RC

c) i=(1-e -t/RC )

d) i= (1-e t/RC )

Answer: b

Explanation: Applying KVL to the given circuit, we get:

i=i 0 e -t/RC = e -t/100

i=10 e -t/100 .

10. What is the voltage in the resistor as soon as the switch is closed at t=0.

basic-electrical-engineering-questions-answers-transients-cr-networks-q10

a) 0V

b) Infinity

c) 220V

d) Insufficient information provided

Answer: c

Explanation: As soon as the switch is closed at t=0, there is no charge in the capacitor, hence the voltage across the capacitor is zero and all the 220V voltage is the voltage across the resistor.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Energy Stored in a Charged Capacitor”.

1. Work done in charging a capacitor is ____________

a) QV

b) 1 ⁄ 2 QV

c) 2QV

d) QV 2

Answer: b

Explanation: We know that work done= Q 2 /2C.

Substituting C as Q/V, we get work done = Q/2V.

2. Energy stored in 2000mF capacitor charged to a potential difference of 10V is?

a) 100J

b) 200J

c) 300J

d) 400J

Answer: a

Explanation: From the expression:

WD = CV 2 /2 = 100J.

3. When do we get maximum energy from a set of capacitors?

a) When they are connected in parallel

b) When they are connected in series

c) Both in series and parallel

d) Insufficient information provided

Answer: a

Explanation: We get maximum energy when capacitors are connected in parallel because the equivalent capacitance is larger than the largest individual capacitance when connected in parallel. The relation between capacitance and energy is:

Energy=CV 2 /2, hence as the capacitance increases, the energy stored in it also increases.

4. If the charge stored in a capacitor is 4C and the value of capacitance is 2F, calculate the energy stored in it.

a) 2J

b) 4J

c) 8J

d) 16J

Answer: b

Explanation: The expression for finding the value of energy is:

U=Q 2 /2C = 4*4/ = 4J.

5. If the charge in a capacitor is 4C and the energy stored in it is 4J, find the value of capacitance.

a) 2F

b) 4F

c) 8F

d) 16F

Answer: a

Explanation: The expression for finding the value of energy is:

U=Q 2 /2C.

Substituting the values of U and Q, we get C=2F.

6. If the charge in a capacitor is 4C and the energy stored in it is 4J, calculate the voltage across its plates.

a) 2V

b) 4V

c) 8V

d) 16V

Answer: a

Explanation: The expression for finding the value of energy is:

U=Q 2 /2C.

Substituting the values of U and Q, we get C=2F.

V=Q/C, hence V=4/2=2V.

7. Calculate the energy in the 2F capacitor.

basic-electrical-engineering-questions-answers-energy-charged-capacitor-q7

a) 8.6kJ

b) 64kJ

c) 64J

d) 6.4kJ

Answer: d

Explanation: From the expression:

WD= CV 2 /2 = 2*80 2 /2=6400J=6.4kJ.

8. Calculate the energy in the 4F capacitor.

basic-electrical-engineering-questions-answers-energy-charged-capacitor-q7

a) 128kJ

b) 1.28kJ

c) 12.8kJ

d) 1280J

Answer: c

Explanation: From the expression:

WD = CV 2 /2 = 4*80 2 /2 = 12800J = 12.8kJ.

9. Calculate the energy stored in the combination of the capacitors.

basic-electrical-engineering-questions-answers-energy-charged-capacitor-q7

a) 192kJ

b) 1.92kJ

c) 19.2kJ

d) 1920J

Answer: c

Explanation: The equivalent capacitance is: Ceq=4+2=6F.

From the expression:

WD = CV 2 /2 = 6*80 2 /2 = 19200J = 19.2kJ.

This set of Basic Electrical Engineering Questions and Answers for Experienced people focuses on “Force of Attraction Between Oppositely Charged Plates”.

1. Which among the following is the correct expression for force between the plates of a parallel plate capacitor?

a) F=epsilon*A* 2 /2

b) F=epsilon*A* 2 /3

c) F=epsilon  2 /2

d) F=epsilon  2 /3

Answer: a

Explanation: The force is proportional to the square of the potential gradient and the area. Hence the force F=epsilon*A* 2 /2.

2. When the area of cross section of the plate increases, what happens to the force between the plates?

a) Increases

b) Decreases

c) Remains the same

d) Becomes zero

Answer: a

Explanation: The force of attraction between the two plates of the capacitor is directly proportional to the area of cross section of the plates, hence an area of cross section increases, the force of attraction also increases.

3. When the potential gradient increases, what happens to the force between the plates?

a) Increases

b) Decreases

c) Remains the same

d) becomes zero

Answer: a

Explanation: The force of attraction between the two plates of the capacitor is directly proportional to the square of potential gradient, hence as a potential gradient increases, the force of attraction also increases.

4. In which of the following mediums, will the force of attraction between the plates of a capacitor be greater?

a) Air

b) Water

c) Does not depend on the medium

d) Cannot be determined

Answer: b

Explanation: The absolute permittivity of water is greater than that of air. The expression relating F and epsilon is F=epsilon*A* 2 /2. From this expression, we can see that as epsilon increases, the force of attraction also increases.

5. A metal parallel plate capacitor has 100mm diameter and the distance between the plates is 1mm. The capacitor is placed in air. Calculate the force on each plate if the potential difference between the plates is 1kV.

a) 350N

b) 0.035kN

c) 0.035N

d) 3.35kN

Answer: c

Explanation: From the given data:

A=pi*d 2 /4=0.007854m 2

Potential gradient = V/x = 10 6 V/m

F=epsilon*A* 2 /2

Therefore, F=0.035N.

6. A metal parallel plate capacitor has 100mm diameter and the distance between the plates is ‘a’ mm. The capacitor is placed in air. Force on each plate is 0.035N and the potential difference between the plates is 1kV. Find ‘a’.

a) 1m

b) 1cm

c) 10cm

d) 1mm

Answer: d

Explanation: From the given data:

A=pi*d 2 /4=0.007854m 2

Potential gradient = V/x = 1000/a

F=epsilon*A* 2 /2

Substituting the given values, we find a=1mm.

7. A metal parallel plate capacitor has ‘a’mm diameter and the distance between the plates is 1mm. The capacitor is placed in air. Force on each plate is 0.035N and the potential difference between the plates is 1kV. Find ‘a’.

a) 10mm

b) 100mm

c) 1000m

d) 1000cm

Answer: b

Explanation: From the given data:

A=pi*d 2 /4=pi*a 2 /4

Potential gradient = V/x = 10 6 V/m

F=epsilon*A* 2 /2

Substituting the given values, we get d=100mm.

8. A metal parallel plate capacitor has 100mm diameter and the distance between the plates is 1mm. The capacitor is placed in air. Calculate the potential difference between the plates if the force on each plate is 0.035N.

a) 1kV

b) 1V

c) 2kV

d) 2V

Answer: a

Explanation: From the given data:

A=pi*d 2 /4=0.007854m 2

Potential gradient = V/x = 1000*V

F=epsilon*A* 2 /2

Substituting the given values in the above expression, we get V=1kV.

9. What happens to the force of attraction between the capacitors when the potential difference between the plates decreases?

a) Increases

b) Decreases

c) Remains the same

d) Becomes zero

Answer: b

Explanation: The force of attraction between the two plates of the capacitor is directly proportional to the square of the potential difference between the plates, hence as the potential difference decreases, the force of attraction also decreases.

10. What happens to the force of attraction between the capacitors when the distance of separation between the plates increases?

a) Increases

b) Decreases

c) Remains the same

d) Becomes zero

Answer: b

Explanation: The force of attraction between the two plates of the capacitor is inversely proportional to the square of the distance between the plates, hence as distance increases, the force of attraction decreases.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Dielectric Strength”.

1. The unit for dielectric strength is ____________

a) V/m 2

b) MV/m 2

c) MV/m

d) Vm

Answer: c

Explanation: Dielectric strength is the potential gradient required to cause a breakdown in the material. Potential gradient is the ratio of voltage and length, its unit is MV/m.

2. If the Voltage increases, what happens to dielectric strength?

a) Increases

b) Decreases

c) Remains the same

d) Becomes zero

Answer: a

Explanation: Dielectric strength is the potential gradient required to cause a breakdown in the material. Potential gradient is the ratio of voltage and length. Hence as potential increases, dielectric strength also increases.

3. If the potential difference in a material is 4MV and the thickness of the material is 2m, calculate the dielectric strength.

a) 2MV/m

b) 4MV/m

c) 6MV/m

d) 8MV/m

Answer: a

Explanation: Dielectric strength is the potential gradient required to cause a breakdown in the material. Potential gradient is the ratio of voltage and thickness.

Dielectric strength= V/t= 4/2= 2MV/m.

4. If the dielectric strength of a material is 4MV/m and its potential difference is 28MV, calculate the thickness of the material.

a) 4m

b) 7m

c) 5m

d) 11m

Answer: b

Explanation: Dielectric strength is the potential gradient required to cause a breakdown in the material. Potential gradient is the ratio of voltage and thickness.

V/dielectric strength= t= 28/4=7m.

5. If the thickness of the material increases, what happens to the dielectric strength?

a) Increases

b) Decreases

c) Remains the same

d) Becomes zero

Answer: b

Explanation: Dielectric strength is the potential gradient required to cause a breakdown in the material. Potential gradient is the ratio of voltage and thickness. Hence as thickness increases, dielectric strength decreases.

6. The thickness of a material having dielectric strength 10MV/m is 5m, calculate the potential difference.

a) 2MV

b) 10MV

c) 50MV

d) 100MV

Answer: c

Explanation: Dielectric strength is the potential gradient required to cause a breakdown in the material. Potential gradient is the ratio of voltage and thickness.

V=t*dielectric strength= 5*10=50MV.

7. Which medium has the highest dielectric strength?

a) Water

b) Mica

c) Air

d) Glass

Answer: c

Explanation: The better material is to prevent electrical conductivity, higher the dielectric strength. And the air is the best insulator so it has high dielectric strength.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Leakage and Conduction Currents in Capacitors”.

1. Leakage in capacitors is primarily caused by _________

a) Transistors

b) Resistors

c) Inductors

d) DC motors

Answer: a

Explanation: Leakage is primarily caused due to electronic devices, such as transistors, connected to the capacitors. Transistors conduct a small amount of current even when they are turned off, hence they are responsible for leakage current.

2. What is the conduction current when a capacitor is fully charged?

a) Infinity

b) Zero

c) 100A

d) 1000A

Answer: b

Explanation: When a capacitor is fully charged, there is no conduction of electrons from one plate of the capacitor to another, hence there is no conduction current and conduction current is equal to zero.

3. The flow of electrons in dielectric is due to _________

a) Conduction

b) Potential difference

c) Breakdown

d) Resistance

Answer: c

Explanation: There is, under normal circumstance, no flow of electrons in a dielectric since a dielectric is basically an insulator. Hence, there is a flow of electrons in a dielectric only at breakdown voltage.

4. The flow of electrons which does not pass through the battery is known as ________

a) Displacement current

b) Leakage current

c) Either displacement or leakage current

d) Neither displacement nor leakage current

Answer: a

Explanation: Displacement current is the flow of electrons from the positive plate of the capacitor to the negative plate of the capacitor, not through the battery. Hence the type of current which flows without passing through the battery is displacement current.

5. The free electrons in practical dielectrics is due to _________

a) There are no free electrons

b) Conductors

c) Impurities

d) Both conductors and impurities

Answer: c

Explanation: Ideally, dielectrics are insulators and do not contain any free electrons. But no dielectric is a perfect dielectric, hence the free electrons are due to impurities present in each dielectric.

6. The current in conductors connecting the voltage source to the plates of a capacitor is ______

a) Conduction current

b) Leakage current

c) Charging current

d) Zero

Answer: c

Explanation: The current in conductors connecting the voltage source to the plates of a capacitor is the charging current and not the conduction or leakage current.

7. What is the type of current where the electrons actually move?

a) Displacement current

b) Conduction current

c) Both conduction and displacement current

d) Neither conduction nor displacement current

Answer: b

Explanation: Conduction current is the current caused by the actual flow of electrons and displacement current is the current where no charge carriers are involved.

8. What is the type of current caused due to variations in the field?

a) Displacement current

b) Conduction current

c) Both conduction and displacement current

d) Neither conduction nor displacement current

Answer: a

Explanation: Displacement current is the current where no charge carriers are involved. It is caused due to variations in the electric field.

9. Under normal conditions capacitors have _________

a) Displacement current

b) Conduction current

c) Both conduction and displacement current

d) Neither conduction nor displacement current

Answer: a

Explanation: Under normal conditions capacitors contain an insulating material called dielectric sandwiched between the plates of the capacitor. Since insulators can carry only an electric field but not moving carriers, therefore normally a capacitor has displacement current and not conduction current.

10. If a large amount of voltage is applied to a capacitor, what is the current that flows through it?

a) Displacement current

b) Conduction current

c) Both conduction and displacement current

d) Neither conduction nor displacement current

Answer: b

Explanation: When a large amount of voltage is applied between the plates of a capacitor, the dielectric between the plates does not behave as an insulator anymore and starts conducting and conduction currents flow through it.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Displacement Current in a Dielectric”.

1. The current in conductors connecting the voltage source to the plates of a capacitor is _______

a) Conduction current

b) Leakage current

c) Charging current

d) Displacement current

Answer: c

Explanation: The current in conductors connecting the voltage source to the plates of a capacitor is the charging current and not the conduction or leakage current.

2. Under normal conditions capacitors have _______

a) Displacement current

b) Conduction current

c) Both conduction and displacement current

d) Neither conduction nor displacement current

Answer: a

Explanation: Under normal conditions capacitors contain an insulating material called dielectric sandwiched between the plates of the capacitor. Since insulators can carry only an electric field but not moving carriers, therefore normally a capacitor has displacement current and not conduction current.

3. What is the unit for displacement current?

a) No unit

b) Ampere

c) Coulomb

d) Ampere/coulomb

Answer: b

Explanation: Displacement current is a type of current and hence it has the same unit as that of current that is ampere.

4. Displacement current depends on ___________

a) Moving charges

b) Change in time

c) Both moving charges and change in time

d) Neither moving charges nor change in time

Answer: b

Explanation: Displacement current is the current which arises due to variations in the field. Hence, it does not depend on the moving charges but it changes with time which causes variation in the field.

5. Magnetic fields between the parallel plates of a capacitor are due to?

a) Displacement current

b) Conduction current

c) Both conduction and displacement current

d) Neither conduction nor displacement current

Answer: a

Explanation: Displacement current is the current which arises due to variations in the field. Change in the field results in the formation of magnetic fields. Hence displacement currents lead to magnetic field between the plates of a capacitor.

6. The free electrons in practical dielectrics are due to ________

a) There are no free electrons

b) Conductors

c) Impurities

d) Displacement currents

Answer: c

Explanation: Ideally, dielectrics are insulators and do not contain any free electrons. But no dielectric is a perfect dielectric, hence the free electrons are due to impurities present in each dielectric.

7. The flow of electrons which does not pass through the battery is known as ____________

a) Conduction current

b) Leakage current

c) Charging current

d) Displacement current

Answer: a

Explanation: Conduction current is the flow of electrons from the positive plate of the capacitor to the negative plate of the capacitor, not through the battery. Hence the type of current which flows without passing through the battery is conduction current.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Types of Capacitor and Capacitance”.

1. Paper capacitor is a type of _________

a) Fixed capacitor

b) Variable capacitor

c) Either fixed or variable depending on its usage

d) Neither fixed nor variable

Answer: a

Explanation: Paper capacitors are fixed capacitors because, like fixed capacitors, its capacitance value remains constant. In paper capacitors, paper is used as the dielectric.

2. A capacitor using chemical reactions to store charge is _______

a) Paper capacitor

b) Ceramic capacitor

c) Polyester capacitor

d) Electrolyte capacitor

Answer: d

Explanation: Electrolyte capacitors use chemical processes like pyrolysis to store charge between its plates.

3. Which, among the following, is the odd one out?

a) Ceramic capacitor

b) Electrolyte capacitor

c) Tuning capacitor

d) Paper capacitor

Answer: c

Explanation: Ceramic capacitor, electrolyte capacitor and paper capacitor are fixed capacitors whereas tuning capacitors is a variable capacitor, hence it is the odd one out.

4. In a variable capacitor, capacitance can be varied by ______

a) Turning the rotatable plates in or out

b) Sliding the rotatable plates

c) Changing the plates

d) Changing the material of plates

Answer: a

Explanation: As the plates are rotated, the area of the plates between which the field exists, will vary. Capacitance depends on area, hence as area varies, capacitance also varies.

5. The simplest kind of capacitor is ________

a) Ceramic capacitor

b) Electrolyte capacitor

c) Tuning capacitor

d) Paper capacitor

Answer: d

Explanation: The paper capacitor consists of two strips of aluminium foil separated by sheets of waxed paper. This whole setup is rolled up into the form of a cylinder. Since the materials requires for its construction are easily available, it is the simplest kind of capacitor.

6. Capacitor preferred when there is high frequency in the circuits is _______

a) Electrolyte capacitor

b) Mica capacitor

c) Air capacitor

d) Glass capacitor

Answer: b

Explanation: Mica capacitors are preferred for high frequency circuits because they have low ohmic losses and less reactance.

7. The type of capacitors used in communication transmitters are?

a) Electrolyte capacitor

b) Variable capacitor

c) Air capacitor

d) Glass capacitor

Answer: b

Explanation: Variable capacitor is used to tune all the circuits to same frequency i.e. resonance frequency so they are used in communication transmitters.

8. Which capacitors relatively costly?

a) Electrolyte capacitor

b) Mica capacitor

c) Air capacitor

d) Glass capacitor

Answer: b

Explanation: Mica capacitors are relatively expensive because it consists either of alternate layers of mica and metal foil clamped tightly together, or of thin films of silver on the two sides of a mica sheet. Silver is an expensive metal, hence mica capacitors are expensive.

9. ____________ capacitors usually have a colour code to find its value.

a) Electrolyte capacitor

b) Variable capacitor

c) Polyester capacitor

d) Glass capacitor

Answer: c

Explanation: Polyester capacitors usually come with a colour code because they are very small and their values cannot be printed on its body.

10. ______________ capacitors have a high leakage voltage.

a) Electrolyte capacitor

b) Variable capacitor

c) Air capacitor

d) Polyester capacitor

Answer: d

Explanation: Polyester capacitors can operate at high voltages, that is, a few thousand volts and the leakage resistance is high, that is, usually 100 M.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Magnetic Field”.

1. What is the magnetic field outside a solenoid?

a) Infinity

b) Half the value of the field inside

c) Double the value of the field inside

d) Zero

Answer: d

Explanation: There are no magnetic lines of force outside a solenoid, hence the magnetic field outside a solenoid is zero.

2. Which, among the following qualities, is not affected by the magnetic field?

a) Moving charge

b) Change in magnetic flux

c) Current flowing in a conductor

d) Stationary charge

Answer: d

Explanation: A stationary charge is not affected by a magnetic field because stationary charges do not have any velocity. Magnetic field cannot occur in a particle having zero velocity.

3. When a charged particle moves at right angles to the magnetic field, the variable quantity is?

a) Momentum

b) Speed

c) Energy

d) Moment of inertia

Answer: a

Explanation: When a charged particle moves perpendicular to the field, its speed remains the same whereas its velocity keeps on changing. Momentum is the product of the mass of the particle and the velocity if the particle, hence since velocity varies, momentum also varies.

4. If the flow of electric current is parallel to the magnetic field, the force will be ______

a) Zero

b) Infinity

c) Maximum

d) Half the original value

Answer: a

Explanation: Force is a cross product. A cross product involves the sine of the angle between them. If two quantities are parallel to each other, the angle between them is zero. Sin is zero, hence force is zero.

5. The ratio of magnetic force to electric force on a charged particle getting undeflected in a field is?

a) 1

b) 0

c) 2

d) 4

Answer: a

Explanation: When a charged particle is undeflected in a field, the magnitude of the magnetic force and electric force acting on the particle is the same, hence the ratio is 1.

6. What is the strength of magnetic field known as ________

a) Flux

b) Density

c) Magnetic strength

d) Magnetic flux density

Answer: d

Explanation: Strength of magnetic field is also known as magnetic flux density. It is the amount of magnetic field lines crossing unit area.

7. Weakest force in nature is?

a) Electric force

b) Gravitational force

c) Weak force

d) Magnetic force

Answer: b

Explanation: Gravitational force is the weakest force in nature as it does not bind anything strongly with its help.

8. How can a magnetic field be produced?

a) Using a permanent magnet

b) Electric current

c) Using a temporary magnet

d) Using a permanent magnet or electric current

Answer: d

Explanation: An electric current as well as the permanent magnet produces a magnetic field whereas a temporary magnet fails to do so.

9. Can we see magnetic flux lines?

a) Yes

b) No

c) Depends on the strength of the field

d) Only when the field strength is very large

Answer: b

Explanation: No, we cannot see magnetic flux lines as the “lines of magnetic flux” is purely an imaginary concept to understand the magnetic field clearly.

10. Magnetic Field lines move from _______

a) North to south

b) South to north

c) West to east

d) East to west

Answer: a

Explanation: Magnetic field lines originate at the north pole and terminate at the south pole of the magnet.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Direction of Magnetic Field”.

1. Field lines move from __________

a) North to south

b) South to north

c) West to east

d) East to west

Answer: a

Explanation: Magnetic field lines originate at the north pole and terminate at the south pole of the magnet.

2. Magnetic field lines ___________ at the north pole.

a) Emerge

b) Converge

c) Neither emerge nor converge

d) Either emerge or converge

Answer: a

Explanation: Magnetic field lines emerge at the north pole. Field lines seem to emerge at the north pole because they originate at the north pole.

3. Magnetic field lines ___________ at the south pole.

a) Emerge

b) Converge

c) Neither emerge nor converge

d) Either emerge or converge

Answer: b

Explanation: Magnetic field lines converge at the south pole. Field lines seem to converge at the south pole because they end at the south pole.

4. Which of the following is used to determine the direction of magnetic field in a current carrying conductor?

a) Left hand thumb rule

b) Right hand thumb rule

c) Right hand palm rule

d) Left hand palm rule

Answer: b

Explanation: The right hand thumb rule determines the direction of a magnetic field in a current carrying conductor. The rule states that when we align our right thumb in the direction of the current and curl our fingers around it, the direction of our fingers is the direction of the magnetic field.

5. According to Flemming’s left hand rule, the index finger denotes?

a) Direction of magnetic field

b) Direction of current

c) Direction of force

d) Direction of force as well as current

Answer: a

Explanation: According to Flemming’s left hand rule, the index finger denotes the direction of the magnetic field, the thumb denoted the direction of force and the middle finger denoted the direction of the current.

6. According to Flemming’s left hand rule, the middle finger denotes?

a) Direction of magnetic field

b) Direction of current

c) Direction of force

d) Direction of force as well as current

Answer: b

Explanation: According to Flemming’s left hand rule, the index finger denotes the direction of the magnetic field, the thumb denoted the direction of force and the middle finger denoted the direction of the current.

7. According to Flemming’s left hand rule, the thumb denotes?

a) Direction of magnetic field

b) Direction of current

c) Direction of force

d) Direction of force as well as current

Answer: c

Explanation: According to Flemming’s left hand rule, the index finger denotes the direction of the magnetic field, the thumb denoted the direction of force and the middle finger denoted the direction of the current.

8. The relation between the direction of force and the direction of magnetic field is _________

a) Same direction

b) Opposite direction

c) Perpendicular

d) Unrelated

Answer: c

Explanation: When a conductor carries a certain value of current, the force developed in the conductor, the current in the conductor and the magnetic field in the conductor are mutually perpendicular to each other.

9. The relation between the direction of current and the direction of magnetic field is ________

a) Same direction

b) Opposite direction

c) Perpendicular

d) Unrelated

Answer: c

Explanation: When a conductor carries a certain value of current, the force developed in the conductor, the current in the conductor and the magnetic field in the conductor are mutually perpendicular to each other.

10. The relation between the direction of current and the direction of the force is _________

a) Same direction

b) Opposite direction

c) Perpendicular

d) Unrelated

Answer: c

Explanation: When a conductor carries a certain value of current, the force developed in the conductor, the current in the conductor and the magnetic field in the conductor are mutually perpendicular to each other.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Characteristics of Lines of Magnetic Flux”.

1. Magnetic field lines seek the path of __________ resistance.

a) Maximum

b) Minimum

c) Infinite

d) Zero

Answer: b

Explanation: Magnetic field lines will always seek the path of least resistance. It does not seek the path of zero resistance because, in practical scenarios, zero resistance is not possible.

2. Magnetic field lines form _________ loops from pole to pole.

a) Open

b) Closed

c) Branched

d) Either closed or branched

Answer: b

Explanation: Magnetic field lines form closed loops from pole to pole. There is no discontinuity in the magnetic flux lines.

3. Do magnetic flux lines intersect?

a) Yes

b) No

c) Depends on the situation

d) Cannot be determined

Answer: b

Explanation: Magnetic field lines do not cross each other because if they cross each other it means that there are two different directions of a magnetic field in that region and that is not possible.

4. Inside the magnet, field lines travel from?

a) North to south

b) South to north

c) West to east

d) East to west

Answer: b

Explanation: Magnetic field lines originate at the north pole and terminate at the south pole of the magnet and magnetic field lines form closed loop so inside the magnet, they move from south to north.

5. Lines of magnetic flux which are parallel and in the same direction __________ each other.

a) Attract

b) Repel

c) Intersect

d) Cancel

Answer: b

Explanation: Lines of magnetic flux which are parallel to each other and in the same direction repel each other because they tend to act as like poles and like poles repel each other.

6. More the number of magnetic flux lines _______ is the force of the magnet.

a) Greater

b) Lesser

c) Either greater or lesser

d) Neither greater nor lesser

Answer: a

Explanation: More the number of magnetic flux lines, greater is the force of the magnet. This is because the magnetic flux lines denote the strength of the field of the magnet.

7. Magnetic field is strong when____________

a) magnetic field lines are closer

b) magnetic field lines are farther

c) magnetic field lines are longer

d) magnetic field lines are thicker

Answer: a

Explanation: Magnetic field is strong where magnetic field lines are closer and weak where magnetic field lines are farther.

This set of Basic Electrical Engineering Interview Questions and Answers for Experienced people focuses on “Magnetic Field Due to an Electric Current”.

1. The force existing between two infinite parallel conductors is inversely proportional to ________

a) Radius of the conductors

b) Current in one of the conductors

c) The product of the current in the two conductors

d) The distance between the two conductors

Answer: d

Explanation: When current is flowing in two different conductors, the force between the two conductors is directly proportional to the product of the current in the two conductors and inversely proportional to the distance between the two conductors.

2. When the distance of operation between the two conductors increases, what happens to the force between the two conductors?

a) Increases

b) Decreases

c) Remains the same

d) Becomes zero

Answer: b

Explanation: When the distance of separation increases, the force between the two conductors decrease because the force between two conductors is inversely proportional to the distance of separation between them.

3. Magnetic field at a point d distance away from long wire due to electric current i in it is ____________

a) µ 0 i/2r

b) µ 0 i/r

c) µ 0 i/2πr

d) µ 0 i/πr

Answer: c

Explanation: Magnetic field at distance d from long wire with carrying current i is given by-

B = µ 0 i/2πr.

4. The force per unit length existing between two infinite parallel conductors is given by ___________

a) µ 0 i 1 i 2 /2πd

b) µ 0 i 1 i 2 /2d

c) µ 0 i 1 /2πdi 2

d) µ 0 i 1 i 2 /d

Answer: a

Explanation: Force of infinite conductor is given by Bil if B, i, l are mutually perpendicular

F/l = Bi = µ 0 i 1 i 2 /2πd if d is separation of two infinite parallel conductors.

5. If the radius of the current carrying conductor increases, what is the effect on the force?

a) increases

b) decreases

c) remain same

d) become zero

Answer: c

Explanation: Force of the infinite conductor is given by Bil if B, i, l are mutually perpendicular

F/l = Bi = µ 0 i 1 i 2 /2πd if d is separation of two infinite parallel conductors. It doesn’t depend on the radius of conductor so force remains the same.

6. If current of conductor increases, what is the effect on the force?

a) increases

b) decreases

c) remain same

d) become zero

Answer: a

Explanation: Force of infinite conductor is given by Bil if B, i, l are mutually perpendicular

F/l = Bi = µ 0 i 1 i 2 /2πd if d is separation of two infinite parallel conductors. So, as current increases force increases.

7. If length of current carrying conductor increases, what is the effect on the force?

a) increases

b) decreases

c) remain same

d) become zero

Answer: a

Explanation: Force of infinite conductor is given by Bil if B, i, l are mutually perpendicular

F/l = Bi = µ 0 i 1 i 2 /2πd if d is a separation of two infinite parallel conductors. So, as length of conductor increases force increases.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Magnetic Field of a Solenoid”.

1. The magnetic field strength of a solenoid can be increased by inserting which of the following materials as the core?

a) Copper

b) Silver

c) Iron

d) Aluminium

Answer: c

Explanation: The Magnetic field of a solenoid increases when we insert an iron core because iron is a ferromagnetic material and ferromagnetic materials help in increasing the magnetic property.

2. If a coil is wound around a steel core and electric current is passed through the coil, the steel core acts as a?

a) Electromagnet

b) Permanent magnet

c) Neither electromagnet nor permanent magnet

d) Either electromagnet or permanent magnet

Answer: b

Explanation: When a coil is wound around a steel core, the steel core behaves like a permanent magnet because it is a ferromagnetic material and once it becomes magnetic it does not lose its magnetic property.

3. What is the formula for the magnetic field due to a solenoid?

a) μnI

b) μn 2

c) μNI

d) μN 2 I 2

Answer: a

Explanation: The magnetic field due to a solenoid is:

B= μnI, where μ is the permeability, n is the number of turns per unit length and I is the current in the solenoid.

4. What happens to the magnetic field in the solenoid when the number of turns increases?

a) Increases

b) Decreases

c) Remains constant

d) Becomes zero

Answer: a

Explanation: The magnetic field of a solenoid is directly proportional to the number of turns in it. Hence as the number of turns increases, the magnetic field also increases.

5. What happens to the magnetic field in the solenoid when the current increases?

a) Increases

b) Decreases

c) Remains constant

d) Becomes zero

Answer: a

Explanation: The magnetic field of a solenoid is directly proportional to the current in it. Hence as the current increases, the magnetic field also increases.

6. What happens to the magnetic field in the solenoid when the length of the solenoid increases?

a) Increases

b) Decreases

c) Remains constant

d) Becomes zero

Answer: b

Explanation: The magnetic field of a solenoid is inversely proportional to the length. Hence as the length increases, the magnetic field decreases.

7. The current in a solenoid is 30A, the number of turns per unit length is 500 turns per metre. Calculate the magnetic field if the core is air.

a) 18.84T

b) 18.84mT

c) 1.84T

d) 1.84mT

Answer: b

Explanation: The magnetic field in a solenoid is given by:

B=μnI

Substituting the values in the given values in the equation, B=18.84mT.

8. The magnetic field of the solenoid is 18.84mT, the number of turns per unit length is 500 turns per metre. Calculate the current if the core is air.

a) 300A

b) 30A

c) 3A

d) 300mA

Answer: b

Explanation: The magnetic field in a solenoid is given by:

B=μnI

Substituting the values in the given values in the equation, I=30A.

9. The magnetic field of the solenoid is 18.84mT, the current is 30A. Calculate the number of turns per unit length if the core is air.

a) 1500 turns/m

b) 1000 turns/m

c) 500 turns /m

d) 2000 turns/m

Answer: c

Explanation: The magnetic field in a solenoid is given by:

B=μnI

Substituting the values in the given values in the equation n=500 turns/m.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Force on a Current Carrying Conductor”.

1. What is the expression for force in a current carrying conductor?

a) F=K/r 2

b) F=Kq/r 2

c) F=Kq 1 q 2 /r 2

d) F=Kq 1 q 2 /r

Answer: c

Explanation: The force in a current carrying conductor is directly proportional to the product of the two charges and inversely proportional to the square of the distance between them. Hence F=Kq 1 q 2 /r 2 , where K is the constant of proportionality.

2. Force in a conductor is__________ to the product of the charges.

a) Directly proportional

b) Inversely proportional

c) Not related

d) Cannot be determined

Answer: a

Explanation: The force in a current carrying conductor is directly proportional to the product of the two charges and inversely proportional to the square of the distance between them.

3. Force in a conductor is __________ to the square of the distance between the charges.

a) Directly proportional

b) Inversely proportional

c) Not related

d) Cannot be determined

Answer: b

Explanation: The force in a current carrying conductor is directly proportional to the product of the two charges and inversely proportional to the square of the distance between them.

4. Calculate the force between two charges having magnitude 3nC and 2nC separated by a distance of 2micro m.

a) 13.5N

b) 13.5kN

c) 1.35N

d) 1.35kN

Answer: b

Explanation: From the expression:

F=Kq 1 q 2 /r 2 , the value of K being 9*10 9 , we get F=13.5kN.

5. If the flow of electric current is parallel to the magnetic field, the force will be?

a) Zero

b) Infinity

c) Maximum

d) Half the original value

Answer: a

Explanation: Force is a cross product. A cross product involves the sine of the angle between them. If two quantities are parallel to each other, the angle between them is zero. Sin is zero, hence force is zero.

6. The ratio of magnetic force to electric force on a charged particle getting undeflected in a field is ______

a) 1

b) 0

c) 2

d) 4

Answer: a

Explanation: When a charged particle is undeflected in a field, the magnitude of the magnetic force and electric force acting on the particle is the same, hence the ratio is 1.

7. Weakest force in nature is __________

a) Electric force

b) Gravitational force

c) Weak force

d) Magnetic force

Answer: a

Explanation: Gravitational force is the weakest force in nature as it does not bind anything strongly with its help.

8. The relation between the direction of force and the direction of magnetic field is __________

a) Same direction

b) Opposite direction

c) Perpendicular

d) Unrelated

Answer: c

Explanation: When a conductor carries a certain value of current, the force developed in the conductor, the current in the conductor and the magnetic field in the conductor are mutually perpendicular to each other.

9. The relation between the direction of current and the direction of the force is ________

a) Same direction

b) Opposite direction

c) Perpendicular

d) Unrelated

Answer: c

Explanation: When a conductor carries a certain value of current, the force developed in the conductor, the current in the conductor and the magnetic field in the conductor are mutually perpendicular to each other.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Force Determination”.

1. Which among the following, is the correct expression for force in a current carrying conductor if magnetic field is perpendicular to it?

a) F=Bi

b) F=B 2 il

c) F=Bil

d) F=Bl 2

Answer: c

Explanation: The correct expression for force in a current carrying conductor in a magnetic field perpendicular to it is F=Bil, where B is the magnetic field, i is the current in the conductor and l is the length of the conductor.

2. When the current in the current carrying conductor increases, what happens to the force in the conductor which is at right angles to the magnetic field?

a) Increases

b) Decreases

c) Remains the same

d) Becomes zero

Answer: a

Explanation: The force at right angles to the magnetic field of a current carrying conductor increases when the current increases because it is directly proportional to the force.

3. When the length of the conductor in the current carrying conductor increases, what happens to the force in the conductor which is at right angles to the magnetic field?

a) Increases

b) Decreases

c) Remains the same

d) Becomes zero

Answer: a

Explanation: The force at right angles to the magnetic field of a current carrying conductor increases when the length of the conductor increases because it is directly proportional to the force.

4. When the magnetic field intensity in the current carrying conductor increases, what happens to the force in the conductor which is at right angles to the magnetic field?

a) Increases

b) Decreases

c) Remains the same

d) Becomes zero

Answer: a

Explanation: The force at right angles to the magnetic field of a current carrying conductor increases when the magnetic field intensity increases because it is directly proportional to the force.

5. The unit for force in a current carrying conductor is _________

a) Tesla*Ampere*metre

b) Tesla

c) Ampere/metre

d) Ampere*metre

Answer: a

Explanation: F=B*i*l. So,unit of force=unit of B * unit of i * unit of l = Tesla*Ampere*metre.

6. If net force is zero on a particle in magnetic field what is relation between velocity and magnetic field?

a) v=E*B

b) v=E/B

c) v=B/E

d) v=1/

Answer: b

Explanation: Since net force is zero on the particle.

Electric force = Magnetic force

QE = QvB => E=vB or v=E/B.

7. If the intensity of the magnetic field is 100T, the length of the conductor is 10m and the magnitude of force perpendicular to the magnetic field is 10kN,calculate the current in the conductor.

a) 100 A

b) 100 mA

c) 10 A

d) 10 mA

Answer: c

Explanation: The formula for calculating the value of the force which is perpendicular to the magnetic field is:

F=Bil

Substituting the values from the question, we get i = 10A.

8. If the intensity of the magnetic field perpendicular to current carrying conductor is 100T, the length of the conductor is 3m and the current in the conductor is 10A, calculate the magnitude of force perpendicular to the electric field.

a) 300N

b) 30N

c) 30kN

d) 3kN

Answer: d

Explanation: The formula for calculating the value of the force which is perpendicular to the magnetic field is:

F=Bil

Substituting the values from the question, we get F=3kN.

9. Force in current carrying conductor placed in magnetic field is ___________ of il and B.

a) dot product

b) scalar product

c) cross product

d) vector addition

Answer: c

Explanation: Force in a current carrying conductor is given by cross product of il and B.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Electromagnetic Induction”.

1. An E.M.F. can be induced by _________

a) Change in the magnetic field only

b) Change in the area of cross section only

c) Change in angle between magnetic field and area only

d) Change in the magnetic field, area or angle between them

Answer: d

Explanation: emf=-dϕ/dt. We know ϕ flux is the dot product of magnetic field vector and area vector.

ϕ=BAcos, hence if either of the three, that is, magnetic field, area or angle changes, the emf will change, flux changes due to which emf can be induced.

2. What happens to the current in a coil while accelerating a magnet inside it?

a) Increases

b) Decreases

c) Remains constant

d) Reverses

Answer: a

Explanation: A change in the magnetic field induces an emf. When there is an emf, there has to be current. Hence, when the magnet is moved inside a coil, the current in it increases.

3. What is the consequence of motor effect?

a) Current

b) Voltage

c) Electromagnetic induction

d) EMF

Answer: c

Explanation: Motor effect is when a current carrying conductor in a magnetic field experiences a force, hence its consequence is electromagnetic induction.

4. The total number of magnetic field lines passing through an area is termed as?

a) Voltage

b) EMF

c) Magnetic flux

d) Magnetic flux density

Answer: c

Explanation: Number of magnetic field lines passing through an area is magnetic flux.

5. The formula for induced emf if magnetic field, length and velocity of conductor all are mutually perpendicular is __________

a) emf=B 2 l

b) emf=Bil

c) emf=Blv

d) emf=B 2 v

Answer: c

Explanation: The formula for induced emf is emf=Blv, where B is the magnetic field, l is the length of the conductor and v is the velocity with which it is moving in the magnetic field and all three quantities are mutually perpendicular to each other.

6. If a conductor 0.2m long moves with a velocity of 0.3m/s in a magnetic field of 5T, calculate the emf induced if magnetic field, velocity and length of conductor are mutually perpendicular to each other.

a) 0.3V

b) 0.03V

c) 30V

d) 3V

Answer: a

Explanation: The formula for induced emf is: emf=Blv if B,l,v are perpendicular to each other. Substituting the values of B, l and v from the question, we get emf=0.3V.

7. Find the length of a conductor which is moving with a velocity 0.4m/s in a magnetic field of 8T, inducing an emf of 20V if magnetic field, velocity and length of conductor are mutually perpendicular to each other.

a) 50m

b) 5m

c) 6.25m

d) 0.5m

Answer: c

Explanation: The formula for induced emf is: emf=Blv if B, l, v are perpendicular to each other. Substituting the values of B, emf and v from the question, we get l=6.25m.

8. Find the strength of the magnetic field in a conductor 0.5m long moving with a velocity of 10m/s, inducing an emf of 20V if magnetic field, velocity and length of conductor are mutually perpendicular to each other.

a) 1T

b) 2T

c) 3T

d) 4T

Answer: d

Explanation: The formula for induced emf is: emf=Blv if B, l, v are perpendicular to each other. Substituting the values of l, emf and v from the question, we get B=4T.

9. What does emf stand for?

a) Electronic magnetic force

b) Electromotive force

c) Electromagnetic force

d) Electromated force

Answer: b

Explanation: Emf stands for electromotive force. It is the voltage developed by any source of electrical energy.

10. What is emf?

a) Force

b) Voltage

c) Current

d) Flux

Answer: b

Explanation: Electromotive force is not actually a force. It is basically a voltage. It is the voltage developed by any source of electrical energy.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Direction of Induced EMF”.

1. According to Faraday’s laws of electromagnetic induction, an emf is induced in a conductor whenever?

a) The conductor is perpendicular to the magnetic field

b) Lies in the magnetic field

c) Cuts magnetic lines of flux

d) Moves parallel to the magnetic field

Answer: c

Explanation: An emf is induced, according to Faraday’s laws of electromagnetic induction, whenever the conductor in a magnetic field cuts the magnetic lines of flux.

2. Direction of induced emf is determined by __________

a) Fleming’s left hand rule

b) Fleming’s right hand rule

c) Faraday’s law

d) Right hand thumb rule

Answer: b

Explanation: Fleming’s left hand rule stated that if the index finger points toward magnetic flux, the thumb towards the motion of the conductor, then the middle finger points towards the induced emf.

3. “The direction of an induced e.m.f. is always such that it tends to set up a current opposing the motion or the change of flux responsible for inducing that e.m.f.”, this is the statement for?

a) Fleming’s left hand rule

b) Fleming’s right hand rule

c) Faraday’s law

d) Lenz’s law

Answer: d

Explanation: The above statement is that of Lenz’s law. It is used to determine the direction of the induced emf.

4. According to Fleming’s right hand rule, the thumb points towards?

a) Current

b) E.M.F.

c) Motion of the conductor

d) Magnetic flux

Answer: c

Explanation: Fleming’s left hand rule stated that if the index finger points toward magnetic flux, the thumb towards the motion of the conductor, then the middle finger points towards the induced emf.

5. According to Fleming’s right hand rule, the index finger points towards?

a) Current

b) E.M.F.

c) Motion of the conductor

d) Magnetic flux

Answer: d

Explanation: Fleming’s left hand rule stated that if the index finger points towards magnetic flux, the thumb towards the motion of the conductor, then the middle finger points towards the induced emf.

6. According to Fleming’s right hand rule, the middle finger points towards?

a) Current

b) E.M.F.

c) Motion of the conductor

d) Magnetic flux

Answer: b

Explanation: Fleming’s left hand rule stated that if the index finger points towards magnetic flux, the thumb towards the motion of the conductor, then the middle finger points towards the induced emf.

7. The relation between the direction of induced emf and the direction of motion of the conductor is?

a) Parallel

b) Equal

c) Not related

d) Perpendicular

Answer: d

Explanation: According to Fleming’s right hand rule, the induced emf, the motion of the conductor and the magnetic flux are mutually perpendicular.

8. The relation between the direction of induced emf and the direction of magnetic flux is _______

a) Parallel

b) Equal

c) Not related

d) Perpendicular

Answer: d

Explanation: According to Fleming’s right hand rule, the induced emf, the motion of the conductor and the magnetic flux are mutually perpendicular.

9. The relation between the direction of magnetic flux and the direction of motion of the conductor is _______

a) Parallel

b) Equal

c) Not related

d) Perpendicular

Answer: d

Explanation: According to Fleming’s right hand rule, the induced emf, the motion of the conductor and the magnetic flux are mutually perpendicular.

This set of Basic Electrical Engineering test focuses on “Magnitude of the Generated or Induced EMF”.

1. Which, among the following, is the formula for induced emf?

a) e=dϕ /dt

b) e=dt/dϕ

c) e=t*ϕ

d) e=t 2 ϕ

Answer: a

Explanation: The formula for induced emf is e = dϕ /dt because the induced emf is the flux linkage per unit time.

2. According to _________________ induced emf is equal to rate of change of magnetic flux.

a) Newton’s law

b) Lenz law

c) Faraday’s law

d) Coulomb’s law

Answer: c

Explanation: According to Faraday law of electromagnetic induction, induced emf is equal to rate of change of magnetic flux.

3. The emf induced in a coil having N turns is?

a) e=ϕ/t

b) e=N*ϕ/t

c) e=N*ϕ*t

d) e=N 2 *ϕ*t

Answer: b

Explanation: The emf induced in a coil having N turns is, e=N*ϕ/t. This is because, the emf in a single coil is the flux linkage per unit time, that is, phi/t. Hence the flux induced in N turns is N*ϕ/t.

4. According to ____________________ induced emf oppose the cause due to which they are produced.

a) Newton’s law

b) Lenz law

c) Faraday’s law

d) Coulomb’s law

Answer: b

Explanation: According to Lenz law, emf is induced in such a way that it opposes the cause due to which it is produced.

5. North pole induces __________

a) Clockwise current

b) Anti-clockwise current

c) Zero current

d) Infinite current

Answer: b

Explanation: A north pole will always induce an anti-clockwise current whereas a south pole will always induce a clockwise current due to electromagnetic theory.

6. What is the principle of the transformer?

a) Gauss law

b) Coulomb’s law

c) Electromagnetic induction

d) Ampere’s law

Answer: c

Explanation: In transformer, flux in secondary coil change due to current in the primary coil and hence current get induced in secondary coil.

7. Voltage induced in secondary coil of transformer is given by__________________

a) N P *V P /N S

b) N S *V P /N P

c) (N P /V P )*N S

d) N P /(V P *N S )

Answer: b

Explanation: In transformer, V P /V S = N P /N S

So, V S = N S *V P /N P .

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Magnetomotive Force and Magnetic Field Strength”.

1. What is a permeable substance?

a) Any good conductor

b) Any bad conductor

c) Any strong magnet

d) Any substance through which the magnetic lines of force can pass easily

Answer: d

Explanation: A permeable substance is one through which the magnetic lines of force can pass easily.

2. Materials having good retentivity are?

a) Strong magnets

b) Weak magnets

c) Temporary magnets

d) Permanent magnets

Answer: d

Explanation: Materials with good retentivity are permanent magnets because they can retain magnetism even when no external magnetic field present.

3. Magnetic field exists along which of the following?

a) Moving charges

b) Stationary charges

c) Copper

d) Iron

Answer: a

Explanation: Moving charges have a magnetic field associated with them because they have magnetic flux lines associated with it.

4. Magnetomotive force is equal to__________________

a) current * number of turns

b) current / number of turns

c) current / number of turns per unit length

d) current * number of turns per unit length

Answer: a

Explanation: MMF is ability to produce flux and is equal to product of current flowing and number of turns.

5. Unit of MMF is ______________

a) A/m

b) A-m

c) A

d) unitless

Answer: c

Explanation: MMF is equal to the product of current flowing and number of turns.

unit of Magnetomotive force = unit of current = A or ampere.

6. When a bar magnet is broken into two pieces, which of the following are true?

a) The magnet loses its magnetism

b) The magnet has only north pole left

c) The magnet has only south pole left

d) The magnet turns into two new bar magnets

Answer: d

Explanation: When a bar magnet is broken into two pieces, it forms two different bar magnets. This happens because the broken pieces of the magnet form a separate north and south pole for itself as monopoles do not exist.

7. When an electric current flows into the page, what is the direction of the magnetic field?

a) Clockwise

b) Anti-clockwise

c) Cannot be determined

d) Parallel to the current

Answer: a

Explanation: When the current flows into the page, the magnetic field is clockwise because of the right hand thumb rule, we orient our thumb into the page and our fingers curl in the clockwise direction.

8. When an electric current flows out of the page, what is the direction of the magnetic field?

a) Clockwise

b) Anti-clockwise

c) Cannot be determined

d) Parallel to the current

Answer: b

Explanation: When the current flows out of the page, the magnetic field is anti-clockwise because of the right hand thumb rule, we orient our thumb out of the page and our fingers curl in the anti-clockwise direction.

9. The relation between the direction of current and the direction of magnetic field is?

a) Same direction

b) Opposite direction

c) Perpendicular

d) Unrelated

Answer: c

Explanation: When a conductor carries a certain value of current, the force developed in the conductor, the current in the conductor and the magnetic field in the conductor are mutually perpendicular to each other.

This set of Basic Electrical Engineering Quiz focuses on “Permeability of Free Space or Magnetic Constant”.

1. Permeability of free space is also known as _________

a) Magnetic constant

b) Electric constant

c) Electrostatic constant

d) Magnetostatic constant

Answer: a

Explanation: The permeability of free space is also known as the magnetic constant. The permittivity of free space is the electrostatic constant.

2. A substance whose permeability is less than the permeability of free space is?

a) Diamagnetic

b) Paramagnetic

c) Ferromagnetic

d) Not a magnetic substance

Answer: a

Explanation: A diamagnetic material creates a magnetic field opposing that of the external magnetic field and it repels the external magnetic field. Hence its permeability is less than that of free space.

3. Which, among the following, have negative susceptibility?

a) Diamagnetic

b) Paramagnetic

c) Ferromagnetic

d) Not a magnetic substance

Answer: a

Explanation: Magnetic susceptibility is the degree of magnetisation of a material in response to the external magnetic field. Diamagnetic substances repel the magnetic field and hence have negative susceptibility.

4. Which, among the following, have positive susceptibility?

a) Diamagnetic

b) Paramagnetic

c) Ferromagnetic

d) Both paramagnetic and ferromagnetic

Answer: d

Explanation: Magnetic susceptibility is the degree of magnetisation of a material in response to the external magnetic field. Both paramagnetic and ferromagnetic materials have positive susceptibility as they get magnetise when placed in external magnetic field.

5. A substance whose permeability is more than the permeability of free space is?

a) Diamagnetic

b) Paramagnetic

c) Ferromagnetic

d) Both paramagnetic and ferromagnetic

Answer: d

Explanation: Relative permeability=1+Magnetic susceptibility

Since both paramagnetic and ferromagnetic materials have positive susceptibility, their relative permeability is greater than unity i.e. their permeability is more than the permeability of free space.

6. The unit for permeability of free space is?

a) Henry

b) Henry-m

c) Henry/m

d) Henry/m 2

Answer: c

Explanation: Using formula of self inductance of solenoid, L=µ 0 n 2 A/l

We get unit of µ 0 = Henry/m.

7. Which among the following, is a correct expression for µ 0 .

a) µ 0 =BH

b) µ 0 =B/H

c) µ 0 =H/B

d) µ 0 =BH 2

Answer: b

Explanation: Magnetic permeability is the constant of proportionality between the magnetic flux density and magnetic field strength of a given medium. Hence µ 0 =B/H.

8. Calculate the magnetic flux density if the magnetic field strength is 2A/m.

a) 4*pi*10 -7 T

b) 8*pi*10 -7 T

c) 10*pi*10 -7 T

d) 12*pi*10 -7 T

Answer: b

Explanation: We know that:

µ 0 =B/H

Substituting the value of H from the question, we get B = 8*pi*10 -7 T.

9. Calculate the magnetic field strength if the magnetic flux density is 4*piT.

a) 10 -7 /16*pi 2 A/m

b) 10 -7 A/m

c) 10 7 A/m

d) 10 -7 A

Answer: c

Explanation: We know that:

µ 0 =B/H

Substituting the value of B from the question, we get H = 10 7 A/m.

10. Calculate the magnetic flux density if the magnetic field strength is 3A/m.

a) 4*pi*10 -7 T

b) 8*pi*10 -7 T

c) 10*pi*10 -7 T

d) 12*pi*10 -7 T

Answer: d

Explanation: We know that:

µ 0 =B/H

Substituting the value of H from the question, we get B = 12*pi*10 -7 T.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Relative Permeability”.

1. What is the unit for relative permeability?

a) H-m

b) H/m

c) H 2 /m

d) No unit

Answer: d

Explanation: Relative permeability is the ratio of permeability of the material to the permeability of free space. Since it is a ratio, it does not have any units.

2. Which of the following expressions is correct with respect to relative permeability?

a) B = µ r µ 0 /H

b) B = µ r µ 0 H

c) B = µ r /µ 0 H

d) B = µ r µ 0 /H 2

Answer: b

Explanation: For a non magnetic material: B=µ 0 H. Hence for a material having relative permeability µ r , B=µ r µ 0 H.

3. A substance whose relative permeability is less than the permeability of free space is?

a) Diamagnetic

b) Paramagnetic

c) Ferromagnetic

d) Not a magnetic substance

Answer: a

Explanation: A diamagnetic material creates a magnetic field opposing that of the external magnetic field and it repels the external magnetic field. Hence its relative permeability is less than that of the free space.

4. A substance whose relative permeability is more than the permeability of free space is?

a) Diamagnetic

b) Paramagnetic

c) Ferromagnetic

d) Both paramagnetic and ferromagnetic

Answer: d

Explanation: Relative permeability=1+Magnetic susceptibility

Since both paramagnetic and ferromagnetic materials have positive susceptibility, their relative permeability is greater than unity i.e. their permeability is more than the permeability of free space.

5. Diamagnetic substances have relative permeability_____________

a) Greater than 1

b) Less than 1

c) Equal to 1

d) Zero

Answer: b

Explanation: A diamagnetic material creates a magnetic field opposing that of the external magnetic field and it repels the external magnetic field. Hence it has relative permeability less than 1.

6. Paramagnetic substances have relative permeability_____________

a) Greater than 1

b) Less than 1

c) Equal to 1

d) Zero

Answer: a

Explanation: A paramagnetic material creates a magnetic field which is weakly attracted to that of the external magnetic field. Hence it has relative permeability greater than 1.

7. As the temperature increases up to the Curie temperature, the relative susceptibility of ferromagnetic materials?

a) Increases

b) Decreases

c) Remains constant

d) Becomes zero

Answer: a

Explanation: The susceptibility of ferromagnetic materials increases with the increase in temperature and then it decreases when the temperature goes beyond the Curie temperature.

8. As the temperature increases beyond the Curie temperature, the relative susceptibility of ferromagnetic materials?

a) Increases

b) Decreases

c) Remains constant

d) Becomes zero

Answer: b

Explanation: The susceptibility of ferromagnetic materials increases with the increase in temperature and then it decreases when the temperature goes beyond the curie temperature.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Reluctance”.

1. Reciprocal of reluctance is __________

a) Permeance

b) Susceptibility

c) Resistance

d) Conductance

Answer: a

Explanation: The reciprocal of reactance is permeance. It is the ability of a material to allow the passage of magnetic lines of flux.

2. Reluctance is ________________ to the length of the material.

a) Directly proportional

b) Inversely proportional

c) Not related

d) Reluctance is ________________ to the length of the material.

Answer: a

Explanation: The formula for reluctance is:

S = l/µ 0 µ r *A.

From the formula, we can see that reluctance is directly proportional to the length of the material.

3. Reluctance is ________________ to the area of cross section the material.

a) Directly proportional

b) Inversely proportional

c) Not related

d) Equal

Answer: b

Explanation: The formula for reluctance is:

S = l/(µ 0 µ r *A).

From the formula, we can see that reluctance is inversely proportional to the area of cross section of the material.

4. When the length of the material increases, what happens to reluctance?

a) Increases

b) Decreases

c) Remains the same

d) Becomes zero

Answer: a

Explanation: Reluctance is directly proportional to the length of the material hence as length increases, reluctance also increases.

5. When the area of cross section of the material increases, what happens to reluctance?

a) Increases

b) Decreases

c) Remains the same

d) Becomes zero

Answer: b

Explanation: Reluctance is inversely proportional to the area of cross section of the material hence as area increases, reluctance decreases.

6. Unit of reluctance is?

a) AWb

b) A 2 /Wb

c) Wb/A

d) A/Wb

Answer: d

Explanation: Reluctance is magnetomotive force per unit flux,

So unit of reluctance = unit of MMF / unit of magnetic flux = A/Wb.

7. The electrical equivalent of reluctance is?

a) Resistance

b) Inductance

c) Capacitance

d) Conductance

Answer: a

Explanation: Resistance is the opposition to the flow of charge, similarly reluctance is the opposition to the flow of magnetic flux.

8. As the magnetic field strength increases, reluctance?

a) Increases

b) Decreases

c) Remains the same

d) Becomes zero

Answer: a

Explanation: Reluctance is directly proportional to the strength of the magnetic field, hence as the strength of magnetic field increases, the reluctance increases.

9. As the magnetic flux density increases, the reluctance _____________

a) Increases

b) Decreases

c) Remains the same

d) Becomes zero

Answer: b

Explanation: Reluctance is inversely proportional to the magnetic flux density, hence as magnetic flux density increases, reluctance decreases.

10. Calculate the reluctance when the magnetomotive force is 10A turns and the flux is 5Wb.

a) 0.5A/Wb

b) 5A/Wb

c) 10A/Wb

d) 2A/Wb

Answer: d

Explanation: We know that:

F=ϕ*S

Substituting the given values from the question:

S=2A/Wb.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Ohm’s Law for a Magnetic Circuit”.

1. Ohm’s law for magnetic circuits is _________

a) F=ϕS

b) F=ϕ/S

c) F=ϕ 2 S

d) F=ϕ/S 2

Answer: a

Explanation: Ohm’s law for magnetic circuits states that the MMF is directly proportional to the magnetic flux where reluctance is the constant of proportionality.

2. What happens to the MMF when the magnetic flux decreases?

a) Increases

b) Decreases

c) Remains constant

d) Becomes zero

Answer: b

Explanation: Ohm’s law for the magnetic circuit’s states that the MMF is directly proportional to the magnetic flux hence as the magnetic flux decreases, the MMF also decreases.

3. Calculate the MMF when the magnetic flux is 5Wb and the reluctance is 3A/Wb.

a) 10At

b) 10N

c) 15N

d) 15At

Answer: d

Explanation: We know that:

F=ϕS

Substituting the given values from the question, we get MMF = 15At.

4. A ring having a cross-sectional area of 500 mm 2 , a circumference of 400 mm and ϕ=800microWb has a coil of 200 turns wound around it. Calculate the flux density of the ring.

a) 1.6T

b) 2.6T

c) 3.6T

d) 4.6T

Answer: a

Explanation: ϕ=BA => Flux density B = ϕ/A

Substituting the values, we get B=1.6T.

5. A ring having a cross-sectional area of 500 mm 2 , a circumference of 400 mm and ϕ=800microWb has a coil of 200 turns wound around it. Relative permeability of ring is 380. Calculate the reluctance.

a) 1.68 * 10 -4 A/Wb

b) 1.68 * 10 4 A/Wb

c) 1.68 * 10 6 A/Wb

d) 1.68 * 10 -6 A/Wb

Answer: c

Explanation: Reluctance=l/µ = l/(µ r µ 0 *A)

Substituting the values, we get Reluctance=1.68*10 6 A/Wb.

6. A ring having a cross-sectional area of 500 mm 2 , a circumference of 400 mm and ϕ=800microWb has a coil of 200 turns wound around it. Relative permeability of ring is 380. Calculate the magnetomotive force.

a) 1442At

b) 1342At

c) 1432At

d) 1344At

Answer: d

Explanation: Reluctance=l/µ = l/(µ r µ 0 *A)

Substituting the values, we get Reluctance S=1.68*106 A/Wb.

F=ϕS

Substituting the given values, we get F = 1344At.

7. A ring having a cross-sectional area of 500 mm 2 , a circumference of 400 mm and ϕ=800microWb has a coil of 200 turns wound around it. Relative permeability of ring is 380. Calculate the magnetising current.

a) 6.7A

b) 7.7A

c) 7.6

d) 6.1A

Answer: a

Explanation: Reluctance = l/µ = l/(µ r µ 0 *A)

Substituting the values, we get Reluctance S=1.68*106 A/Wb.

F=ϕS Substituting the given values, we get F = 1344At.

I=F/N Substituting the values from the question, we get I=6.7A.

8. Can we apply Kirchhoff’s law to magnetic circuits?

a) Yes

b) No

c) Depends on the circuit

d) Insufficient information provided

Answer: a

Explanation: Magnetic circuits have an equivalent to the potential difference of electric circuits. This is the magnetic potential difference which allows us to apply Kirchhoff’s laws to magnetic circuit analysis.

9. What is MMF?

a) Magnetic Machine Force

b) Magnetomotive Force

c) Magnetic Motion Force

d) Magnetomotion Force

Answer: b

Explanation: MMF stands for magnetomotive force. Actually, it is not a force. It is analogous to potential in electric field.

10. The equivalent of the current I in magnetic ohm’s law is?

a) Flux

b) Reluctance

c) MMF

d) Resistance

Answer: a

Explanation: The equivalent of current in magnetic ohm’s law is flux as:

V=IR is equivalent to F=ϕS.

This set of Basic Electrical Engineering MCQs focuses on “Determination of the B/H Characteristic”.

1. The B/H characteristics can be determined using _______

a) Ammeter

b) Fluxmeter

c) Voltmeter

d) Multimeter

Answer: b

Explanation: The fluxmeter is an electronic display instrument used to measure the magnetic flux of permanent magnets hence it can be used to determine B/H characteristics.

2. The B/H curve can be used to determine?

a) Iron loss

b) Hysteresis loss

c) Voltage loss

d) Eddy current loss

Answer: b

Explanation: Hysteresis loss is basically a heat loss due to the reversal of magnetisation of the transformer core whenever it is subjected to a changing magnetic field. It can be determined using the B/H curve.

3. The B/H ratio is not constant for ________

a) Diamagnetic materials

b) Ferromagnetic materials

c) Paramagnetic materials

d) Non-magnetic materials

Answer: b

Explanation: As the magnetizing field increases, the relative permeability increases, reaches a maximum, and then decreases. Due to varying permeability, B/H ratio is not constant for ferromagnetic materials.

4. When using a fluxmeter, if the flux changes from Φ to -Φ, what happens to the current?

a) Becomes zero

b) Becomes infinity

c) Remains the same

d) Reverses

Answer: d

Explanation: When the flux changes from Φ to -Φ, the current direction will change as the direction of flux is changing.

5. Why is the coil of a ballistic galvanometer wound on a non- metallic former?

a) To minimise damping when high resistance is connected in series

b) To maximise damping when high resistance is connected in series

c) To minimise damping when high resistance is connected in parallel

d) To maximise damping when high resistance is connected in parallel

Answer: a

Explanation: The coil of ma ballistic galvanometer is wound on a nonmagnetic former in order to minimise damping when high resistance is connected in series.

6. The ballistic galvanometer is usually lightly damped so that ________

a) It can oscillate

b) It will remain stable

c) Amplitude of the first swing is very large

d) Amplitude of the first swing is very small

Answer: c

Explanation: The ballistic galvanometer is usually lightly damped so that the amplitude of its first swing is very large.

7. PMMC instruments can be used as a fluxmeters by _______

a) Using a low resistance shunt

b) Removing the control spring

c) Making the control springs having a large moment of inertia

d) Using a high resistance in series

Answer: b

Explanation: A PMMC or a permanent moving magnet coil instrument can be used as a fluxmeter by removing the control spring.

8. Hysteresis loss is determined from _______

a) B/H curve

b) H/B curve

c) BH curve

d) B 2 H curve

Answer: c

Explanation: Hysteresis loss is basically a heat loss due to the reversal of magnetisation of the transformer core whenever it is subjected to a changing magnetic field. It can be determined using the B/H curve.

9. What is a PMMC instrument?

a) Permanent moving magnet coil instrument

b) Permanent machine magnet coil instrument

c) Permanent moving machine coil instrument

d) Premature moving magnet coil instrument

Answer: a

Explanation: A PMMC instrument is a permanent moving magnet coil instrument. It uses two magnets to create a stationary magnetic field.

10. B/H curve shows the relationship between?

a) Magnetic field strength and magnetic flux

b) Magnetic field strength and magnetic flux density

c) Current and magnetic flux density

d) Voltage and magnetic flux density

Answer: b

Explanation: The B/H curve shows the relation between magnetic field strength and magnetic flux density.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Inductive and Non-Inductive Circuits”.

1. In case of Inductive circuit, Frequency is ______________ to the inductance.

a) Directly proportional

b) Inversely proportional

c) Unrelated

d) Much greater than

Answer: b

Explanation: The formula for frequency in an inductive circuit is:

X L =2*π*f*L.

Therefore: f is inversely proportional to L.

2. In case of Inductive circuit, Frequency is ______________ to the current.

a) Directly proportional

b) Inversely proportional

c) Unrelated

d) Much greater than

Answer: b

Explanation: The formula for frequency in an inductive circuit is:

X L =2*π*f*L => i=V/

Therefore: f is inversely proportional to i.

3. In an inductive circuit, when the XL value increases, the circuit power factor?

a) Increases

b) Decreases

c) Remains the same

d) Becomes zero

Answer: b

Explanation: tan ϕ = X L /R and Power factor=cos ϕ

As X L increases, tan ϕ increases, ϕ increases, cos ϕ decreases and hence power factor decreases.

4. If the current and voltage are 90 degrees out of phase, the power factor will be?

a) 0

b) Infinity

c) 1

d) Insufficient information provided

Answer: a

Explanation: The power factor is the cosine of the angle between the voltage and the current. If the angle between the voltage and the current is 90, then cos90=0. Hence, the power factor is zero.

5. In a pure inductive circuit, the power factor is __________

a) Maximum

b) Minimum

c) 0

d) Infinity

Answer: c

Explanation: In a pure inductive circuit, the current is lagging by 90 degrees from the voltage. The power factor is the cosine of the angle between the voltage and the current. If the angle between the voltage and current is 90, then cos90=0. Hence, the power factor is zero.

6. If the power factor is 1/10 and the value of impedance is 20 ohm, calculate the resistance in the circuit.

a) 1 ohm

b) 2 ohm

c) 3 ohm

d) 4 ohm

Answer: b

Explanation: We know that:

cos=R/Z

R=Z cos = 20/10 = 2 ohm.

7. If the resistance in a circuit is 2 ohm and the impedance is 20 ohm, calculate the power factor.

a) 1/10

b) 1/20

c) 1/30

d) 1/40

Answer: a

Explanation: We know that:

cos=R/Z = 2/20 = 1/10 ohm.

8. If tan ϕ = 10 and the resistance is 2 ohm, calculate the inductive reactance.

a) 10 ohm

b) 20 ohm

c) 30 ohm

d) 40 ohm

Answer: b

Explanation: We know that:

tan=XL/R

From the given question, we find that the inductive reactance is 20 ohm.

9. What is the unit for inductive reactance?

a) Henry

b) Ohm

c) Farad

d) Volts

Answer: b

Explanation: Inductive reactance is nothing but the impedance. Impedance is the AC equivalent of resistance, hence the unit for inductive reactance is ohm.

10. An induced emf is said to be ________

a) Inductive

b) Capacitive

c) Resistive

d) Cannot be determined

Answer: a

Explanation: Any circuit in which a change of current is accompanied by a change of flux, and therefore by an induced emf, is said to be inductive.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers focuses on “Inductance in Terms of Flux Linkages Per Ampere”.

1. Among the following, which is the right formula for inductance?

a) L=emf*t/I

b) L=emf/t*I

c) L=emf*I/t

d) L=emf*t*I

Answer: a

Explanation: The average emf induced is proportional to the current per unit time, the constant of proportionality being L. Hence emf=LI/t. Making L the subject of the formula, we get L=emf*t/I.

2. Among the following, which is the right formula for inductance of N turns?

a) L=et/Ni

b) L=N*i *e*t

c) L=Ni/et

d) L=N/iet

Answer: a

Explanation: We know that:

emf=NLi/t

Inductance = L = et/N.

3. For a coil having a magnetic circuit of constant reluctance, the flux is ___________ to the current.

a) Directly proportional

b) Inversely proportional

c) Not related

d) Very large compared to

Answer: a

Explanation: For a coil having a magnetic circuit of constant reluctance, the flux is directly proportional to the current.

4. For a coil having a magnetic circuit of constant reluctance, if the flux increases, what happens to the current?

a) Increases

b) Decreases

c) Remains constant

d) Becomes zero

Answer: a

Explanation: For a coil having a magnetic circuit of constant reluctance, the flux is directly proportional to the current. Hence as the flux increases, the current also increases.

5. The unit for inductance is ___________

a) Ohm

b) Henry

c) A/m

d) A/s

Answer: b

Explanation: The unit of induction is named after a famous scientist Joseph Henry who independently discovered electromagnetic induction.

6. If either the inductance or the rate of change of current is doubled, the induced e.m.f?

a) Remains constant

b) Becomes zero

c) Doubles

d) Becomes half

Answer: c

Explanation: If either the inductance or the rate of change of current is doubled, the induced e.m.f. becomes double because of emf=LI/t.

7. If the current changes from 5A to 3A in 2 seconds and the inductance is 10H, calculate the emf.

a) 5V

b) 10V

c) 15V

d) 20V

Answer: b

Explanation: We know that:

emf=L/t

Substituting the values from the question, we get emf=10V.

8. If the current changes from 5A to 3A in x sec and inductance is 10H. The emf is 10V, calculate the value of x.

a) 2s

b) 3s

c) 4s

d) 5s

Answer: a

Explanation: We know that:

emf=L/t

Substituting the values from the question, we get x=2s.

9. If the current changes from 3A to 5A in 2s and the emf is 10V. Calculate the inductance.

a) 10H

b) 20H

c) 30H

d) 40H

Answer: a

Explanation: If the current changes from 5A to 3A in 2s and the emf is 10V. Calculate the inductance.

This set of Basic Electrical Engineering online test focuses on “Factors Determining the Inductance of a Coil”.

1. As the number of turns in the coil increases, what happens to the inductance of the coil?

a) Increases

b) Decreases

c) Remains the same

d) Becomes zero

Answer: a

Explanation: Inductance is directly proportional to the square of the number of turns in the coil, hence as the number of turns increases, inductance also increases.

2. What happens to the inductance when the magnetic field strength decreases?

a) Increases

b) Decreases

c) Remains the same

d) Becomes zero

Answer: b

Explanation: Inductance is directly proportional to the magnetic field strength in the coil, hence as the magnetic field strength decreases, inductance decreases.

3. What happens to the inductance when the current in the coil becomes double its original value?

a) Becomes half

b) Becomes four times

c) Becomes double

d) Remains same

Answer: d

Explanation: ϕ is directly proportional to i.

Φ=Li where L is constant of proportionality

So, when current get double ϕ also becomes double keeping L same.

4. When the coil is wrapped around a ferromagnetic core, why is it difficult to determine the inductance?

a) The variation of flux is no longer proportional to the variation of current

b) Current does not exist in the coil

c) Flux does not exist in the coil

d) The value of current is too large to measure

Answer: a

Explanation: When a coil is wrapped around a ferromagnetic core, it is difficult to determine the inductance because the variation of flux is no longer proportional to the variation of the current.

5. What happens to the inductance as the area of the cross section of the coil increases?

a) Increases

b) Decreases

c) Remains the same

d) Becomes zero

Answer: a

Explanation: L=µ 0 *N 2 *A/l, hence as the area of cross section A increases, the inductance also increases.

6. What happens to the inductance as the length of the magnetic circuit increases?

a) Increases

b) Decreases

c) Remains the same

d) Becomes zero

Answer: b

Explanation: L=µ 0 *N 2 *A/l, hence as the length of the magnetic circuit l increases, the inductance decreases.

7. If the current changes from 20A to 10A in 5 seconds and the value of inductance is 1H, calculate the emf induced.

a) 8V

b) 6V

c) 4V

d) 2V

Answer: d

Explanation: We know that:

emf=L/t

Substituting the values from the question we get emf=2V.

This set of Basic Electrical Engineering online quiz focuses on “Ferromagnetic Cored Inductor in a DC Circuit”.

1. When a ferromagnetic core is inserted into an inductor, what happens to the flux linkage?

a) Increases

b) Decreases

c) Remains the same

d) Becomes zero

Answer: a

Explanation: When a ferromagnetic core is introduced into an inductor, its flux increases because the number of magnetic field lines increases due to the introduction of magnetic field within the coil.

2. What happens to the current when a ferromagnetic material is introduced within an inductor?

a) Increases

b) Decreases

c) Remains the same

d) Becomes zero

Answer: c

Explanation: When a ferromagnetic is introduced within an inductor, the current remains fairly constant. This is because the current does not depend on the magnetic field.

3. What is the relation between the flux and the magnetizing current when a ferromagnetic core is introduced within the inductor?

a) Directly proportional

b) Inversely proportional

c) Not proportional

d) Current is double of flux

Answer: c

Explanation: When a ferromagnetic core is introduced within an inductor the flux changes rapidly, whereas the current changes at the same pace. Hence the two are not proportional.

4. What happens to the effective inductance when a ferromagnetic core is introduced?

a) Increases

b) Decreases

c) Remains the same

d) Becomes zero

Answer: a

Explanation: The effective inductance increases when a ferromagnetic core is introduced within an inductor because of the flux increases. Inductance varies directly with the flux hence it increases.

5. A laminated steel ring is wound with 200 turns. When the magnetizing current varies between 5 and 7 A, the magnetic flux varies between 760 and 800 Wb. Calculate the inductance of the coil.

a) 40H

b) 4 H

c) 4000H

d) 0.004 H

Answer: c

Explanation: From the formula of incremental inductance, we know that:

L=*Number of turns

Substituting the values from the given question, we get L = 4000 H.

6. Calculate the number of turns in an inductor with a ferromagnetic core when the inductance is 4000 H, the current changes from 5A to 7A and the flux changes from 760 to 800 Wb.

a) 100

b) 200

c) 300

d) 400

Answer: b

Explanation: From the formula of incremental inductance, we know that:

L=*Number of turns

Substituting the values from the given question, we get N=200.

7. Calculate the change in current in an inductor having inductance 4000H, number of turns is 200 and the flux changes from 760 to 800 Wb.

a) 2A

b) 4A

c) 6A

d) 8A

Answer: a

Explanation: From the formula of incremental inductance, we know that:

L=*Number of turns

Substituting the values from the given question, we get a change in current = 2A.

8. Calculate the initial current in an inductor having inductance 4000 H, number of turns is 200 and the flux changes from 760 to 800 Wb. Current changes to 7A.

a) 10A

b) 2A

c) 3A

d) 5A

Answer: d

Explanation: From the formula of incremental inductance, we know that:

L=*Number of turns

Substituting the values from the given question, we get change in current = 2A.

Change in current = final current- initial current.

2=7-initial current.

Initial current = 5A.

9. Calculate the change in flux of an inductor having inductance 4000 H, number of turns is 200 and the current changes from 5A to 7A.

a) 20 Wb

b) 40 Wb

c) 60 Wb

d) 80 Wb

Answer: b

Explanation: From the formula of incremental inductance, we know that:

L=*Number of turns

Substituting the values from the given question, we get change in flux = 40 Wb.

10. Calculate the final flux in an inductor having inductance 4000 H, number of turns is 200 and the current changes from 5A to 7A. The initial flux is 760 Wb.

a) 200 Wb

b) 400 Wb

c) 600 Wb

d) 800 Wb

Answer: d

Explanation: From the formula of incremental inductance, we know that:

L=*Number of turns

Substituting the values from the given question, we get a change in flux = 40 Wb.

Change in flux = final flux- initial flux.

Thus final flux = 800 Wb.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Growth in an Inductive Circuit”.

1. In a pure inductive circuit, the power factor is?

a) Maximum

b) Minimum

c) 0

d) Infinity

Answer: c

Explanation: In a pure inductive circuit, the current is lagging by 90 degrees from the voltage. The power factor is the cosine of the angle between the voltage and the current. If the angle between the voltage and the current is 90, then cos90=0. Hence, the power factor is zero.

2. Among the following, which is the right formula for growth in an inductive circuit?

a) VL=V(1-e -tR/L )

b) VL=(e -tR/L )

c) VL=(1-e -tR/L )

d) VL=V(e -tR/L )

Answer: d

Explanation: The correct formula for growth in an inductive circuit is VL=V(e -tR/L ). As the time increases, voltage decreases.

3. The charging time constant of a circuit consisting of an inductor is the time taken for the voltage in the inductor to become __________ % of the initial voltage.

a) 33

b) 63

c) 37

d) 36

Answer: c

Explanation: We know that: V=V 0 (1-e -tR/L ).

When time constant=t, we have: V=V 0 (1-e -1 ) = 0.37*V 0 .

Hence the time constant is the time taken for the charge in an inductive circuit to become 0.37 times its initial charge.

4. What is the time constant of an inductive circuit?

a) LR

b) R/L

c) 1/LR

d) L/R

Answer: d

Explanation: The time constant in an inductive circuit is the time taken for the voltage across the inductor to become 63 percent of its initial value. It is given by Time constant = L/R.

5. Calculate the time constant of an inductive circuit having resistance 5 ohm and inductance 10H.

a) 2s

b) 4s

c) 5s

d)10s

Answer: a

Explanation: We know that: Time constant = L/R

Substituting the values from the given question, we get a time constant = 2s.

6. Calculate the resistance in an inductive circuit whose time constant is 2s and the inductance is 10H.

a) 7ohm

b) 10ohm

c) 2ohm

d) 5ohm

Answer: d

Explanation: We know that: Time constant = L/R

Substituting the values from the given question, we get R=5ohm.

7. Calculate the inductance in an inductive circuit whose time constant is 2s and the resistance is 5 ohm.

a) 10H

b) 20H

c) 5H

d) 15H

Answer: a

Explanation: We know that: Time constant = L/R

Substituting the values from the given question, we get L=10H.

8. The charging time constant of a circuit consisting of an inductor is the time taken for the current in the inductor to become __________% of the initial current.

a) 33

b) 63

c) 37

d) 36

Answer: b

Explanation: We know that: i=i 0 (1-e -tR/L ).

When t=L/R, we have: i=i 0 (1-e -1 ) = 0.63*i 0 .

Hence the time constant is the time taken for the current in an inductive circuit to become 0.63 times its initial current.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Analysis of Growth & Decay”.

1. What is the total applied voltage in an inductive circuit?

a) V=Ri+Ldi/dt

b) V=Ri+di/dt

c) V=i+Ldi/dt

d) V=R+Ldi/dt

Answer: a

Explanation: The total voltage in an inductive circuit is the sum of the voltage due to the resistor which is Ri and the voltage due to the inductor which is Ldi/dt. Hence V=Ri+Ldi/dt.

2. What is Helmholtz equation?

a) i=I(e Rt/L )

b) i=I(1-e -Rt/L )

c) i=I(1+e -Rt/L )

d) i=I(e -Rt/L )

Answer: b

Explanation: Helmholtz equation is an equation which gives the formula for the growth in an inductive circuit. Hence the Helmholtz formula is: i=I(1-e -Rt/L ).

3. A coil has a resistance of 4 ohms and an inductance of 2H. It is connected to a 20V dc supply. Calculate the initial value of the current in the circuit.

a) 5A

b) 10A

c) 0 A

d) 20A

Answer: c

Explanation: Initially, inductor behave as open circuit for dc current so, i=0.

4. A coil has a resistance of 4 ohms and an inductance of 2H. It is connected to a 20V dc supply. Calculate the final value of the current in the circuit.

a) 5A

b) 10A

c) 15A

d) 20A

Answer: a

Explanation: The final value of the current in the circuit is:

I=V/R = 5A.

5. A coil has a resistance of 4 ohms and an inductance of 2H. It is connected to a 20V dc supply. Calculate the value of current 1s after the switch is closed.

a) 5.44A

b) 4.32A

c) 6.56A

d) 2.34A

Answer: b

Explanation: We know that:

i=I(1-e Rt/L )

I=V/R=5A

Substituting the remaining values from the given question, we get i=4.32A.

6. A coil has a resistance of 4 ohms and an inductance of 2H. It is connected to a 20V dc supply. Calculate the value of voltage 1s after the switch is closed.

a) 5.4V

b) 10.8V

c) 0 V

d) 2.7V

Answer: d

Explanation: V=V 0 e -Rt/L

V=20e -2 =2.7V.

7. Among the following, which is the right formula for decay in an inductive circuit?

a) i=I(1-e -t /time constant)

b) i=I(1-e t /time constant)

c) i=(1-e -t /time constant)

d) i=I(e -t /time constant)

Answer: d

Explanation: The correct formula for decay in an inductive circuit is i=I. As the time increases, the current in the inductor decreases, the voltage also increases.

8. The discharging time constant of a circuit consisting of an inductor is the time taken for the voltage in the inductor to become __________ % of the initial voltage.

a) 33

b) 63

c) 37

d) 36

Answer: c

Explanation: We know that: V=V 0 (e -tR/L ).

When t=L/R, we have: V=V 0 (e -1 ) = 0.37*Vsub>0.

Hence the time constant is the time taken for the voltage in an inductive circuit to become 0.37 times its initial voltage.

9. A coil has a resistance of 4 ohms and an inductance of 2H. It is connected to a 20V dc supply. Calculate the initial value of the voltage across the inductor.

a) 5V

b) 10V

c) 0 V

d) 20V

Answer: d

Explanation: Initially, inductor behave as open circuit for dc current so, V = V0 = 20V i.e. same as voltage source.

10. A coil has a resistance of 4 ohms and an inductance of 2H. It is connected to a 20V dc supply. Calculate the final value of the voltage across the inductor.

a) 5V

b) 10V

c) 0 V

d) 20V

Answer: c

Explanation: At steady state, inductor behaves as a short circuit for dc current so, V=0

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Transients in LR Networks”.

1. An RL network is one which consists of ____________

a) Resistor and capacitor in parallel

b) Resistor and capacitor in series

c) Resistor and inductor in parallel

d) Resistor and inductor in series

Answer: d

Explanation: An R-L network is a network which consists of a resistor which is connected in series to an inductor.

2. If the switch is opened at t=0, what is the current in the circuit?

basic-electrical-engineering-questions-answers-transients-lr-networks-q4

a) 0A

b) 1A

c) 2A

d) 3A

Answer: c

Explanation: Initially when switch was closed,current in the inductor was 60/30=2A.

Current in inductor doesn’t change suddenly so when switch is opened, current in inductor remains same i.e. 2A.

3. In an RL series circuit, when the switch is closed and the circuit is complete, what is the response?

a) Response does not vary with time

b) Decays with time

c) Increases with time

d) First increases, then decrease

Answer: b

Explanation: In an RL series circuit, the response decays with time because according to the equation, there is an exponential decrease in the response.

4. If the switch is closed at t=0, what is the current in the circuit?

basic-electrical-engineering-questions-answers-transients-lr-networks-q4

a) 0A

b) 10A

c) 20A

d) 30A

Answer: a

Explanation: Initially, when the switch is open, the current in the circuit is 0. As soon as the switch is closed at t=0 + , the inductor acts as an open circuit, hence the current in the circuit is zero. Since the current in the circuit is zero, there is no voltage drop across the resistor and hence voltage across the inductor is equal to the supply voltage, i.e. 60V.

5. What is the voltage across the inductor at t=0?

basic-electrical-engineering-questions-answers-transients-lr-networks-q4

a) 0V

b) 20V

c) 60V

d) 58V

Answer: c

Explanation: Initially, when the switch is open, the current in the circuit is 0. As soon as the switch is closes at t=0+, the inductor acts as an open circuit, hence the current in the circuit is zero. Since the current in the circuit is zero, there is no voltage drop across the resistor and the voltage across the inductor is equal to the supply voltage, which is equal to 60V.

6. What is the expression for current in the given circuit?

basic-electrical-engineering-questions-answers-transients-lr-networks-q4

a) i=2(e -2t )A

b) i=2(1-e -2t )A

c) i=2(e 2t )A

d) i=2(1+e -2t )A

Answer: b

Explanation: Applying KVL in above circuit, we get

60-30i-15di/dt =0

i=2(1-e -2t )A

7. What is the expression for voltage in the given circuit?

basic-electrical-engineering-questions-answers-transients-lr-networks-q4

a) V=60e -0.5t

b) V=30e -0.5t

c) V=60e -2t

d) V=30e -2t

Answer: c

Explanation: Applying KVL in above circuit, we get

60-30i-15di/dt = 0

i=2(1-e -2t )A

di/dt = 4e -2t

V=Ldi/dt=15*4e- -2t =60e -2t .

8. At steady state, the current in the inductor is?

a) Maximum

b) Minimum

c) Zero

d) Infinity

Answer: a

Explanation: At steady state maximum current flows in the inductor because it acts as an open circuit.

9. Initially, when the switch in a series RL circuit is closed, the inductor acts as?

a) Open circuit

b) Short circuit

c) Resistor

d) Capacitor

Answer: a

Explanation: Before switch is closed, current in inductor is zero. When the switch in a series RL circuit is closed, current in the inductor remains zero since current in inductor doesn’t change suddenly. So, the inductor acts as an open circuit.

10. Initially, when the switch in a series RL circuit is closed, the current in the inductor is?

a) Maximum

b) Minimum

c) Zero

d) Infinity

Answer: c

Explanation: Initially, when the switch in a series RL circuit is closed, the inductor acts as an open circuit. Current in an open circuit is zero, hence the inductor current is zero.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Energy Stored in an Inductor”.

1. If the current in a coil having a constant inductance of L henrys grows at a uniform rate, what is the value of the average current?

a) I

b) I/2

c) I/4

d) 0

Answer: b

Explanation: The average current is the average of the current which flows in the inductor. Hence it is I/2.

2. What is the power in the magnetic field if the current in a coil has a constant inductance of L henrys grows at a uniform rate?

a) LI/2t

b) LI 2 /2t

c) L/2It

d) L/2I 2 t

Answer: b

Explanation: EMF induced in the coil is Ldi/dt.

Energy=ei*dt=Lidi/dt*dt=Li 2 /2.

Power=Energy/Time = Li 2 /2t.

3. What is the energy stored in the magnetic field if the current in a coil has a constant inductance of L henrys grows at a uniform rate?

a) LI/2

b) LI 2 /2

c) L/2I

d) L/2I 2

Answer: b

Explanation: EMF induced in the coil is Ldi/dt.

Energy=ei*dt=Lidi/dt*dt=Li 2 /2.

4. Find the average current in an inductor if the total current in the inductor is 26A.

a) 10A

b) 26A

c) 13A

d) 5A

Answer: c

Explanation: Average current = I/2.

Substituting the value of I from the equation, average current = 13A.

5. Calculate the power in an inductive circuit if the inductance is 10H, the current flowing in the inductor is 2A in 4s.

a) 50W

b) 4W

c) 5W

d) 10W

Answer: c

Explanation: The expression for power in an inductive circuit is:

P = LI 2 /2t

Substituting the values from the given question, we get P=5W.

6. Calculate the value of stored energy in an inductor if the value of inductance is 20H and 4A of current flows through it.

a) 220J

b) 150J

c) 190J

d) 160J

Answer: d

Explanation: The expression for energy in an inductor is

E = LI 2 /2. Substituting the values from the given question, we get E = 160J.

7. Calculate the emf induced in an inductor if the inductance is 10H and the current is 2A in 4s.

a) 2.5V

b) 1.5V

c) 3.5V

d) 5V

Answer: d

Explanation: The expression for emf in an inductive circuit is:

emf = LdI/dt

Substituting the values from the given question, we get emf = 5V.

8. Calculate the value of emf in an inductor if the value of inductance is 15H and an average current of 5A flows through it in 10s.

a) 15V

b) 7.5V

c) 10V

d) 5.5V

Answer: b

Explanation: The expression for emf in an inductive circuit is:

emf = LdI/dt

Substituting the values from the given question, we get emf = 7.5V.

9. Calculate the current in an inductor if the energy stored is 160J and the inductance is 20H.

a) 1A

b) 2A

c) 3A

d) 4A

Answer: d

Explanation: The expression for energy in an inductor is:

E = LI 2 /2

Substituting the values from the given question, we get I=4A.

10. Find the time taken for the current in an inductor to change to 2A from 0A if the power in the inductor is 5W. The value of inductance is 10H.

a) 1s

b) 2s

c) 3s

d) 4s

Answer: d

Explanation: The expression for power in an inductive circuit is:

P = LI 2 /2t

Substituting the values from the given question, we get t=4s.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Mutual Inductance”.

1. The phenomenon due to which there is an induced current in one coil due to the current in a neighbouring coil is?

a) Electromagnetism

b) Susceptance

c) Mutual inductance

d) Steady current

Answer: c

Explanation: When there is a current in a coil, due to the magnetic field caused by the current there is current induced in the neighbouring coil as well. This is known as mutual inductance.

2. If the current in one coil becomes steady, the current in neighbouring coil is?

a) Zero

b) Infinity

c) Doubles

d) Halves

Answer: a

Explanation: A current is induced when there is changing magnetic flux. Hence the induced current in neighbouring coil is zero when the current is steady.

3. If the current in one coil is steady, what happens to the mutual inductance?

a) Zero

b) Infinity

c) Doubles

d) Halves

Answer: a

Explanation: A current is induced when there is changing magnetic flux. The induced current in neighbouring coil is zero when the current is steady. So, mutual inductance is zero.

4. What is the SI unit of mutual inductance?

a) Ohm

b) Henry

c) Volt

d) Siemens

Answer: b

Explanation: Mutual inductance is the inductance between the two neighbouring coils. Since it is a type of inductance, its unit is that of inductance, that is, henry.

5. Which, among the following, is the correct expression for mutual inductance?

a) M=N2φ2/I2

b) M=N2φ2/I1

c) M=N1φ2/I2

d) M=N1φ1/I1

Answer: b

Explanation: Mutual inductance is the product of the number of turns in one coil and the flux linkages of that coil, divided by the current in the other coil. Hence M=N2φ2/I1 is the correct expression.

6. If the flux linkage in coil 1 is 3Wb and it has 500 turns and the current in coil 2 is 2A, calculate the mutual inductance.

a) 750H

b) 500H

c) 450H

d) 900H

Answer: a

Explanation: We know that mutual inductance is the product of the number of turns in one coil and the flux linkages of that coil, divided by the current in the other coil.

M=3*500/2=750H.

7. The flux linkage in coil 1 is 3Wb and it has x turns and the current in coil 2 is 2A, calculate the value of x if the mutual inductance is 750H.

a) 300

b) 400

c) 500

d) 700

Answer: c

Explanation: We know that mutual inductance is the product of the number of turns in one coil and the flux linkages of that coil, divided by the current in the other coil.

N=750*2/3 = 500 turns.

8. The flux linkage in coil 1 is x and it has 500 turns and the current in coil 2 is 2A, calculate the value of x if the mutual inductance is 750H.

a) 1Wb

b) 2Wb

c) 3Wb

d) 4Wb

Answer: c

Explanation: We know that mutual inductance is the product of the number of turns in one coil and the flux linkages of that coil, divided by the current in the other coil.

φ=750*2/500 = 3Wb.

9. The flux linkage in coil 1 is 3 Wb and it has 500 turns and the current in coil 2 is xA, calculate the value of x if the mutual inductance is 750H.

a) 1A

b) 2A

c) 3A

d) 4A

Answer: b

Explanation: We know that mutual inductance is the product of the number of turns in one coil and the flux linkages of that coil, divided by the current in the other coil.

I=3*500/750 = 2A.

10. Practical application of mutual inductance is ____________

a) DC generator

b) AC generator

c) Transformer

d) Capacitor

Answer: c

Explanation: A transformer is a device made of two or more inductors, one of which is powered by AC, inducing an AC voltage across the second inductor.

11. The types of inductors are ____________

a) Fixed and variable

b) Only fixed

c) Only variable

d) Neither fixed nor variable

Answer: a

Explanation: The two types of inductors are fixed and variable inductors. Fixed inductors are those whose inductance value cannot be changed. Variable inductors are those whose values can be changed as and when required.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Coupling Coefficient”.

1. What is the coupling coefficient when all the flux of coil 1 links with coil 2?

a) 0

b) 100

c) 1

d) Insufficient information provided

Answer: c

Explanation: When all the flux of coil 1 links with coil 2 it is known as an ideal coupling where the coupling coefficient is 1.

2. What is the coupling coefficient when there is ideal coupling?

a) 0

b) 100

c) 1

d) Insufficient information provided

Answer: c

Explanation: When all the flux of coil 1 links with coil 2 it is known as an ideal coupling where the coupling coefficient is 1.

3. Can the coupling coefficient practically ever be equal to 1?

a) Yes

b) No

c) Depends on the current in coil 1

d) Depends on the current in coil 2

Answer: b

Explanation: All the flux of coil 1 can never link with coil 2. Loss occurs practically due to which coupling coefficient cannot be equal to 1.

4. Mutual inductance between two coupled coils depend on?

a) Amount of flux linkage

b) Rate of change of flux linkage

c) Rate of change of current

d) Flux density

Answer: b

Explanation: Faraday’s law of induction states that the magnitude of the induced EMF is the product of the number of turns of the coil and the rate of change of flux linkage in it. Hence, the mutual inductance depends on the rate of change of flux linkage.

5. Which, among the following, is the correct formula to find the coupling coefficient?

a) k=M/sqrt

b) k=M/sqrt(L1 2 )

c) k=M/sqrt(L2 2 )

d) k=M/

Answer: a

Explanation: The correct formula for the coupling coefficient is k=M/sqrt. Where L1 and L2 are the inductance values of the first and second coil respectively and M is the mutual inductance.

6. What happens to the coupling coefficient when the flux linkage of coil 1 and coil 2 increases?

a) Increases

b) Decreases

c) Remains the same

d) Becomes zero

Answer: a

Explanation: When the flux linkage of coil 1 and coil 2 increases, its mutual inductance increases. The coupling coefficient is directly proportional to the mutual inductance hence as mutual inductance increases, the coupling coefficient increases.

7. What is the SI unit of coupling coefficient?

a) H

b) H -1

c) No unit

d) H 2

Answer: c

Explanation: The expression to find mutual inductance is k=M/sqrt= H/sqrt= 1. Therefore it does not have any unit.

8. Find the coupling coefficient if the Mutual inductance is 20H, the inductance of coil 1 is 2H and the inductance of coil 2 is 8H.

a) 5

b) 20

c) 2

d) 8

Answer: a

Explanation: we know that:

k=M/sqrt

Substituting the values from the question, we get k=5.

9. Find the value of x if the Mutual inductance is x H, the inductance of coil 1 is 2H and the inductance of coil 2 is 8H. The coupling coefficient is 5.

a) 10H

b) 20H

c) 16H

d) 15H

Answer: b

Explanation: we know that:

k=M/sqrt

Substituting the values from the question, we get M=20H.

10. Find the value of x if the Mutual inductance is 20H, the inductance of coil 1 is xH and the inductance of coil 2 is 8H. The coupling coefficient is 5.

a) 2H

b) 4H

c) 6H

d) 8H

Answer: a

Explanation: we know that:

k=M/sqrt

Substituting the values from the question, we get L1=2H.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Coils Connected in Series”.

1. What is the equivalent inductance when inductors are connected in series?

a) Sum of all the individual inductances

b) Product of all the individual inductances

c) Sum of the reciprocal of all the individual inductances

d) Product of the reciprocal of all the individual inductances

Answer: a

Explanation: When inductances are connected in series, the equivalent inductance is equal to the sum of all the individual inductance values.

2. When inductances are connected in series, the equivalent inductance is ____________ the largest individual inductance.

a) Greater than

b) Less than

c) Equal to

d) Not related to

Answer: a

Explanation: When inductances are connected in series, the equivalent inductance is equal to the sum of all the individual inductance values. Hence the equivalent inductance is greater than the largest individual inductance.

3. Three inductors having inductance values 3H, 4H and 5H are connected in series, calculate the equivalent inductance.

a) 10H

b) 12H

c) 3H

d) 5H

Answer: b

Explanation: When inductances are connected in series, the equivalent inductance is equal to the sum of all the individual inductance values.

Hence Leq= L1+L2+L3= 12H.

4. Calculate the equivalent inductance between A and B.

basic-electrical-engineering-questions-answers-coils-connected-series-q4

a) 30H

b) 54H

c) 44H

d) 60H

Answer: c

Explanation: The 4 inductors are connected in series, hence their equivalent inductance is:

Leq=L1+L2+L3+L4=44H.

5. When inductors are connected in series, the voltage across each inductor is _________

a) Equal

b) Different

c) Zero

d) Infinity

Answer: b

Explanation: In a series circuit, the current across all elements remain the same and the total voltage of the circuit is the sum of the voltages across all the elements. The voltage across each inductor in series is different.

6. In a series circuit, which of the parameters remain constant across all circuit elements such as resistor, capacitor, inductor etc?

a) Voltage

b) Current

c) Both voltage and current

d) Neither voltage nor current

Answer: b

Explanation: In a series circuit, the current across all elements remain the same and the total voltage of the circuit is the sum of the voltages across all the elements.

7. Find voltage across 2H inductor.

basic-electrical-engineering-questions-answers-coils-connected-series-q7

a) 2V

b) 10V

c) 12V

d) 20V

Answer: a

Explanation: e=Ldi/dt

L eq =2+10+12+20=44H

di/dt=44/44 = 1 A/s.

Voltage across 2H inductor = 2*di/dt = 2*1=2V.

8. Find voltage across 10H inductor.

basic-electrical-engineering-questions-answers-coils-connected-series-q7

a) 2V

b) 10V

c) 12V

d) 20V

Answer: b

Explanation: e=Ldi/dt

L eq =2+10+12+20=44H

di/dt=44/44 = 1 A/s.

Voltage across 10H inductor = 10*di/dt = 10*1=10V.

9. Find voltage across 12H inductor.

basic-electrical-engineering-questions-answers-coils-connected-series-q7

a) 2V

b) 10V

c) 12V

d) 20V

Answer: c

Explanation: e=Ldi/dt

L eq =2+10+12+20=44H

di/dt=44/44 = 1 A/s.

Voltage across 12H inductor = 12*di/dt = 12*1=12V.

10. Find voltage across 20H inductor.

basic-electrical-engineering-questions-answers-coils-connected-series-q7

a) 2V

b) 10V

c) 12V

d) 20V

Answer: d

Explanation: e=Ldi/dt

L eq =2+10+12+20=44H

di/dt=44/44 = 1 A/s.

Voltage across 20H inductor = 20*di/dt = 20*1=20V.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Generation of an Alternating EMF”.

1. Which, among the following, is the correct expression for alternating emf generated?

a) e=2Blvsin

b) e=2B 2 lvsin

c) e=Blvsin

d) e=4Blvsin

Answer: c

Explanation: The correct expression for alternating emf generated is e=Blvsin. Where B stands for magnetic field density, l is the length of each of the parallel sides v is the velocity with which the conductor is moved and θ is the angle between the velocity and the length.

2. What should theta be in order to get maximum emf?

a) 0 0

b) 90 0

c) 180 0

d) 45 0

Answer: b

Explanation: The value of θ should be 90 0 in order to get maximum emf because e = Blvsin and sin is maximum when θ is 90 0 .

3. Calculate the maximum emf when the velocity is 10m/s, the length is 3m and the magnetic field density is 5T.

a) 150V

b) 100V

c) 300V

d) 0V

Answer: a

Explanation: We know that: e max =Bvl

Substituting the values from the given question, we get e=150V.

4. When a coil is rotated in a magnetic field, the emf induced in it?

a) Is maximum

b) Is minimum

c) Continuously varies

d) Remains constant

Answer: c

Explanation: When a coil is rotated in a magnetic field, cross sectional area varies due to which the number of flux lines crossing it varies, which causes the emf to vary continuously.

5. emf is zero if the angle between velocity and length is _____

a) 0 0

b) 90 0

c) 270 0

d) 45 0

Answer: a

Explanation: If the angle between velocity and length is zero, sinθ=0

So, e=Bvlsinθ = 0.

6. In an A.C. generator, increase in number of turns in the coil _________

a) Increases emf

b) Decreases emf

c) Makes the emf zero

d) Maintains the emf at a constant value

Answer: a

Explanation: In an A.C. generator, the emf increases as the number of turns in the coil increases because the emf is directly proportional to the number of turns.

7. The number of cycles that occur in one second is termed as ___________

a) Waveform

b) Frequency

c) Amplitude

d) Period

Answer: b

Explanation: The number of cycles that occur in one second is known as the frequency. It is the reciprocal of the time period.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Waveform Terms and Definitions”.

1. The variation of a quantity such as voltage or current shown on a graph is known as ___________

a) Waveform

b) Peak value

c) Instantaneous value

d) Period

Answer: a

Explanation: The variation of a quantity, which is voltage or current in this case, shown on a graph with the x-axis as time is known as a waveform.

2. What is the duration of one cycle known as _________

a) Waveform

b) Peak value

c) Instantaneous value

d) Period

Answer: d

Explanation: The duration of one cycle is known as a period. A function which repeats the same waveform at equal intervals of time is known as a periodic function.

3. The repetition of a variable quantity, recurring at equal intervals, is known as ___________

a) Waveform

b) Instantaneous value

c) Cycle

d) Period

Answer: c

Explanation: Each repetition of a variable quantity, recurring at equal intervals, is termed as a cycle.

4. The value of a given waveform at any instant time is termed as ___________

a) Waveform

b) Instantaneous value

c) Cycle

d) Period

Answer: b

Explanation: Instantaneous value is the value of the waveform at that instant. Hence the value of a given waveform at any instant time is termed as instantaneous value.

5. The maximum instantaneous value measured from zero value is known as?

a) Peak value

b) Peak to peak value

c) Cycle

d) Period

Answer: a

Explanation: The maximum instantaneous value measured from the zero value is termed as the peak value.

6. The maximum variation between the maximum positive and the maximum negative value is known as?

a) Peak value

b) Peak to peak value

c) Cycle

d) Period

Answer: b

Explanation: The maximum variation between the maximum positive instantaneous value and the maximum negative instantaneous value is the peak-to-peak value.

7. What is the correct relation between the peak value and peak to peak value for a sinusoidal waveform?

a) Vp=4Vp-p

b) Vp=Vp-p

c) Vp-p=2Vp

d) Vp=2Vp-p

Answer: c

Explanation: The maximum variation between the maximum positive instantaneous value and the maximum negative instantaneous value is the peak-to-peak value. For a sinusoidal waveform, it is twice the peak value. Hence Vp-p=2Vp.

8. If the peak to peak voltage is 10V, calculate the peak voltage.

a) 10V

b) 2V

c) 4V

d) 5V

Answer: d

Explanation: Vp-p=2Vp

Substituting the values from the question, we get Vp=5V.

9. If the peak voltage is 9V, calculate the peak to peak voltage.

a) 9V

b) 20V

c) 18V

d) 12V

Answer: c

Explanation: Vp-p=2Vp

Substituting the values from the question, we get Vp-p= 18V.

This set of Basic Electrical Engineering Question Bank focuses on “Relationship between Frequency, Speed and Number of Pole Pairs”.

1. The waveform of the emf generated undergoes one complete cycle when?

a) Conductors move past north pole

b) Conductors move past south pole

c) Conductors move past north and south poles

d) Conductors are stationary

Answer: c

Explanation: The waveform of the e.m.f. generated in an a.c. generator undergoes one complete cycle of variation when the conductors move past an N and an S pole.

2. When is the shape of the negative half of the emf waveform equal to the positive half?

a) When the conductors move past north pole

b) When conductors move past south pole

c) When conductors move past both north and south pole

d) When conductors are stationary

Answer: c

Explanation: The waveform of the e.m.f. generated in an a.c. generator undergoes one complete cycle of variation when the conductors move past an N and an S pole and the shape of the wave over the negative half is exactly the same as that over the positive half.

3. Which is the correct formula for frequency in an ac generator?

a) f=p*n

b) f=p/n

c) f=n/p

d) f=n 2 p

Answer: a

Explanation: The frequency in an ac generator is p*n, where p is pairs of poles and speed is n revolutions per second.

4. What will happen to the frequency if the number of revolutions increases?

a) Increases

b) Decreases

c) Remains the same

d) Becomes zero

Answer: a

Explanation: We know that:

f=p*n, therefore, as n increases, f also increases.

Hence frequency increases if number of revolutions increases.

5. What happens to the frequency if the number of pairs of poles increases?

a) Increases

b) Decreases

c) Remains the same

d) Becomes zero

Answer: a

Explanation: We know that:

f=p*n, therefore, as p increases, f also increases.

Hence frequency increases if number of pair of poles increases.

6. Calculate the frequency if the number of revolutions is 300 and the paired poles are 50.

a) 15kHz

b) 150kHz

c) 1500kHz

d) 150Hz

Answer: a

Explanation: We know that f=p*n

f=50*300=15000 Hz = 15kHz.

7. Calculate the number of revolutions if the frequency is 15kHz and the paired poles are 50.

a) 100

b) 200

c) 300

d) 400

Answer: c

Explanation: We know that f=p*n

f=15kHz=15000Hz, p=50

15000=50*n => n=15000/50=300.

8. Calculate the number of paired poles if the frequency id 15kHz and the number of revolutions is 300.

a) 10

b) 30

c) 50

d) 70

Answer: c

Explanation: We know that f=p*n

f=15kHz=15000Hz, n=300

15000=p*300 => p=15000/300=50.

9. What is the frequency of a two pole machine having n=50?

a) 100Hz

b) 200Hz

c) 50Hz

d) 25Hz

Answer: c

Explanation: For a two pole machine, p=1.

f=pn = 1*50 = 50Hz.

10. What is the minimum number of poles that a machine must have __________

a) 1

b) 2

c) 4

d) 10

Answer: b

Explanation: The minimum number of poles that a machine must have is 2 because a machine must have at least one pair of poles = 2 poles.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Average and RMS Values of an Alternating Current”.

1. Find the average value of current when the current that are equidistant are 4A, 5A and 6A.

a) 5A

b) 6A

c) 15A

d) 10A

Answer: a

Explanation: The average value of current is the sum of all the currents divided by the number of currents. Therefore average current = /3=5A.

2. What is the current found by finding the current in an equidistant region and dividing by n?

a) RMS current

b) Average current

c) Instantaneous current

d) Total current

Answer: b

Explanation: The average value of the current is the sum of all the currents divided by the number of currents.

3. RMS stands for ________

a) Root Mean Square

b) Root Mean Sum

c) Root Maximum sum

d) Root Minimum Sum

Answer: a

Explanation: RMS stands for Root Mean Square. This value of current is obtained by squaring all the current values, finding the average and then finding the square root.

4. What is the type of current obtained by finding the square of the currents and then finding their average and then fining the square root?

a) RMS current

b) Average current

c) Instantaneous current

d) Total current

Answer: a

Explanation: RMS stands for Root Mean Square. This value of current is obtained by squaring all the current values, finding the average and then finding the square root.

5. __________ current is found by dividing the area enclosed by the half cycle by the length of the base of the half cycle.

a) RMS current

b) Average current

c) Instantaneous current

d) Total current

Answer: b

Explanation: The average value of current is the sum of all the currents divided by the number of currents. Hence it can also be found by dividing the area enclosed by the half cycle by the length of the base of the half cycle.

6. What is the effective value of current?

a) RMS current

b) Average current

c) Instantaneous current

d) Total current

Answer: a

Explanation: RMS current is also known as the effective current. RMS stands for Root Mean Square. This value of current is obtained by squaring all the current values, finding the average and then finding the square root.

7. In a sinusoidal wave, average current is always _______ rms current.

a) Greater than

b) Less than

c) Equal to

d) Not related

Answer: b

Explanation: The average value of current is the sum of all the currents divided by the number of currents whereas RMS current is obtained by squaring all the current values, finding the average and then finding the square root. Hence RMS current is greater than average current.

8. For a rectangular wave, average current is ______ rms current.

a) Greater than

b) Less than

c) Equal to

d) Not related

Answer: c

Explanation: The rms value is always greater than the average except for a rectangular wave, in which the heating effect remains constant so that the average and the rms values are the same.

9. Peak value divided by the rms value gives us?

a) Peak factor

b) Crest factor

c) Both peak and crest factor

d) Neither peak nor crest factor

Answer: c

Explanation: Peak and crest factor both mean the same thing. Hence the peak value divided by the rms value gives us the peak or crest factor.

10. Calculate the crest factor if the peak value of current is 10A and the rms value is 2A.

a) 5

b) 10

c) 5A

d) 10A

Answer: a

Explanation: We know that:

Crest factor = Peak value/RMS value.

Substituting the values from the given question, we get crest factor=5.

This set of Basic Electrical Engineering Questions and Answers for Entrance exams focuses on “Average and RMS Values of Sinusoidal & Non-Sinusoidal Currents and Voltages”.

1. If maximum value of current is 5√2 A, what will be the value of RMS current?

a) 10 A

b) 5 A

c) 15 A

d) 25 A

Answer: b

Explanation: We know, value of RMS current =value of max current/√2

Substituting the value of max current we get, rms current = 5A.

2. If Im is the maximum value of a sinusoidal voltage, what is the instantaneous value?

a) i=Im/2

b) i=Imsinθ

c) i=Imcosθ

d) i=Imsinθ or i=Imcosθ

Answer: d

Explanation: The instantaneous value of a sinusoidal varying current is i=Imsinθ or i=Imcosθ where Im is the maximum value of current.

3. Average value of current over a half cycle is?

a) 0.67Im

b) 0.33Im

c) 6.7Im

d) 3.3Im

Answer: a

Explanation: Average current = ∫ 0 π idθ/π = ∫ 0 π Imsinθ dθ/π = 2Im/π =0.67 Im.

4. What is the correct expression for the rms value of current?

a) Irms=Im/2

b) Irms=Im/√2

c) Irms=Im/4

d) Irms=Im

Answer: b

Explanation: I rms 2 = ∫ 0 π dθ i 2 /2π = Im 2 /2

I rms =Im/√2.

5. Average value of current over a full cycle is?

a) 0.67Im

b) 0

c) 6.7Im

d) 3.3Im

Answer: b

Explanation: Average of sine or cosine over a period is zero so, average value of current over full cycle is zero.

6. What is the correct expression for the form factor?

a) I rms * I av

b) I rms / I av

c) I rms + I av

d) I rms – I av

Answer: b

Explanation: The correct expression for form factor is I rms /I av where I rms is the rms value of the current and I av is the average current.

7. For a direct current, the rms current is ________ the mean current.

a) Greater than

b) Less than

c) Equal to

d) Not related to

Answer: c

Explanation: For a direct current, the mean current value is the same as that of the rms current.

8. For a direct current, the rms voltage is ________ the mean voltage.

a) Greater than

b) Less than

c) Equal to

d) Not related to

Answer: c

Explanation: For a direct current, the mean voltage value is the same as that of the rms voltage.

9. What is the value of the form factor for sinusoidal current?

a) π/2

b) π/4

c) 2π

d) π/√2

Answer: a

Explanation: For sinusoidal current, I rms =Im/√2

I av =√2 Im/π

So, form factor = I rms /I av = π/2.

10. If the maximum value of the current is 5√2 A, what will be the value of the average current?

a) 10/π A

b) 5/π A

c) 15/π A

d) 25/π A

Answer: a

Explanation: We know, the value of the average current = value of max current *√2 /π

Substituting the value of max current we get, rms current = 10/π A.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Representation of an Alternating Quantity by a Phasor”.

1. For addition and subtraction of phasors, we use the _________ form.

a) Rectangular

b) Polar

c) Either rectangular or polar

d) Neither rectangular nor polar

Answer: a

Explanation: For addition and subtraction of phasors, we use the rectangular form because in the rectangular form we can only add the real part and the complex part separately to get the total value.

2. For multiplication and division of phasors, we use ____________ form.

a) Rectangular

b) Polar

c) Either rectangular or polar

d) Neither rectangular nor polar

Answer: b

Explanation: For multiplication and division of phasors, we use the polar form because in the polar form we just multiply or divide the values and add or subtract the angles.

3. If a voltage of 2+5j and another voltage of 3+ 6j flows through two different resistors, connected in series, in a circuit, find the total voltage in the circuit.

a) 2+5j V

b) 3+6j V

c) 5+11j V

d) 5+10j V

Answer: c

Explanation: The total voltage in the circuit is the sum of the two voltages where we add the real parts and imaginary parts separately.

Therefore, Vtotal= 5+11j V.

4. Find the total current in the circuit if two currents of 4+5j flow in the circuit.

a) 4+5j A

b) 4A

c) 5A

d) 8+10j A

Answer: d

Explanation: The total current in the circuit is the sum of the two currents where we add the real parts and imaginary parts separately.

Therefore, Itotal= 8+10j A.

5. What is the correct expression of ω?

a) ω=2π

b) ω=2πf

c) ω=πf

d) ω=2f 2

Answer: b

Explanation: The correct expression for ω is ω=2πf where f is the frequency of the alternating voltage or current.

6. Find the value of ω if the frequency is 5Hz?

a) 3.14 rad/s

b) 31.4 rad/s

c) 34 rad/s

d) 341 rad/s

Answer: b

Explanation: The expression for ω is ω=2*π*f.

Substituting the value of f from the question, we get ω=31.4 rad/s.

7. When one sine wave passes through the zero following the other, it is _________

a) Leading

b) Lagging

c) Neither leading nor lagging

d) Either leading or lagging

Answer: b

Explanation: The sine wave is said to lag because it passes though zero following the other, hence it crosses zero after the first wave, therefore it is said to lag.

8. A phasor has frozen at 30 degrees, find the value of the phase angle.

a) 30 degrees

b) 60 degrees

c) 120 degrees

d) 180 degrees

Answer: a

Explanation: The value of the phase angle is the value at which the phasor stops or freezes. Here, it freezes at 30 degree, hence the phase angle is 30 degrees.

9. The time axis of an AC phasor represents?

a) Time

b) Phase angle

c) Voltage

d) Current

Answer: b

Explanation: The time axis while measuring an AC sinusoidal voltage or current represents the phase angle when converting it to a phasor.

10. The length of the phasor represents?

a) Magnitude of the quantity

b) Direction of the quantity

c) Neither magnitude nor direction

d) Either magnitude or direction

Answer: a

Explanation: The length of the phasor arrow represents the magnitude of the quantity, whereas the angle between the phasor and the reference represents the phase angle.

This set of Basic Electrical Engineering Questions and Answers for Aptitude test focuses on “Phasor Diagrams Drawn with r.m.s. Values Instead of Maximum Values”.

1. Ammeters and voltmeters are calibrated to read?

a) RMS value

b) Peak value

c) Average value

d) Instantaneous value

Answer: a

Explanation: Ammeters and voltmeters are calibrated to read the rms value because the rms value is the most accurate than average value.

2. The rms value is _________ times he maximum value

a) 1.414

b) 0.5

c) 2

d) 0.707

Answer: d

Explanation: We know that the rms value is 1/√2 times the maximum value, hence the rms value is 0.707 times the maximum value.

3. The rms value is 0.707 times the _________ value.

a) Peak

b) Instantaneous

c) Average

d) DC

Answer: a

Explanation: We know that the rms value is 1/√2 times the maximum value, hence the rms value is 0.707 times the maximum value.

4. If the phasors are drawn to represent the maximum values instead of the rms values, what would happen to the phase angle between quantities?

a) Increases

b) Decreases

c) Remains constant

d) Becomes zero

Answer: c

Explanation: When phasors are drawn representing the maximum values instead of the rms value, the shape of the diagram remains unaltered and hence the phase angle remains the same.

5. Usually phasor diagrams are drawn representing?

a) RMS value

b) Peak value

c) Average value

d) Instantaneous value

Answer: a

Explanation: Ammeters and voltmeters are calibrated to read the rms value, hence the phasors are drawn representing the rms values.

6. If two current phasors, having magnitude 12A and 5A intersect at an angle of 90 degrees, calculate the resultant current.

a) 13 A

b) 10 A

c) 6 A

d) 5 A

Answer: a

Explanation: Using the parallelogram law of addition, I 2 = I 1 2 + I 2 2 + 2I 1 I 2 cosθ

I=13 A.

7. If two current phasors, having magnitude 5A and 10A intersect at an angle of 60 degrees, calculate the resultant current.

a) 12.23 A

b) 12.54 A

c) 13.23 A

d) 14.24 A

Answer: c

Explanation: Resultant current can be found using I 2 = I 1 2 + I 2 2 + 2I 1 I 2 cosθ

Substituting the values, we get I=13.23 A.

8. The instantaneous values of two alternating voltages are given as _________

v1=60sinθ and v2=40sin. Find the instantaneous sum.

a) 87.2 sin V

b) 87.2 sin V

c) 87.2 sin V

d) 87.2 cos V

Answer: a

Explanation: Horizontal component of v1 = 40V

Vertical component of v1=0V

Horizontal component of v2=60cos600

Vertical component of v2=60sin600

Resultant horizontal component=60cos600 + 40 = 70V

Resultant vertical component = 30√3 V

Resultant = 87.2V

tan = 30√3 / 70 => ϕ=36.50

Therefore sum = 87.2 sin V.

9. The instantaneous values of two alternating voltages are given as:

v1=60sinθ and v2=40sin. Find the instantaneous difference.

a) 53 sin V

b) 53 sin V

c) 53 sin V

d) 53 cos V

Answer: b

Explanation: Horizontal component of v1 = 40V

Vertical component of v1=0V

Horizontal component of v2=-60cos600

Vertical component of v2=-60sin600

Resultant horizontal component=40-30 = 10V

Resultant vertical component = -30√3 V

Resultant v = 53 V

tan = 30√3 / 10 => ϕ=79.50

Therefore sum = 53 sin  V.

10. The resultant of two alternating sinusoidal voltages or currents can be found using ___________

a) Triangular law

b) Parallelogram law

c) Either triangular or parallelogram law

d) Neither triangular nor parallelogram law

Answer: b

Explanation: The resultant current can be found by using the parallelogram law of addition I 2 = I 1 2 + I 2 2 + 2I 1 I 2 cosθ.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Alternating Current in a Resistive & Inductive Circuit”.

1. Instantaneous voltage is the product of resistance and _____________ current in a resistive circuit.

a) Instantaneous

b) Average

c) RMS

d) Peak

Answer: a

Explanation: V=IR. So, V=iR

Instantaneous voltage is the product of resistance and instantaneous current in a resistive circuit.

2. Find the value of the instantaneous voltage if the resistance is 2 ohm and the instantaneous current in the circuit is 5A.

a) 5V

b) 2V

c) 10V

d) 2.5V

Answer: c

Explanation: We know that,

v=iR, substituting the given values from the question, we get v=10V.

3. The power for a purely resistive circuit is zero when?

a) Current is zero

b) Voltage is zero

c) Both current and voltage are zero

d) Either current or voltage is zero

Answer: d

Explanation: P=VIcosϕ Power in a circuit is the product of voltage, current and the cosine of the phase angle. Phase angle is 00 for purely resistive circuit so, P=VI. Hence if either voltage or current is zero, the power is zero.

4. The correct expression for the instantaneous current if instantaneous voltage is Vm in a resistive circuit is?

a) 1A

b) 2A

c) 3A

d) 4A

Answer: b

Explanation: We know that:V=Vm

Since i=V/R, we can write, i=Vm/R.

5. Calculate the resistance in the circuit if the rms voltage is 20V and the rms current is 2A.

a) 2 ohm

b) 5 ohm

c) 10 ohm

d) 20 ohm

Answer: c

Explanation: We know that:

R=V/I

Substituting the given values from the question, we get R=10 ohm.

6. The correct expression for the instantaneous current in a resistive circuit is?

a) i=Vm/R

b) i=Vm/R

c) i=V/R

d) i=V/R

Answer: a

Explanation: The instantaneous voltage can be written in terms of the maximum voltage in the following manner:

v=Vm

Since i=v/R, we can write, i=Vm/R.

7. Can ohm’s law be applied in an ac circuit?

a) Yes

b) No

c) Depends on the rms current

d) Depends on the rms voltage

Answer: a

Explanation: Ohm’s law can be applied in ac as well as dc circuits. It can be applied in ac circuits because the condition V=IR holds true even in ac circuits.

8. The correct expression for the instantaneous current if instantaneous voltage is Vm in an inductive circuit is?

a) i = Vm/X L

b) i = Vm/X L

c) i = -Vm/X L

d) i = -Vm/X L

Answer: d

Explanation: V=Vm*sint

I=V/X L = -Vm/X L .

This set of Basic Electrical Engineering Question Paper focuses on “Mechanical Analogy, Current and Voltage of an Inductive Circuit”.

1. Inductor does not allow sudden changes in?

a) Voltage

b) Current

c) Resistance

d) Inductance

Answer: b

Explanation: The inductor does not allow sudden changes in current because if current changes in the inductor occur in zero time, the voltage becomes zero which is not possible.

2. Inductance is _____________________ to number of turns in the coil.

a) directly proportional

b) inversely proportional

c) equal

d) not related

Answer: a

Explanation: L=µ 0 N 2 A/l

Inductance is directly proportional to number of turns in the coil.

3. Choke involve use of _____________

a) Resistor

b) Capacitor

c) Inductor

d) Transistor

Answer: c

Explanation: Choke is a type of coil so it involves use of inductor. Capacitors cannot be used in choke coil.

4. What is the value of current in an inductive circuit when there is no applied voltage?

a) Minimum

b) Maximum

c) Zero

d) Cannot be determined

Answer: b

Explanation: The current in an inductive circuit is maximum when there is no voltage applied because the coils of the inductor store electric flux.

5. What is the current in an inductive circuit when the applied voltage is maximum?

a) Infinity

b) Maximum

c) Zero

d) Cannot be determined

Answer: c

Explanation: The current in an inductive circuit is zero or minimum when the value of the applied voltage is maximum.

6. In an inductive circuit, the voltage_______ the current?

a) Leads

b) Lags

c) Is greater than

d) Is less than

Answer: a

Explanation: In a pure inductive circuit the voltage leads the current and the current lags the voltage by a phase difference of 90 degrees.

7. In an inductive circuit, the current________ the voltage?

a) Leads

b) Lags

c) Is greater than

d) Is less than

Answer: b

Explanation: In a pure inductive circuit the voltage leads the current and the current lags the voltage by a phase difference of 90 degrees.

8. In which device inductor cannot be used?

a) filter circuit

b) transformer

c) choke

d) dielectric

Answer: d

Explanation: Inductor has wide number of applications.

It is used in LR filter circuits, transformer and choke coil.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Resistance and Inductance in Series”.

1. A resistance of 7 ohm is connected in series with an inductance of 31.8mH. The circuit is connected to a 100V 50Hz sinusoidal supply. Calculate the current in the circuit.

a) 2.2A

b) 4.2A

c) 6.2A

d) 8.2A

Answer: d

Explanation: X L =2*π*f*L = 10 ohm. Z 2 =(R 2 +X L 2 )

Therefore the total impedance Z = 12.2ohm.

V=IZ, therefore I=V/Z=100/12.2 = 8.2A.

2. A resistance of 7 ohm is connected in series with an inductance of 31.8mH. The circuit is connected to a 100V 50Hz sinusoidal supply. Calculate the phase difference.

a) -55.1

b) 55.1

c) 66.1

d) -66.1

Answer: a

Explanation: φ=tan -1 =55.1

Since this is an inductive circuit, the current will lag, hence φ= -55.1.

3. A resistance of 7 ohm is connected in series with an inductance of 31.8mH. The circuit is connected to a 100V 50Hz sinusoidal supply. Calculate the voltage across the resistor.

a) 31.8V

b) 57.4V

c) 67.3V

d) 78.2V

Answer: b

Explanation: X L =2*π*f*L = 10 ohm. Z 2 =(R 2 +X L 2 )

Therefore, the total impedance Z = 12.2ohm.

V=IZ, therefore I=V/Z=100/12.2 = 8.2A. Voltage across resistor = 8.2*7 = 57.4V.

4. A resistance of 7 ohm is connected in series with an inductance of 31.8mH. The circuit is connected to a 100V 50Hz sinusoidal supply. Calculate the voltage across the inductor.

a) 52V

b) 82V

c) 65V

d) 76V

Answer: b

Explanation: X L =2*π*f*L = 10 ohm. Z 2 =(R 2 +X L 2 )

Therefore, the total impedance Z = 12.2ohm.

V=IZ, therefore I=V/Z=100/12.2 = 8.2A. Voltage across inductor = 8.2*10 = 82V.

5. A resistance of 7 ohm is connected in series with an inductance of 31.8mH. The circuit is connected to a x V 50Hz sinusoidal supply. The current in the circuit is 8.2A. Calculate the value of x.

a) 10V

b) 50V

c) 100V

d) 120V

Answer: c

Explanation: X L =2*π*f*L= 10 ohm. Z 2 =(R 2 +X L 2 )

Therefore, the total impedance Z = 12.2ohm.

V=IZ, therefore V = 12.2*8.2 = 100V.

6. Which, among the following, is the correct expression for φ.

a) φ=tan -1 

b) φ=tan -1 

c) φ=tan -1 

d) φ=cos -1 

Answer: a

Explanation: From the impedance triangle, we get tanφ= XL/R.

Hence φ=tan -1 .

7. For an RL circuit, the phase angle is always ________

a) Positive

b) Negative

c) 0

d) 90

Answer: b

Explanation: For a series resistance and inductance circuit the phase angle is always a negative value because the current will always lag the voltage.

8. What is φ in terms of voltage?

a) φ=cos -1 V/VR

b) φ=cos -1 V*VR

c) φ=cos -1 VR/V

d) φ=tan -1 V/VR

Answer: c

Explanation: From the voltage triangle, we get cosφ= VR/V.

Hence φ=cos -1 VR/V.

9. What is sinϕ from impedance triangle?

a) X L /R

b) X L /Z

c) R/Z

d) Z/R

Answer: b

Explanation: In Impedance triangle, Base is R, Hypotenuse is Z, Height is X L .

So, sinϕ = X L /Z.

This set of Basic Electrical Engineering Problems focuses on “Alternating Current in a Capacitive Circuit”.

1. What is the resonance frequency of ac circuit?

a) 1/√LC

b) √

c) √LC

d) LC

Answer: a

Explanation: At resonance, X L =X C

ωL=1/ωC

ω=1/√LC.

2. What is impedance at resonance?

a) maximum

b) minimum

c) zero

d) cannot be determined

Answer: b

Explanation: At resonance, X L =X C

Z 2 =R 2 +(X L -X C ) 2

Z=R So Z is minimum at resonance.

3. What is the value of impedance at resonance?

a) X L

b) X C

c) R

d) 0

Answer: c

Explanation: At resonance, X L =X C

Z 2 =R 2 +(X L -X C ) 2

Z=R So Z is minimum at resonance.

4. What is φ in terms of voltage?

a) φ=cos -1 V/V R

b) φ=cos -1 V*V R

c) φ=cos -1 V R /V

d) φ=tan -1 V/V R

Answer: c

Explanation: Form the voltage triangle, we get cosφ= V R /V.

Hence φ=cos -1 V R /V.

5. What is tanϕ for RC circuit?

a) X C /R

b) X L /R

c) R/Z

d) Z/R

Answer: a

Explanation: From the impedance triangle, height gives capacitive reactance and base gives resistance.

tanϕ=X C /R.

6. What is the resonance condition?

a) When X L >X C

b) When X L <X C

c) When X L =X C

d) When X C =infinity

Answer: c

Explanation: The current is in phase with the voltage when the capacitive reactance is in equal to the inductive reactance. This is known as resonance condition.

7. What is the frequency in resonance condition?

a) Minimum

b) Maximum

c) Cannot be determined

d) Zero

Answer: b

Explanation: At resonance condition, the frequency is maximum since the inductive reactance is equal to the capacitive reactance. X L =X C .

This set of Basic Electrical Engineering Assessment Questions and Answers focuses on “Current and Voltage in a Capacitive Circuit”.

1. Can capacitor fully charge using alternating current?

a) yes

b) no

c) may or may not

d) depend on value of capacitance

Answer: a

Explanation: No, the capacitor cannot be fully charged using alternating current because as soon as the capacitor charges, the alternating current reverses its polarity thereby discharging it.

2. What is the resistance offered by a capacitor?

a) Susceptance

b) Conductance

c) Admittance

d) Reactance

Answer: d

Explanation: Resistance offered to alternating current by inductor or capacitor is known as reactance which is equivalent to resistance of resistor.

3. The combination of resistance and reactance known as ___________

a) Susceptance

b) Impedance

c) Conductance

d) Admittance

Answer: b

Explanation: The combination of resistance and reactance is known as impedance. It is equivalent resistance of an RLC alternating circuit.

4. What is the relation between reactance, resistance and impedance?

a) Z=R+jX

b) Z=R+X

c) Z=R-X

d) Z=R-jX

Answer: a

Explanation: The combination of resistance and reactance is known as impedance. Z=R+jX

Where Z is impedance, R is resistance and X is reactance.

5. What is the real part of the impedance of RLC circuit?

a) Resistance

b) Conductance

c) Admittance

d) Reactance

Answer: a

Explanation: The combination of resistance and reactance is known as impedance. Z=R+jX where Z is impedance, R is resistance and X is reactance. R is real part of Z.

6. What is imaginary part of the impedance of RLC circuit?

a) Resistance

b) Conductance

c) Admittance

d) Reactance

Answer: d

Explanation: The combination of resistance and reactance is known as impedance. Z=R+jX where Z is impedance, R is resistance and X is reactance. X is imaginary part of Z.

7. Which type of current can be stored in a capacitor?

a) Alternating current

b) Direct current

c) Both alternating current and direct current

d) Neither alternating current nor direct current

Answer: b

Explanation: Only direct current can be stored in the capacitor. Capacitor cannot be fully charged using alternating current because as soon as the capacitor charges, alternating current reverses its polarity thereby discharging it. So, we cannot store ac current in capacitor.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Resistance and Capacitance in Series”.

1. If in an alternating current circuit, resistance is 5 ohm, capacitive reactance is 12 ohm, what is the impedance?

a) 5 ohm

b) 10 ohm

c) 12 ohm

d) 13 ohm

Answer: d

Explanation: R=5Ω, X C =12Ω

Z 2 =R 2 +X C 2 Substituting the values we get,

Z 2 = 5 2 + 12 2

Z 2 = 169

Z=13 Ω.

2. If in an alternating current circuit, impedance is 26 ohm, capacitive reactance is 24 ohm, what is the resistance?

a) 25 ohm

b) 10 ohm

c) 12 ohm

d) 23 ohm

Answer: b

Explanation: Z=26Ω, X C =24Ω

Z 2 =R 2 +X C 2 Substituting the values we get,

26 2 = R 2 + 24 2

676 = R 2 + 576

R 2 = 100

R=10 Ω.

3. If in an alternating current circuit, capacitance of 30 µF is connected to a supply of 200V,50Hz. Find the current in the circuit.

a) 1.38 A

b) 1.89 A

c) 1.74 A

d) 0.89 A

Answer: a

Explanation: X C =1/=106.1

I=V/X C =200/106.1=1.89 A.

4. If in an alternating current circuit, capacitance C is connected to a supply of 200V,50Hz. Current in the circuit is 1.89 A. Find the capacitance C.

a) 30 µF

b) 20 µF

c) 10 µF

d) 15 µF

Answer: a

Explanation: X C = V/I = 200/1.89 = 106.1 Ω. X C =1/ Substituting the values we get C = 30 µF.

5. In ac circuit, resistance 5 ohm is connected with capacitor having capacitive reactance 12 ohm. Supply of 260 V is connected to the circuit. Calculate the current in the circuit.

a) 40 A

b) 10 A

c) 20 A

d) 30 A

Answer: c

Explanation: Z 2 =R 2 +X C 2 Substituting the values we get Z 2 = 5 2 + 12 2

Z 2 = 169

Z=13 Ω.

I=V/Z = 260/13=20 A.

6. In ac circuit, resistance 5 ohm is connected with capacitor having capacitive reactance 12 ohm. Supply of 260 V is connected to the circuit. Calculate the voltage across resistance.

a) 300 V

b) 200 V

c) 240 V

d) 100 V

Answer: d

Explanation: Z 2 =R 2 +X C 2 Substituting the values we get

Z 2 = 5 2 + 12 2

Z 2 = 169

Z=13 Ω.

I = V/Z = 260/13=20 A.

V R =iR=20*5=100 V.

7. In ac circuit, resistance 5 ohm is connected with a capacitor having capacitive reactance 12 ohm. Supply of 260 V is connected to the circuit. Calculate the voltage across a capacitor.

a) 300 V

b) 200 V

c) 240 V

d) 100 V

Answer: c

Explanation: Z 2 =R 2 +X C 2 Substituting the values we get

Z 2 = 5 2 + 12 2

Z 2 = 169

Z=13 Ω.

I = V/Z = 260/13=20 A.

V C =iX C =20*12=240 V.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Alternating Current in an RLC Circuit”.

1. Find the total voltage applied in a series RLC circuit when i=3mA, V L =30V, V C =18V and R=1000 ohms.

a) 3.95V

b) 51V

c) 32.67V

d) 6.67V

Answer: b

Explanation: Total voltage= V R +V L +V C .

V R =1000*3*10 -3 =3V. Therefore, total voltage = 30+18+3=51V.

2. In an RLC circuit, which of the following is always used as a vector reference?

a) Voltage

b) Resistance

c) Impedance

d) Current

Answer: a

Explanation: In an RLC circuit, the voltage is always used as a reference and according to the phase of the voltage, the phase of the other parameters is decided.

3. In an RLC circuit, the power factor is always ____________

a) Positive

b) Negative

c) Depends on the circuit

d) Zero

Answer: c

Explanation: In an RLC series circuit, the power factor depends on the number of resistors and inductors in the circuit, hence it depends on the circuit.

4. In an RLC series phasor, we start drawing the phasor from which quantity?

a) Voltage

b) Resistance

c) Impedance

d) Current

Answer: d

Explanation: In an RLC series phasor diagram, we start drawing the phasor from the quantity which is common to all three components, that is the current.

5. What is the correct expression for the phase angle in an RLC series circuit?

a) φ=tan -1 (X L -X C )/R

b) φ=tan -1 (X L +X C )/R

c) φ=tan(X L -X C )/R

d) φ=tan -1 (X L -X C )

Answer: a

Explanation: from the impedance triangle we get tanφ=(X L -X C )/R.

Hence φ=tan -1 (X L -X C )/R.

6. When is tanφ positive?

a) When inductive reactance is less than capacitive reactance

b) When inductive reactance is greater than capacitive reactance

c) When inductive reactance is equal to capacitive reactance

d) When inductive reactance is zero

Answer: b

Explanation: tanφ is positive when inductive reactance is greater than capacitive reactance because current will lag the voltage.

7. When is tanφ negative?

a) When inductive reactance is less than capacitive reactance

b) When inductive reactance is greater than capacitive reactance

c) When inductive reactance is equal to capacitive reactance

d) When inductive reactance is zero

Answer: a

Explanation: tanφ is negative when inductive reactance is less than capacitive reactance because current will lead the voltage.

8. Which of the following is not ac waveform?

a) sinusoidal

b) square

c) constant

d) triangular

Answer: c

Explanation: AC waveforms are those which periodically changes polarity with time. Sinusoidal wave, square wave, triangular wave change their polarity at regular intervals so they are ac wavefoms. Constant wave doesnot change its polarity so it is not an ac waveform.

9. What is not a frequency for ac current?

a) 50 Hz

b) 55 Hz

c) 0Hz

d) 60 Hz

Answer: c

Explanation: DC current is a type of constant current so it has frequency of zero hertz. So, AC current can have rest other frequencies other than zero.

10. Which type of ac waveform is given in figure?

basic-electrical-engineering-questions-answers-alternating-current-rlc-circuit-q10

a) sinusoidal

b) triangular

c) square

d) complex waveform

Answer: a

Explanation: The figure depicts ac waveform of sinusoidal nature changing its polarity after regular intervals sinusoidally.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Kirchhoff’s Laws and Network Solution”.

1. In an AC circuit, resistance 50 Ω, inductance 0.3 H and capacitance 15 μF is connected to an AC voltage source 25 V, 50 Hz. Determine the inductive reactance in the circuit.

a) 36 ohm

b) 95 ohm

c) 125 ohm

d) 140 ohm

Answer: b

Explanation: X L =2πfL f=50Hz and L=0.3H

X L =2π = 94.25 ohm.

2. In an AC circuit, resistance 50 Ω, inductance 0.3 H and capacitance 15 μF is connected to an AC voltage source 25 V, 50 Hz. Determine the capacitive reactance in the circuit.

a) 316 ohm

b) 195 ohm

c) 124 ohm

d) 212 ohm

Answer: d

Explanation: X C =1/ f=50Hz and C=15 μF

X C =1/(2π*50*15*10 -6 ) = 212.21 ohm.

3. In an AC circuit, resistance 50 Ω, inductance 0.3 H and capacitance 15 μF is connected to an AC voltage source 25 V, 50 Hz. Determine the impedance in the circuit.

a) 110 ohm

b) 100 ohm

c) 125 ohm

d) 140 ohm

Answer: c

Explanation: Z 2 = R 2 +(X L -X C ) 2 , X L =2πfL, X C =1/ f=50Hz and L=0.3H and C=15 μF

X L =2π= 94.25 ohm .

X C =1/(2π*50*15*10 -6 ) = 212.21 ohm.

Z 2 =50 2 + 2 =15625

Z = 125 ohm.

4. In an AC circuit, resistance 50 Ω, inductance 0.3 H and capacitance 15 μF is connected to an AC voltage source 25 V, 50 Hz. Determine the current in the circuit.

a) 0.01 A

b) 0.2 A

c) 0.02 A

d) 0.002 A

Answer: b

Explanation: Z 2 = R 2 +(X L -X C ) 2 , X L =2πfL, X C =1/ f=50Hz and L=0.3H and C=15 μF

X L =2π= 94.25 ohm .

X C =1/(2π*50*15*10 -6 ) = 212.21 ohm.

Z 2 =50 2 + 2 =15625

Z = 125 ohm.

Current in the circuit i=V/Z=25/125=0.2 A.

5. Find the value of the source current from the following circuit.

basic-electrical-engineering-questions-answers-kirchhoffs-law-network-solution-q5

a) 2.54A

b) 6.67A

c) 3.35A

d) 7.65A

Answer: a

Explanation: I 3 =A

V 2 =I 3 R==V

I2=V2/Xc= A

I1 =I2 +I3 =+=A

I1=(3 2 +1.5 2 ) 1/2 = 3.35A.

6. Find the value of the source voltage from the following circuit.

basic-electrical-engineering-questions-answers-kirchhoffs-law-network-solution-q5

a) 49.2V

b) 34.6V

c) 65.2V

d) 25.6V

Answer: a

Explanation: I3 =A

V2 =I3R==V

I2=V2/Xc= A

I1 =I2 +I3 =+=A

I1=(3 2 +1.5 2 ) 1/2 = 3.35A.

V1 =I1 =V

E=V1 +V2 =V

E=(39 2 +30 2 ) 1/2 = 49.2V.

7. In an AC circuit, resistance 50 Ω, inductance 0.3 H and capacitance 15 μF is connected to an AC voltage source 25 V, 50 Hz. Determine the phase difference between current and voltage.

a) 670

b) 540

c) 470

d) 770

Answer: a

Explanation: X L =2πfL, X C =1/ f=50Hz and L=0.3H and C=15 μF

X L =2π = 94.25 ohm.

X C =1/(2π*50*15*10 -6 ) = 212.21 ohm.

tanϕ = |(X L -X C )|/R = /50 = 2.3592

ϕ=67 0 .

8. What value of direct current must flow through a resistor to produce the same heating power as an alternating current with a peak value of 3.5 A?

a) 1.5 A

b) 2.5 A

c) 3.5 A

d) 4.5 A

Answer: b

Explanation: Power in dc circuit = Power in ac circuit 

I DC 2 R = I RMS 2 R

I DC =I RMS

I RMS =I 0 /√2 = 3.5/√2 = 2.5 A.

I DC =2.5 A.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Frequency Variation in a Series RLC Circuit”.

1. If the resonant frequency in a series RLC circuit is 50kHz along with a bandwidth of 1kHz, find the quality factor.

a) 5

b) 50

c) 100

d) 500

Answer: b

Explanation: We know that Quality factor is equal to the resonant frequency divided by the bandwidth.

Q=f res /Bandwidth = 50/1 = 50.

2. What is the SI unit for quality factor?

a) Hz

b) kHz

c) MHz

d) No unit

Answer: d

Explanation: We know that Quality factor is equal to the resonant frequency divided by the bandwidth. It is one frequency divided by another hence it has no unit.

3. What happens to the quality factor when the bandwidth increases?

a) Increases

b) Decreases

c) Remains the same

d) Becomes zero

Answer: b

Explanation: Q=f res /Bandwidth

Quality factor is inversely proportional to bandwidth. So, if bandwidth increases quality factor decreases.

4. What happens to the quality factor when resonant frequency increases?

a) Increases

b) Decreases

c) Remains the same

d) Becomes zero

Answer: a

Explanation: Q=f res /Bandwidth

Quality factor is directly proportional to resonant frequency. So, if resonant frequency increases quality factor increases.

5. Resonance frequency occurs when __________________

a) X L =X C

b) X L >X C

c) X L <X C

d) Cannot be determined

Answer: a

Explanation: The frequency of a system is said to be resonating when the value of the capacitive reactance and the inductive reactance is the same.

6. The current leads the supply voltage in a series RLC circuit has its frequency _________ the resonant frequency.

a) Above

b) Below

c) Equal to

d) Cannot be determined

Answer: b

Explanation: Current is leading the voltage indicates capacitor dominating circuit. X C >X L => 1/ > ωL => ω<1/√LC

So, frequency less than resonant frequency.

7. What is the power factor of a series RLC circuit under resonance condition?

a) 0

b) 1

c) Infinity

d) 100

Answer: b

Explanation: The power factor for a series RLC circuit in resonance condition is always unity because the current is in phase with the voltage under resonance condition.

Φ=0 0 => cos ϕ = 1 i.e. power factor = 1.

8. The current lags the supply voltage in a series RLC circuit has its frequency _________ the resonant frequency.

a) Above

b) Below

c) Equal to

d) Cannot be determined

Answer: a

Explanation: Current is lagging the voltage indicates inductor dominating circuit. X C < X L => 1/ < ωL => ω > 1/√LC

So, frequency more than resonant frequency.

9. What is the correct formula for quality factor?

a) Q=BW*fr

b) Q=BW/fr

c) Q=fr/BW

d) Q=fr2

Answer: c

Explanation: The correct formula for quality factor is Q=fr/BW, where fr is the resonant frequency, BW is the bandwidth frequency and Q is the quality factor.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Quality Factor”.

1. Quality factor is also known as _________

a) Voltage magnification

b) Current magnification

c) Resistance magnification

d) Impedance magnification

Answer: a

Explanation: Quality factor is also known as voltage magnification because the voltage across the capacitor or inductor in resonance condition is equal to Q times the source voltage.

2. At resonance condition, the voltage across the capacitor and inductor is _________ the source voltage.

a) Greater than

b) Less than

c) Equal to

d) Much less than

Answer: a

Explanation: In resonance condition, the voltage across the capacitor and inductor is greater than the source voltage because the voltage across the capacitor or inductor in resonance condition is equal to Q times the source voltage.

3. What is the voltage across the capacitor when the source voltage is 100V and the Q factor is 10?

a) 100V

b) 10V

c) 1000V

d) 0V

Answer: c

Explanation: We know that voltage across the capacitor in resonance condition is equal to Q times the source voltage.

Q=V C /V S where V C is capacitive voltage and V S is source voltage.

10=V C /100

V C =1000 V.

4. Find the Q factor when the voltage across the capacitor is 1000V and the source voltage is 100V.

a) 10

b) 20

c) 30

d) 40

Answer: a

Explanation: We know that voltage across the capacitor in resonance condition is equal to Q times the source voltage.

Q=V C /V S where V C is capacitive voltage and V S is source voltage. Q=1000/100 = 10 V.

5. Find the source voltage when the voltage across the capacitor is 1000V and the Q factor is 10.

a) 10V

b) 200V

c) 100V

d) 90V

Answer: c

Explanation: We know that voltage across the capacitor in resonance condition is equal to Q times the source voltage.

Q=V C /V S where V C is capacitive voltage and V S is source voltage. 10=1000/V S

V S =100 V.

6. What is the voltage across the inductor when the source voltage is 200V and the Q factor is 10?

a) 100V

b) 20V

c) 2000V

d) 0V

Answer: c

Explanation: We know that voltage across the capacitor in resonance condition is equal to Q times the source voltage.

Q=V L /V S where V L is inductive voltage and V S is source voltage. 10=V L /200 => V L = 2000 V.

7. Find the Q factor when the voltage across the inductor is 2000V and the source voltage is 100V.

a) 10

b) 20

c) 30

d) 40

Answer: b

Explanation: We know that voltage across the capacitor in resonance condition is equal to Q times the source voltage.

Q=V L /V S where V L is inductive voltage and V S is source voltage. Q=2000/100=20.

8. Find the source voltage when the voltage across the inductor is 2000V and the Q factor is 20.

a) 10V

b) 200V

c) 100V

d) 90V

Answer: c

Explanation: We know that voltage across the capacitor in resonance condition is equal to Q times the source voltage.

Q=V L /V S where V L is inductive voltage and V S is source voltage.

20=2000/V S

V S =100 V.

9. What happens to the voltage across the capacitor when the Q factor increases?

a) Increases

b) Decreases

c) Remains the same

d) Becomes zero

Answer: a

Explanation: We know that voltage across the capacitor in resonance condition is equal to Q times the source voltage. Hence as the Q factor increases, the voltage across the capacitor also increases.

10. What happens to the voltage across the inductor when the Q factor decreases?

a) Increases

b) Decreases

c) Remains the same

d) Becomes zero

Answer: b

Explanation: We know that voltage across the inductor in resonance condition is equal to Q times the source voltage. Hence as the Q factor decreases, the voltage across the inductor also decreases.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Oscillation of Energy at Resonance”.

1. The energy stored in the capacitor is of _________ nature.

a) Electrostatic

b) Magnetic

c) Neither electrostatic nor magnetic

d) Either electrostatic or magnetic

Answer: a

Explanation: Since capacitor stores charge in between the plates and energy associated with static charge is of electrostatic nature, so we can say energy stored in the capacitor is of electrostatic nature.

2. The energy stored in the inductor is of _________ nature.

a) Electrostatic

b) Magnetic

c) Neither electrostatic nor magnetic

d) Either electrostatic or magnetic

Answer: b

Explanation: Since inductor stores current which involves moving charge and energy associated with moving charge is of magnetic nature so we can say energy stored in the inductor is of magnetic nature.

3. At resonance, the circuit appears __________

a) Inductive

b) Capacitive

c) Either inductive or capacitive

d) Resistive

Answer: d

Explanation: At resonance, the circuit appears resistive because the capacitive and inductive energies are equal to each other.

4. At resonance, the capacitive energy is ___________ inductive energy.

a) Greater than

b) Less than

c) Equal to

d) Depends on the circuit

Answer: c

Explanation: At resonance, energy stored in the capacitor is equal to energy stored in the inductor because capacitive reactance and inductive reactance are equal at resonance. So, at resonance, capacitive energy is equal to inductive energy.

5. At resonance, electrostatic energy is ___________ the magnetic energy.

a) Greater than

b) Less than

c) Equal to

d) Depends on the circuit

Answer: c

Explanation: At resonance, energy stored in the capacitor is equal to energy stored in the inductor because capacitive reactance and inductive reactance are equal at resonance. The capacitor stores electrostatic energy and the inductor stores magnetic energy hence they are equal.

6. The maximum magnetic energy stored in an inductor at any instance is?

a) E=LI m 2 /2

b) E=LI m /2

c) E=LI m 2

d) E=LI m 2 *2

Answer: a

Explanation: At any instant, the magnetic energy stored in an inductor is E=LI m 2 /2, where I m is the maximum current and L is the value of the inductor.

7. The maximum electrostatic energy stored in a capacitor at any instance is?

a) CV m 2

b) 1/2*CV m 2

c) CV m

d) CV m /2

Answer: b

Explanation: The maximum electrostatic energy stored in a capacitor at any instance is 1/2*CV m 2 , where C is the capacitance value and V m is the peak voltage.

8. Q is the ratio of?

a) Active power to reactive power

b) Reactive power to active power

c) Reactive power to average power

d) Reactive power to capacitive power

Answer: c

Explanation: Q is the ratio of the reactive power to the average power. The reactive power is due to the inductance or capacitance and the average power is due to the resistance.

9. Find the value of Q if the reactive power is 10W and the average power is 5W.

a) 10

b) 5

c) 2

d) 1

Answer: c

Explanation: Q is the ratio of the reactive power to the average power.

Q = Reactive power / Average power = 10/5 = 2.

10. Find the reactive power when the average power is 5W and Q=2.

a) 10W

b) 5W

c) 2W

d) 1W

Answer: a

Explanation: Q is the ratio of the reactive power to the average power.

Q = Reactive power / Average power

2 = Reactive power / 5

Reactive Power = 2*5 = 10W.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Bandwidth”.

1. The SI unit for bandwidth is?

a) Hz

b) Watt

c) kHz

d) kW

Answer: a

Explanation: The SI unit for bandwidth is Hz. Hertz is the SI unit because bandwidth is basically frequency and the unit for frequency is Hz.

2. At bandwidth frequency range, the value of the current I is?

a) I=Im/2

b) I=Im 2

c) I=Im

d) I=Im/√2

Answer: d

Explanation: At the bandwidth frequency range, the value of the current is equal to the maximum value of current divided by √2.

3. At bandwidth frequency range, the value of the voltage V is?

a) V=Vm/2

b) V=Vm 2

c) V=Vm

d) V=Vm/√2

Answer: d

Explanation: At the bandwidth frequency range, the value of the voltage is equal to the maximum value of voltage divided by √2.

4. At resonance, bandwidth includes the frequency range that allows _____ percent of the maximum current to flow.

a) 33.33

b) 66.67

c) 50

d) 70.7

Answer: d

Explanation: At resonance, bandwidth includes the frequency range that allows 70.2 percent of the maximum current to flow. This is because in the bandwidth frequency range, the value of the current is equal to the maximum value of current divided by √2.

5. At resonance, bandwidth includes the frequency range that allows _____ percent of the maximum voltage to flow.

a) 33.33

b) 66.67

c) 50

d) 70.7

Answer: d

Explanation: At resonance, bandwidth includes the frequency range that allows 70.2 percent of the maximum voltage to flow. This is because in the bandwidth frequency range, the value of the voltage is equal to the maximum value of voltage divided by √2.

6. Find the value of current in the bandwidth range when the maximum value of current is 50A.

a) 56.65A

b) 35.36A

c) 45.34A

d) 78.76A

Answer: b

Explanation: At the bandwidth frequency range, the value of the current is equal to the maximum value of current divided by √2. Hence I =50/√2= 35.36A.

7. Find the value of voltage in the bandwidth range when the maximum value of voltage is 100 V.

a) 56.65 V

b) 35.36 V

c) 45.34 V

d) 70.72 V

Answer: d

Explanation: At the bandwidth frequency range, the value of the voltage is equal to the maximum value of voltage divided by √2. Hence V =100/√2= 70.72V.

8. Bandwidth is the difference of_____________________ frequencies.

a) half power

b) full power

c) double power

d) wattless

Answer: a

Explanation: Current for the end frequencies of bandwidth is 1/√2 times the maximum current. So, power at the end frequencies of bandwidth is half the maximum power. So, bandwidth is the difference of half power frequencies.

9. For a sharp resonance, bandwidth is ______________

a) low

b) high

c) zero

d) infinity

Answer: a

Explanation: For sharp resonance quality factor is high and the quality factor is inversely proportional to bandwidth so bandwidth is low for sharp resonance.

10. Current is maximum at __________ frequency of bandwidth.

a) left end

b) middle

c) right end

d) all end

Answer: b

Explanation: Current will be maximum at a frequency which is at the middle of bandwidth.

On both sides, it decreases and is 1/√2 times the maximum current at the ends of bandwidth.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Selectivity”.

1. Shape of the resonance curve depends upon the?

a) Q-factor

b) Voltage

c) Current

d) Either voltage or current

Answer: a

Explanation: The shape of the resonance curve depends on the Q factor because of the equation:

Q=Resonance frequency / Bandwidth. Sharp resonance means high quality factor.

2. A circuit is said to be selective if it has a _____ peak and ____ bandwidth.

a) Blunt, narrow

b) Sharp, narrow

c) Sharp, broad

d) Blunt, broad

Answer: b

Explanation: For a circuit to be selective, it should have high quality factor. And we know that for high quality factor, resonance frequency should be high and bandwidth should be narrow.

3. What is the Q factor of a selective circuit?

a) Very low

b) Very high

c) Zero

d) Infinity

Answer: b

Explanation: For a circuit to be selective, it should have high quality factor. It should have a sharp peak with narrow bandwidth.

4. In selective circuits, higher the Q factor _________ the peak.

a) Sharper

b) Blunter

c) Neither sharper nor blunter

d) Either sharper or blunter

Answer: a

Explanation: Q=Resonance frequency / Bandwidth.

Higher the quality factor, sharper the peak of resonance curve.

5. Q is a measure of _________

a) Resonance

b) Bandwidth

c) Selectivity

d) Either resonance or bandwidth

Answer: c

Explanation: For a circuit to be selective, it should have a high quality factor. It should have a sharp peak with narrow bandwidth.

6. In selective circuits, the resonant frequency lies in the ________ of the bandwidth frequency range.

a) Beginning

b) End

c) Midpoint

d) Cannot be determined

Answer: c

Explanation: In selective circuits, the resonant frequency lies in the midpoint of the bandwidth frequency range.

7. In order for high selectivity, the resistance must be?

a) Small

b) Large

c) Negative

d) Positive

Answer: a

Explanation: For high selectivity, the Q factor should be large and for Q factor to be large, the resistance would be small because Q is inversely proportional to the resistance.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Voltages in a Series RLC Circuit”.

1. In a series RLC circuit, the phase difference between the voltage across the capacitor and the voltage across the resistor is?

a) 0 0

b) 90 0

c) 180 0

d) 360 0

Answer: b

Explanation: In a series RLC circuit, voltage across capacitor lag the current by 90 0 and voltage across resistor is in phase with current so, the phase difference between the voltage across the capacitor and the voltage across the resistor is 90 0 .

2. In a series RLC circuit, the phase difference between the voltage across the inductor and the voltage across the resistor is?

a) 0 0

b) 90 0

c) 180 0

d) 360 0

Answer: b

Explanation: In a series RLC circuit, voltage across inductor lead the current by 90 0 and voltage across resistor is in phase with current so, the phase difference between the voltage across the inductor and the voltage across the resistor is 90 0 .

3. In a series RLC circuit, the phase difference between the voltage across the capacitor and the voltage across the inductor is?

a) 0 0

b) 90 0

c) 180 0

d) 360 0

Answer: c

Explanation: In a series RLC circuit, voltage across inductor lead the current by 90 0 and voltage across capacitor lag the current by 900 so, the phase difference between the voltage across the inductor and the voltage across the capacitor is 180 0 .

4. In a series RLC circuit, the phase difference between the voltage across the resistor and the current in the circuit is?

a) 0 0

b) 90 0

c) 180 0

d) 360 0

Answer: a

Explanation: In a series RLC circuit, the phase difference between the voltage across the resistor and the current in the circuit is 0 degrees because they are in phase.

5. In a series RLC circuit, the phase difference between the voltage across the capacitor and the current in the circuit is?

a) 0 0

b) 90 0

c) 180 0

d) 360 0

Answer: b

Explanation: In a series RLC circuit, voltage across capacitor lag the current by 90 0 so, the phase difference between the voltage across the capacitor and current is 90 0 .

6. In a series RLC circuit, the phase difference between the voltage across the inductor and the current in the circuit is?

a) 0 0

b) 90 0

c) 180 0

d) 360 0

Answer: b

Explanation: In a series RLC circuit, voltage across inductor lead the current by 90 0 so, the phase difference between the voltage across the inductor and the current is 90 0 .

7. The current in the inductor lags the voltage in a series RLC circuit ___________ resonant frequency.

a) Above

b) Below

c) Equal to

d) Depends on the circuit

Answer: a

Explanation: The current in the inductor lags the voltage in a series RLC circuit if circuit is inductive dominant i.e. if X L > X C ωL > 1/ωC => ω > 1/√LC => ω > ω 0 . So, the current in the inductor lags the voltage in a series RLC circuit above the resonant frequency.

8. The current in the capacitor leads the voltage in a series RLC circuit ___________ resonant frequency.

a) Above

b) Below

c) Equal to

d) Depends on the circuit

Answer: b

Explanation: The current in the capacitor leads the voltage in a series RLC circuit if circuit is capacitive dominant i.e. if X L < X C

ωL < 1/ωC => ω < 1/√LC => ω < ω 0 . So, the current in the capacitor leads the voltage in a series RLC circuit below the resonant frequency.

9. The current in the inductor ___________ the voltage in a series RLC circuit above the resonant frequency.

a) Leads

b) Lags

c) Equal to

d) Depends on the circuit

Answer: b

Explanation: ω > ω 0 => ω > 1/√LC

=> ωL > 1/ωC => X L > X C

The circuit is inductive dominant so, the current in the inductor lags the voltage in a series RLC circuit above the resonant frequency.

10. The current in the capacitor ___________ the voltage in a series RLC circuit below the resonant frequency.

a) Leads

b) Lags

c) Equal to

d) Depends on the circuit

Answer: a

Explanation: ω < ω 0 => ω < 1/√LC

=> ωL < 1/ωC => X L < X C The circuit is capacitive dominant so, the current in the capacitor leads the voltage in a series RLC circuit above the resonant frequency.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “The Current in a Series RLC Circuit”.

1. In a series RLC circuit, the phase difference between the current in the capacitor and the current in the resistor is?

a) 0 0

b) 90 0

c) 180 0

d) 360 0

Answer: a

Explanation: In a series RLC circuit, the phase difference between the current in the capacitor and the current in the resistor is 0 0 because same current flows in the capacitor as well as the resistor.

2. In a series RLC circuit, the phase difference between the current in the inductor and the current in the resistor is?

a) 0 0

b) 90 0

c) 180 0

d) 360 0

Answer: a

Explanation: In a series RLC circuit, the phase difference between the current in the inductor and the current in the resistor is 0 0 because same current flows in the inductor as well as the resistor.

3. In a series RLC circuit, the phase difference between the current in the capacitor and the current in the inductor is?

a) 0 0

b) 90 0

c) 180 0

d) 360 0

Answer: a

Explanation: In a series RLC circuit, the phase difference between the current in the inductor and the current in the capacitor is 0 0 because same current flows in the inductor as well as the capacitor.

4. In a series RLC circuit, the phase difference between the current in the circuit and the voltage across the resistor is?

a) 0 0

b) 90 0

c) 180 0

d) 360 0

Answer: a

Explanation: In a series RLC circuit, the phase difference between the voltage across the resistor and the current in the circuit is 0 0 because they are in phase.

5. In a series RLC circuit, the phase difference between the current in the circuit and the voltage across the capacitor is?

a) 0 0

b) 90 0

c) 180 0

d) 360 0

Answer: b

Explanation: In a series RLC circuit, voltage across capacitor lags the current in the circuit by 90 0 so, the phase difference between the voltage across the capacitor and the current in the circuit is 90 0 .

6. _________ the resonant frequency, the current in the inductor lags the voltage in a series RLC circuit.

a) Above

b) Below

c) Equal to

d) Depends on the circuit

Answer: a

Explanation: The current in the inductor lags the voltage in a series RLC circuit if a circuit is inductive dominant i.e. if X L > X C

ω L > 1/ω C => ω > 1/√LC => ω > ω 0 .

So, the current in the inductor lags the voltage in a series RLC circuit above the resonant frequency.

7. _________ the resonant frequency, the current in the capacitor leads the voltage in a series RLC circuit.

a) Above

b) Below

c) Equal to

d) Depends on the circuit

Answer: b

Explanation: The current in the capacitor leads the voltage in a series RLC circuit if circuit is capacitive dominant i.e.i.e. if X L < X C

ω L < 1/ω C => ω < 1/√LC => ω < ω 0 .

So, the current in the capacitor leads the voltage in a series RLC circuit below the resonant frequency.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Basic AC Parallel Circuits”.

1. In a parallel circuit, we consider _____________ instead of impedance.

a) Resistance

b) Capacitance

c) Inductance

d) Admittance

Answer: d

Explanation: In a parallel circuit, we consider admittance instead of impedance, where admittance is the reciprocal of impedance.

2. In a parallel circuit, we consider admittance instead of _________

a) Resistance

b) Capacitance

c) Inductance

d) Impedance

Answer: d

Explanation: In a parallel circuit, we consider admittance instead of impedance, where admittance is the reciprocal of impedance.

3. Which, among the following is the correct expression for impedance?

a) Z=Y

b) Z=1/Y

c) Z=Y 2

d) Z=1/Y 2

Answer: b

Explanation: We know that impedance is the reciprocal of admittance, hence the correct expression for impedance is: Z=1/Y.

4. Which, among the following is the correct expression for admittance?

a) Y=Z

b) Y=1/Z

c) Y=Z 2

d) Y=1/Z 2

Answer: b

Explanation: We know that admittance is the reciprocal of impedance, hence the correct expression for admittance is: Y=1/Z.

5. What is the unit of admittance?

a) ohm

b) henry

c) farad

d) ohm -1

Answer: d

Explanation: The unit for admittance is ohm -1 because the unit of impedance is ohm and admittance is the reciprocal of impedance.

6. As the impedance increases, the admittance ____________

a) Increases

b) Decreases

c) Remains the same

d) Becomes zero

Answer: b

Explanation: As the impedance increases, the admittance decreases because admittance is equal to 1/impedance.

7. if the impedance of a system is 4 ohm, calculate its admittance.

a) 0.25 ohm -1

b) 4 ohm -1

c) 25 ohm -1

d) 0.4 ohm -1

Answer: a

Explanation: We know that: Y=1/Z.

Substituting the value of Z from the question, we get Y = 1/4 = 0.25 => Y= 0.25 ohm -1 .

8. The admittance of a system is 10 ohm -1 , calculate its impedance.

a) 10 ohm

b) 0.1 ohm

c) 1 ohm

d) 1.1 ohm

Answer: b

Explanation: We know that: Z=1/Y.

Z = 1/10 = 0.1 => Z = 0.1 ohm.

9. In A parallel circuit, with any number of impedances, The voltage across each impedance is?

a) equal

b) divided equally

c) divided proportionaly

d) zero

Answer: a

Explanation: In parallel circuits, the current across the circuits vary whereas the voltage remains the same. So, voltage across each impedance is equal in parallel circuit.

10. In a parallel circuit, current in each impedance is_____________

a) equal

b) different

c) zero

d) infinite

Answer: b

Explanation: In parallel circuits, the current across the circuits vary whereas the voltage remains the same. So, current in each impedance is different.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Simple Parallel Circuits”.

1. From the given circuit, find the value of I R .

basic-electrical-engineering-questions-answers-simple-parallel-circuits-q1

a) 0

b) V/I

c) V/R

d) Cannot be determined

Answer: c

Explanation: In the given circuit, the voltage across the resistor is the same as the source voltage as they are connected in parallel. The current in the resistor is I R hence I R =V/R.

2. What is the relation between I R and V in the following circuit?

basic-electrical-engineering-questions-answers-simple-parallel-circuits-q1

a) I R leads V

b) I R lags V

c) I R and V are in phase

d) No relation

Answer: c

Explanation: In the following circuit I R and V are in phase because I R is the current in the resistor and the current in the resistor is always in phase with the voltage across it.

3. What is the expression for the current in the inductor from the following circuit?

basic-electrical-engineering-questions-answers-simple-parallel-circuits-q3

a) V/I

b) V/X L

c) 0

d) Cannot be determined

Answer: b

Explanation: In the given circuit, the voltage across the inductor is the same as the source voltage as they are connected in parallel. The current in the inductor is I L hence I L =V/X L .

4. What is the phase relation between I L and V from the following circuit?

basic-electrical-engineering-questions-answers-simple-parallel-circuits-q3

a) I L lags V

b) I L leads V

c) I L and V are in phase

d) No relation

Answer: a

Explanation: IL is the current across the inductor and we know that the current across the inductor always lags the voltage across it. Hence IL lags V.

5. Find the expression for the current I from the given circuit.

basic-electrical-engineering-questions-answers-simple-parallel-circuits-q1

a) I=I C

b) I=I R

c) I=I C +I R

d) I=0

Answer: c

Explanation: I is the total current in the circuit. Since this is a parallel connection, the total current in the circuit is equal to the sum of the currents in each branch of the circuit. Hence I=I C +I R .

6. Find the total current if I C =2A and I R =5A.

basic-electrical-engineering-questions-answers-simple-parallel-circuits-q1

a) 3A

b) -3A

c) 7A

d) 10A

Answer: c

Explanation: I is the total current in the circuit. Since this is a parallel connection, the total current in the circuit is equal to the sum of the currents in each branch of the circuit. Hence I=I C +I R .

I=2+5=7A.

7. Find the value of I R if I=10A and I C =8A.

basic-electrical-engineering-questions-answers-simple-parallel-circuits-q1

a) 5A

b) 18A

c) 12A

d) 2A

Answer: d

Explanation: I is the total current in the circuit. Since this is a parallel connection, the total current in the circuit is equal to the sum of the currents in each branch of the circuit. Hence I=I C +I R .

10=8+I R => I R =2A.

8. Find the value of IL if I C =10A and I R =6A.

basic-electrical-engineering-questions-answers-simple-parallel-circuits-q1

a) 4A

b) 18A

c) 12A

d) 2A

Answer: a

Explanation: I is the total current in the circuit. Since this is a parallel connection, the total current in the circuit is equal to the sum of the currents in each branch of the circuit. Hence I=I C +I R .

10=I C +6 => I C =4A.

9. What is the expression for the current in the capacitor from the following circuit?

basic-electrical-engineering-questions-answers-simple-parallel-circuits-q1

a) V/C

b) V/I

c) 0

d) V/X C

Answer: d

Explanation: In the given circuit, the voltage across the capacitor is the same as the source voltage as they are connected in parallel. The current in the capacitor is I C hence I C =V/X C .

10. What is the phase relation between I C and V from the following circuit?

basic-electrical-engineering-questions-answers-simple-parallel-circuits-q1

a) I C lags V

b) I C leads V

c) I C and V are in phase

d) No relation

Answer: b

Explanation: I C is the current across the capacitor and we know that the current across the capacitor always leads the voltage across it. Hence I C leads V.

This set of Basic Electrical Engineering Multiple Choice Questions & Answers  focuses on “Parallel Impedance Circuits”.

1. In an impedance parallel network, the reactive component will ____________ the voltage by 90 degrees.

a) Lead

b) Lag

c) Either lead or lag

d) Depends on the circuit

Answer: c

Explanation: In an impedance parallel network the reactive component will either lead or lag the voltage by 90 degrees.

2. In an impedance parallel network, the reactive component will either lead or lag the voltage by _________ degrees.

a) 0

b) 90

c) 45

d) 180

Answer: b

Explanation: In an impedance parallel network the reactive component will either lead or lag the voltage by 90 degrees.

3. In an impedance parallel network, the reactive component will either lead or lag the ________ by 90 degrees.

a) Voltage

b) Current

c) Either voltage or current

d) Cannot be determined

Answer: a

Explanation: In an impedance parallel network the reactive component will either lead or lag the voltage by 90 degrees.

4. The reactive component in an impedance parallel circuit leads the voltage when the current _________ the voltage.

a) Leads

b) Lags

c) Either leads or lags

d) Cannot be determined

Answer: a

Explanation: The reactive component in an impedance parallel circuit leads the voltage when the current leads the voltage.

5. The active component in an impedance parallel circuit will __________ the voltage.

a) Leads

b) Lags

c) Be in phase with

d) Either leads or lags

Answer: c

Explanation: The active component in an impedance parallel network will always be in phase with the voltage in the circuit.

6. The phase difference between the active component of an impedance parallel circuit and the voltage in the network is __________

a) 0

b) 90

c) 180

d) 360

Answer: a

Explanation: The active component in an impedance parallel network will always be in phase with the voltage in the circuit. Hence the phase difference is 0.

7. The quadrature component is also known as?

a) Active component

b) Reactive component

c) Either active or reactive component

d) Neither active nor reactive component

Answer: b

Explanation: The quadrature component is also known as the reactive component because the reactive component forms a quadrature with the voltage.

8. Find the expression for the current I from the given circuit.

basic-electrical-engineering-questions-answers-parallel-impedance-circuits-q8

a) I=I L

b) I=I R

c) I=I L +I R

d) I=0

Answer: c

Explanation: I is the total current in the circuit. Since this is a parallel connection, the total current in the circuit is equal to the sum of the currents in each branch of the circuit. Hence I=I R +I L .

9. Find the value of I R if I=10A and I L =8A.

basic-electrical-engineering-questions-answers-parallel-impedance-circuits-q8

a) 5A

b) 18A

c) 12A

d) 2A

Answer: d

Explanation: We know that I=I R +I L .

10=I R +8 => I R =2A.

10. Find the total current if I L =2A and I R =8A.

basic-electrical-engineering-questions-answers-parallel-impedance-circuits-q8

a) 3A

b) -3A

c) 7A

d) 10A

Answer: d

Explanation: We know that I=I R +I L .

I=8+2=10A.