Kirchhoff’s Voltage Law (KVL) MCQ – Objective Question Answer for Kirchhoff’s Voltage Law Quiz

11. 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

Current = Charge/Time.

Here charge = 50c

time = 5seconds,

so current = 50/5 = 10A.

 

12. The voltages developed across the 3 Ω and 2 Ω resistors shown in the figure are 6 V and 2 V respectively, with the polarity as marked. What is the power (in Watt) across the resistance having a voltage of 5 V?

The voltages developed across the 3 Ω and 2 Ω resistors shown in the figure are 6 V and 2 V respectively, with the polarity as marked. What is the power (in Watt) across the resistance having a voltage of 5 V?

  1. 5
  2. 7
  3. 10
  4. 14

Answer.1. 5 A

Let current through 3 Ω be I1 and through 2 Ω is I2 and current through 5 V is I ampere respectively as shown below

I1 = 6/3 = 2 A

I2 = 2/2 = 1 A

Apply KCL at N/W N2,

I + I2 = I1

I + 1 = 2

I = 1 A

P = 5 × 1 = 5 W

 

13. KCL deals with the conservation of?

A. Momentum
B. Mass
C. Potential Energy
D. Charge

Answer: D

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.

 

14. Find the voltage V0 across a 4 Ω resistor in the following circuit.

Find the voltage V0 across 4 Ω resistor in the following circuit.

  1. 20 V
  2. 12 V
  3. 24 V
  4. 18 V

Answer.3. 24 V

Apply KCL at node ‘a’

i= 0.5i0 + 3

0.5i0 = 3

i0 = 6 A

∴v0 = 4i0 = 4× 6 = 24 V

 

15. KCL is applied at _________

A. Loop
B. Node
C. Both loop and node
D. Neither loop nor node

Answer: B

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.

 

16. Find the value of current I using KCL

Find the value of current I using KCL

  1. 2 A
  2. 3 A
  3. 1 A
  4. 5 A

Answer.3 .1 A

KCL at Node A

I1 + 1A = 3A

I1 = 2 A

KCL at node B

I2 + 2A = 6A

I2 = 4 A

KCL at node C

I3 = 4A + 2A

I3 = 6A

KCL at node D

I = 6A – 5A = 1 A

 

17. 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

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

 

18. Consider the following two statements

1. Kirchhoffs junction law follows from conservation of charge.

2. Kirchhoffs loop law follows the conservative nature of the electric field.

  1. Both 1 and 2 are correct
  2. 2 is correct but 1 is wrong
  3. 1 is correct but 2 are wrong
  4. Both 1 and 2 are wrong

Answer.2

  • Kirchhoff’s junction law states that the current entering the junction is equal to the current leaving the junction, which indirectly means the charge entering the junction is equal charge leaving the junction. From the statement, it is clear that Kirchhoff’s junction law follows from the conservation of charge. Therefore statement 1 is correct.
  • Kirchhoff’s loop law states that the net potential drop in a loop is Zero Σ V = 0, which means potential changes around a closed loop is zero.
  • Voltage is the measure of energy per unit charge in this context KVL can be restated as the total energy gained per unit charge must be equal to the amount of energy lost per unit charge (ie: the Law of conservation of Energy). Therefore statement 2 is correct.

 

19. Find the value of the current I by using KCL?

Find the value of the current I by using KCL?

A. 8A
B. 7A
C. 6A
D. 5A

Answer: A

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.

At the junction, I-2+3-4-5 = 0.

Hence I = 8A.

 

20. If two capacitors having a capacitance of 5 μF and 10 μF respectively are connected in series across a 200 V supply, find the potential difference across each capacitor.

  1. 50 V, 100 V
  2. 133.33 V, 66.66 V
  3. 200 V, 100 V
  4. 100 V, 200 V

Answer.2. 133.33 V, 66.66 V

Given,

C1 = 5 μF
C2 = 10 μF
V = 200 V

From the above concept, the voltage across C1 is given as,

VC1 = V × C2/(C1 + C2)

= 200 × 10 (10 +15) = 133.33 V

And the voltage across C2 is given as,

VC2 = V × C1/(C1 + C2)

= 200 × 5 (10 +15) = 66.66 V

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