Kirchhoff’s Voltage Law MCQ [Free PDF] – Objective Question Answer for Kirchhoff’s Voltage Law Quiz

11. Kirchhoff’s voltage law is based on the principle of conservation of _________

A. energy
B. momentum
C. mass
D. charge

Answer: A

KVL is based on the law of conservation of energy. Kirchhoff’s 2nd law states that the sum of the potential drop across all the components in a loop must be zero. Thus Kirchhoff’s 2nd law is based on the conservation of energy.

 

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

A. KCL
B. Ohm’s law
C. KVL
D. None of the mentioned

Answer: C

Kirchhoff’s voltage law is based on the principle of conservation of energy. This requires that the total work done in taking a unit positive charge around a closed path and ending up at the original point is zero. KVL can be best suited for circuits with more number of loops.

 

13. Mathematically, Kirchhoff’s Voltage law can be ________

A. ∑_(k=0)n(V) = 0
B. V2∑_(k=0)n(V) = 0
C. V∑_(k=0)n(V) = 0
D. None of the mentioned

Answer: A

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

Kirchhoff’s voltage law is based on the principle of conservation of energy. This requires that the total work done in taking a unit positive charge around a closed path and ending up at the original point is zero.

 

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

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

A. 10V
B. 12V
C. 14V
D. 16V

Answer: B. 12 V

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 s

ince they are connected in parallel.

v = 12V.

 

15. In the circuit shown in the figure, find the current flowing through the 8 Ω resistance.

In the circuit shown in the figure, find the current flowing through the 8 Ω resistance.

  1. 0.25 A
  2. 0.50 A
  3. 0.75 A
  4. 0.10 A

Answer.1. 0.25A

Let voltage across the 8 Ω resistance is ‘V’ volt.

∴ Current across the 8 Ω is given by

I = V/8

Now by applying KCL at the node  we get

\({{V – 5} \over 2}+{{V +3} \over 4}+{{V } \over 8}=0\)

4V – 20 + 2V + 6 + V = 0

V = 14/7

Now current flowing through the 8 Ω resistance is

I = 2/8

I = 0.25 A

 

16. Calculate the current A by using Kirchhoff’s current law

Calculate the current A by using Kirchhoff's current law

A. 5A
B. 10A
C. 15A
D. 20A

Answer: C

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.

 

17. In the figure shown, the current 𝑖 (in ampere) is __________

In the figure shown, the current 𝑖 (in ampere) is

  1. -1 Amp
  2. 5 Amp
  3. 2 Amp
  4. -2 Amp

Answer.1. -1 Amp

Apply KCL at node V1, we get:

In the figure shown, the current 𝑖 (in ampere) is

\(\frac{{{{\rm{V}}_1} – 0}}{1} + \frac{{{{\rm{V}}_1} – 8}}{1} + \frac{{{{\rm{V}}_1} – 0}}{1} + \frac{{{{\rm{V}}_1} – 8}}{1} = 0\)

4V1 – 16 = 0

V1 = 4 V

Again, applying KCL, we can write:

\({\rm{i}} + \frac{{\left( {0 – {{\rm{V}}_1}} \right)}}{1} + 5 = 0 \)

i = V1 − 5 = 4 − 5 = −1 Amp

 

18. By using Kirchhoff’s current law calculate the current across the 20-ohm resistor.

By using Kirchhoff's current law calculate the current across the 20-ohm resistor.

A. 20A
B. 1A
C. 0.67A
D. 0.33A

Answer: D

Assume a lower terminal of 20 ohms 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 = (20/3)/20

= 1/3 = 0.33V.

 

19. The total charge q(t), in the coulombs, that enters the terminal of an element is:

\(q(t) = \left\{ {\begin{array}{*{20}{c}} {0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,t < 0}\\ {2t\,\,\,\,\,\,\,\,\,\,\,\,0 \le t \le 2}\\ {3 + {e^{ – 2(t – 2)}}\,\,t > 2} \end{array}} \right.\)

Determine the current at t = 5 s.

  1. 0 A
  2. 2 A
  3. -2e-6 A
  4. 3 + e-6 A

Answer.3.

Electric current, i = Rate of transfer of electric charge.

i(t) = dQ/dt

Calculation:

t = 5 s so, equation 3rd is consider.

\(i = \frac{{dQ}}{{dt}} = \frac{d}{{dt}}\left( {3 + {e^{ – 2\left( {t – 2} \right)}}} \right)\)

 

\(i = {e^{ – 2\left( {t – 2} \right)}}\frac{d}{{dt}}\left[ { – 2\left( {t – 2} \right)} \right]\)

 

\(i = {e^{ – 2\left( {t – 2} \right)}}\left( { – 2} \right)\)

 

\(i = – 2{e^{ – 2\left( {t – 2} \right)}}\)

Put the value of t = 5, then we get,

i = −2e−6A

 

20. Calculate the value of I3, if I1 = 2A and I2 = 3A by applying Kirchhoff’s current law

Calculate the value of I3, if I1= 2A and I2=3A by applying Kirchhoff's current law

A. -5A
B. 5A
C. 1A
D. -1A

Answer: A

According to KCl, I1+I2+I3 = 0.

Hence I3 = -(I1+I2) = -5A.

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