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

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

a) energy

b) momentum

c) mass

d) charge

View Answer

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

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

a) KCL

b) Ohm’s law

c) KVL

d) None of the mentioned

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

3. Determine the unknown voltage drop in the circuit below

a) 11V

b) 10V

c) 19V

d) 5V

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Explanation: By applying KVL to the loop, we get,

2 + 1 + V + 3 + 5 – 30 = 0.

4. Determine V in the circuit

a) 28.8V

b) 34.4v

c) -28.8V

d) 28V

View Answer

Explanation: First, apply KCL to the second node i.e, i6 + I = i8 . Next, apply KVL to the first loop i.e, -20 + V2 + V5 =0. But V2 = 7*2=14V. Obtain V5 and I5=V5/5. Again apply KVL to central loop i.e, -V5 + V6 + V = 0.

Apply KCL to the first node again, seek the value of V6 and finally we get V.

5. Find V and I in the circuit

a) -39V , -4.875A

b) 39V , -4.875A

c) -39v , 4.875a

d) 39V , 4.875A

View Answer

Explanation: By applying KVL to the loop, we get,

+30+9-V = 0. then, I=V/8.

6. Mathematically, Kirchhoff’s Voltage law can be as

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

View Answer

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

7. Determine the value of V and the power supplied by the independent current source

a) 20V , 300mw

b) 27V , 498mW

c) 26.6v , 532mW

d) 25V , 322mW

View Answer

Explanation: Apply KCL to the node, we get i – 2i1 – 0.02 – i1 = 0. Next, apply ohm’s law to each resistor, i=v/4000 and i1= -v/6000 and substitute in the above equation. For the power supplied by the independent source P = V*0.02A.

8. Determine V in the circuit

a) -11.6V

b) 23.2V

c) -23.2V

d) 11.6V

View Answer

Explanation: First, apply KCL to the second node i.e, i8 + I = i10 . Next, apply KVL to the first loop i.e, -20 + V i

_{1}+ V6 =0. But V4 = 4*4=16V. Obtain V6 and I6=V6/6. Again apply KVL to central loop i.e, -V6 + V8 + V = 0.

Apply KCL to the first node again, seek the value of V8 and finally we get V.

9. Find V and I in the circuit

a) 19V, 0.0633A

b) -19V, 0.0633A

c) 19V, -0.0633A

d) -19V,- 0.0633A

View Answer

Explanation: By applying KVL to the loop, we get,

-12-7+V = 0. then, I=V/300.

**Sanfoundry Global Education & Learning Series – Network Theory.**

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