This set of Network Theory Multiple Choice Questions & Answers (MCQs) focuses on “Supermesh Analysis”.

1.Consider the circuit shown below. Find the current I_{1} (A).

a) 1

b) 1.33

c) 1.66

d) 2

View Answer

Explanation: Applying Super mesh analysis, the equations will be I

_{2}-I

_{1}=2 -10+2I

_{1}+I

_{2}+4=0. On solving the above equations, I

_{1}=1.33A.

2. Find the current I_{2} (A) in the circuit shown in the question 1.

a) 1.33

b) 2.33

c) 3.33

d) 4.33

View Answer

Explanation: Applying Super mesh analysis, the equations will be I

_{2}-I

_{1}=2

-10+2I

_{1}+I

_{2}+4=0. On solving the above equations, I

_{2}=3.33A.

3. Consider the circuit shown in the figure. Find the current I_{1} (A).

a) -1

b) -2

c) -3

d) -4

View Answer

Explanation: Applying Super mesh analysis, the equations will be I

_{1}+I

_{1}+10+I

_{2}+I

_{2}=0. I

_{1}+I

_{2}=-5. I

_{2}-I

_{1}=1. On solving, I

_{1}=-3A.

4. Find the current I_{2} (A) in the figure shown in the question 3.

a) -2

b) -1

c) 2

d) 1

View Answer

Explanation: Applying Super mesh analysis, the equations will be I

_{1}+I

_{1}+10+I

_{2}+I

_{2}=0. I

_{1}+I

_{2}=-5. I

_{2}-I

_{1}=1. On solving, I

_{2}=-2A.

5. Find the power (W) supplied by the voltage source in the figure shown in the question 3.

a) 0

b) 1

c) 2

d) 3

View Answer

Explanation: I

_{3}-I

_{2}=2. As I

_{2}=-2A, I

_{3}=0A. Th term power is the product of voltage and current. So, power supplied by source= 10×0=0W.

6. Find the current i_{1} in the circuit shown below.

a) 8

b) 9

c) 10

d) 11

View Answer

Explanation: The current in the first loop is equal to 10A. So the current i

_{1}in the circuit is i

_{1}= 10A.

7. Find the current i_{2} in the circuit shown in the question 6.

a) 6.27

b) 7.27

c) 8.27

d) 9.27

View Answer

Explanation: For 2nd loop, 10 + 2(i

_{2}-i

_{3}) + 3(i

_{2}-i

_{1}) =0. For 3rd loop, i

_{3}+ 2(i

_{3}-i

_{2})=10. As i

_{1}=10A, On solving above equations, we get i

_{2}=7.27A.

8. Find the current i_{3} in the circuit shown in the question 6.

a) 8.18

b) 9.18

c) 10.18

d) 8.8

View Answer

Explanation: For 2nd loop, 10 +2(i

_{2}-i

_{3}) +3(i

_{2}-i

_{1}) =0. For 3rd loop, i

_{3}+2(i

_{3}-i

_{2})=10. As i

_{1}=10A, On solving above equations, we get i

_{3}=8.18A.

9. Find the current I_{1} in the circuit shown below.

a) 8

b) -8

c) 9

d) -9

View Answer

Explanation: Applying Super Mesh analysis, (10+5)I

_{1}– 10(I

_{2}) – 5(I

_{3}) =50. 2(I

_{2}) + I

_{3}+ 5(I

_{3}-I

_{1}) + 10(I

_{2}-I

_{1}) =0. I

_{2}– I

_{3}= 2. On solving above equations, we get I

_{1}=-8A.

10. Find the current I_{2} in the circuit shown in the question 9.

a) 5.3

b) -5.3

c) 7.3

d) -7.3

View Answer

Explanation: Applying Super Mesh analysis, (10+5)I

_{1}-10(I

_{2})-5(I

_{3}) =50. 2(I

_{2}) + I

_{3}+ 5(I

_{3}-I

_{1}) + 10(I

_{2}-I

_{1}) =0. I

_{2}– I

_{3}=2. On solving above equations, we get I

_{2}=-7.3A.

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

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