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

1. Mesh analysis is applicable for non planar networks also.

a) true

b) false

View Answer

Explanation: Mesh analysis is applicable only for planar networks. A circuit is said to be planar if it can be drawn on a plane surface without crossovers.

2. A mesh is a loop which contains ____ number of loops within it.

a) 1

b) 2

c) 3

d) no loop

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Explanation: A loop is a closed path. A mesh is defined as a loop which does not contain any other loops within it.

3. Consider the circuit shown below. The number mesh equations that can be formed are?

a) 1

b) 2

c) 3

d) 4

View Answer

Explanation: We know if there are n loops in the circuit, n mesh equations can be formed. So as there are 2 loops in the circuit. So 2 mesh equations can be formed.

4. In the figure shown in the question 2, the current through loop 1 be I_{1} and through the loop 2 be I_{2}, then the current flowing through the resistor R_{2} will be?

a) I_{1}

b) I_{2}

c) I_{1}-I_{2}

d) I_{1}+I_{2}

View Answer

Explanation: Through the resistor R

_{2}both the currents I

_{1}, I

_{2}are flowing. So the current through R

_{2}will be I

_{1}-I

_{2}.

5. If there are 5 branches and 4 nodes in graph, then the number of mesh equations that can be formed are?

a) 2

b) 4

c) 6

d) 8

View Answer

Explanation: Number of mesh equations= B-(N-1). Given number of branches = 5 and number of nodes = 4. So Number of mesh equations = 5-(4-1) =2.

6. Consider the circuit shown in the figure. Find voltage V_{x}.

a) 1

b) 1.25

c) 1.5

d) 1.75

View Answer

Explanation: Consider current I

_{1}(CW) in the loop 1 and I

_{2}(ACW) in the loop 2. So, the equations will be V

_{x}+I

_{2}-I

_{1}=0. I

_{1}=5/2=2.5A. I

_{2}=4Vx/4= V

_{x}. V

_{x}+V

_{x}-2.5=0. V

_{x}=1.25V.

7. Consider the circuit shown below. Find the current I_{1}.

a) 3.3

b) 4.3

c) 5.3

d) 6.3

View Answer

Explanation: According to mesh analysis, (1+3+6)I

_{1}– 3(I

_{2}) – 6(I

_{3}) =10

-3(I

_{1}) + (2+5+3)I

_{2}=4 -6(I

_{1}) + 10(I

_{3}) = -4 +20 On solving the above equations, I

_{1}=4.3A.

8. Find the current I_{2} (A) in the figure shown in the question 7.

a) 1.7

b) 2.6

c) 3.6

d) 4.6

View Answer

Explanation: According to mesh analysis, (1+3+6)I

_{1}– 3(I

_{2}) – 6(I

_{3}) = 10. -3(I

_{1}) + (2+5+3)I

_{2}= 4. -6(I

_{1}1) + 10(I

_{3}) = -4 + 20 On solving the above equations, I

_{2}=1.7A.

9. Find the current I_{3} (A) in the figure shown in the question 7.

a) 4

b) 4.7

c) 5

d) 5.7

View Answer

Explanation: According to mesh analysis, (1+3+6)I

_{1}– 3(I

_{2}) – 6(I

_{3}) = 10. -3(I

_{1}) + (2+5+3)I

_{2}= 4. -6(I

_{1}) + 10(I

_{3}) = -4 + 20. On solving the above equations, I

_{3}= 4.7A.

10. Find current through R_{2} resistor.

a) 3

b) 3.25

c) 3.5

d) 3.75

View Answer

Explanation: Applying mesh analysis, 5(I

_{1}) + 2(I

_{1}-I

_{2}) = 10. 10(I

_{2}) + 2(I

_{2}-I

_{1}) + 40 = 0. On solving, I

_{1}= 0.5A, I

_{2}= -3.25A. So current through R

_{2}resistor is 0.5-(-3.25) = 3.75 A.

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

To practice all areas of Network Theory, __here is complete set of 1000+ Multiple Choice Questions and Answers__.