# Network Theory Questions and Answers – Mesh Analysis

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

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

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

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 below, the current through loop 1 be I1 and through the loop 2 be I2, then the current flowing through the resistor R2 will be?

a) I1
b) I2
c) I1-I2
d) I1+I2

Explanation: Through the resistor R2 both the currents I1, I2 are flowing. So the current through R2 will be I1-I2.

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

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.
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6. Consider the circuit shown in the figure. Find voltage Vx.

a) 1
b) 1.25
c) 1.5
d) 1.75

Explanation: Consider current I1 (CW) in the loop 1 and I2 (ACW) in the loop 2. So, the equations will be Vx+I2-I1=0. I1=5/2=2.5A. I2=4Vx/4= Vx. Vx+Vx-2.5=0. Vx = 1.25V.

7. Consider the circuit shown below. Find the current I1 (A).

a) 3.32
b) 3.78
c) 5.33
d) 6.38

Explanation: According to mesh analysis,
(1+3+6)I1 – 3(I2) – 6(I3) = 10
-3(I1) + (2+5+3)I2 = 4
-6(I1) + 10(I3) = – 4 + 20
On solving the above equations, we get I1 = 3.78A.

8. Consider the following figure. Find the current I2 (A).

a) 1.5
b) 2.6
c) 3.6
d) 4.6

Explanation: According to mesh analysis,
(1+3+6)I1 – 3(I2) – 6(I3) = 10
-3(I1) + (2+5+3)I2 = 4
-6(I1) + 10(I3) = – 4 + 20
On solving the above equations, we get I2 = 1.53A.

9. Consider the following figure. Find the current I3 (A).

a) 4.34
b) 3.86
c) 5.45
d) 5.72

Explanation: According to mesh analysis,
(1+3+6)I1 – 3(I2) – 6(I3) = 10
-3(I1) + (2+5+3)I2 = 4
-6(I1) + 10(I3) = – 4 + 20
On solving the above equations, we get I3 = 3.86A.

10. Find current through R2 resistor.

a) 3
b) 3.25
c) 3.5
d) 3.75

Explanation: Applying mesh analysis, 5(I1) + 2(I1-I2) = 10. 10(I2) + 2(I2-I1) + 40 = 0. On solving, I1 = 0.5A, I2 = -3.25A. So current through R2 resistor is 0.5-(-3.25) = 3.75 A.

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