# Network Theory Questions and Answers – Nodal Analysis

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

1. If there are 8 nodes in network, we can get ____ number of equations in the nodal analysis.
a) 9
b) 8
c) 7
d) 6

Explanation: Number of equations = N-1 = 7. So as there are 8 nodes in network, we can get 7 number of equations in the nodal analysis.

2. Nodal analysis can be applied for non planar networks also.
a) true
b) false

Explanation: Nodal analysis is applicable for both planar and non planar networks. Each node in a circuit can be assigned a number or a letter.

3. In nodal analysis how many nodes are taken as reference nodes?
a) 1
b) 2
c) 3
d) 4

Explanation: In nodal analysis only one node is taken as reference node. And the node voltage is the voltage of a given node with respect to one particular node called the reference node.

4. Find the voltage at node P in the following figure.

a) 8V
b) 9V
c) 10V
d) 11V

Explanation: I1 = (4-V)/2, I2 = (V+6)/3. The nodal equation at node P will be I1+3=I2. On solving, V=9V.

5. Find the resistor value R1(Ω) in the figure shown below.

a) 10
b) 11
c) 12
d) 13

Explanation: 10=(V1-V2)/14+(V1-V3)/R1. From the circuit, V1=100V, V2=15×2=30V, V3=40V. On solving, R1=12Ω.
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6. Find the value of the resistor R2 (Ω) in the circuit shown below.

a) 5
b) 6
c) 7
d) 8

Explanation: As V1=100V, V2=15×2=30V, V3=40V. (V1-V2)/14+(V1-V3)/R2=15. On solving we get R2 = 6Ω.

7. Find the voltage (V) at node 1 in the circuit shown.

a) 5.32
b) 6.32
c) 7.32
d) 8.32

Explanation: At node 1, (1/1+1/2+1/3)V1-(1/3)V2 = 10/1. At node 2, -(1/3)V1+(1/3+1/6+1/5)V2 = 2/5+5/6. On solving above equations, we get V1=6.32V.

8. Find the voltage (V) at node 2 in the circuit shown below.

a) 2.7
b) 3.7
c) 4.7
d) 5.7

Explanation: At node 1, (1/1+1/2+1/3)V1-(1/3)V2 = 10/1. At node 2, -(1/3)V1+(1/3+1/6+1/5)V2 = 2/5+5/6. On solving above equations, we get V2=4.7V.

9. Find the voltage at node 1 of the circuit shown below.

a) 32.7
b) 33.7
c) 34.7
d) 35.7

Explanation: Applying Kirchhoff’s current law at node 1, 10 = V1/10+(V1-V2)/3. At node 2, (V2-V1)/3+V2/5+(V2-10)/1=0. On solving the above equations, we get V1=33.7V.

10. Find the voltage at node 2 of the circuit shown below.

a) 13
b) 14
c) 15
d) 16

Explanation: Applying Kirchhoff’s current law at node 1, 10 = V1/10+(V1-V2)/3. At node 2, (V2-V1)/3+V2/5+(V2-10)/1=0. On solving the above equations, we get V2=14V.

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