# 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 N nodes in a circuit, then the number of nodal equations that can be formed are?
a) N+1
b) N
c) N-1
d) N-2

Explanation: If there are N nodes in a circuit, then the number of nodal equations that can be formed are N-1. Number of nodal equations = N-1.

2. In the network shown below, find the voltage at node ‘a’.

a) 5.22∠104.5⁰
b) 5.22∠-104.5⁰
c) 6.22∠104.5⁰
d) 6.22∠-104.5⁰

Explanation: Applying nodal analysis at node ‘a’, (Va-10∠0o)/j6+Va/(-j6)+(Va-Vb)/3=0. Applying nodal analysis at node ‘b’, (Vb-Va)/3+Vb/j4+Vb/j1=0. Solving the above equations we get, Va = 5.22∠-104.5⁰V.

3. Determine the voltage at node ‘b’ in the circuit shown below.

a) -1.34∠-180⁰
b) 1.34∠-180⁰
c) -0.34∠-180⁰
d) 0.34∠-180⁰

Explanation: Applying nodal analysis at node ‘a’, (Va-10∠0o)/j6+Va/(-j6)+(Va-Vb)/3=0. Applying nodal analysis at node ‘b’, (Vb-Va)/3+Vb/j4+Vb/j1=0. Solving the above equations we get, Vb = -1.34∠-180⁰V.

4. In the circuit shown below we get a nodal equation as (1/3+1/j4-1/j6)Va—(-1/j6)Vb=x. Find the value of ‘x‘.

a) (5∠0o)/3
b) – (5∠0o)/3
c) (10∠0o)/3
d) – (10∠0o)/3

Explanation: The general equations are YaaVa+YabVb = I1, YbaVa+YbbVb = I2. We get Yaa=1/3+1/j4+1/(-j6) and the self admittance at node a is the sum of admittances connected to node a. Yab=-(1/(-j6)). I1 = (10∠0o)/3=x.

5. Find the value of ‘y’ in the equation –(-1/j6)Va+(1/5+1/j5-1/j6)Vb=y obtained from the following circuit.

a) (10∠30o)/5
b) -(10∠30o)/5
c) (5∠30o)/5
d) (-5∠30o)/5

Explanation: We got Ybb=1/5+1/j5-1/j6 and Yab=–(-1/j6). The mutual admittance between node b and a is the sum of the admittances between nodes b and a. I2=-(10∠30o)/5=y.
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6. In the circuit shown below find the power dissipated by 2Ω resistor.

a) 16.24
b) 17.24
c) 18.24
d) 19.24

Explanation: Applying nodal analysis at node ‘a’, (Va-20∠30o)/3+Va/(-j4)+Va/(2+j5)=0. On solving, Va = 16.27∠18.91⁰. Current in 2Ω resistor I2 = Va/(2+j5) = (16.27∠18.91o)/(5.38∠68.19o)=3.02∠-49.28o. Power dissipated in 2Ω resistor P=I22 R=3.022×2 = 18.24W.

7. In the circuit shown below. Determine the power dissipated in 3Ω resistor.

a) 7.77
b) 8.77
c) 9.77
d) 10.77

Explanation: Current in 3Ω resistor I3 = (-20∠30o+16.27∠18.91o)/3=1.71∠-112o. Power dissipated in 3Ω resistor P=I32 R=1.712×3=8.77W.

8. In the following circuit. Find the power output of the source.

a) 27
b) 28
c) 29
d) 30

Explanation: Total power delivered by the source is the product of voltage and current and is given by power output of the source VIcosφ = 20 x 1.71cos142⁰ = 26.95W.

9. For the circuit shown below, find the voltage across the resistance RL if RL is infinite.

a) 3
b) 2
c) 1
d) 0

Explanation: If RL is infinite, the voltage across it will be 0. So the voltage across the resistance RL if RL is infinite is zero.

10. Find the voltage Vab in the circuit shown below.

a) 21.66∠-45.02⁰
b) 20.66∠-45.02⁰
c) 21.66∠45.02⁰
d) 20.66∠45.02⁰

Explanation: Applying nodal analysis at node ‘a’, (Va-20∠0o)/(3+2)+(Va-20∠90o)/(j4+3)=0. On solving, we get Va = 21.66∠45.02⁰V.

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