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

View Answer

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

View Answer

Explanation: Applying nodal analysis at node ‘a’, (V

_{a}-10∠0

^{o})/j6+V

_{a}/(-j6)+(V

_{a}-Vb)/3=0. Applying nodal analysis at node ‘b’, (V

_{b}-V

_{a})/3+V

_{b}/j4+V

_{b}/j1=0. Solving the above equations we get, V

_{a}= 5.22∠-104.5⁰V.

3. Determine the voltage at node ‘b’ in the circuit shown in the question 2.

a) -1.34∠-180⁰

b) 1.34∠-180⁰

c) -0.34∠-180⁰

d) 0.34∠-180⁰

View Answer

Explanation: Applying nodal analysis at node ‘a’, (V

_{a}-10∠0

^{o})/j6+V

_{a}/(-j6)+(V

_{a}-V

_{b})/3=0. Applying nodal analysis at node ‘b’, (V

_{b}-V

_{a})/3+V

_{b}/j4+V

_{b}/j1=0. Solving the above equations we get, V

_{b}= -1.34∠-180⁰V.

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

a) (5∠0^{o})/3

b) – (5∠0^{o})/3

c) (10∠0^{o})/3

d) – (10∠0^{o})/3

View Answer

Explanation: The general equations are Y

_{aa}V

_{a}+Y

_{ab}V

_{b}= I

_{1}, Y

_{ba}V

_{a}+Y

_{bb}V

_{b}= I

_{2}. We get Y

_{aa}=1/3+1/j4+1/(-j6) and the self admittance at node a is the sum of admittances connected to node a. Y

_{ab}=-(1/(-j6)). I

_{1}= (10∠0

^{o})/3=x.

5. Find the value of ‘y’ in the equation –(-1/j6)V_{a}+(1/5+1/j5-1/j6)V_{b}=y obtained from the circuit shown in the question 4.

a) (10∠30^{o})/5

b) -(10∠30^{o})/5

c) (5∠30^{o})/5

d) (-5∠30^{o})/5

View Answer

Explanation: We got Y

_{bb}=1/5+1/j5-1/j6 and Y

_{ab}=–(-1/j6). The mutual admittance between node b and a is the sum of the admittances between nodes b and a. I

_{2}=-(10∠30

^{o})/5=y.

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

View Answer

Explanation: Applying nodal analysis at node ‘a’, (Va-20∠〖30

^{o})/3+Va/(-j4)+Va/(2+j5)=0. On solving, Va = 16.27∠18.91⁰. Current in 2Ω resistor I

_{2}= Va/(2+j5)=(16.27∠18.91

^{o})/(5.38∠68.19

^{o})=3.02∠-49.28

^{o}. Power dissipated in 2Ω resistor P=I

_{2}

^{2}R=3.02

^{2}×2= 18.24W.

7. In the circuit shown in the question 6 determine the power dissipated in 3Ω resistor.

a) 7.77

b) 8.77

c) 9.77

d) 10.77

View Answer

Explanation: Current in 3Ω resistor I

_{3}= (-20∠30

^{o}+16.27∠18.91

^{o})/3=1.71∠-112

^{o}. Power dissipated in 3Ω resistor P=I

_{3}

^{2}R=1.71

^{2}×3=8.77W.

8. In the circuit shown in the question 6 find the power output of the source.

a) 27

b) 28

c) 29

d) 30

View Answer

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 R_{L} if R_{L} is infinite.

a) 3

b) 2

c) 1

d) 0

View Answer

Explanation: If R

_{L}is infinite, the voltage across it will be 0. So the voltage across the resistance R

_{L}if R

_{L}is infinite is zero.

10. Find the voltage V_{ab} in the circuit shown question 9.

a) 21.66∠-45.02⁰

b) 20.66∠-45.02⁰

c) 21.66∠45.02⁰

d) 20.66∠45.02⁰

View Answer

Explanation: Applying nodal analysis at node ‘a’, (Va-20∠0

^{o})/(3+2)+(Va-20∠90

^{o})/(j4+3)=0. On solving, we get Va = 21.66∠45.02⁰V.

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