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

1. Superposition theorem states that the response in any element is the ____________ of the responses that can be expected to flow if each source acts independently of other sources.

a) algebraic sum

b) vector sum

c) multiplication

d) subtraction

View Answer

Explanation: The superposition theorem is used to analyse ac circuits containing more than one source. Superposition theorem states that the response in any element is the vector sum of the responses that can be expected to flow if each source acts independently of other sources.

2. Superposition theorem is valid for only linear systems.

a) true

b) false

View Answer

Explanation: Superposition theorem is valid for only linear systems. Superposition theorem is not valid for non-linear systems. In a network containing complex impedance, all quantities must be treated as complex numbers.

3. Determine the current through (2+j5) Ω impedance considering 50∠0⁰ voltage source.

a) 6.42∠77.47⁰

b) 6.42∠-77.47⁰

c) 5.42∠77.47⁰

d) 5.42∠-77.47⁰

View Answer

Explanation: According to the superposition theorem the current due to the 50∠0

^{o}source is I

_{1}with the current source 20∠30⁰ A short-circuited. I

_{1}= (50∠0

^{o})/(2+j4+j5) = 5.42∠-77.47

^{o}A.

4. Find the voltage across (2+j5) Ω impedance considering 50∠0⁰ voltage source.

a) 30.16∠-9.28⁰

b) 30.16∠9.28⁰

c) 29.16∠-9.28⁰

d) 29.16∠9.28⁰

View Answer

Explanation: Voltage across (2+j5) Ω impedance considering 50∠0⁰ voltage source is V

_{1}= 5.42∠-77.47

^{o}(2+j5) = 29.16∠-9.28

^{o}V.

5. Find the current through (2+j5) Ω impedance considering 20∠30⁰ voltage source.

a) 8.68∠-42.53⁰

b) 8.68∠42.53⁰

c) 7.68∠42.53⁰

d) 7.68∠-42.53⁰

View Answer

Explanation: The current through (2+j5) Ω impedance considering 20∠30⁰ voltage source is I

_{2}= 20∠30

^{o}×j4/(2+j9) = 8.68∠42.53

^{o}A.

6. Determine the voltage across (2+j5) Ω impedance considering 20∠30⁰ voltage source.

a) 45.69∠-110.72⁰

b) 45.69∠110.72⁰

c) 46.69∠-110.72⁰

d) 46.69∠110.72⁰

View Answer

Explanation: The voltage across (2+j5) Ω impedance considering 20∠30⁰ voltage source is V

_{2}= 8.68∠42.53

^{o}(2+j5) = 46.69∠110.72⁰V.

7. Find the voltage across (2+j5) Ω impedance using Superposition theorem.

a) 40.85∠72.53⁰

b) 40.85∠-72.53⁰

c) 41.85∠72.53⁰

d) 41.85∠-72.53⁰

View Answer

Explanation: The voltage across (2+j5) Ω impedance using Superposition theorem is the sum of the voltages V

_{1}and V

_{2}. => V = V

_{1}+V

_{2}= 29.16∠-9.28

^{o}+46.69∠110.72

^{o}=40.85∠72.53

^{o}V.

8. Determine the voltage V_{ab} considering the source 50∠0⁰V.

a) 50∠0⁰

b) 4∠0⁰

c) 54∠0⁰

d) 46∠0⁰

View Answer

Explanation: Let source 50∠0⁰ act on the circuit and set the source 4∠0⁰A equal to zero. If the current source is zero, it becomes open-circuited. Then the voltage across ‘ab’ is 50∠0⁰.

9. Determine the voltage V_{ab} considering the source 4∠0⁰A in the circuit shown above.

a) 46∠0⁰

b) 4∠0⁰

c) 54∠0⁰

d) 50∠0⁰

View Answer

Explanation: Set the voltage source 50∠0⁰V to zero, and is short-circuited. So, the voltage drop across ‘ab’ is zero.

10. Find the voltage V_{ab} in the circuit shown above using Superposition theorem.

a) 4∠0⁰

b) 50∠0⁰

c) 54∠0⁰

d) 46∠0⁰

View Answer

Explanation: The total voltage is the sum of the two voltages V

_{1}and V

_{2}=> V

_{ab}= V

_{1}+V

_{2}= 50∠0⁰V.

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

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