250+ TOP MCQs on Advanced Problems on Superposition Theorem and Answers

Network Theory Multiple Choice Questions on “Advanced Problems on Superposition Theorem”.

1. The Superposition Theorem is not applicable for _________
A. Power calculation
B. Voltage calculation
C. Current Calculation
D. Both Voltage and Current calculation

Answer: A
Clarification: The Superposition Theorem is not applicable for Power calculation because for power, the calculations involve either the product of voltage and current or the square of current or the square of the voltage thus making them non-linear operations. Hence they cannot be calculated using Superposition Theorem.

2. The current I in the circuit given below is ________

A. (frac{1}{4}) A
B. (frac{5}{4}) A
C. (frac{3}{4}) A
D. (frac{1}{2}) A

Answer: B
Clarification: Using Superposition Theorem, we can write
I = (frac{24}{3+2} – frac{72}{3+2} – frac{3I}{3+2} )
Or, I = 2 – (frac{3I}{3+2})
Or, I = 2 – (frac{3I}{5})
∴ I = (frac{5}{4}) A.

3. The value of the voltage V₂ at the node in the circuit given below is __________

A. 2.745 V
B. 1.393 V
C. -1.393 V
D. -2.745 V

Answer: C
Clarification: The datum node is the lower branch.
By superposition, the current i is given by, i = 2 (frac{7}{15+5+7} + frac{3}{15+5+7} + 4I frac{7+15}{15+5+7})
Or, I = (frac{17}{27} + frac{88}{27I})
∴ The solution for i yields, I = (frac{frac{17}{27}}{1 – frac{88}{27}} = -frac{17}{61}) A
∴ V₂ = 5I = -1.393 V.

4. The value of the voltage V₁ at the node in the circuit given below is __________

A. 11.148 V
B. 10.989 V
C. 20.151 V
D. 25.148 V

Answer: A
Clarification: The datum node is the lower branch.
By superposition, the current i is given by, i = 2 (frac{7}{15+5+7} + frac{3}{15+5+7} + 4I frac{7+15}{15+5+7})
Or, I = (frac{17}{27} + frac{88}{27I})
∴ The solution for i yields, I = (frac{frac{17}{27}}{1 – frac{88}{27}} = -frac{17}{61}) A
∴ V₁ = V₂ – (4I – i) 15 = 11.148 V.

5. The value of the current I in the circuit given below, is ___________

A. 0.7 A
B. 0.5 A
C. 1 A
D. 3.5 A

Answer: A
Clarification: Using Superposition theorem, we can write,
I = (frac{30}{6+4+2} + 3frac{4}{6+4+2} – 8ifrac{6}{6+4+2})
Or, I = (frac{42}{12}) – 4I
Or, I = (frac{frac{42}{12}}{1+4}) = 0.7 A.

6. In superposition theorem when we consider one voltage source, all the other voltage sources are ___________
A. Shorted
B. Removed
C. Undisturbed
D. Opened

Answer: A
Clarification: To determine the contribution of each individual source in case of Superposition Theorem, we short circuit all the other voltage sources (which are independent) and open circuit all the independent current sources (which are independent).

7. In the figure given below, the value of Resistance R by Superposition Theorem is ___________

A. 10
B. 20
C. 30
D. 40

Answer: B
Clarification: Using Superposition theorem, we get,
(frac{V_P-100}{10} + frac{V_P}{10}) + 2 = 0
Or, 2V_P – 100 +20= 0
∴ V_P = 80/2 = 40V
∴ R = 20Ω.

8. In the circuit given below, the value of voltage V_X using Superposition Theorem is _________

A. 5(I_O – 3)
B. 2.5(I_O – 3)
C. 4.5(3 – I_O)
D. 2.5(3 – I_O)

Answer: D
Clarification: Using Superposition Theorem the voltage V_X is given as,
V_X = (3-I_O) (2||40) + 5V_X (frac{2}{40+2})
Or, V_X = (frac{80}{42}(3 – I_O) + frac{10}{42}V_X)
Or, V_X = (frac{frac{80}{42}}{1-frac{10}{42}}(3 – I_O)) = 2.5(3 – I_O).

9. In the circuit given below, the value of voltage V_O using Superposition Theorem is _________

A. −30 – 10I_O
B. 30 + 10I_O
C. −30 + 10I_O
D. +30 – 10I_O

Answer: C
Clarification: Using Superposition Theorem the voltage V_X is given as,
V_X = (3-I_O) (2||40) + 5V_X (frac{2}{40+2})
Or, V_X = (frac{80}{42}(3 – I_O) + frac{10}{42})V_X
Or, V_X = (frac{frac{80}{42}}{1-frac{10}{42}})(3 – I_O) = 2.5(3 – I_O)
∴ V_O = V_X – 5V_X = −30 + 10I_O.

10. The value of the voltage V_O in the circuit given below is __________

A. 20 V
B. 30 V
C. 40 V
D. 50 V

Answer: B
Clarification: Using Superposition Theorem, the current IB is given by,
I_B = (frac{70}{4||20+2||10} × frac{20}{4+20} + frac{50}{10+4||20||2} × frac{20||2}{4+20||2} – frac{2I_B}{20||2+4||10} × frac{10}{4+10})
Or, I_B = (frac{35}{3} + frac{25}{18} – frac{11}{36}) I_B
Now, Solving for I_B, we get, I_B = (frac{frac{35}{3}+frac{25}{18}}{1+frac{11}{36}}) = 10 A
Now, V_O = 70 − 4i_b
∴ V_O = 30 V.

11. The value of voltage V_O in the circuit given below is _____________

A. 24 V
B. 48 V
C. 12 V
D. 4 V

Answer: A
Clarification: Using Superposition Theorem, we get,
V_∆ = −0.4V_∆ × 10 + 5 × 10
Solving for V_∆, we get, V_∆ = (frac{5 × 10}{1+0.4×10}) = 10 V
Again by using Superposition Theorem, we get, I_∆ = (frac{10}{5+20} – 0.4V_∆ frac{20}{5+20})
Or, I_∆ = (frac{10}{25} – 0.4V_∆ frac{20}{25} = -frac{70}{25}) A
Thus, V_O is given by, V_O = 10 − 5I_∆
∴ V_O = 24 V.

12. In the circuit given below, the value of V in terms of V_S and I_S using Superposition Theorem is ___________

A. V = (frac{2}{7} I_S – frac{4}{7} V_S)
B. V = (frac{2}{7} V_S + frac{4}{7} I_S)
C. V = (frac{2}{7} V_S – frac{4}{7} I_S)
D. V = (frac{2}{7} I_S – frac{4}{7} V_S)

Answer: C
Clarification: By superposition, the current I is given by I = (frac{V_S}{5} − frac{2}{5} × I_S − frac{3}{5} × 3i)
This can be solved for I to obtain, I = (frac{V_S}{14} – frac{I_S}{7})
Now, by Superposition Theorem, V = (frac{V_S}{5} − frac{2}{5} × I_S − frac{3}{5} × 3i)
Or, V = (frac{V_S}{5} − frac{2}{5} × I_S − frac{3}{5} × 3(frac{V_S}{14} – frac{I_S}{7}))
∴ V = (frac{2}{7} V_S – frac{4}{7} I_S).

13. In the circuit given below, the value of V_X due to the 10 V source is ____________

A. 1 V
B. 2 V
C. 3 V
D. 4 V

Answer: B
Clarification: Due to the effect of the 10 V source, we short the 16 V source and open the 3A and 15 A sources. From the resulting circuit, we can calculate the value of V.
∴ V = (frac{10 ×20}{80+20}) = 2 V.

14. In the following circuit, when R = 0 Ω, the current IR equals to 10 A. The maximum power will be?

A. 50 W
B. 100 W
C. 200 W
D. 400 W

Answer: A
Clarification: Using Superposition Theorem, we get, I = 10 A and R = 2 Ω
∴ P_max = ((frac{10}{2}))² × 2
= 5 × 5 × 2 = 50 W.

15. The circuit shown in figure has a load equivalent to _________

A. (frac{4}{3}) Ω
B. (frac{8}{3}) Ω
C. 4 Ω
D. 2 Ω

Answer: B
Clarification: Using Superposition Theorem, in the given circuit, (frac{V}{4} + frac{V-2I}{2}) = I
Or, (frac{3V-4I}{4}) = I
Or, 3V = 8I
∴ (frac{V}{I} = frac{8}{3}) Ω.