250+ TOP MCQs on DC Response of an R-L-C Circuit and Answers

Network Theory Multiple Choice Questions on “DC Response of an R-L-C Circuit”.

1. For an R-L-C circuit, we get [D – (K₁ + K₂)][D – (K₁ – K₂)] i = 0. If K₂ is positive, then the curve will be?
A. damped
B. over damped
C. under damped
D. critically damped
Answer: B
Clarification: For an R-L-C circuit, we get [D – (K₁ + K₂)][D – (K₁ – K₂)] i = 0. If K₂ is positive, then the curve will be over damped response.

2. If the roots of an equation are real and unequal, then the response will be?
A. critically damped
B. under damped
C. over damped
D. damped
Answer: C
Clarification: If the roots of an equation are real and unequal, then the response will be over damped response. Over damped response of a system is defined as the system returns (exponentially decays) to equilibrium without oscillating.

3. If the roots of an equation are complex conjugate, then the response will be?
A. over damped
B. critically damped
C. damped
D. under damped
Answer: D
Clarification: If the roots of an equation are complex conjugate, then the response will be under damped response. Damping is an influence within or upon an oscillatory system that has the effect of reducing, restricting or preventing its oscillations.

4. If the roots of an equation are real and equal, then the response will be?
A. over damped
B. damped
C. critically damped
D. under damped
Answer: C
Clarification: If the roots of an equation are real and equal, then the response will be critically damped response. For a critically damped system, the system returns to equilibrium as quickly as possible without oscillating.

5. The circuit shown in the figure consists of resistance, capacitance and inductance in series with a 100V source when the switch is closed at t = 0. Find the equation obtained from the circuit in terms of current.
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A. 100 = 20i + 0.05 (frac{di}{dt} + frac{1}{20 times 10^{-6}} int idt)
B. 100 = 20i – 0.05 (frac{di}{dt} + frac{1}{20 times 10^{-6}} int idt)
C. 100 = 20i + 0.05 (frac{di}{dt} – frac{1}{20 times 10^{-6}} int idt)
D. 100 = 20i – 0.05 (frac{di}{dt} – frac{1}{20 times 10^{-6}} int idt)
Answer: A
Clarification: At t = 0, switch S is closed when the 100V source is applied to the circuit and results in the following differential equation.
100 = 20i + 0.05 (frac{di}{dt} + frac{1}{20 times 10^{-6}} int idt)

6. Replacing the differentiation with D₁, D₂ in the equation 100 = 20i + 0.05 (frac{di}{dt} + frac{1}{20 times 10^{-6}} int idt). Find the values of D₁, D₂.
A. 200±j979.8
B. -200±j979.8
C. 100±j979.8
D. -100±j979.8
Answer: B
Clarification: Let the roots of the characteristic equation are denoted by D₁, D₂. So on differentiating the equation 100 = 20i + 0.05 (frac{di}{dt} + frac{1}{20 times 10^{-6}} int idt), we get D₁ = -200+j979.8, D₂ = -200-j979.8.

7. The expression of current from the circuit shown below.
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A. i=e^-200t [c₁ cos979.8t+c₂ 979.8t]A
B. i=e^200t [c₁ cos979.8t-c₂ 979.8t]A
C. i=e^-200t [c₁ cos979.8t-c₂ 979.8t]A
D. i=e^200t [c₁ cos979.8t+c₂ 979.8t]A
Answer: A
Clarification: The expression of current from the circuit will be i = e^K₁t[c₁cosK₁t + c₂sinK₂t]. So, i=e^-200t [c₁ cos979.8t+c₂ 979.8t]A.

8. At time t = 0, the value of current in the circuit shown below.

A. 1
B. 2
C. 3
D. 0