# Sinusoidal Response of an R-L-C Circuit MCQ’s

This set of Network Theory Multiple Choice Questions & Answers (MCQs) focuses on “Sinusoidal Response of an R-L-C Circuit”.

1. In the sinusoidal response of R-L-C circuit, the complementary function of the solution of i is?

a) i_{c} = c_{1} e^{(K1+K2)t} + c_{1} e^{(K1-K2)t}

b) i_{c} = c_{1} e^{(K1-K2)t} + c_{1} e^{(K1-K2)t}

c) i_{c} = c_{1} e^{(K1+K2)t} + c_{1} e^{(K2-K1)t}

d) i_{c} = c_{1} e^{(K1+K2)t} + c_{1} e^{(K1+K2)t}

2. The complete solution of the current in the sinusoidal response of R-L-C circuit is?

a) i = c_{1} e^{(K1+K2)t} + c_{1} e^{(K1-K2)t} – V/√(R^{2}+(1/ωC-ωL)^{2}) cos(ωt+θ+tan^{-1})((1/ωC-ωL)/R))

b) i = c_{1} e^{(K1+K2)t} + c_{1} e^{(K1-K2)t }– V/√(R^{2}+(1/ωC-ωL)^{2}) cos(ωt+θ-tan^{-1})((1/ωC-ωL)/R))

c) i = c_{1} e^{(K1+K2)t} + c_{1} e^{(K1-K2)t} + V/√(R^{2}+(1/ωC-ωL)^{2}) cos(ωt+θ+tan^{-1})((1/ωC-ωL)/R))

d) i = c_{1} e^{(K1+K2)t} + c_{1} e^{(K1-K2)t }+ V/√(R^{2}+(1/ωC-ωL)^{2}) cos(ωt+θ-tan^{-1})((1/ωC-ωL)/R))

3. The particular current obtained from the solution of i in the sinusoidal response of R-L-C circuit is?

a) i_{p} = V/√(R^{2}+(1/ωC+ωL)^{2}) cos(ωt+θ+tan^{-1})((1/ωC+ωL)/R))

b) i_{p} = V/√(R^{2}+(1/ωC-ωL)^{2}) cos(ωt+θ+tan^{-1})((1/ωC-ωL)/R))

c) i_{p} = V/√(R^{2}+(1/ωC+ωL)^{2}) cos(ωt+θ+tan^{-1})((1/ωC-ωL)/R))

d) i_{p} = V/√(R^{2}+(1/ωC-ωL)^{2}) cos(ωt+θ+tan^{-1})((1/ωC+ωL)/R))

4. In the circuit shown below, the switch is closed at t = 0. Applied voltage is v (t) = 400cos (500t + π/4). Resistance R = 15Ω, inductance L = 0.2H and capacitance = 3 µF. Find the roots of the characteristic equation.

a) -38.5±j1290

b) 38.5±j1290

c) 37.5±j1290

d) -37.5±j1290

5. In the circuit shown below, the switch is closed at t = 0. Applied voltage is v (t) = 400cos (500t + π/4). Resistance R = 15Ω, inductance L = 0.2H and capacitance = 3 µF. Find the particular solution.

a) i_{p} = 0.6cos(500t + π/4 + 88.5⁰)

b) i_{p} = 0.6cos(500t + π/4 + 89.5⁰)

c) i_{p} = 0.7cos(500t + π/4 + 89.5⁰)

d) i_{p} = 0.7cos(500t + π/4 + 88.5⁰)

6. The value of the c_{1} in the following equation is?

i = e^{-37.5t}(c_{1}cos1290t + c_{2}sin1290t) + 0.7cos(500t + π/4 + 88.5⁰).

a) -0.5

b) 0.5

c) 0.6

d) -0.6

7. In the circuit shown below, the switch is closed at t = 0. Applied voltage is v (t) = 400cos (500t + π/4). Resistance R = 15Ω, inductance L = 0.2H and capacitance = 3 µF. Find the complementary current.

a) i_{c} = e^{-37.5t}(c_{1}cos1290t + c_{2}sin1290t)

b) i_{c} = e^{-37.5t}(c_{1}cos1290t – c_{2}sin1290t)

c) i_{c} = e^{37.5t}(c_{1}cos1290t – c_{2}sin1290t)

d) i_{c} = e^{37.5t}(c_{1}cos1290t + c_{2}sin1290t)

8. In the circuit shown below, the switch is closed at t = 0. Applied voltage is v (t) = 400cos (500t + π/4). Resistance R = 15Ω, inductance L = 0.2H and capacitance = 3 µF. Find the complete solution of current.

a) i = e^{-37.5t}(c_{1}cos1290t + c_{2}sin1290t) + 0.7cos(500t + π/4 + 88.5⁰)

b) i = e^{-37.5t}(c_{1}cos1290t + c_{2}sin1290t) + 0.7cos(500t – π/4 + 88.5⁰)

c) i = e^{-37.5t}(c_{1}cos1290t + c_{2}sin1290t) – 0.7cos(500t – π/4 + 88.5⁰)

d) i = e^{-37.5t}(c_{1}cos1290t + c_{2}sin1290t) – 0.7cos(500t + π/4 + 88.5⁰)

9. The value of the c_{2} in the following equation is?

i = e^{-37.5t}(c_{1}cos1290t + c_{2}sin1290t) + 0.7cos(500t + π/4 + 88.5⁰).

a) 2.3

b) -2.3

c) 1.3

d) -1.3

10. The complete solution of current obtained by substituting the values of c_{1} and c_{2} in the following equation is?

i = e^{-37.5t}(c_{1}cos1290t + c_{2}sin1290t) + 0.7cos(500t + π/4 + 88.5⁰).

a) i = e^{-37.5t}(0.49cos1290t – 1.3sin1290t) + 0.7cos(500t + 133.5⁰)

b) i = e^{-37.5t}(0.49cos1290t – 1.3sin1290t) – 0.7cos(500t + 133.5⁰)

c) i = e^{-37.5t}(0.49cos1290t + 1.3sin1290t) – 0.7cos(500t + 133.5⁰)

d) i = e^{-37.5t}(0.49cos1290t + 1.3sin1290t) + 0.7cos(500t + 133.5⁰)

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