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

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

1. The particular current obtained from the solution of i in the sinusoidal response of R-L-C circuit is?
A. ip = V/√(R2+(1/ωC+ωL)2) cos⁡(ωt+θ+tan-1)⁡((1/ωC+ωL)/R))
B. ip = V/√(R2+(1/ωC-ωL)2) cos⁡(ωt+θ+tan-1)⁡((1/ωC-ωL)/R))
C. ip = V/√(R2+(1/ωC+ωL)2) cos⁡(ωt+θ+tan-1)⁡((1/ωC-ωL)/R))
D. ip = V/√(R2+(1/ωC-ωL)2) cos⁡(ωt+θ+tan-1)⁡((1/ωC+ωL)/R))

Answer: B
Clarification: The characteristic equation consists of two parts, viz. complementary function and particular integral. The particular integral is ip = V/√(R2+(1/ωC-ωL)2) cos⁡(ωt+θ+tan-1)⁡((1/ωC-ωL)/R)).

2. In the sinusoidal response of R-L-C circuit, the complementary function of the solution of i is?
A. ic = c1 e(K1+K2)t + c1 e(K1-K2)t
B. ic = c1 e(K1-K2)t + c1 e(K1-K2)t
C. ic = c1 e(K1+K2)t + c1 e(K2-K1)t
D. ic = c1 e(K1+K2)t + c1 e(K1+K2)t

Answer: A
Clarification: From the R-L circuit, we get the characteristic equation as
(D2+R/L D+1/LC.=0. The complementary function of the solution i is ic = c1 e(K1+K2)t + c1 e(K1-K2)t.

3. The complete solution of the current in the sinusoidal response of R-L-C circuit is?
A. i = c1 e(K1+K2)t + c1 e(K1-K2)t – V/√(R2+(1/ωC-ωL)2) cos⁡(ωt+θ+tan-1)⁡((1/ωC-ωL)/R))
B. i = c1 e(K1+K2)t + c1 e(K1-K2)t – V/√(R2+(1/ωC-ωL)2) cos⁡(ωt+θ-tan-1)⁡((1/ωC-ωL)/R))
C. i = c1 e(K1+K2)t + c1 e(K1-K2)t + V/√(R2+(1/ωC-ωL)2) cos⁡(ωt+θ+tan-1)⁡((1/ωC-ωL)/R))
D. i = c1 e(K1+K2)t + c1 e(K1-K2)t + V/√(R2+(1/ωC-ωL)2) cos⁡(ωt+θ-tan-1)⁡((1/ωC-ωL)/R))

Answer: C
Clarification: The complete solution for the current becomes i = c1 e(K1+K2)t + c1 e(K1-K2)t + V/√(R2+(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

Answer: D
Clarification: By applying Kirchhoff’s voltage law to the circuit,
On differentiating the above equation and on solving, we get roots of the characteristic equation as -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 complementary current.
A. ic = e-37.5t(c1cos1290t + c2sin1290t)
B. ic = e-37.5t(c1cos1290t – c2sin1290t)
C. ic = e37.5t(c1cos1290t – c2sin1290t)
D. ic = e37.5t(c1cos1290t + c2sin1290t)

Answer: A
Clarification: The roots of the charactesistic equation are D1 = -37.5+j1290 and D2 = -37.5-j1290. The complementary current obtained is ic = e-37.5t(c1cos1290t + c2sin1290t).

6. 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. ip = 0.6cos(500t + π/4 + 88.5⁰)
B. ip = 0.6cos(500t + π/4 + 89.5⁰)
C. ip = 0.7cos(500t + π/4 + 89.5⁰)
D. ip = 0.7cos(500t + π/4 + 88.5⁰)

Answer: D
Clarification: Particular solution is ip = V/√(R2+(1/ωC-ωL)2) cos⁡(ωt+θ+tan-1)⁡((1/ωC-ωL)/R)). ip = 0.7cos(500t + π/4 + 88.5⁰).

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 complete solution of current.

A. i = e-37.5t(c1cos1290t + c2sin1290t) + 0.7cos(500t + π/4 + 88.5⁰)
B. i = e-37.5t(c1cos1290t + c2sin1290t) + 0.7cos(500t – π/4 + 88.5⁰)
C. i = e-37.5t(c1cos1290t + c2sin1290t) – 0.7cos(500t – π/4 + 88.5⁰)
D. i = e-37.5t(c1cos1290t + c2sin1290t) – 0.7cos(500t + π/4 + 88.5⁰)

Answer: A
Clarification: The complete solution is the sum of the complementary function and the particular integral. So i = e-37.5t(c1cos1290t + c2sin1290t) + 0.7cos(500t + π/4 + 88.5⁰).

8. The value of the c1 in the following equation is?
i = e-37.5t(c1cos1290t + c2sin1290t) + 0.7cos(500t + π/4 + 88.5⁰).
A. -0.5
B. 0.5
C. 0.6
D. -0.6

Answer: B
Clarification: At t = 0 that is initially the current flowing through the circuit is zero that is i = 0. So, c1 = -0.71cos (133.5⁰) = 0.49.

9. The value of the c2 in the following equation is?
i = e-37.5t(c1cos1290t + c2sin1290t) + 0.7cos(500t + π/4 + 88.5⁰).
A. 2.3
B. -2.3
C. 1.3
D. -1.3

Answer: C
Clarification: Differentiating the current equation, we have di/dt = e-37.5t (-1290c1sin1290t + 1290c2cos1290t) – 37.5e-37.5t(c1cos1290t+c2sin1290t) – 0.71x500sin(500t+45o+88.5o). At t = 0, di/dt = 1414. On solving, we get c2 = 1.31.

10. The complete solution of current obtained by substituting the values of c1 and c2 in the following equation is?
i = e-37.5t(c1cos1290t + c2sin1290t) + 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)

Answer: D
Clarification: The complete solution is the sum of the complementary function and the particular integral. So i = e-37.5t(0.49cos1290t + 1.3sin1290t) + 0.7cos(500t + 133.5⁰).

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