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. i_p = V/√(R²+(1/ωC+ωL)²) cos⁡(ωt+θ+tan^-1)⁡((1/ωC+ωL)/R))
B. i_p = V/√(R²+(1/ωC-ωL)²) cos⁡(ωt+θ+tan^-1)⁡((1/ωC-ωL)/R))
C. i_p = V/√(R²+(1/ωC+ωL)²) cos⁡(ωt+θ+tan^-1)⁡((1/ωC-ωL)/R))
D. i_p = 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 i_p = 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. i_c = c₁ e^(K₁+K₂)t + c₁ e^(K₁-K₂)t
B. i_c = c₁ e^(K₁-K₂)t + c₁ e^(K₁-K₂)t
C. i_c = c₁ e^(K₁+K₂)t + c₁ e^(K₂-K₁)t
D. i_c = 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 i_c = 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

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. i_c = e^-37.5t(c₁cos1290t + c₂sin1290t)
B. i_c = e^-37.5t(c₁cos1290t – c₂sin1290t)
C. i_c = e^37.5t(c₁cos1290t – c₂sin1290t)
D. i_c = e^37.5t(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 i_c = 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. 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)

Answer: D
Clarification: Particular solution is i_p = V/√(R²+(1/ωC-ωL)²) cos⁡(ωt+θ+tan^-1)⁡((1/ωC-ωL)/R)). i_p = 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+45^o+88.5^o). 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).