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

Network Theory online test on “Sinusoidal Response of an R-C Circuit”.

1. In the sinusoidal response of R-C circuit, the complementary function of the solution of i is?
A. i_c = ce^-t/RC
B. i_c = ce^t/RC
C. i_c = ce^-t/RC
D. i_c = ce^t/RC

Answer: A
Clarification: From the R-c circuit, we get the characteristic equation as (D+1/RC.i=-Vω/R sin⁡(ωt+θ). The complementary function of the solution i is i_c = ce^-t/RC.

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

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

3. The value of ‘c’ in complementary function of ‘i’ is?
A. c = V/R cosθ+V/√(R²+(1/(ωC.)²) cos⁡(θ+tan^-1(1/ωRC.)
B. c = V/R cosθ+V/√(R²+(1/(ωC.)²) cos⁡(θ-tan^-1(1/ωRC.)
C. c = V/R cosθ-V/√(R²+(1/(ωC.)²) cos⁡(θ-tan^-1(1/ωRC.)
D. c = V/R cosθ-V/√(R²+(1/(ωC.)²) cos⁡(θ+tan^-1(1/ωRC.)

Answer: D
Clarification: Since the capacitor does not allow sudden changes in voltages, at t = 0, i = V/R cosθ. So, c = V/R cosθ-V/√(R²+(1/(ωC.)²) cos⁡(θ+tan^-1(1/ωRC.).

4. The complete solution of the current in the sinusoidal response of R-C circuit is?
A. i = e^-t/RC[V/R cosθ+V/√(R²+(1/(ωC.)²) cos⁡(θ+tan^-1(1/ωRC.)+V/√(R²+(1/ωC.²) cos⁡(ωt+θ+tan^-1(1/ωRC.]
B. i = e^-t/RC[V/R cosθ-V/√(R²+(1/ωC.²) cos⁡(θ+tan^-1(1/ωRC.)-V/√(R²+(1/ωC.²) cos⁡(ωt+θ+tan^-1(1/ωRC.]
C. i = e^-t/RC[V/R cosθ+V/√(R²+(1/ωC.²) cos⁡(θ+tan^-1(1/ωRC.)-V/√(R²+(1/ωC.²) cos⁡(ωt+θ+tan^-1(1/ωRC.]
D. i = e^-t/RC[V/R cosθ-V/√(R²+(1/(ωC.)²) cos⁡(θ+tan^-1(1/ωRC.)+V/√(R²+(1/ωC.²) cos⁡(ωt+θ+tan^-1(1/ωRC.]

Answer: D
Clarification: The complete solution for the current becomes i = e^-t/RC[V/R cosθ-V/√(R²+(1/(ωC.)²) cos⁡(θ+tan^-1(1/ωRC.)+V/√(R²+(1/ωC.²) cos⁡(ωt+θ+tan^-1(1/ωRC.).

5. In the circuit shown below, the switch is closed at t = 0, applied voltage is v (t) = 50cos (102t+π/4), resistance R = 10Ω and capacitance C = 1µF. The complementary function of the solution of ‘i’ is?

A. i_c = c exp (-t/10^-10)
B. i_c = c exp(-t/10¹⁰)
C. i_c = c exp (-t/10^-5)
D. i_c = c exp (-t/10⁵)

6. In the circuit shown below, the switch is closed at t = 0, applied voltage is v (t) = 50cos (102t+π/4), resistance R = 10Ω and capacitance C = 1µF. The particular integral of the solution of ‘i_p’ is?

A. i_p = (4.99×10^-3) cos⁡(100t+π/4-89.94^o)
B. i_p = (4.99×10^-3) cos⁡(100t-π/4-89.94^o)
C. i_p = (4.99×10^-3) cos⁡(100t-π/4+89.94^o)
D. i_p = (4.99×10^-3) cos⁡(100t+π/4+89.94^o)

Answer: D
Clarification: Assuming particular integral as i_p = A cos (ωt + θ) + B sin (ωt + θ)
we get i_p = V/√(R²+(1/ωC.²) cos⁡(ωt+θ-tan^-1(1/ωRC.)
where ω = 1000 rad/sec, θ = π/4, C = 1µF, R = 10Ω. On substituting, we get i_p = (4.99×10^-3) cos⁡(100t+π/4+89.94^o).

7. In the circuit shown below, the switch is closed at t = 0, applied voltage is v (t) = 50cos (102t+π/4), resistance R = 10Ω and capacitance C = 1µF. The current flowing in the circuit at t = 0 is?

A. 1.53
B. 2.53
C. 3.53
D. 4.53

Answer: C
Clarification: At t = 0 that is initially current flowing through the circuit is i = V/R cosθ = (50/10)cos(π/4) = 3.53A.

8. In the circuit shown below, the switch is closed at t = 0, applied voltage is v (t) = 50cos (102t+π/4), resistance R = 10Ω and capacitance C = 1µF. The complete solution of ‘i’ is?

A. i = c exp (-t/10^-5) – (4.99×10^-3) cos⁡(100t+π/2+89.94^o)
B. i = c exp (-t/10^-5) + (4.99×10^-3) cos⁡(100t+π/2+89.94^o)
C. i = -c exp(-t/10^-5) + (4.99×10^-3) cos⁡(100t+π/2+89.94^o)
D. i = -c exp(-t/10^-5) – (4.99×10^-3) cos⁡(100t+π/2+89.94^o)

Answer: B
Clarification: The complete solution for the current is the sum of the complementary function and the particular integral. The complete solution for the current becomes i = c exp (-t/10^-5) + (4.99×10^-3) cos⁡(100t+π/2+89.94^o).

9. In the circuit shown below, the switch is closed at t = 0, applied voltage is v (t) = 50cos (102t+π/4), resistance R = 10Ω and capacitance C = 1µF. The value of c in the complementary function of ‘i’ is?

A. c = (3.53-4.99×10^-3) cos⁡(π/4+89.94^o)
B. c = (3.53+4.99×10^-3) cos⁡(π/4+89.94^o)
C. c = (3.53+4.99×10^-3) cos⁡(π/4-89.94^o)
D. c = (3.53-4.99×10^-3) cos⁡(π/4-89.94^o)

Answer: A
Clarification: At t = 0, the current flowing through the circuit is 3.53A. So, c = (3.53-4.99×10^-3) cos⁡(π/4+89.94^o).

10. In the circuit shown below, the switch is closed at t = 0, applied voltage is v (t) = 50cos (102t+π/4), resistance R = 10Ω and capacitance C = 1µF. The complete solution of ‘i’ is?

A. i = [(3.53-4.99×10^-3)cos⁡(π/4+89.94^o)] exp⁡(-t/0.00001)+4.99×10^-3) cos⁡(100t+π/2+89.94^o)
B. i = [(3.53+4.99×10^-3)cos⁡(π/4+89.94^o)] exp⁡(-t/0.00001)+4.99×10^-3) cos⁡(100t+π/2+89.94^o)
C. i = [(3.53+4.99×10^-3)cos⁡(π/4+89.94^o)] exp⁡(-t/0.00001)-4.99×10^-3) cos⁡(100t+π/2+89.94^o)
D. i = [(3.53-4.99×10^-3)cos⁡(π/4+89.94^o)] exp⁡(-t/0.00001)-4.99×10^-3) cos⁡(100t+π/2+89.94^o)

Answer: A
Clarification: The complete solution for the current is the sum of the complementary function and the particular integral. So, i = [(3.53-4.99×10^-3)cos⁡(π/4+89.94^o)] exp⁡(-t/0.00001)+4.99×10^-3) cos⁡(100t+π/2+89.94^o).