Network Theory Questions and Answers – Sinusoidal Response of an R-C Circuit

This set of Network Theory Multiple Choice Questions & Answers (MCQs) focuses 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) ic = ce-t/RC
b) ic = cet/RC
c) ic = ce-t/RC
d) ic = cet/RC
View Answer

Answer: a
Explanation: 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 ic = ce-t/RC.

2. The particular current obtained from the solution of i in the sinusoidal response of R-C circuit is?
a) ip = V/√(R2+(1/ωC)2) cos⁡(ωt+θ+tan-1(1/ωRC))
b) ip = -V/√(R2+(1/ωC)2) cos⁡(ωt+θ-tan-1(1/ωRC))
c) ip = V/√(R2+(1/ωC)2) cos⁡(ωt+θ-tan-1(1/ωRC))
d) ip = -V/√(R2+(1/ωC)2) cos⁡(ωt+θ+tan-1(1/ωRC))
View Answer

Answer: a
Explanation: The characteristic equation consists of two parts, viz. complementary function and particular integral. The particular integral is ip = 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))
View Answer

Answer: d
Explanation: 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)).
advertisement

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)]
View Answer

Answer: d
Explanation: 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?
The complementary function of the solution of ‘i’ is ic = c exp (-t/10-5)
a) ic = c exp (-t/10-10)
b) ic = c exp(-t/1010)
c) ic = c exp (-t/10-5)
d) ic = c exp (-t/105)
View Answer

Answer: c
Explanation: By applying Kirchhoff’s voltage law to the circuit, we have
By applying Kirchhoff’s voltage law to the circuit (D+1/10-5) )i=-500sin⁡(1000t+π/4)
(D+1/10-5) )i=-500sin⁡(1000t+π/4). The complementary function is ic = c exp (-t/10-5).
Free 30-Day C++ Certification Bootcamp is Live. Join Now!

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 ‘ip’ is?
The complementary function of the solution of ‘i’ is ic = c exp (-t/10-5)
a) ip = (4.99×10-3) cos⁡(100t+π/4-89.94o)
b) ip = (4.99×10-3) cos⁡(100t-π/4-89.94o)
c) ip = (4.99×10-3) cos⁡(100t-π/4+89.94o)
d) ip = (4.99×10-3) cos⁡(100t+π/4+89.94o)
View Answer

Answer: d
Explanation: Assuming particular integral as ip = A cos (ωt + θ) + B sin (ωt + θ)
we get ip = 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 ip = (4.99×10-3) cos⁡(100t+π/4+89.94o).

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?
The complementary function of the solution of ‘i’ is ic = c exp (-t/10-5)
a) 1.53
b) 2.53
c) 3.53
d) 4.53
View Answer

Answer: c
Explanation: 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?
The complementary function of the solution of ‘i’ is ic = c exp (-t/10-5)
a) i = c exp (-t/10-5) – (4.99×10-3) cos⁡(100t+π/2+89.94o)
b) i = c exp (-t/10-5) + (4.99×10-3) cos⁡(100t+π/2+89.94o)
c) i = -c exp(-t/10-5) + (4.99×10-3) cos⁡(100t+π/2+89.94o)
d) i = -c exp(-t/10-5) – (4.99×10-3) cos⁡(100t+π/2+89.94o)
View Answer

Answer: b
Explanation: 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+89.94o).

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?
The complementary function of the solution of ‘i’ is ic = c exp (-t/10-5)
a) c = (3.53-4.99×10-3) cos⁡(π/4+89.94o)
b) c = (3.53+4.99×10-3) cos⁡(π/4+89.94o)
c) c = (3.53+4.99×10-3) cos⁡(π/4-89.94o)
d) c = (3.53-4.99×10-3) cos⁡(π/4-89.94o)
View Answer

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

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?
The complementary function of the solution of ‘i’ is ic = c exp (-t/10-5)
a) i = [(3.53-4.99×10-3)cos⁡(π/4+89.94o)] exp⁡(-t/0.00001)+4.99×10-3) cos⁡(100t+π/2+89.94o)
b) i = [(3.53+4.99×10-3)cos⁡(π/4+89.94o)] exp⁡(-t/0.00001)+4.99×10-3) cos⁡(100t+π/2+89.94o)
c) i = [(3.53+4.99×10-3)cos⁡(π/4+89.94o)] exp⁡(-t/0.00001)-4.99×10-3) cos⁡(100t+π/2+89.94o)
d) i = [(3.53-4.99×10-3)cos⁡(π/4+89.94o)] exp⁡(-t/0.00001)-4.99×10-3) cos⁡(100t+π/2+89.94o)
View Answer

Answer: a
Explanation: 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+89.94o)] exp⁡(-t/0.00001)+4.99×10-3) cos⁡(100t+π/2+89.94o).

Sanfoundry Global Education & Learning Series – Network Theory.

To practice all areas of Network Theory, here is complete set of 1000+ Multiple Choice Questions and Answers.

advertisement
advertisement
Subscribe to our Newsletters (Subject-wise). Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

Youtube | Telegram | LinkedIn | Instagram | Facebook | Twitter | Pinterest
Manish Bhojasia - Founder & CTO at Sanfoundry
I’m Manish - Founder and CTO at Sanfoundry. I’ve been working in tech for over 25 years, with deep focus on Linux kernel, SAN technologies, Advanced C, Full Stack and Scalable website designs.

You can connect with me on LinkedIn, watch my Youtube Masterclasses, or join my Telegram tech discussions.

If you’re in your 40s–60s and exploring new directions in your career, I also offer mentoring. Learn more here.