Network Theory Online Test

This set of Network Theory online test 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) i_c = ce^-t/RC
b) i_c = ce^t/RC
c) i_c = ce^-t/RC
d) i_c = ce^t/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 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))
View Answer

Answer: a
Explanation: 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))
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)).

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?

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)
View Answer

Answer: c
Explanation: By applying Kirchhoff’s voltage law to the circuit, we have

(D+1/10^-5) )i=-500sin⁡(1000t+π/4). The complementary function is i_c = c exp (-t/10^-5).

6. The particular integral of the solution of ‘i’ from the information provided in the question 5.
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)
View Answer

Answer: d
Explanation: 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. The current flowing in the circuit at t = 0 in the question 5 is?
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. The complete solution of ‘i’ from the information provided in the question 5.
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 )
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+89.94^o ).

9. The value of c in the complementary function of ‘i’ in the question 5 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)
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+89.94^o ).

10. The complete solution of ‘i’ in the question 5 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 )
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+89.94^o)] exp⁡(-t/0.00001)+4.99×10^-3) cos⁡(100t+π/2+89.94^o).

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