This set of Network Theory Multiple Choice Questions & Answers (MCQs) focuses on “Sinusoidal Response of an R-L Circuit”.

1. In the sinusoidal response of R-L circuit, the complementary function of the solution of i is?

a) i_{c} = ce^{-t(R/L)}

b) i_{c} = ce^{t(RL)}

c) i_{c} = ce^{-t(RL)}

d) i_{c} = ce^{t(R/L)}

View Answer

Explanation: From the R-L circuit, we get the characteristic equation as (D+R/L)i=V/L cos(ωt+θ). The complementary function of the solution i is i

_{c}= ce

^{-t(R/L)}.

2. The particular current obtained from the solution of i in the sinusoidal response of R-L circuit is?

a) i_{p} = V/√(R^{2}+(ωL)^{2}) cos(ωt+θ+tan^{-1}(ωL/R))

b) i_{p} = V/√(R^{2}+(ωL)^{2}) cos(ωt+θ-tan^{-1}(ωL/R))

c) i_{p} = V/√(R^{2}+(ωL)^{2}) cos(ωt-θ+tan^{-1}(ωL/R))

d) i_{p} = V/√(R^{2}+(ωL)^{2}) cos(ωt-θ+tan^{-1}(ωL/R))

View Answer

Explanation: The characteristic equation consists of two parts, viz. complementary function and particular integral. The particular integral is i

_{p}= V/√(R

^{2}+(ωL)

^{2}) cos(ωt+θ-tan

^{-1}(ωL/R)).

3. The value of ‘c’ in complementary function of ‘i’ is?

a) c = -V/√(R^{2}+(ωL)^{2}) cos(θ+tan^{-1}(ωL/R))

b) c = -V/√(R^{2}+(ωL)^{2}) cos(θ-tan^{-1}(ωL/R))

c) c = V/√(R^{2}+(ωL)^{2}) cos(θ+tan^{-1}(ωL/R))

d) c = V/√(R^{2}+(ωL)^{2}) cos(θ-tan^{-1}(ωL/R))

View Answer

Explanation: Since the inductor does not allow sudden changes in currents, at t = 0, i = 0. So, c = -V/√(R

^{2}+(ωL)

^{2}) cos(θ-tan

^{-1}(ωL/R)).

4. The complete solution of the current in the sinusoidal response of R-L circuit is?

a) i = e^{-t(R/L)}[V/√(R^{2}+(ωL)^{2}) cos(θ-tan^{-1})(ωL/R))]+V/√(R^{2}+(ωL)^{2}) cos(ωt+θ-tan^{-1})(ωL/R))

b) i = e^{-t(R/L)}[-V/√(R^{2}+(ωL)^{2}) cos(θ-tan^{-1})(ωL/R))]-V/√(R^{2}+(ωL)^{2}) cos(ωt+θ-tan^{-1})(ωL/R))

c) i = e^{-t(R/L)}[V/√(R^{2}+(ωL)^{2}) cos(θ-tan^{-1})(ωL/R))]-V/√(R^{2}+(ωL)^{2}) cos(ωt+θ-tan^{-1})(ωL/R))

d) i = e^{-t(R/L)}[-V/√(R^{2}+(ωL)^{2}) cos(θ-tan^{-1})(ωL/R))]+V/√(R^{2}+(ωL)^{2}) cos(ωt+θ-tan^{-1})(ωL/R))

View Answer

Explanation: The complete solution for the current becomes i = e

^{-t(R/L)}[-V/√(R

^{2}+(ωL)

^{2}) cos(θ-tan

^{-1})(ωL/R))]+V/√(R

^{2}+(ωL)

^{2})cos(ωt+θ-tan

^{-1})(ωL/R)).

5. In the circuit shown below, the switch is closed at t = 0, applied voltage is v (t) = 100cos (103t+π/2), resistance R = 20Ω and inductance L = 0.1H. The complementary function of the solution of ‘i’ is?

a) i_{c} = ce^{-100t}

b) i_{c} = ce^{100t}

c) i_{c} = ce^{-200t}

d) i_{c} = ce^{200t}

View Answer

Explanation: By applying Kirchhoff’s voltage law to the circuit, we have 20i+0.1di/dt=100cos(10

^{3}t+π/2) => (D+200)i=1000cos(1000t+π/2). The complementary function is i

_{c}= ce

^{-200t}.

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

a) i_{p} = 0.98cos(1000t+π/2-78.6^{o})

b) i_{p} = 0.98cos(1000t-π/2-78.6^{o})

c) i_{p} = 0.98cos(1000t-π/2+78.6^{o})

d) i_{p} = 0.98cos(1000t+π/2+78.6^{o})

View Answer

Explanation: Assuming particular integral as i

_{p}= A cos (ωt + θ) + B sin(ωt + θ). We get i

_{p}= V/√(R

^{2}+(ωL)

^{2}) cos(ωt+θ-tan

^{-1}(ωL/R)) where ω = 1000 rad/sec, V = 100V, θ = π/2, L = 0.1H, R = 20Ω. On substituting, we get i

_{p}= 0.98cos(1000t+π/2-78.6

^{o}).

7. The complete solution of ‘i’ from the information provided in the question 5.

a) i = ce^{-200t} + 0.98cos(1000t-π/2-78.6^{o})

b) i = ce^{-200t} + 0.98cos(1000t+π/2-78.6^{o})

c) i = ce^{-200t} + 0.98cos(1000t+π/2+78.6^{o})

d) i = ce^{-200t} + 0.98cos(1000t-π/2+78.6^{o})

View Answer

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 = ce

^{-200t}+ 0.98cos(1000t+π/2-78.6

^{o}).

8. The current flowing through the circuit at t = 0 in the circuit shown in the question 5 is?

a) 1

b) 2

c) 3

d) 0

View Answer

Explanation: At t = 0 that is initially the current flowing through the circuit is zero that is i = 0. So, i = 0.

9. The value of c in the complementary function of ‘i’ in the question 5 is?

a) c = -0.98cos(π/2-78.6^{o})

b) c = -0.98cos(π/2+78.6^{o})

c) c = 0.98cos(π/2+78.6^{o})

d) c = 0.98cos(π/2-78.6^{o})

View Answer

Explanation: At t = 0, the current flowing through the circuit is zero. Placing i = 0 in the current equation we get c = -0.98cos(π/2-78.6

^{o}).

10. The complete solution of ‘i’ in the question 5 is?

a) i = [-0.98 cos(π/2-78.6^{o})] exp(-200t)+0.98cos(1000t+π/2-78.6^{o})

b) i = [-0.98 cos(π/2-78.6^{o})] exp(-200t)-0.98cos(1000t+π/2-78.6^{o})

c) i = [0.98 cos(π/2-78.6^{o})] exp(-200t)-0.98cos(1000t+π/2-78.6^{o})

d) i = [0.98 cos(π/2-78.6^{o})] exp(-200t)+0.98cos(1000t+π/2-78.6^{o})

View Answer

Explanation: The complete solution for the current is the sum of the complementary function and the particular integral.

So, i = [-0.98 cos(π/2-78.6

^{o})] exp(-200t)+0.98cos(1000t+π/2-78.6

^{o}).

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