Explore topic-wise MCQs in Network Theory.

This section includes 17 Mcqs, each offering curated multiple-choice questions to sharpen your Network Theory knowledge and support exam preparation. Choose a topic below to get started.

1.

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» E.
Discussion
2.

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» D. -1.3
Discussion
3.

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» C. 0.6
Discussion
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 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» B. i = e-37.5t(c1cos1290t + c2sin1290t) + 0.7cos(500t – π/4 + 88.5⁰)
Discussion
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 particular solution.

A. ip = 0.6cos(500t + π/4 + 88.5⁰)
B. ip = 0.6cos(500t + π/4 + 89.5⁰)
C. ip = 0.7cos(500t + π/4 + 89.5⁰)
D. ip = 0.7cos(500t + π/4 + 88.5⁰)
Answer» E.
Discussion
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 complementary current.

A. ic = e-37.5t(c1cos1290t + c2sin1290t)
B. ic = e-37.5t(c1cos1290t – c2sin1290t)
C. ic = e37.5t(c1cos1290t – c2sin1290t)
D. ic = e37.5t(c1cos1290t + c2sin1290t)
Answer» B. ic = e-37.5t(c1cos1290t – c2sin1290t)
Discussion
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 roots of the characteristic equation.

A. -38.5±j1290
B. 38.5±j1290
C. 37.5±j1290
D. -37.5±j1290
Answer» E.
Discussion
8.

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

A. ic = c1 e(K1+K2)t + c1 e(K1-K2)t
B. ic = c1 e(K1-K2)t + c1 e(K1-K2)t
C. ic = c1 e(K1+K2)t + c1 e(K2-K1)t
D. ic = c1 e(K1+K2)t + c1 e(K1+K2)t
Answer» B. ic = c1 e(K1-K2)t + c1 e(K1-K2)t
Discussion
9.

THE_COMPLETE_SOLUTION_OF_CURRENT_OBTAINED_BY_SUBSTITUTING_THE_VALUES_OF_C1_AND_C2_IS??$

A. i = e<sup>-37.5t</sup>(0.49cos1290t – 1.3sin1290t) + 0.7cos(500t + 133.5⁰)
B. i = e<sup>-37.5t</sup>(0.49cos1290t – 1.3sin1290t) – 0.7cos(500t + 133.5⁰)
C. i = e<sup>-37.5t</sup>(0.49cos1290t + 1.3sin1290t) – 0.7cos(500t + 133.5⁰)
D. i = e<sup>-37.5t</sup>(0.49cos1290t + 1.3sin1290t) + 0.7cos(500t + 133.5‚Äö√Ö‚àû)
Answer» E.
Discussion
10.

THE_VALUE_OF_THE_C2_OBTAINED_IN_THE_COMPLETE_SOLUTION_OF_QUESTION_7.?$

A. 2.3
B. -2.3
C. 1.3
D. -1.3
Answer» D. -1.3
Discussion
11.

The value of the c1 obtained in the complete solution of question 7?

A. -0.5
B. 0.5
C. 0.6
D. -0.6
Answer» C. 0.6
Discussion
12.

The complete solution of current from the information provided in the question 4.

A. i = e<sup>-37.5t</sup>(c<sub>1</sub>cos1290t + c<sub>2</sub>sin1290t) + 0.7cos(500t + π/4 + 88.5⁰)
B. i = e<sup>-37.5t</sup>(c<sub>1</sub>cos1290t + c<sub>2</sub>sin1290t) + 0.7cos(500t – π/4 + 88.5⁰)
C. i = e<sup>-37.5t</sup>(c<sub>1</sub>cos1290t + c<sub>2</sub>sin1290t) – 0.7cos(500t – π/4 + 88.5⁰)
D. i = e<sup>-37.5t</sup>(c<sub>1</sub>cos1290t + c<sub>2</sub>sin1290t) – 0.7cos(500t + π/4 + 88.5⁰)
Answer» B. i = e<sup>-37.5t</sup>(c<sub>1</sub>cos1290t + c<sub>2</sub>sin1290t) + 0.7cos(500t ‚Äö√Ñ√∂‚àö√ë‚àö¬® ‚âà√¨‚àö√ë/4 + 88.5‚Äö√Ñ√∂‚àö√ñ‚Äö√†√ª)
Discussion
13.

The particular solution from the information provided in the question 4.

A. i<sub>p</sub> = 0.6cos(500t + π/4 + 88.5⁰)
B. i<sub>p</sub> = 0.6cos(500t + π/4 + 89.5⁰)
C. i<sub>p</sub> = 0.7cos(500t + π/4 + 89.5⁰)
D. i<sub>p</sub> = 0.7cos(500t + π/4 + 88.5⁰)
Answer» E.
Discussion
14.

Find the complementary current from the information provided in the question 4.

A. i<sub>c</sub> = e<sup>-37.5t</sup>(c<sub>1</sub>cos1290t + c<sub>2</sub>sin1290t)
B. i<sub>c</sub> = e<sup>-37.5t</sup>(c<sub>1</sub>cos1290t – c<sub>2</sub>sin1290t)
C. i<sub>c</sub> = e<sup>37.5t</sup>(c<sub>1</sub>cos1290t – c<sub>2</sub>sin1290t)
D. i<sub>c</sub> = e<sup>37.5t</sup>(c<sub>1</sub>cos1290t + c<sub>2</sub>sin1290t)
Answer» B. i<sub>c</sub> = e<sup>-37.5t</sup>(c<sub>1</sub>cos1290t ‚Äö√Ñ√∂‚àö√ë‚àö¬® c<sub>2</sub>sin1290t)
Discussion
15.

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

A. i = c<sub>1</sub> e<sup>(K<sub>1</sub>+K<sub>2</sub>)t</sup> + c<sub>1</sub> e<sup>(K<sub>1</sub>-K<sub>2</sub>)t</sup> – V/√(R<sup>2</sup>+(1/ωC-ωL)<sup>2</sup> ) cos⁡(ωt+θ+tan<sup>-1</sup>)⁡((1/ωC-ωL)/R))
B. i = c<sub>1</sub> e<sup>(K<sub>1</sub>+K<sub>2</sub>)t</sup> + c<sub>1</sub> e<sup>(K<sub>1</sub>-K<sub>2</sub>)t </sup> – V/√(R<sup>2</sup>+(1/ωC-ωL)<sup>2</sup> ) cos⁡(ωt+θ-tan<sup>-1</sup>)⁡((1/ωC-ωL)/R))
C. i = c<sub>1</sub> e<sup>(K<sub>1</sub>+K<sub>2</sub>)t</sup> + c<sub>1</sub> e<sup>(K<sub>1</sub>-K<sub>2</sub>)t</sup> + V/√(R<sup>2</sup>+(1/ωC-ωL)<sup>2</sup> ) cos⁡(ωt+θ+tan<sup>-1</sup>)⁡((1/ωC-ωL)/R))
D. i = c<sub>1</sub> e<sup>(K<sub>1</sub>+K<sub>2</sub>)t</sup> + c<sub>1</sub> e<sup>(K<sub>1</sub>-K<sub>2</sub>)t </sup> + V/√(R<sup>2</sup>+(1/ωC-ωL)<sup>2</sup> ) cos⁡(ωt+θ-tan<sup>-1</sup>)⁡((1/ωC-ωL)/R))
Answer» D. i = c<sub>1</sub> e<sup>(K<sub>1</sub>+K<sub>2</sub>)t</sup> + c<sub>1</sub> e<sup>(K<sub>1</sub>-K<sub>2</sub>)t </sup> + V/‚Äö√Ñ√∂‚àö‚Ć‚àö‚àÇ(R<sup>2</sup>+(1/‚âà√¨‚àö¬¢C-‚âà√¨‚àö¬¢L)<sup>2</sup> ) cos‚Äö√Ñ√∂‚àö√ñ¬¨‚àû(‚âà√¨‚àö¬¢t+‚âà√≠‚Äö√†√®-tan<sup>-1</sup>)‚Äö√Ñ√∂‚àö√ñ¬¨‚àû((1/‚âà√¨‚àö¬¢C-‚âà√¨‚àö¬¢L)/R))
Discussion
16.

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

A. i<sub>c</sub> = c<sub>1</sub> e<sup>(K<sub>1</sub>+K<sub>2</sub>)t</sup> + c<sub>1</sub> e<sup>(K<sub>1</sub>-K<sub>2</sub>)t</sup>
B. i<sub>c</sub> = c<sub>1</sub> e<sup>(K<sub>1</sub>-K<sub>2</sub>)t</sup> + c<sub>1</sub> e<sup>(K<sub>1</sub>-K<sub>2</sub>)t</sup>
C. i<sub>c</sub> = c<sub>1</sub> e<sup>(K<sub>1</sub>+K<sub>2</sub>)t</sup> + c<sub>1</sub> e<sup>(K<sub>2</sub>-K<sub>1</sub>)t</sup>
D. i<sub>c</sub> = c<sub>1</sub> e<sup>(K<sub>1</sub>+K<sub>2</sub>)t</sup> +c<sub>1</sub>e<sup>(K<sub>1</sub>+K<sub>2</sub>)t</sup>
Answer» B. i<sub>c</sub> = c<sub>1</sub> e<sup>(K<sub>1</sub>-K<sub>2</sub>)t</sup> + c<sub>1</sub> e<sup>(K<sub>1</sub>-K<sub>2</sub>)t</sup>
Discussion
17.

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

A. i<sub>p</sub> = V/√(R<sup>2</sup>+(1/ωC+ωL)<sup>2</sup> ) cos⁡(ωt+θ+tan<sup>-1</sup>)⁡((1/ωC+ωL)/R))
B. i<sub>p</sub> = V/√(R<sup>2</sup>+(1/ωC-ωL)<sup>2</sup> ) cos⁡(ωt+θ+tan<sup>-1</sup>)⁡((1/ωC-ωL)/R))
C. i<sub>p</sub> = V/√(R<sup>2</sup>+(1/ωC+ωL)<sup>2</sup> ) cos⁡(ωt+θ+tan<sup>-1</sup>)⁡((1/ωC-ωL)/R))
D. i<sub>p</sub> = V/√(R<sup>2</sup>+(1/ωC-ωL)<sup>2</sup> ) cos⁡(ωt+θ+tan<sup>-1</sup>)⁡((1/ωC+ωL)/R))
Answer» C. i<sub>p</sub> = V/‚Äö√Ñ√∂‚àö‚Ć‚àö‚àÇ(R<sup>2</sup>+(1/‚âà√¨‚àö¬¢C+‚âà√¨‚àö¬¢L)<sup>2</sup> ) cos‚Äö√Ñ√∂‚àö√ñ¬¨‚àû(‚âà√¨‚àö¬¢t+‚âà√≠‚Äö√†√®+tan<sup>-1</sup>)‚Äö√Ñ√∂‚àö√ñ¬¨‚àû((1/‚âà√¨‚àö¬¢C-‚âà√¨‚àö¬¢L)/R))
Discussion