Explore topic-wise MCQs in Network Theory.

This section includes 16 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 current equation obtained from the circuit shown below is?

A. i=e-200t (1.04 sin979.8t)A
B. i=e-200t (2.04 sin979.8t)A
C. i=e-200t (3.04 sin979.8t)A
D. i=e-200t (4.04 sin979.8t)A
Answer» C. i=e-200t (3.04 sin979.8t)A
Discussion
2.

The voltage across the inductor at t = 0 in the circuit shown below.

A. 50
B. 100
C. 150
D. 200
Answer» C. 150
Discussion
3.

At time t = 0, the value of current in the circuit shown below.

A. 1
B. 2
C. 3
D. 0
Answer» E.
Discussion
4.

The expression of current from the circuit shown below.

A. i=e-200t [c1 cos979.8t+c2 979.8t]A
B. i=e200t [c1 cos979.8t-c2 979.8t]A
C. i=e-200t [c1 cos979.8t-c2 979.8t]A
D. i=e200t [c1 cos979.8t+c2 979.8t]A
Answer» B. i=e200t [c1 cos979.8t-c2 979.8t]A
Discussion
5.

Replacing the differentiation with D1, D2 in the equation 100 = 20i + 0.05 \(\frac{di}{dt} + \frac{1}{20 \times 10^{-6}} \int idt\). Find the values of D1, D2.

A. 200±j979.8
B. -200±j979.8
C. 100±j979.8
D. -100±j979.8
Answer» C. 100±j979.8
Discussion
6.

The circuit shown in the figure consists of resistance, capacitance and inductance in series with a 100V source when the switch is closed at t = 0. Find the equation obtained from the circuit in terms of current.

A. 100 = 20i + 0.05 \(\frac{di}{dt} + \frac{1}{20 \times 10^{-6}} \int idt\)
B. 100 = 20i – 0.05 \(\frac{di}{dt} + \frac{1}{20 \times 10^{-6}} \int idt\)
C. 100 = 20i + 0.05 \(\frac{di}{dt} – \frac{1}{20 \times 10^{-6}} \int idt\)
D. 100 = 20i – 0.05 \(\frac{di}{dt} – \frac{1}{20 \times 10^{-6}} \int idt\)
Answer» B. 100 = 20i – 0.05 \(\frac{di}{dt} + \frac{1}{20 \times 10^{-6}} \int idt\)
Discussion
7.

If the roots of an equation are real and equal, then the response will be?

A. over damped
B. damped
C. critically damped
D. under damped
Answer» D. under damped
Discussion
8.

For an R-L-C circuit, we get [D – (K1 + K2)][D – (K1 – K2)] i = 0. If K2 is positive, then the curve will be?

A. damped
B. over damped
C. under damped
D. critically damped
Answer» C. under damped
Discussion
9.

THE_CURRENT_EQUATION_OBTAINED_FROM_THE_CIRCUIT_SHOWN_IN_THE_QUESTION_5.?$

A. i=e<sup>-200t</sup> (1.04 sin979.8t)A
B. i=e<sup>-200t</sup> (2.04 sin979.8t)A
C. i=e<sup>-200t</sup> (3.04 sin979.8t)A
D. i=e<sup>-200t</sup> (4.04 sin979.8t)A
Answer» C. i=e<sup>-200t</sup> (3.04 sin979.8t)A
Discussion
10.

The voltage across the inductor at t = 0 in the circuit shown in the question 5?

A. 50
B. 100
C. 150
D. 200
Answer» C. 150
Discussion
11.

At time t = 0, the value of current in the circuit shown in the question 5?

A. 1
B. 2
C. 3
D. 0
Answer» E.
Discussion
12.

The expression of current from the circuit shown in the question 5.

A. i=e<sup>-200t</sup> [c<sub>1</sub> cos979.8t+c<sub>2</sub> 979.8t]A
B. i=e<sup>200t</sup> [c<sub>1</sub> cos979.8t-c<sub>2</sub> 979.8t]A
C. i=e<sup>-200t</sup> [c<sub>1</sub> cos979.8t-c<sub>2</sub> 979.8t]A
D. i=e<sup>200t</sup> [c<sub>1</sub> cos979.8t+c<sub>2</sub> 979.8t]A
Answer» B. i=e<sup>200t</sup> [c<sub>1</sub> cos979.8t-c<sub>2</sub> 979.8t]A
Discussion
13.

Replacing the differentiation with D1, D2 in the equation obtained from the question 5. Find the values of D1, D2.

A. 200±j979.8
B. -200±j979.8
C. 100±j979.8
D. -100±j979.8
Answer» C. 100¬¨¬®¬¨¬±j979.8
Discussion
14.

If the roots of an equation are complex conjugate, then the response will be?

A. over damped
B. critically damped
C. damped
D. under damped
Answer» E.
Discussion
15.

If the roots of an equation are real and unequal, then the response will be?

A. critically damped
B. under damped
C. over damped
D. damped
Answer» D. damped
Discussion
16.

For an R-L-C circuit, we get [D – (K1 + K2)][D – (K1 – K2)] i = 0. If K2 is positive, then the curve will be?

A. damped
B. over damped
C. under damped
D. critically damped
Answer» C. under damped
Discussion