Test: Transient Response of DC & AC Networks - 1 | 10 Questions MCQ Test Network Theory (Electric Circuits) Electrical Engineering (EE)

QUESTION: 1

The natural response of an RLC circuit is described by the differential equation

The v(t) is

A.
10(1 + t)e^-t V
B.
10(1 - t)e^-t V
C.
10 e^-t V
D.
10 te^-t V

Solution:

QUESTION: 2

The differential equation for the circuit shown in fig. P1.6.2. is

A.
B.
C.
D.

Solution:

QUESTION: 3

The differential equation for the circuit shown in fig. P1.6.3 is

A.
B.
C.
D.

Solution:

QUESTION: 4

In the circuit of fig. P.1.6.4 vs = 0 for t > 0. The initial condition are v(0) = 6 V and dv(0) dt = -3000 Vs. The v(t) for t > 0 is

A.
-2 e^-100^t + 8e^-400t V
B.
6 e^-100t+ 8e^-400t V
C.
6 e^-100t - 8e^-400t V
D.
None of the above

Solution:

A + B = 6, -100 A - 400B = -3000 ⇒ B = 8, A = -2

QUESTION: 5

The circuit shown in fig. P1.6.5 has been open for a long time before closing at t = 0. The initial condition is v(0) = 2 V. The v(t) for t > is

A.
5 e^-y - 7e^-3t V
B.
7 e^-t - 5e^-3t V
C.
-e^{-t +} 3e^-3t V
D.
3e^-t - e^-3t V

Solution:

The characteristic equation is
After putting the values, s² +4s + 3 = 0

QUESTION: 6

Circuit is shown in fig. P.1.6. Initial conditions are i₁(0) = i₂(0) = 11A

Q. i₁ (1 s) = ?

A.
0.78 A
B.
1.46 A
C.
2.56 A
D.
3.62 A

Solution:

In differential equation putting t = 0 and solving

QUESTION: 7

Circuit is shown in fig. P.1.6. Initial conditions are i₁(0) = i₂(0) = 11A

Q. i₂ (1 s) = ?

A.
0.78 A
B.
1.46 A
C.
2.56 A
D.
3.62 A

Solution:

QUESTION: 8

v_C(t) = ? for t > 0

A.
4e ^{-1000 t} - e^-2000t V
B.
( 3 + 6000 t)e^{-2000 t} V
C.
2 e^{-1000 t} - e^-2000t V
D.
( 3 - 6000 t)e^{-2000 t}V

Solution:

QUESTION: 9

The circuit shown in fig. P1.6.9 is in steady state with switch open. At t = 0 the switch is closed. The output voltage v_C(t) for t > 0 is

A.
-9 e^{-400 t}+12e^-300t
B.
e^{-400 t}[ 3cos 300t + 4 sin 300t]
C.
e^{-300 t}[ 3cos 400t + 4 sin 300t]
D.
e^{-300 t}[ 3cos 400t + 2.25 sin 300t]

Solution:

QUESTION: 10

The switch of the circuit shown in fig. P1.6.10 is opened at t = 0 after long time. The v(t) , for t > 0 is

A.
4e^-2tsin 2t V
B.
-4e^-2tsin 2t V
C.
4e^-2tcos 2t V
D.
-4e^-2tcos 2t V

Solution:

Test: Transient Response of DC & AC Networks - 1

10 Questions MCQ Test Network Theory (Electric Circuits) | Test: Transient Response of DC & AC Networks - 1

The natural response of an RLC circuit is described by the differential equation

The v(t) is

The differential equation for the circuit shown in fig. P1.6.2. is

The differential equation for the circuit shown in fig. P1.6.3 is

Recommended Content

Electromagnetic System

Overview: Generation, Transmission & Distribution of Electric Power

Fermi Level

Modern Power Systems

Syllabus: Electrical Engineering, GATE 2022

Recommended Tests

Test: Operating Characteristics of DC Generators

Test: Power Systems- 1

Test: Combinational Logic Circuits

Test: Electrical & Electronic Measurements- 1

Test: Introduction to Electrical & Electronics Measurements

Download free EduRev App

Similar Content

Transient Response

AC Transients

AC Transient Free Condition, Network Theory, Electrical Engineering, GATE

AC Transient Free Condition, Network Theory, Electrical Engineering, GATE

Related tests