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This mock test of Test: First Order RL & RC Circuits- 2 for Electrical Engineering (EE) helps you for every Electrical Engineering (EE) entrance exam.
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QUESTION: 1

Laplace transform analysis gives

Solution:

QUESTION: 2

The Laplace transform of the derivative of the signal f(t) = e^{-t} u(t) is

Solution:

The derivative of function f(t) is

f'(t) = e^{-t}δ (t) + u (t) (-e^{-t})

= -e^{-t} u(t) + e^{-t}δ(t)

QUESTION: 3

For a series R - C circuit excited by a dc voltage of 20 V, and with time-constant τ seconds, the voltage across the capacitor at time t = τ is given by

Solution:

V_{c}(t) = V(1 - e^{-t/τ}) volts

V = 20 volt (given)

At t = τ, V_{c}|_{t = τ} = 20 (1 - e^{-1}) volts

QUESTION: 4

The Laplace-transform equivalent of a given network will have 7/5 F capacitor replaced by 5

Solution:

QUESTION: 5

An RC circuit has a capacitor C = 2 μF in series with a resistance R = 1 MΩ The time of 6 secs will be equal to

Solution:

Time constant of series RC circuit is

τ = RC = 1 x 10^{6} x 2 x 10^{-6}

= 2 secs

Hence, time of 6 seconds will be equal to 3τ = 3 times of time constant.

QUESTION: 6

The Laplace transform of the gate function shown below is

Solution:

From the given waveform, we have

∴

QUESTION: 7

Assuming initial condition to be zero, the current i(t) in the circuit shown below is

Solution:

The given circuit in Laplace domain is shown below.

Here

(Using current divider theorem)

or, i(t) = 0.25 e^{-0-5t} u(t)

QUESTION: 8

A 10 volts step voltage is applied across a RC series circuit at t = 0 having R = 100 Ω, and C = 100 μF.

The values of i(t) and di(t)/dt at t = 0 are respectively given by

Solution:

At t = 0, capacitor acts as short-circuit.

So,

Time constant of given circuit is

τ = RC = 100 x 100 x 10^{-6}

= 10^{-2} sec = 0.01 sec

With zero initial condition, current in the series RC circuit is

Hence,

di/dt = -10 e^{-100t}

i(0^{+}) = 0.1 A

QUESTION: 9

If the input voltage V_{1}(t) = 10 e^{-2t} V then, the output voltage V_{2}(t) for the circuit shown below is

Solution:

Given circuit is a low-pass filter having transfer function

When, V_{1}(t) = 10e^{-2t}

QUESTION: 10

In the circuit shown below, the switch 'S' is opened at t = 0. Prior to that, switch was closed.

The current i(t) is at t = 0^{+ }is

Solution:

For t = 0^{-}, switch Sis closed. Hence, in steady state capacitor will act as open circuit.

Thus, V_{0}(0^{-}) = V_{C}(0^{+}) = 2 volts

The circuit at t = 0^{+} is shown below where,

cananitor ants as a constant voltage source.

Using source transformation,

and

Hence,

QUESTION: 11

A rectangular pulse of duration T and magnitude A has the Laplace transform

Solution:

∴

QUESTION: 12

The time constant of the circuit shown below is

Solution:

R_{eg} across inductor

(5 A C.S open circuited)

∴ Time constant of RL circuit is

QUESTION: 13

Assertion (A): Laplace transform is preferred for solving networks involving higher order differential equations.

Reason (R): The classical method for solving differential equations of higher order is quite cumbersome.

Solution:

QUESTION: 14

Assertion (A): In a purely capacitive circuit, the current wave is more distorted than the voltage wave.

Reason (R): The harmonics in the current wave are increased in proportion to their frequency numbers.

Solution:

QUESTION: 15

The time constant of the circuit shown below is

Solution:

For finding time constant of the circuit 5 A current source is open circuited and 2 V voltage source is short circuited.

C_{eq }= C

∴

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