Test: The RLC Circuits

# Test: The RLC Circuits

Test Description

## 20 Questions MCQ Test GATE Electrical Engineering (EE) 2023 Mock Test Series | Test: The RLC Circuits

Test: The RLC Circuits for Railways 2022 is part of GATE Electrical Engineering (EE) 2023 Mock Test Series preparation. The Test: The RLC Circuits questions and answers have been prepared according to the Railways exam syllabus.The Test: The RLC Circuits MCQs are made for Railways 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: The RLC Circuits below.
Solutions of Test: The RLC Circuits questions in English are available as part of our GATE Electrical Engineering (EE) 2023 Mock Test Series for Railways & Test: The RLC Circuits solutions in Hindi for GATE Electrical Engineering (EE) 2023 Mock Test Series course. Download more important topics, notes, lectures and mock test series for Railways Exam by signing up for free. Attempt Test: The RLC Circuits | 20 questions in 60 minutes | Mock test for Railways preparation | Free important questions MCQ to study GATE Electrical Engineering (EE) 2023 Mock Test Series for Railways Exam | Download free PDF with solutions
 1 Crore+ students have signed up on EduRev. Have you?
Test: The RLC Circuits - Question 1

### The natural response of an RLC circuit is described by the differential equation The v(t) is

Detailed Solution for Test: The RLC Circuits - Question 1

S2 + 2s + 1 = 0 ⇒ s = -1, -1,
v(t) = (A1 + A2t)e-t
v(0) = 10V,
A1 = A2 = 10

Test: The RLC Circuits - Question 2

### The differential equation for the circuit shown in fig

Detailed Solution for Test: The RLC Circuits - Question 2

Test: The RLC Circuits - Question 3

### In the circuit of fig. v∞ = 0 for t > 0. The initial condition are v(0) = 6V and dv(0) /dt =-3000 V s. The v(t) for t > 0 is

Detailed Solution for Test: The RLC Circuits - Question 3

⇒

⇒

Test: The RLC Circuits - Question 4

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) = 2V. The v(t) for t > is

Detailed Solution for Test: The RLC Circuits - Question 4

The characteristic equation is
After putting the values,
v(t) = Ae -t + Be-3t

Test: The RLC Circuits - Question 5

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

i1 (1s) = ?

Detailed Solution for Test: The RLC Circuits - Question 5

In differential equation putting t = 0 and sovling

Test: The RLC Circuits - Question 6

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

i2 (1 s)= ?

Detailed Solution for Test: The RLC Circuits - Question 6

C = -1 and D = 12

Test: The RLC Circuits - Question 7

v(t ) =? for t > 0

Detailed Solution for Test: The RLC Circuits - Question 7

Test: The RLC Circuits - Question 8

The circuit shown in fig is in steady state with switch open. At t = 0 the switch is closed. Theoutput voltage vt (c) for t > 0 is

Detailed Solution for Test: The RLC Circuits - Question 8

Test: The RLC Circuits - Question 9

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

Detailed Solution for Test: The RLC Circuits - Question 9

A2 = -4

Test: The RLC Circuits - Question 10

In the circuit of fig.the switch is opened at t = 0 after long time. The current iL(t) for t > 0 is

Detailed Solution for Test: The RLC Circuits - Question 10

Test: The RLC Circuits - Question 11

In the circuit shown in fig.all initial condition are zero.

If is (t) = 1 A, then the inductor current iL(t) is

Detailed Solution for Test: The RLC Circuits - Question 11

Test: The RLC Circuits - Question 12

In the circuit shown in fig. all initialcondition are zero

If is(t) = 0.5t A, then iL(t) is

Detailed Solution for Test: The RLC Circuits - Question 12

Trying iL (t)= At+ B,

Test: The RLC Circuits - Question 13

In the circuit of fig. switch is moved from position a to b at t =  0. The iL(t) for t > 0 is

Detailed Solution for Test: The RLC Circuits - Question 13

α = ωo critically damped
v(t) = 12 + (A + Bt)e-5t
0 = 12 + A, 150 = -5A + B A = -12, B = 90
v(t) =12 + (90t -12)e-5t
iL(t) = 0.02(-5) e-5t(90t -12) +0.02(90)e-5t = (3 -9t)e-5t

Test: The RLC Circuits - Question 14

In the circuit shown in fig. a steady state has been established before switch closed. The i(t) for t > 0 is

Detailed Solution for Test: The RLC Circuits - Question 14

Test: The RLC Circuits - Question 15

The switch is closed after long time in the circuit of fig. The v(t) for t > 0 is

Detailed Solution for Test: The RLC Circuits - Question 15

Test: The RLC Circuits - Question 16

i(t) = ?

Detailed Solution for Test: The RLC Circuits - Question 16

Test: The RLC Circuits - Question 17

In the circuit of fig. i(0) = 1A and v(0) = 0. The current i(t) for t > 0 is

Detailed Solution for Test: The RLC Circuits - Question 17

Test: The RLC Circuits - Question 18

In the circuit of fig. a steady state has been established before switch closed. The vo  (t) for t >0 is

Detailed Solution for Test: The RLC Circuits - Question 18

α = Wo, So critically damped respones
s = -10, -10

Test: The RLC Circuits - Question 19

Calculate the quality factor Q for an RLC circuit having R = 10 Ω, C = 30μF, and L = 27mH.

Detailed Solution for Test: The RLC Circuits - Question 19

CONCEPT:

RLC CIRCUIT:

•  An RLC circuit is an electrical circuit consisting of an inductor (L), Capacitor (C), Resistor (R) it can be connected either parallel or series.
• When the LCR circuit is set to resonate (XL = XC), the resonant frequency is expressed as

⇒f=12π1LC

• Quality factor is

⇒Q=ω0LR=1RLC

Where, XL & XC = Impedance of inductor and capacitor, L, R & C = Inductance, resistance, and capacitance, f = frequency and, ω0 = angular resonance frequency

CALCULATION:

Given that: R = 10 Ω, C = 30 μF = 30 × 10-6 F, and L = 27mH = 27 × 10-3 H

From the above discussion,

⇒Q=ω0LR=1RLC

⇒Q=11027×10−330××10−6

⇒Q=3×1010=3

• Hence option (1) is correct.
Test: The RLC Circuits - Question 20

In the circuit of fig. a steady state has been established before switch closed. The i(t) for t > 0 is

Detailed Solution for Test: The RLC Circuits - Question 20

α = Wo, critically damped response

s = -2, -2
i(t) = (A + Bt)e-2t, A = -2

At t = 0. ⇒ B = -2

## GATE Electrical Engineering (EE) 2023 Mock Test Series

20 docs|208 tests
 Use Code STAYHOME200 and get INR 200 additional OFF Use Coupon Code
Information about Test: The RLC Circuits Page
In this test you can find the Exam questions for Test: The RLC Circuits solved & explained in the simplest way possible. Besides giving Questions and answers for Test: The RLC Circuits, EduRev gives you an ample number of Online tests for practice

## GATE Electrical Engineering (EE) 2023 Mock Test Series

20 docs|208 tests