Test: The RLC Circuits - Electrical Engineering (EE) MCQ

# Test: The RLC Circuits - Electrical Engineering (EE) MCQ

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## 20 Questions MCQ Test GATE Electrical Engineering (EE) Mock Test Series 2025 - Test: The RLC Circuits

Test: The RLC Circuits for Electrical Engineering (EE) 2024 is part of GATE Electrical Engineering (EE) Mock Test Series 2025 preparation. The Test: The RLC Circuits questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: The RLC Circuits MCQs are made for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: The RLC Circuits below.
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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

### Find the current in the circuit which supplies an active power of 600 watts and reactive power of 800 VAR, if the voltage applied is 200 V (RMS value).

Detailed Solution for Test: The RLC Circuits - Question 2

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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

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 19

α = Wo, critically damped response

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

At t = 0. ⇒ B = -2

Test: The RLC Circuits - Question 20

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 20

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.

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## GATE Electrical Engineering (EE) Mock Test Series 2025

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