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Test: The RLC Circuits


20 Questions MCQ Test RRB JE for Electrical Engineering | Test: The RLC Circuits


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Attempt Test: The RLC Circuits | 20 questions in 60 minutes | Mock test for Railways preparation | Free important questions MCQ to study RRB JE for Electrical Engineering for Railways Exam | Download free PDF with solutions
QUESTION: 1

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

The v(t) is

Solution:

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

QUESTION: 2

The differential equation for the circuit shown in fig.

Solution:


108vs(t) = v"(t) + 3000v'(t) + 1.02v(t)

 

QUESTION: 3

The differential equation for the circuit shown in fig

Solution:





QUESTION: 4

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
Solution:


⇒ 

⇒ 


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) = 2V. The v(t) for t > is
Solution:

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


 

QUESTION: 6

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

i1 (1s) = ?
Solution:







In differential equation putting t = 0 and sovling

QUESTION: 7

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

i2 (1 s)= ?
Solution:



C = -1 and D = 12

QUESTION: 8

vc (t ) =? for t > 0
Solution:





 
 

QUESTION: 9

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
Solution:






QUESTION: 10

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



A2 = -4

QUESTION: 11

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






QUESTION: 12

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

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





QUESTION: 13

In the circuit shown in fig. all initialcondition are zero

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

Trying iL (t)= At+ B,



QUESTION: 14

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




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

QUESTION: 15

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

QUESTION: 16

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


QUESTION: 17

i(t) = ?
Solution:

QUESTION: 18

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

 

QUESTION: 19

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

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

QUESTION: 20

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



α = Wo, critically damped response

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

At t = 0. ⇒ B = -2

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