Test: The RLC Circuits

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This mock test of Test: The RLC Circuits for Railways helps you for every Railways entrance exam. This contains 20 Multiple Choice Questions for Railways Test: The RLC Circuits (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: The RLC Circuits quiz give you a good mix of easy questions and tough questions. Railways students definitely take this Test: The RLC Circuits exercise for a better result in the exam. You can find other Test: The RLC Circuits extra questions, long questions & short questions for Railways on EduRev as well by searching above.

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.
10e^-t V
D.
10te^-t V

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.

A.
v"+3000v'(t) + 1.02 x 10⁸v(t) = 10⁸v_s (t)
B.
v"+1000v'(t) + 1.02 x 10⁸v(t) = 10⁸v_s (t)
C.
D.

Solution:

10⁸v_s(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

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

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

A.
5e^-t - 7e ^-3t V
B.
7e^-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,
v(t) = Ae ^-t + Be^-3t,

QUESTION: 6

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

i₁ (1s) = ?

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

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)= ?

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

Solution:

C = -1 and D = 12

QUESTION: 8

v_c(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 v_t (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 i_L(t) for t > 0 is

Solution:

QUESTION: 12

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

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

A.
1 A
B.
t A
C.
t + 1 A
D.
0 A

Solution:

QUESTION: 13

In the circuit shown in fig. all initialcondition are zero

If i_s(t) = 0.5t A, then i_L(t) is

A.
B.
C.
D.
2t + 3250 A

Solution:

Trying i_L (t)= At+ B,

QUESTION: 14

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

Solution:

α = ω_o 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
i_L(t) = 0.02(-5) e^-5t(90t -12) +0.02(90)e^-5t = (3 -9t)e^-5t