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QUESTION: 1

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

The v(t) is

Solution:

S^{2} + 2s + 1 = 0 ⇒ s = -1, -1,

v(t) = (A_{1} + A_{2}t)e^{-t}

v(0) = 10V,

A_{1} = A_{2} = 10

QUESTION: 2

The differential equation for the circuit shown in fig.

Solution:

10^{8}v_{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

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 i_{1}(0) = i_{2}(0) =11A

i_{1} (1s) = ?

Solution:

In differential equation putting t = 0 and sovling

QUESTION: 7

Circuit is shown in fig. P.1.6. Initial conditions are (0)i_{1=}i_{2}(0)=11A

i_{2} (1 s)= ?

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:

A_{2} = -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

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

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}

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 v_{o} (t) for t >0 is

Solution:

α = W_{o}, 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 = W_{o}, critically damped response

s = -2, -2

i(t) = (A + Bt)e^{-2t}, A = -2

At t = 0. ⇒ B = -2

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