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QUESTION: 1

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

The v(t) is

Solution:

QUESTION: 2

The differential equation for the circuit shown in fig. P1.6.2. is

Solution:

QUESTION: 3

The differential equation for the circuit shown in fig. P1.6.3 is

Solution:

QUESTION: 4

In the circuit of fig. P.1.6.4 vs = 0 for t > 0. The initial condition are v(0) = 6 V and dv(0) dt = -3000 Vs. The v(t) for t > 0 is

Solution:

A + B = 6, -100 A - 400B = -3000 ⇒ B = 8, A = -2

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) = 2 V. The v(t) for t > is

Solution:

The characteristic equation is

After putting the values, s^{2} +4s + 3 = 0

QUESTION: 6

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

Q. i_{1} (1 s) = ?

Solution:

In differential equation putting t = 0 and solving

QUESTION: 7

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

Q. i_{2} (1 s) = ?

Solution:

QUESTION: 8

v_{C}(t) = ? for t > 0

Solution:

QUESTION: 9

The circuit shown in fig. P1.6.9 is in steady state with switch open. At t = 0 the switch is closed. The output voltage v_{C}(t) for t > 0 is

Solution:

QUESTION: 10

The switch of the circuit shown in fig. P1.6.10 is opened at t = 0 after long time. The v(t) , for t > 0 is

Solution:

### Transient Response

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