Transient and Steady State Analysis

8/20/2019 Transient and Steady State Analysis

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Transient and Steady State Analysis

1.

A 10 Ω resistor, a 1 H inductor and 1 µF capacitor are connected in parallel. The

combination is driven by a unit step current. Under the steady state condition, the source

current flows through.

(a) the resistor

(b) the inductor

(c) the capacitor only

(d) all the three elements

[GATE 1989: 2 Marks]

Soln. Under steady state condition, the capacitor is open circuit and inductor short

circuit. The source current flows through inductor.

Option (a)

2. If the Laplace transform of the voltage across a capacitor of value of ½ F is = 13 1 The value of the current through the capacitor at t = 0+ is

(a) 0 A

(b) 2 A

(c) (1/2) A

(d) 1 A

[GATE 1989: 2 Marks]

Soln. Impedance of capacitor

= = Current through the capacitor

=

= = + = →


2/16

= → = → =

Option (c)

3. A ramp voltage, v(t) = 100 t volts, is applied to an RC differentiating circuit with R = 5

kΩ and C = 4 µF. The maximum output voltage is

(a) 0.2 volt

(b) 2.0 volts

(c) 10.0 volts

(d) 50.0 volts

[GATE 1994: 1 Mark]

Soln. The output of an RC differentiating circuit

= = × × × −

= × − ×

= Option (b)

4. In the figure, the switch was closed for a long time before opening at t = 0. The voltage

Vx at t = 0+ is


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20

2.5A

5H

20

Vx- +

(a)

25 V(b) 50 V

(c)

-50 V(d) 0 V

[GATE 2002 :1 Mark]

Soln. When the switch was closed for a long time, the steady state is reached and inductor

is short circuit.

− = . The current cannot change instantaneously in a inductor

− = + = .

At = + inductor can be replaced by a current source of 2.5A. The equivalentcircuit for the same is drawn.

20

2.5A

20

V -+

2.5A


4/16

= × . =

= is of opposite polarity

= Option (c)

5. For the R-L circuit shown in the figure, the input voltage Vi(t) = u(t). Plot the current i(t)

i(t)V1(t)

1H

2

Soln. = =

= [ ] = −

= , =

=

= ∞, ∞ = = . = , = −


5/16

= . = .

= .

0.316

0.500

i (t)

t1/2

6. The circuit shown in the figure has initial current iL(0-) = 1 A through the inductor and an

initial voltage VC (0-) = -1 V across the capacitor. For input v(t) = u(t) find the laplace

transform of the current i(t) for ≥ 0

i(t)V1(t)

1H

+

-

1F

+

-

1

[GATE 2004 :2 Marks]


6/16

Soln. Writing KVL

= ∫

−

Taking Laplace transform on both sides

= + −

Where + is the initial current and is the initial voltage of the capacitor.

= = [ ]

=

7. A square pulse of 3 volts amplitude is applied to C-R circuit shown in the figure. The

capacitor is initially uncharged. The output voltage V2 at time t = 2 sec is


7/16

Vi

2 sect

V1 V2

+ +

- -

1k

0. 1 µF

3

(a) 3 V

(b) -3 V

(c) 4 V

(d) -4 V

[GATE : 2005 2 Marks]

Soln. Time constant = × × . × − = . × − = .

Steady state will be reached in time ≥ time constant (0.5 m sec). The capacitorgets charge to + 3 volts

=

Option (b)

8. A 2 mH inductor with some initial current is in figure. Where s is the laplace transform

variable. The value of initial current is.

+

-

I(s)

0.002s

1 mV


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(a) 0.5 A

(b) 2.0 A

(c) 1.0 A

(d) 0.0 A

[GATE : 2006 1 Mark]

Soln. Voltage across inductor L

= = + + =

+ = = . Option (a)

9. In the figure shown below, assume that all the capacitors are initially uncharged. If Vi (t)

= 10 u(t) Volts, V0(t) is given by

+ +

- -

1 K

V1(t)4 µF

4K 1 µF V0(t)

(a) 8e-t/0.004 Volts

(b) 8 (1 – e-t/0.004) Volts

(c) 8u(t) Volts

(d) 8 Volts


Soln. Let

= ‖ = = × −


9/16

+ +

- -

1 K

V1(t)

4 µF

4 K 1 µF V0(t)

R 1

C1

R 2

C2

= ‖

= = =

= = .

= . × = Option (c)

10. In the following circuit, the switch S is closed at t = 0. The rate of change of current

0+


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i(t)IS R S

S

R

L

(a) 0

(b)

(c) +

(d) ∞ [GATE : 2008 1 Mark]

Soln. Drawing the equivalent circuit as the switch is closed

i(t)

IS

R S R

LDC I(S) R S

=

= +, = + + Since current can not change instantaneously in a inductor so + = − =

+ =

Option (b)

11. For t > 0, the voltage across the capacitor is:


11/16

DCδ (t) VC (t)

1 H 1

1 F

(a) √ 3 −√ −

(b) − √ 3 √ 3 √ 3 (c)

√ 3 −√ 3 (d)

√ 3 −√ 3

[GATE: 2008 2 Marks]

Soln. Writing KVL

=

∫

Taking laplace transform on both sides = [ ] = =

= =

=

= √ − =


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= √ √

√

= √ − √ Option(c)

12. The switch in the circuit shown was on position a for a long time, and is moved to

position b at time t = 0. Find the current i(t) for t > 0

DC100 V

10k

0.2µF

0.5µF 0.3µF

5 k

a b

i(t)


Soln. Total capacitance : C

=

.

.

= =


13/16

= = . The switch is in position a for a long a for a long time, the capacitor C is fully

charged to 100 volts.

When switch is in position b, the equivalent circuit for the same is

1 / CS

I(S)

DC

R 5 K

100 / SV (0+) / S

= ⁄

= =

= − ⁄ = − = −

13. The time domain behavior of an RL circuit is represented by

= 1

−

For an initial current of 0 = the steady state value of the current is given by(a) → (b) →

(c) → 1 (d) = 1



14/16

Soln. = − Taking Laplace transform

+ =

+ =

= + The steady sate value of current is

→ = →

=

→ = → =

Option (a)


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14. In the circuit shown below, the initial charge on the capacitor is 2.5 mC, with the voltage

polarity as indicated. The switch is closed at time t = 0. The current i(t) at a time t after

the switch is closed is

DC

10

100 V

i(t)

-

+50 µF

(a)

= 15 exp 2 × 1 0

3

(b)

= 5 exp 2 × 1 03

(c)

= 10exp 2 × 1 0

3

(d)

= 5 exp 2 × 1 03


Soln. Initial charge = 2.5 mc

Initial voltage on capacitor = . × − × − = Net voltage

= =

= − ⁄ Where = × × − = × −

= − ×⁄

= (−× )

Option (a)

15. In the following figure C1 and C2 are ideal capacitors. C1 had been charged to 12V before

the ideal switch S is closed at t = 0. The current i(t) for all t is


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S t = 0

C1 C2i(t)

(a) Zero

(b) A step function

(c) An exponentially decaying function

(d) An impulse function


Soln. Since there is no resistance so time constant is zero. This means as the switch is

closed, C2 will get charge. Charging and discharge time constant is zero. Sudden

change of voltage exists only if impulse of current passes through it.

Option (d)

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Transient and Steady State Analysis