Transient/Steady State Analysis of RLC Circuits to DC Input

ECE - 2006

1. In the figure shown, assume that all the capacitors are initially uncharged.

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

(A) 8e Volts (B) 8( e) Volts (C) 8 u(t) Volts

(D) 8 Volts ECE - 2007

2. In the circuit shown, Vc is 0 volts at t = 0 sec. For t>0, the capacitor current iC (t), where t is in seconds, is given by

(A) 0.50 exp ( 25t)mA (B) 0.25 exp ( 25t)mA (C) 0.50 exp ( 12.5 t)mA (D) 0.25 exp ( 6.25 t)mA ECE - 2008

3. The circuit shown in the figure is used to charge the capacitor C alternately from two current sources as indicated. The switches S1 and S2 are mechanically coupled and connected as follows

For 2nT ≤ t< (2n+ )T,

(n = 0, , 2…) S to P and S2 to P2 For (2n+ )T ≤ t< (2n+2)T, (n = 0, , 2…) S to Q and S2 to Q2

Assume that the capacitor has zero initial charge. Given that u(t) is a unit step function, the voltage (t) across the capacitor is given by

(A) ∑( ) tu(t nT)

(B) u(t) + 2 ∑( ) u(t nT) (C) tu(t) + 2 ∑( ) (t nT)

u(t nT)

(D) ∑ 0 e^{( )}+0 e^{( )}] Common Data for Questions 4 and 5:

The following series RLC circuit with zero initial condition is excited by a unit impulse function δ(t)

4. For t > 0, the output voltage (t) is (A) √ (e e^√)

(B) √ te

5. For t > 0, the voltage across the resistor is

(A) √ (e^√ e) (B) e*cos (^√)

√ sin (^√)+

1F 1 H

VC(t) Ω

δ(t) +

(t) F

P2 Q2 P

S S2

10V 20k 4 F VC

+ - 20k

1µF 4 F 4K

Vi(t) +

Vo(t) +

6. In the following circuit, the switch S is closed at t=0. The rate of change of current (0 ) is given by

(A) 0 (B)

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

+ Ri = ( + e^⁄ sint) u(t) For an initial current of i(0) = , the steady state value of the current is given by

(A) i(t) → (B) i(t) → (C) i(t) → (1+B) (D) i(t) → (1+B)

8. The switch in the circuit shown was on position ‘a’ for a long time and is moved to position ‘b’ at time t = 0. The current i(t) for t > 0 is given by

(A) 0.2e u(t)m (B) 20e u(t)m (C) 0.2e u(t)m (D) 20eu(t)m ECE - 2010

9. In the circuit shown, the switch S is open for a long time and is closed at t = 0. The current i(t) for t 0 is

(A) i(t) = 0 0 2 e (B) i(t) = 0 2 e (C) i(t) = 0 0 e (D) i(t) = 0 e ECE - 2011

10. 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

(A) i(t) = exp( 2 0 t) (B) i(t) = exp( 2 0 t) (C) i(t) = 0 exp( 2 0 t) (D) i(t) = exp( 2 0 t)

ECE/EE/IN - 2012

11. In the following figure, C1 and C2 are ideal capacitors. has been charged to 12 V before the ideal switch S is closed at t = 0.

The current i(t) for all t is.

(A) Zero

(B) a step function

S t = 0

i(t)

0 F +

1.5A

0Ω 15mH

0Ω

0Ω S t=0 i(t)

100 V +

0 kΩ a

i(t) 0.2 F

0.3 F 0.5 F

kΩ b

i(t) IS

L S

ECE - 2013

Common Data for Questions 12 and 13:

Consider the following figure

12. The current Is in Amps in the voltage source, and voltage Vs is Volts across the current source respectively , are

(A) , 20 (B) , 0

(A) 2 (B) 3.33

14. In the figure shown, the ideal switch has been open for a long time. If it is closed at t = 0, then the magnitude of the current (in m ) through the 4 k resistor at t = 0 is _______.

15. A series LCR circuit is operated at a frequency different from its resonant frequency. The operating frequency is such that the current leads the supply voltage. The magnitude of current is half the value at resonance. If the values of L, C and R are H, F and , respectively, the operating angular frequency (in rad/s) is ________.

16. In the figure shown, the capacitor is initially uncharged. Which one of the following expressions describes the current I(t) (in mA) for t >0?

( ) (t) = ( e^⁄ ), =2 msec ( ) (t) =

2( e^⁄ ), =2 msec ( ) (t) = ( e^⁄ ), = msec ( ) (t) =

2( e^⁄ ), = msec 17. A series RC circuit is connected to a DC

voltage source at time t = 0. The relation between the source voltage , the resistance R, the capacitance C, and the current i(t) is given below:

= R i(t) + ∫ i(u)du

Which one of the following represents the current i(t)?

i(t)

t 0

( )

t 0

( ) ( )

+ F

R 2k

k R

0 0 F

k 4k k

+ mH

5

2

1

2 A Vs

10V

18. In the circuit shown in the figure, the value of (t) (in volts )for t → is ________

EE - 2006

1. In the circuit shown in the figure, the current source I = 1A, voltage source V = 5V, R = R = R = 1Ω,

= = = 1H, = = 1F.

The currents (in A) through R3 and the voltage source V respectively will be

(A) 1,4 (B) 5,1

2. An ideal capacitor is charged to a voltage Vo and connected at t = 0 across an ideal inductor L. (The circuit now consists of a capacitor and inductor alone). If we let = √ , the voltage across the capacitor at time t > 0 is given by

(A) Vo

(B) Vocos ( t) (C) Vosin ( t) (D) Voe cos ( t) EE - 2007

3. In the circuit shown in figure switch is initially CLOSED and Sw is OPEN. The inductor L carries a current of 10 A and the capacitor is charged to 10 V with polarities as indicated. Sw is initially caps at t = 0 and Sw is OPENED at t = 0.

The current through C and the voltage across L at t = 0⁺ is

(A) 55 A, 4.5 V (B) 5.5 A, 45 V

shown in the network shown below in terms of the voltage Vx and the

C 10V SW1

R2 0Ω

L 10A R1= 0Ω

SW2

+ _

I R

R R

(t)

+ 2i 0u(t)

+ i(t)

t 0

( ) i(t)

t 0

( )

independent source V, is given by

(A) = 1.4 Vx – 3.75I1 + V (B)

= +1.4 Vx – 3.75I1 V (C) = 1.4 Vx + 3.75I1 + V (D) = 1.4 Vx + 3.75I1 V

EE - 2008

Statement for Linked Answer Questions 5 and 6

The current i(t) sketched in the figure flows through an initially uncharged 0.3 nF capacitor.

5. The charge stored in the capacitor at t = 5 µs, will be

(A) 8nC (B) 10nC

6. The capacitor charged upto 5 ms, as per the current profile given in the figure, is connected across an inductor of 0.6 mH.

Then the value of voltage across the capacitor after 1s will approximately be (A) 18.8 V

(B) 23.5 V

will be

(A) ⁄ s (B) 4⁄ s

8. In the figure shown, all elements used are ideal. For time t<0, S remained closed and S open. At t = 0, S is opened and S is closed. If the voltage Vc2 across the capacitor at t = 0 is zero, the voltage across the capacitor combination at t=0⁺ will be

(A) 1V (B) 2 V

Linked Answer Questions 9 and 10

The L-C circuit shown in the figure has an inductance L = 1mH and a capacitance C = 10µF.

9. The initial current through the inductor is zero, while the initial capacitor voltage is 100 V. The switch is closed at t = 0. The current i through the circuit is:

(A) 5 cos( 0 t) A (B) 5 sin( 0 t) A (C) 10cos( 0 t) A (D) 10 sin( 0 t) A

t = 0

100 V 100 V

+ C

S1 S2

3V C1 1F C2 2F

1F 1F

0 1 2 3 4 5 6 7 8 9

i(t) mA

t ( s) 1

2 3 6 5 4

+ +

0 2H

0 2

0 H +

10. The L – C circuit of Q.9 is used to commutate a thyristor. Which is initially carrying a current of 5 A as shown in the figure below. The values and initial conditions of L and C are the same as in Q.9. The switch is closed at t=0. If the forward drop is negligible, the time taken for the device to turn off is

(A) 2 s (B) 6 s

11. The switch in the circuit has been closed for a long time. It is opened at t = 0.

At t = 0 , the current through the 1µF capacitor is

(A) 0A (B) 1A

Common Data For Q.No 12 & Q.No 13 An RLC circuit with relevant data is given below

12. The power dissipated in the resistor R is (A) 0.5 W

(B) 1 W

(A) 2 (B)

√

√ (D) +j2A

EE/IN - 2012

Statement for Linked answer question 14 and 15:

In the circuit shown, the three voltmeter reading = 220V, = 122 V, =136 V

14. The power factor of the load is (A) 0.45

(B) 0.50

15. If RL = 5, the approximate power consumption in the load is

(A) 700W (B) 750W

16. The switch SW shown in the circuit is kept at position ‘ ’ for a long duration t t = 0 the switch is moved to position ‘2’

Assuming , the voltage (t) across the capacitor is

(A) (t) = ( e^⁄) (B) (t) = ( e^⁄) +

(D) (t) = ( ) ( e^⁄) +

17. The voltage across the capacitor, as shown in the figure, is expressed as

(t) = sin( t )

+ sin ( t ) The values of and respectively, are

2 S

~

R oad R

~

R = 0

= √2 4

= √2 4 t =0

5V Ω

1µF 4Ω L

t=0 C 100V

100V 5A 20

(A) 2.0 and 1.98 (B) 2.0 and 4.20

1. In the circuit shown in the following figure, the input voltage vi(t) is constant at 2V for time t s and then it changes to 1 V. The output voltage, v0(t), 2 s after the change will be

(A) – exp ( 2) V (B) – 1 + exp ( 2) V (C) exp ( 2) V (D) 1 – exp ( 2) V IN - 2007

2. In the circuit shown in the figure, the input signal is ( ) = + cos

The steady – state output is expressed as

(t) = P + Q cos( t ). If R = 2, the values of P and Q are

(A) P = 0 and Q = 6/√

(B) P = 0 and Q = 3/√

(D) P = 5 and Q = 3 IN - 2008

Statement for Linked Answer Question 3 and 4:

In the circuit shown below the steady-state is reached with the switch K open subsequently the switch is closed at time t=0.

3. At time t=0 , current I is (A) ( ⁄ )

(B) 0A

4. At time t =0 , is

is (A) 5A / s

(B) ( 0 ⁄ ) S

5. In the dc circuit shown in the adjoining figure, the node voltage V2 at steady state is

(A) 0V (B) 1V

6. In the circuit shown below, the switch, initially at position 1 for a long time, is changed to position 2 at t = 0.

0 F 2k

+ k

20 F

F 0 2

+ +

t = 0

(t) +

Fig.

M 1µF

Vo(t) Vi(t)

+ +

2 1 0 Vi(t)

~

^{0 sin t}

20 sin 0t

(t) F

The current i through the inductor for t 0 is

(A) e (B) + e

7. In the circuit shown below, the current through the inductor is

(A) A (B) A

IN - 2014

8. The circuit shown in figure was at steady state for t < 0 with the switch at position

‘ ’ The switch is thrown to position ‘ ’ at time = 0, . The voltage V (volts) across the 0 resistor at time

t = 0 is _________________

9. The average real power in watts delivered to a load impedance = (4 j2) by an ideal current source

i (t) = 4 sin ( t + 20 ) is ______________

10. In the circuit shown in the figure, initially the capacitor is uncharged. The switch ‘S’

is closed at = 0. Two milliseconds after the switch is closed, the current through the capacitor (in mA) is _____________

11. In the microprocessor controlled measurement scheme shown in the figure, R is the unknown resistance to be measured, while R and known.

is charged from voltage to (by a constant DC voltage source ), once through Rin T seconds and then discharged to . It is again charged from voltage to through R in T seconds.

If T = kT then

(A) R = kR( ) (B) R = kRln ( )

12. capacitor ‘ ’ is to be connected across the terminals ‘ ’ and ‘ ’ as shown in the figure so that the power factor of the parallel combination becomes unity. The value of the capacitance required in F is___________.

j 4 60

0 Hz

0 Hz 0 Hz

~

R analog interface Sense harge

control Sense

ischarge P

t = 0 2 k 2 k

4 F S

i(t)

6 t = 0

0 H

~

0 0 0

+ +

0Ω

10V 4A 1H

t = 0

i 1

Answer Keys and Explanations

ECE

1. [Ans. D]

The transform circuit is shown below

For the given circuit ^{( )}

( )= ^{( )}

( ) ( ) Where (S) =

, Where, R = 4 , = F

(S) =

, Where, R = ,

= 4 F

R = R = 0 4 0 sec = 4 0sec

For t 0u(t) = s = 0 (for dc signal)

( ) ( ) =

v (S) = (S) , v (t) = 0 v(t) for v(t) = 0 for t 0, v (t) = 2. [Ans. A]

At t = 0 Capacitor is short circuit and at t = apacitor is open circuit

So (0 ) =

= 0 m ( ) = 0

Time constant of the circuit = R = 4 F 20 k 20 = 40 ms Using direct formula

(t) = ( ) ( ) (0)]e (t) = 0 (0 0 )e (t) = 0 e m

3. [Ans. C]

The waveform of voltage Vc(t) is shown below.

In mathematical form,

(t) = tu(t) 2(t T)u(t 2T) + 2(t 2T)u(t 2T)………

= tu(t) + 2 ∑( ) (t nT)u(t nT)

4. [Ans. D]

( )

( )= ^⁄

( ⁄ )=

( ) (t) = δ(t) (S) = (S) =

(S + 2⁄ ) + (^√) (t) =

√ e^⁄ Sin (^√t) 5. [Ans. B]

(S) = R (S) = ( S (s)) = S (S) (S) = S

(S + S + ) = ⁽^⁄^{⁄ )}

( ) (√ ⁄ )

(t) = e^⁄ os (^√t)

√ e^⁄ Sin (^√t) 6. [Ans. B]

In the circuit shown,

= 0 = R

(R + R ) R= (R + R ) T = R⁄ = (R + R⁄ )

di dt|

=T=

R⁄(R + R )

(R + R )

⁄ = R

t Vc(t)

-1 T

2 T

(s) +

(s)

7. [Ans. A]

Take L.T of given differential equation with i(0)= and use the L.T pairs:

sin(t) u(t)→ s + e sin(t) u(t)

→

(s + ) + ,

L^{( )} + Ri = ( + esin t) u(t) Take Laplace on both side

[s (s)

R] + R (s) =

s +

(s + ) + r ( s + R) (s) = + + +

( ) (s) = [ + +

( ) ]

( ) According to final value theorem, Steady state value of

i(t) = i(t)

→ = t

→ s (s)] = R 8. [Ans. B]

hen S is in position ‘a’, (0 ) = 00 fter S is moved to b, (for t 0)

= 00 = 0 R = =0 0 2

= 0 6 F

(t) = 00 e^⁄= 00e i(t) = (t)

= 20e u(t)m 9. [Ans. A]

S is open for < ≤ 0

At t = 0 the status of the circuit is shown in Fig. 1

i (0 ) = 0 , i(0 ) = 0 S is closed for 0 ≤ t <

At t = 0 , i (0 ) = i (0 ) = 0 The status of the circuit is shown in Fig. 2

(0 ) =

t t = , the status of the circuit is shown in Fig. 3

i( ) = 0

The given circuit is a first order circuit with time constant, =

R = 0 + ( 0 0) = 0 + = = mH = 0H

= =

= 0sec, = 0 sec General formula:

i(t) = + (F ) ( e) , t 0 i(t) = + (

2 ) ( e)

= 0 0 2 e , t 0 10. [Ans. A]

Q(0 ) = 2 m

(0 ) =⁽ ⁾== 0 For t 0, = 0 = 00 = R = 0 m sec

(t) = ( 00 0 e)u(t) i (t) = d

= 0 0 0 2000 e u(t)

= e u(t)

11. [Ans. D]

When the switch in closed at t = 0

Capacitor C1 will discharge and C2 will get charge since both C1 and C2 are ideal and there is no-resistance in the circuit

0Ω

0.5 A 0Ω

i( ) 0Ω

1.5 A S

0.5 A 0.5 A i ( )

0Ω

i (0 ) i(0 )

0Ω 1.5 A

4 0Ω

0Ω

0Ω 0.75 A 0.75 A

i (0 )

15mH i(0 )

0Ω 1.5 A

S 0.75 A

charging and discharging time constant will be zero.

Thus current will exist like an impulse function.

12. [Ans. D]

Across AB voltage drop is 10V

= 13A, = 20V 2 = ++= 0

= 0 + 2 = 13A 13. [Ans. C]

current through = = 0

14. [Ans. *] Range 1.2 to 1.3

urrent after long time = 0

0k= m At t = 0 the voltage across the capacitor is 10

Where, = = 0 = Now at t= 0

The current is

4m = 2 m

15. [Ans. *] Range 0.45 to 0.47

A series LCR circuit operating frequency is such that current leads the supply voltage so

The current at resonance = = amp R = = H = F

The magnitude of current is the value at resonance

= 2

=√R + (x x ) =

√ + ( )

2 = √ + ( )

( ) =

= √ as ] + √ = 0

= √ + √ + 4 2 = 0 4 rad/sec

(omitting the negative value) 16. [Ans. A]

= 2000i(t) i = cdv

dt = 0 002di dt i= [0 002di

dt+ i(t)] amp = 2di

dt+ 000i(t) + =

2di

dt+ 000i(t) + 2000i(t) = s 2di

dt+ 000i(t) + 2000i(t) = s 2 [s (s) + 000

2 (s)] = s

⁺ F

i(t) i 2k

k i

+ 4k 0 0 F

k 4k k

+ mH

2Ω Ω

10V 2A

+ 20 _

0 +

Ω 0

2 = 0

= 0

2 (s) [s + 000 2 ] =

s (s) =

2s(s + 00)

=2[

s s + 00] 00

=600[

s s + 00] amp

= [

s s + 00] m i(t) = e]

= * e + where is in m sec

= [ e], =2 msec

17. [Ans. A]

= Ri(t) + ∫ i(u)du

= R +

S aplace transform ] =

= is a D.C voltage source, so = where depends on

= S k s(RS + )=

R (s +) i(t) =k

R e

18. [Ans. *] Range 31.24 to 31.26

Since we have to evaluate (t) in steady state the inductor will behave as a short circuit and hence = i

By Nodal analysis at node A 0 + + i = 0 0 + + i = 0

= 0 i =

(t) = i = == 2

1. [Ans. D]

At steady state, inductor acts as short circuit & capacitor acts open circuit

Current through R = 5/1 =5A

Current delivered by 5V source = 5 1

= 4A 2. [Ans. B]

The relevant circuit is shown in fig.

It is a standard LC circuit.

With v (t) = cos( t) or sin( t + 0 ) 3. [Ans. D]

Equivalent circuit at t = 0 is,

By nodal analysis,

0 +^{( )}

= 0 2 0 = 00 = =^{( )} = 4

10A I

0Ω 10V

L 0Ω +

(t) C L

Fig.

2Ω

Ω _ + 5V

(t)

+ 2i 0u(t)

+ ( )

4. [Ans. A]

V-3( + ) 0 = 0 …… (1) 0 = 0.2

+ 0 2 = 0……… (2)

Eliminating I2 for eq(1) and (2) we get

= 4 + 5. [Ans. C]

Charged stored in the capacitor = area under i – t curve

Q = +

=2(2 0) (4 0) +

2(4 + 2) 0 ( 2) 0

= [4 +6

2 ] 0 = n 6. [Ans. D]

apacitor charged upto s, so total charge stored in capacitor = Q = 13nC Voltage across the capacitor before connecting to indictor

= 0

0 0= 4 Voltage across the capacitor at time t

(t)at t = s

(t) = cos t]

t =_√

0 = 2.357 rad =

(t) = 4 cos 0 6

7. [Ans. C]

= F ; R = 6Ω T = R = 4sec

8. [Ans. A]

The status of the circuit at t = 0 is shown in fig. 1.

The status of the circuit at t = 0 is shown in Fig. 2.

= ==

9. [Ans. D]

Initial current through the inductor is zero and capacitor voltage is charged upto to voltage

(0 ) = 00

As current through inductor and voltage across cannot change abrupty.

So, after closing the switch i (0 ) = i (0 ) = 0

And (0 ) = (0 ) = 00 The circuit is s – domain

(s) =

(s +

)

= 00

(s +)

= 00√ ( √ ⁄ s + ( √ ⁄ ) ) I(s

) sL

1/s C

(0 ) s = 00

s 3 V

3 V

B +

Fig. 2 +

V 3 V 3 V

1 F 2 F

B +

Fig. 1 i(t)

mA 4 2

A1 A2 s s 2 s

Taking inverse Laplace transform i(t) = (s)]

= 00√ sin

√ t

00 √

sin (

√ t) i(t) = 10sin (10⁴t)A

10. [Ans. A]

When the switch is closed, I flows through thyristor.

Load current = = Net current through thyrostor i = i

i = sin 0

Let at t=T, circuit get turned off and current i becomes zero

i = 0 sin 0 T = 0 0 sin T =

sin 0 T = 0

0 T = 0 or 0 2 rad T 2 s

11. [Ans. B]

(0 ) = 4 (0 ) = 4 i (0 ) = ⁽ ⁾= A

12. [Ans. B]

Power supplied by the source = cos Where = angle between

= 4 inductor and capacitor do not consume power.

Therefore, power dissipated in R = Power supplied by the source

P = cos

= √2 cos 4

= √2

√2= 13. [Ans. D]

Using KCL,

+ + = 0 =

= √2 √2 = 2 0

= + 2 14. [Ans. A]

Pharor diagram = 220 = 122 V = 136V

⃗⃗⃗ = ⃗⃗⃗ + ⃗⃗⃗

By parallelogram law of addition of vectors

= + + 2 cos by using options, cos = 0.45 15. [Ans. B]

: R =

= cos = 136 x 0.45 [From Q 4]

= 61.2V

P = = 749W = 750W

16. [Ans. D]

At t = 0

Voltage across capacitor = t t 0

+ iR + i R + = 0

+ 2cd

dt R + = 0 Taking Laplace

s + (2Rc) s (0)] + = 0

s + 2Rcs 2R + = 0 ( + 2Rcs) = 2R

I 𝛟

VC(t)

100V 20

i iT

(s) = 2R + 2Rcs

S( + 2Rcs) (s) = 2R

+ 2Rcs+ 2Rc + 2Rcs

(t) (e) = e + e +

* (t) = ( + ) ( e ) + +

17. [Ans. A]

Applying superposition theorem Resolving 0 sin t source

(t) =

√ ()

sin ( 0t + tan() 0) Removing 20 sin 0t voltage source

= +

= + j

= 0 sin( t) ( )

= 0 sin( t) + j

= 0

√ + 2 sin t tan( )]

(t) = 20

√ 00 + sin( 0t ( 0 tan(

0))

+ 0

√ + 2 sin( t tan ) IN

1. [Ans. A]

Input is 2V for t 1s.

Time constant ( ) = = 0 0

= So (0 ) = 2

Laplace Equivalent circuit after the change

(s) = s 2 s R +

= s

(s) = Rsc + (s) = c

+ s= c e

Now voltage across Resister at t = 2s (t) = (t) R

( )= ce R = Rc e _{( )}= e ] 2. [Ans. A]

The circuit is shown in Fig.

(t) = + cos( t)

Due to 5V d.c alone , C is open and no current flows through R,

= 0 and = 0 ue to cos ( t) alone, with

(t) +

Fig.

1/sc

1/s

2/s

I(s) R

M 1µF

Vo(t) Vi(t)

+ +

2 1 0 Vi(t)

H( ) = = R

According to Superposition Theorem v (t) = v(t) + v(t) relevant circuit is shown in Fig. 3

iL (t) andv (t) cannot change

From Fig. 3: Write the Outer loop equation: steady state is shown in Fig.2

Apply voltage division across R1 = 2k

Node voltage, V1= = = V1 is divided between C1 = 0 F and C2 = 20 F

Apply, again voltage division across C1

and C2

+ = 0 0= 6. [Ans. D]

For t < 0, the status of the circuit is shown in Fig. 1 inductor behaves as short circuit after a long time.

i(0 ) =

for t 0, the status of the circuit is shown in Fig. 2

Current through the inductor cannot change instantaneously.

i(0 ) = i(0 ) = initial value (I.V) After a long time inductor behaves as short circuit.

i( ) = 2 = final value (F. V) Time constant = = , = 20

i(t) = + (F ) ( e⁽⁾) for t 0

= + (2 )( e) = 2 e

7. [Ans. C]

From the given circuit, = 0

= j = ( )

8. [Ans. ] t t = 0

T =6

2= = (0 ) At t= 0

= 0 = 0 ] 9. [Ans. 32]

Load impedance = = (4 j2) Ideal current source

i(t) = 4 sin ( t + 20 ) Average real power = i

= (

√ ) 4 Power = 6

2 4 = 2 6

(inductor in steady state) 6

0 +

+ +

0 0

0 ( )

10 10

4 A 1 H

Fig. 2 Fig. 1

10 V 1 H

10. [Ans. *]Range 1.5 to 1.6 at t = 0 , switch is open so

(0 ) =

at (t = )so capacitor is in steady state

( ) = 2k

4k= 2 We know that

(t) = (0 ) v ( )]e+ ( ) = 0 2 ]e+ 2

= (2k

2k) 4

= k 4 = 0 4 0 = 4 0s

= 000

4 = 2 0

(t) = 2 e+ 2 (t) = 2 2 e we know that

i (t) = dv (t) dt i (t) = cd

dt 2 2 e] = 2 c( 2 0)e = 4 F

i (t) = +2 4 0 2 0 e i (t) = 2 0e i (t) = m ]

11. [Ans. C]

Time constant of RC Circuit is = R In case I

Time constant T = R

In case II

Time constant (T ) = R Given T = k T

So R = k R

R = k R]

12. [Ans. *] Range 186 to 188

This is a tank circuit configuration of a Parallel R-L-C circuit.

Resonant frequency of tank circuit f =√

= 4 f =

( ) =

( ) +( ) = R = 4 , = f = 0 H

+= =

=2 0 = F

j 4

60 0 Hz

0 Hz 0 H

0 Hz

( )

In document GQB_IN.pdf (Page 173-191)