An AC source of rms voltage 20 V with internal impedance ZS = (1 + 2j) feeds a load of impedance ZL = (7 + 4J) in the figure below. the reactive power consumed by the load is

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An AC source of rms voltage 20 V with internal impedance ZS = (1 + 2j) feeds a load of impedance ZL = (7 + 4J) in the figure below. the reactive power consumed by the load is ?

Question and Answers given below (with solution) :

  1. 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) = 0.5 – 0.125 e-1000t A

(b) I(t) = 1.5 – 0.125 e-1000t A

(c) I(t) = 0.5 – 0.5 e-1000t A

(d) I(t) = 0.375 e-1000t A

  1. The current I in the circuit shown is

(a) -j1A

(b) j1A

(c) zero

(d) 20 A

  1. In the circuit shown, the power supplied by the voltage source is

(a) zero

(b) 5 W

(c) 10 W

(d) 100 W

  1. An AC source of rms voltage 20 V with internal impedance ZS = (1 + 2j) feeds a load of impedance ZL = (7 + 4J) in the figure below. the reactive power consumed by the load is

(a) 8 VAR

(b) 16 VAR

(c) 28 VAR

(d) 32 VAR

  1. 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) 0.2 e-125tu(t) mA

(b) 20 e-1250t u(t) mA

(c) 0.2 e-1250tu(t) mA

(d) 20 e-1000t u(t) mA

  1. In the circuit shown, what value of RL maximizes the power delivered to RL?

(a) 2.4

(b) 8/3

(c) 4

(d) 6

  1. The thevenin equivalent impedance Zth between the nodes P and Q in the following circuit is

(a) 1

(b) 1 + s + 1/s

(c) 2 + s + 1/s

(d) s2 + s + 1/s2 + 2s + 1

  1. The driving point impedance of the following network is given by Z(s) = 0.2s/s2 + 0.1s + 2 the component values are

(a) L = 5 H, R = 0.5, C = 0 1F

(b) L = 0.1 H, R = 0.5, C = 5 F

(c) L = 5 H, R = 2, C = 0.1 F

(d) L = 0.1 H, R = 2, C = 5 F

  1. In the following circuit the value of open circuit volage and thevenin resistance at terminal ab are

(a) VOC = 100 V, RTH = 1800

(b) VOC = 0 V, RTH = 270

(c) VOC = 100 V, RTH = 90

(d) VOC = 0 V, RTH = 90

  1. In the following linear circuit it is given that VAB = 4 V for RL 10 K and VAB = 1 V for RL = 2 K.

The values of thevenin resistance and voltage for the network N are

(a) 16 k, 30 V

(b) 30 k, 16 V

(c) 3 k, 6 V

(d) 50 k, 30 V

  1. Consider the following linear circuit. three sources have fixed value only IS1 can be varied.

It is given that if IS1 = 3 mA then VOUT = 16 V and if IS1 = 1 mA then VOUT = 8 V. if IS1 = 0.5 mA, then the value of VOUT is

(a) 6 V

(b) 8 V

(c) 4 V

(d) 12 V

  1. For the circuit shown in figure below the value of RTH is

(a) 100

(b) 136.4

(c) 200

(d) 272.8

  1. Consider the following circuits shown below.

The relation betwwwn IA and IB is

(a) IB = IA + 6

(b) IB = IA + 2

(c) IB = 1.5IA

(d) IB = IA

  1. 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 2n T < t <(2n + 1) T, (n = 0,1,2,……..) S1 to P1 and S2 to P2

For (2n + 1) T < t < (2n + 2) T, (n = 0, 1, 2,……) S1 to Q1 and S2 to Q2

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

(a) n – 0 (-1)n tu (t – nT)

(b) u(t) + 2 n = 0 (-1)n u (t – nT)

(c) tu (t) + 2 n = 0b(-1)n (t – nT) U (t – nt)

(d) n = 0 [0.5 – e-(t – 2nt) + 0.5 e-(t – 2nT – T)]

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

  1. For t > 0, the output voltage VC (t) is

(a) 1/3 (e-2/3t – e3/2t)

(b) 2/3 te-1/2t

(c) 2/3 e-1/2 t cos (3/2t)

(d) 2/3 e-1/2t sin (3/2t)

  1. For t > 0. the voltage across the resistor is

(a) 1/3 (e-3/2t – e-1/2t)

(b) e-1/2t [cos (3t/2) – 1/3 sin (3/2)]

(c) 2/3 e-1/2t sin (3t/2)

(d) 2/3 e-1/2t cos (3t/2)

  1. Two series resonant filters are as shown in the figure. let the 3 dB bandwidth of filter 1 be B1 and that of filter 2 be B2 . the value of B1/B2 is

(a) 4

(b) 1

(c) 1/2

(d) 1/4

  1. For the circuit shown in the figure, the thevenin voltage and resistance looking into X-Y are

(a) 4/3 V, 2

(b) 4 V, 2/3

(c) 4/3 V, 2/3

(d) 4V, 2

  1. In the circuit shown VC is 0 volt at = 0 s. for t < 0, the capacitor current IC (t) where t is in second, is given by

(a) 0.50 exp (-25t) mA

(b) 0.25 exp (-25t) mA

(c) 0.50 exp (-12.5t) mA

(d) 0.25 exp (-6.25t) mA

  1. In the AC network shown in the figure, the phasor voltage VAB (in volt) is

(a) zero

(b) 5 < 300

(c) 12.5 <300

(d) 17 <300

  1. In the following circuit capacitor is initially uncharged. At t = 0+ the value of d2vc/dt2 and d3vc/dt3

(a) 2 V/s2, -8 V/S3

(b) -8 V/S2 , 26 V/S3

(c) -2 V/S2, 8 V/S3

(d) 8 V/S2, -26 V/S3

  1. A square pulse of 3 V amplitude is applied to CR circuit shown below. the capacitor is initially uncharged. the output voltage V2 at time t = 2 s is

(a) 3 V

(b) -3 V

(c) 4 V

(d) -4 V

  1. In the given circuit below initially capacitor is uncharged. the VC (t) for t >0 is

(a) (8 – 8e-t) V

(b) (8 + 8e-t) V

(c) 8 V

(d) (-8 + 8et) V

  1. In the circuit given below VIN (t) = 10 u(t) , the current IL (t) is

(a) (-0.01 + 0.01e-5000t)A

(b) 0.1 A

(c) (-0.1 – 0.1e-5000t) A

(d) 0.2 A

  1. In the following circuit the initial values are VC1(0) = 6 V and VC2 (0) = 24 V. the voltage VO(t) for t > 0 is

(a) 38 – 8e -t/1.2 V

(b) 8 + 22e -t/1.2 V

(c) 8 + 22e – 3.75t V

(d) 38 – 8e -t/1.2 V

Answers with solution

  1. (a)

at steady state,

inductor becomes short-circuit.

I(t) at steady state

I(t) = 1.5/3 = 0.5

LEQ = 15 mH

REQ = 5 + 10 = 15

L = LEQ/REQ = 15 x 103/15 = 1/1000

I(t) A – (A – B)e-t/10

= 0.5 – (0.5 – B) e-1000t

= 0.5 (0.5 – 0.375) e-1000t

I(t) = 0.5 – 0.125 e-1000t

when switch is closed for a long time then inductor is short and current through inductor 1.5/2 = 0.75 A

current through inductor remains same after closing the switch at t = 0.

so, current through 10

= 0.75/2 = 0.375

  1. (a)

XEQ = sL + R x 1/sC/ R + 1/sC

= sL + R / 1 + sRC

IO = V/XEQ

I = XC/XC + R IO

= 1/sC/1/sC + R x V/sL (1 + sRC) + R/1 + sRC

= 1/1 + sRC x V/sL (1 + sRC) + R/(1 + sRC)

= V/ sL (1 + sRC) + R

= V/j x 103 x 20 x 10-3 (1 + j x 103 x 50 x 10-6 + 1)

V/20j (1 + j50 x 10-3) + 1

= V/20j – 1 + 1 = 20/20j = – j1A

  1. (a)

let current supplied by voltage source. the current in different branch is indicated in circuit using Kirchhoff’s current law.

applying KVL in outer loop,

10 – (I + 3) x (1 + 1) – (I + 2) x 2 = 0

10 – 2(I + 3) – 2 (I – 2) = 0

I = 0

VI = 0

  1. (b)

current I = V/ZL + ZS = 20 <00/8 + 6J

= 20/82 + 62 = <00/<tan-13/4

= 20/10 < – tan-13/4

2 < – tan-13/4

power consumed by load = |I|2ZL

= 4 (7 + 4J)

= 28 + J16

The reactive power = 16 VAR

  1. (c)

CEQ = 0.8 x 0.2/0.8 + 0.2 = 0.16

when the switch is at position a for a long time then voltage across capacitor

VC(t = 0) = 100 V

as capacitor acts as open-circuit

at t > 0

the discharging current

I(t) = VO/R e-t/rc for t > 0

= 100/5000 e-t/5 x 103 x 0.16 x 10-6

= 0.020 e-125t A

on comparing RL = 0.2

RLC = 1

L = 0.1 H

R = 2

C = 5 F

  1. (c)

Maximum power delivered to RL when load resistance equals to thevenin resistance of circuit.

RL = RTH = VOC/ISC

due to open-circuit VOC = 100 V

ISC = I1 + I2

Applying KVL in lower loop.

100 – 8I1  = 0

I1 = 100/8 = 25/2

VX = – 4I1 = – 4 x 25/2 = – 50 V

100 + VX – 4I2 = 0

I2 = 100 – 50/4 = 25/2

ISE = I1 + I2

= 25/2 + 25/2 = 25

RTH = VOC/ISC = 100/25 4

RL = RTH = 4

  1. (a)

to calculate thevenin resistance, all the current sources get open-circuited and voltage sources short-circuited. writing in impedance of inductor and capacitor.

RTH = (1/S + 1) ||(1 + S)

(1 + S) (1/S + 1)/1 + S + 1/S + 1 = (1/S + 1 + 1 + S)/(1/S + 1 + 1 + S)

RTH = 1

  1. (d)

dividing point impedance = R ||1/sC||sL

= {(R) (1/sC)/R + 1/sC}||sL

(R/1 + sRC) = sRL/S2RLC + sL + R

R/1 + sRC + sL

Z(s) = 0.2s/s2 + 0.1s + 2

  1. (d)

by writing loop equation for the circuit,

VS = VX,IS = IX

IS = Test source to find out thevenin equivalent

VS = 600(I1 – I2) + 300 (I1 – I2) + 900(I1)

VS = (600 + 300 + 900)I1 – 600 I2 – 300 I3

VS = 1800I1– 600I2 – 300I3

I1 = IS,I2 = 0.3 VS

I3 = 3IS + 0.2VS

VS = 1800IS – 600(0.01VS) -300 (3IS + 0.01VS)

VS = 1800IS – 6VS – 900IS – 3VS

10VS = 900IS

VS = 90I2

For thevenin equivalent

VS = RTH IS + VOC

So, thevenin voltage VOC = 0 V

Resistance RTH = 90

  1. (b)

the circuit is shown in figure below.

when RL = 10 K and VAB = 4 V,

Current in circuit

I – VAB/RL = 4/10 = 0.4 mA

thevenin voltage is given by

VTH = I(RTH + RL)

VTH = 0.4 (RTH + 10)

VTH = 0.4 RTH + 4…………..(1)

Simiarly,

RL = 2 K, VAB = 1 V

I = 1/2 = 0.5 mA

VTH = 0.5 (RTH + 2)

VTH = 0.5RTH + 1……………(2)

From eq. (1) and (2)

0.1RTH = 3

RTH = 30 K

VTH = (12 + 4) = 16 V

  1. (a)

for a linear circuit, we can write

VOUT = AIS1 + BVS1 + CIS2 + DIS3

= AIS1 + (BVS1 + CIS3 + DIS3)

VOUT = AIS1 + K

(VS1, IS2 and IS3     are fixed)

16 = A x 3 + K

8 =  A x 1 + K

2A = 8

A = 4

4 + k = 8, K = 4 V

so, the equation for VOUT is

VOUT = 4IS1 + 4

IS1 = 0.5 mA

VOUT = 4 x 0.5 + 4 = 6 V

  1. (a)

The circuit is shown below,

IX = 1 A, VX = VTEST

VTEST = 100(1 – 2IX) + 300 (1 – 2IX – 0.01VX) + 800

VTEST = 1200 – 800IX – 3VTEST

4VTEST = 1200 – 800 = 400

VTEST = 100 V

RTH = VTEST/1 = 100

  1. (c)

in circuit (b) transforming the 3 A source is to 18 V source all sources are 1.5 times of that in circuit (a). hence, IB = 1.5 IA.

  1. (c)

for 2nt < t < (2n + 1) t, n = 0, 1, 2, 3…………….

the capacitor gets charge to its peak value. as the constant current is flowing, the voltage across capacitor is ramp function.

for (2n + 1) T < t (2n + 2) T

the capacitor get discharge from peak value with same slope as for charging to zero value.

  1. (d)

in frequency domain,

VC(S) = t/s/1/s + 1 + s

VI (t) = s(t)

VI(s) = 1

VC(s) = 1/s2 + s + 1

= 2/3 |B/2/(s + 1/2)2 + (3/2)2|

taking inverse laplace transform,

VC(t) = 2/3 sin (3/2 t) .e-1/2t

  1. (b)

in frequency domain

VR(s) = 1. VI(s)/1/s + 1 + s

s/s2 + s + 1

VR(s) = s + 1/2 – 1/2/(s + 1/2)2 + 3/4

=  (s + 1/2)/(s + 1/2)2 + (3/2)2 – 1/2 /(s + 1/2)2 + (3/2)2

taking inverse laplace transform,

VR(t) = e-t/2 cos 3/2 t -1/2 x 2/3 e-t/2 sin 3/2 t

= e-t/2 [cos 3/2 + 1/3 sin 3/2 t]

  1. (d)

for series resonant circuit, 3 dB bandwidth is R/L

B1 = R/L1

B2 = R/L2 = R/L1/4 = 4R/L1

B1/B2 = 1/4

Note  bandwidth of series resonant circuit is independent from value of capacitor.

RTH = VOC/ISC

VTH = VOC

Applying KCL at node A,

2I – VTH/1 + 2 = I + VTH/2 ……………(1)

I = VTH/1

Putting 2VTH – VTH + 2 = VTH + VTH/2

VTH = 4 V

When XY get shorted, 2A current flows through short-circuited path.

RTH = 4/2 = 2

  1. (a)

the capacitor voltage

VC(T) = VC(00) – (VC(00) – VC(0)) e-t/req c

REQ = 20 K || 20 K

= 10 K

VC (00) = 10 x 20/20 + 20 = 5 V

given VC(0) = 0

VC(t) = 5 – (5 – 0) e-t/10 x 4 x 10-6 x 103

VC(t) = 5 (1 – e-25t)

IC(t) = C dVC(t)/dt = 4 x 10-6 d/dt 5 (1 – e-25t)

= 4 x 10-6 x 5 x 25 e-25t

IL(T) = 0.50 e-2.5t mA

  1. (d)

equivalent impedance

= (5 + j3) ||(5 – 3)

= 5 + j3) (5 – j3)/5 + j3 + 5 – j3

= 25 + 9/1 = 3.4

VAB = Current x impedance

= 5 <300 x 3.4

= 17 <300

  1. (b)

at t = 0+, VC(0+) = 0

VC/20 + VC – e-1/10 + 1/20 . dVC/dt = 0 ……………(1)

dVC/dt + 3VC = 2e-t ……………….(2)

at t = 0+ dVC(0+)/dt = 2 v/s

differentiating eq. (1), d2VC/dt2 + 3 dVC/dt = – 2e-t …………..(3)

at t = 0+, d2VC(0+)/dt2 = -2 – 6 = – 8 v/s2

differentiating eq. (ii), d2VC/dt3 + 3 d2VC/dt = 2e-t

at t = 0+, d3vc(0+)/dt2 = 2 + 24 = 26 v/s3

  1. (b)

RC = 0.1 x 10-6 x 103 = 10-4 s

as RC is very small, so steady state will be reached in 2 s.

thus, VC = 3 V and V2 = -VC = – 3 V

  1. (d)

by source transformation, we have equivalent circuit shown below.

thevenin equivalent for above circuit can be obtained as (by putting a test source)

by writing KVL,

VS – IS 0.5 – 9VX + 8 – 500IS = 0

VS = 5IS + 9VX – 8 + 500IS …………….(1)

VX = VS + 8 – 500IS

VS = 5IS + 9[VS + 8 – 500IS] – 8 + 500IS

VS = 1000IS + 64 + 9VS

VS = 125 IS – 8

For thevenin’s equivalent circuit

VS = RTHIS + VTH

VTH = – 8 V, RTH = – 125

The circuit is

VC(t) = VC(00) + [V (0) – V(00)]e-1/rc

VC(t) = – 8 + (0 + 8) e-t/-125 x 8 x 10-3

here, time constant is negative.

VC(t) = – 8 + 8et

  1. (a)

thevenin equivalent across the inductor for t > 0 can be obtained as

by applying node equation,

IS = 0.01VS + (VS – VA + VIN)/100

100IS = 1. VA + VS – VA + VIN

VS = – VIN + 100IS

For thevenin’s equivalent circuit,

VS = VTH + ISRTH

VTH = – VIN = – 10 V.

RTH = 100

t < 0, IL(0) = 0

IL(00) = -10/100 = – 0.1A

IL(t) = 0.1 [1 – e-t] where = time constant

= L/RTH = 20 x 10-3/100

IL(t) = 0.1 [1 – e-5000t]

= (-0.1 + 0.1 e-5000t) A

  1. (a)

let 1/sensitivity = 1/20 = 50 uA

for 0 – 10 V scale, RM = 10 x 20 = 200 k

for 0 – 50 V scale, RM = 50 x 20 = 1 m

for 4 V reading I = 4/10 x 50 = 20 uA

VTH = 20 RTH + 20 x 200 = 4 + 20uRTH……………..(1)

For 5 V reading I = 5/50 x 50 = 5uA

VTH = 5 x RTH + 5 x 1m = 5 + 5RTH ……………….(2)

Solving eqs. (1) and (2)

VTH = 16/3 V. RTH = 200/3 K