Electric Circuit and Fields

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Test Paper-1

electric circuits and fields

Source Book:GATE CLOUD Electric Circuits & Fields (Vol-1 and Vol-2)

Author: RK Kanodia & Ashish Murolia Edition:1st

ISBN: 9788192348360, 9788192348377

Publisher :JHNUJHUNUWALA

Visit us at:www.nodia.co.in

Q. No. 1 - 10 Carry One Mark Each

MCQ 1.1 Consider a network which consists of resistors and voltage sources only. If the

values of all the voltage sources are doubled, then the values of mesh current will be(A) doubled (B) same

(C) halved (D) none of these

SOL 1.1 Option (A) is correct.

From the principal of superposition, doubling the values of voltage source doubles

the mesh currents.

MCQ 1.2 In the circuit below, equivalent capacitance across terminal a-bis measured with

switch Sis open. If the switch is closed, the equivalent capacitance

(A) increases

(B) decreases

(C) remains unchanged

(D) none of these

SOL 1.2 Option (C) is correct.

When the switch is open


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Ceq( ) ( )( )

2 42 4

4 24 2

=+

++

68

68

= + F616

38

= =

When the switch is closed

eqCl ( ) ( )( )( ) ( )( )

F2 2 4 42 2 4 4

4 84 8

38

=+ + +

+ +=

+ =

eqCl Ceq=

MCQ 1.3 What is the time constant of the circuit shown in the figure below ?

(A) 24 s (B) 128 s

(C) 32 s (D) 96 sSOL 1.3 Option (A) is correct.

Time constant, R Ceq eq =

Ceq( )( ) 3 F

C CC C

2 62 6

21 21 2=+

=+

=

Req"Thevenin resistance across Ceq

Req 10 8 || 24 10 6 16 = + = + =^ h 16 24 s

23

#= =

MCQ 1.4 The value of R that satisfies the following circuit to exhibits critically damped

response, is


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(A) 88.4 (B) 176.8

(C) 353.6 (D) 125 k


For critically damped response

0=

RC21

LC

1=

L R C4 2=

5 103

#

R4 10 102 9

# # #=

R 353.6 =

MCQ 1.5 Which of the following is equivalent of the circuit shown in figure

SOL 1.5 Option (B) is correct.

I 10 Aj2

20 090

cc= = (Using source transformation)

Z 2j =

MCQ 1.6 For an ac circuit if

( )v t 112 ( 20 )cos Vt c= +

( )i t 8 ( 40 )cos At c=


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the average power and reactive power absorbed by the circuit respectively are

(A) 224 W, 388 VAR (B) 448 W, 3 VAR88

(C) 448 W, 153 VAR (D) 421W, 153 VAR

SOL 1.6 Option (A) is correct.Phasors, V 112 V20c=

I 8 A40c=

Complex power, S VI21

= )21 112 20 8 40# #c c=

448 60c=

224 387.97 VAj= +

Average Power, P 224 W=

Reactive power Q 388 VAR-

MCQ 1.7 What is the value of Leqin the circuit shown below ?

(A) 25 H (B) 15 H

(C) 5 H (D) 20 H

SOL 1.7 Option (C) is correct.To obtain equivalent inductance, put a test source at the input port

By Writing kVL

Mesh 1: V1 5 5 5j j jI I I1 1 2= + +

V1 10 5j jI I1 2= + ...(i)

Mesh 2: 0 3 2 5j j jI I I2 2 1= + +

0 5 5j jI I1 2= +

or, I1 I2=

Substituting above into equation (i)

V1 10 5( )j jI I1 1= +

V1 5j I1=

Input impedance Zin 5j j LI

Veq

1

1 = = =


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Leq 5 H=

MCQ 1.8 A series RLCcircuit has lower and upper half power frequencies as 100 /secrad

and 110 / secrad , quality factor is(A) 10 (B) 0.1

(C) 10.48 (D) 11


Resonant frequency, 0 1 2 =

104.88 / secrad100 110#= =

Bandwidth, B 2 1 =

110 100 10 /secrad= =

Quality factor, Q . 10.48 / secradB 10

104 880= = =

MCQ 1.9 An inductor of inductance LHenry and with an initial current I0is represented in

s-domain by

(A) an impedance Lsconnected in series with a voltage source LI0 .

(B) an impedance Lsconnected in parallel with a current sourcesI0b l.

(C) an impedance Lsconnected in series with a current source I0.

(D) both (A) and (B).

SOL 1.9 Option (D) is correct.

In the time domain

( )v t ( )

Ldt

di t=

Taking Laplace transform

( )V s [ ( ) ( )]L sI s i 0=

( )LsI s LI 0= ( ( ) )i I0 0=The equivalent circuit in s-domain


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Now, ( )V s ( )LsI s LI 0=

( )I s ( )

LsV s

sI0= +

The equivalent circuit in s-domain

MCQ 1.10 Which parameters are most useful for characterizing the circuit shown in figure

below ?

(A) z-parameter

(B) y-parameter

(C) h-parameter

(D) ABCD-parameter


The z-parameters are defined as

V1 z I z I 11 1 12 2= + ...(i)

V2 z I z I 21 1 22 2= + ...(ii)

From the above equations, the general equivalent network can be obtained as

shown below

From the above circuit equation (i) and (ii) can be obtained directly.

Now, comparing this circuit with given circuit

z11 Z1=

z12 0=


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z21 =

z22 Z2=

Q. No. 11- 21 Carry Two Marks Each

MCQ 1.11 The circulation of cos sinzA a az = + around the edge Lof the wedge given

below is

(A) 1 (B) 1

(C) 0 (D) 3

SOL 1.11 Option (A) is correct

As shown in the figure the integral will be taken in three segments as

dlAL

:# dlA321

:= + +b l###

C C C1 2 3= + +

Along 1st contour d dl a=

, C1 cos d00

2

= =

# 222

= =

Along 2nd contour d dl a = , C2 0dlA : =

Along 3rd contour d dl a=

, C3 cos d0

2

60 =

c =# 14

2

0

2

= =; E dlA

L:# C C C1 2 3= + + 1=


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Alternate method :

Since, we have to direction the integral for the closed contour so we can also

evaluate it by using stokess theorem as

dA lL :# dA S#d= ^ h#now A#d

a

A

a

A

a

A

1 z

z

z

=2

2

2

2

2

2

R

T

SSSS

V

X

WWWW

cos sinz

a a a

1

0

z

z

= 2

2

2

2

2

2

R

T

SSSS

V

X

WWWW

1cos sinz a a a1 0 0 1 0

= + + ^ ^h h6 @

cossin

za az

= +

d d dS az =

So, dA l:# sinA d d d d zS 0

60

0

2

= =c

==^ ^h h# ##

cos2

2

0

2

060

#

= c; 6E @ 1

24 0

21#= b ^l h: D

1=MCQ 1.12 The network function of circuit below is

( )H ( )( )

.jj

jV

V

1 0 014

i

o

= =+

The value of the Cand Ais

(A) 10 F , 6 (B) 5 F , 10

(C) 5 F , 6 (D) 10 F , 10


( )jVC /

/( )

j C

j CjV

2 10 1

1i3

#

=

+^ ^h h (Using voltage division)

( )jVC ( )

j C

jV

1 2 10i

3#

=

+ ...(i)


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( )jVo ( )A jV15k 30k30k

C = + (Using voltage division)

( )A jV

3

2C =

Substituting ( )jVC from equation (i)

( )jVo ( )

( )

j C

A jV

3 1 2 10

2 i3

#

=

+

Transfer function

&( )( )j

j

V

V

i

o

/

j C

A

1 2 10

2 33

# =

+

Comparing with given equation, we have

A3

2 4 6A&= =

2 10 C3# 0.01= C 5= F

MCQ 1.13 In the ideal transformer circuit shown in the figure, average power absorbed by

10 resistor will be

(A) 10 W (B) 20 W

(C) 40 W (D) 5 W


Applying mesh analysis

Mesh 1:


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Complex power S2 P jQ2 2= +

.

.sin cosj200 820 0 81= + 6 @ j20 15= +

Complex power of the source S 36 22.75jS S1 2= + = + 42.59 . kVA32 29c=

S 6 42.59 .V I V 32 29o o c#= = =)

Vo 7.1 . V32 29c=

MCQ 1.15 The circuit shown in figure (A) is reduced to the circuit as shown in figure (B).

In figure (A) it is given that

( )v ts1 200 (400 135 )cos Vt2 c= +

( )v ts2 200 (400 36.87 )cos Vt c= +

( )i ts 8 (400 90 )cos At c= +

The values of ( )v ts and Zeqin figure (B) are

(A) ( ) 416 (400 117.2 )cos Vv t ts c= + , 20 12.5Z jeq = +(B) ( ) 232 (400 56 )cos Vv t ts c= + , 20 40Z jeq =

(C) ( ) 160 (400 32 )cos Vv t ts c= + , 20 40Z jeq = +

(D) ( ) 80 (400 128 )cos Vv t ts c= + , 10 12.5Z jeq = +


400 / secrad=

25mH (400)(25 10 ) 10j j3& # =

50 F ( )( )

50j

j400 50 10

16&

#=

In phasor form Vs1 200 V2 135c= , 200 . VV 36 87s2 c= , 8 AI 90s c=The circuit in frequency domain is shown as below


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Applying the source transformation

Is1 j jV

10 10200 2 135s1 c= = 20 A

10 90200 2 135

2 45c

cc= =

Combining parallel impedance

( )( )

j jj j

10 5010 50

12.5j =

Adding parallel current source

I 8 90s1 c+ 20 2 45 8 90c c= +

( )j j20 1 8= + + 20 28 34.4 . Aj 54 46c= + =

Transforming the current sources into voltage source

Vs3 . . .j34 4 54 46 12 5#c=

. . .34 4 54 46 12 5 90#c c= . .430 12 144 46c=

Adding source and combing resistance

Vs 430. . 200 .12 144 6 36 87c c= +

j j350 250 160 120= + + +


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190 370 Vj= + 415.93 . V117 18c=

( )v ts 416 (400 117.18 )cos Vt c= +

Zeq 20 12.5j = +

MCQ 1.16 In the circuit shown in figure bulb A uses 36 W when lit, bulb B uses 24 W whenlit, and bulb C uses 14.4 W when lit. The additional A bulbs in parallel to this

circuit, that would be required to blow the fuse is

(A) 4 (B) 5

(C) 6 (D) 7


Current through bulb A IA 31236

= = A

Current through bulb B IB 21224

= = A

Current through bulb C IC. 1.2

1214 4

= = A

Current required to blow fuse If 20= A

The excess current which is required to blow the fuse is 20 (3 2 1.2) + + 13.8= A

No. of additional A bulbs nA. 4.63

13 8= =

So 5 additional A bulbs are required.

MCQ 1.17 In the following circuit the current I1is


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(A) 0.12 A (B) 0.24 A

(C) 0.36 A (D) 0.48 A


Vx 500I1= (Using Ohms Law)

Applying KVL in the loop DED

Vy 400( 0.001 )I Vx1= 400( 0.5 ) 200I I I1 1 1= =

Writing KVL in the upper left mesh ABCDEA^ h 180 . .I I V I V 500 100 0 6 100 0 005y y1 1 1= + + + +^ ^h hSubstituting V I200y 1= into above equation, we have

180 500 100( 0.6) 200 100( 0.005 )I I I I V y1 1 1 1= + + + + +

180 900 60 100 0.005 200I I1 1# #= + +

I1 0.12= A

MCQ 1.18 In the following circuit the value of open circuit voltage and Thevenin resistance

at terminals ,a bare

(A) 100 VVoc = , 1800RTh =


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(B) 0 VVoc = , 270RTh =

(C) 100 VVoc = , 90RTh =

(D) 0 VVoc = , 90RTh =

SOL 1.18 Option (D) is correct.To obtain Thevenin resistance put a test source across the terminal a, bas shown.

Vtest Vx= , I Itest x =

By writing loop equation for the circuit

Vtest 600( ) 300( ) 900( )I I I I I 1 2 1 3 1= + +

Vtest (600 300 900) 600 300I I I1 2 3= + +

Vtest 1800 600 300I I I1 2 3= ...(i)

The loop current are given as,

I1 Itest= , 0.3I Vs2 = , and 3 0.2I I Vtest s 3 = +

Substituing theses values into equation (i),

Vtest 1800 600(0.01 ) 300(3 0.01 )I V I V test s test s = + Vtest 1800 6 900 3I V I V test s test s =

10Vtest 900Itest= ,

Vtest 90Itest=

Thevenin resistance

RTh 90IV

test

test = =

Thevenin voltage or open circuit voltage will be zero because there is no independent

source present in the network, i.e. 0Voc = V

MCQ 1.19 The voltage ( )v t shown in the figure below is applied across a 0.5 H inductor,

having initial current of 2 A .


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Which of the following plot is correct for the inductor current ( )i t ?


For an inductor v-irelationship is given by

( )i t ( ) ( )L v d i t 1 t

00

= +#For 0 0.t 5< < ,

( )v 1 V=

( )i t .

( )d i0 51 1 0

t

0= +# 2 2t

0= 6 @ (0) 2 Ai =

(2 2)At=

For . t0 5 1< < ,

( )v 0=


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( )i t ( ) ( . )L

v d i1 0 5

.

t

0 5 = +#

.

( )d

0 5

1 0 1.

t

0 5

= + # (0.5) 2(0.5) 2 1 Ai = = 1 A=

For .t1 1 5< < ,

( )v 1 V=

( )i t ( ) ( )L

v d i1 1

t

1 = +#

.

d0 51 1 1

t

1= # (1) 1 Ai =

2 1t1

= 6 @ t2 2 1= (2 3)At= For . t1 5 2< < ,

( )v 0= ( )i t ( ) ( . )

L v d i

1 1 5.

t

1 5 = +#

L

d1 0 0.

t

1 5= +# ( . ) ( . )i 1 5 2 1 5 3 0= =

0=

We can obtained ( )i t in similar way for entire ( )v t . It is shown as

MCQ 1.20 The circuit shown in the given figure is in steady state with switch Sopen.

If Scloses at t 0= , then current ( )i t for t 0> , will be

(A) a decreasing exponential

(B) an increasing exponential


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(C) oscillatory

(D) constant


( )i t ( ) (0) ( )i i i e /t3 3= + 6 @ , t 0>For t 0< : The switch was open, in steady state inductor acts as a short circuit

( )i 0 4

5k

mA20= =

for t 0> : Again steady state reaches and at t 3= inductor is replaced by a shortcircuit

( )ix 3 0= and ( ) 5 mAi 3 =

So, ( )i t ( )e5 5 5/t

= +

5 0 5 mA= + = , t 0>

MCQ 1.21 In the following circuit what is the value of( )

dtdv tR at t 0= +?

(A) 0 (B) 100 / secV(C) 20 /secV (D) 40 /secV


For t 0< : ( )u t10 0=

So the circuit is


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( )i 0L1 0=

(0 )iL2 0=

For 0t> , ( )u t10 10 volt=

and (0 )iL1+ (0 )i 0L1= =

(0 )iL2+ (0 ) 0iL2= =

Mesh current, ( )i t1 ( )i tL1=

( )i t2 ( )i tL2=

( )v tR ( ) ( )R i t i t 1 2= 6 @

( )dt

dv tR ( ) ( )

Rdt

di tdt

di t1 2= ; E

( )dt

dv 0R+

( ) ( )Rdt

didt

di0 01 2= + +

; E ...(i)Let voltage across inductors are ( )v tL1 and ( )v tL2

( )v t

R [ ( ) ( )]R i t i t

1 2=

( )v 0R+ (0 ) [ ( ) ( )]v R i i 0 0L 1 22= =

+ + +

[ ]R 0 0 0= =

Writing KVL

( ) ( ) ( )i v v10 3 0 0 0L L R1 1 + + + 0=

10 0 ( )v 0 0L1 + 0=

& ( )v 0L1+ 10=

( )v 0L1+

( )L

dtdi 01

=+


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( )

dtdi 01

+

( )

.Lv 0

0 210 50L1= = =

+

( )

dtdi 02

+

( )L

v 00L2= =

+

Substituting values of ( )/di dt 01+ and (0 )/di dt 2

+ into equation (ii)

( )dt

dv 0R + 2[50 0] 100 /secV= =

Statement For Linked Questions 22 and 23 :

Consider a two port network as shown in figure below.

MCQ 1.22 The ABCD-parameter matrix for this network is

(A)5

8

8

13> H (B) 42 64> H(C)

7

4

12

7> H (D) 64 42> HSOL 1.22 Option (C) is correct.

The given network is equivalent to cascade connection of two port network shown

as

For the above network T-equivalent circuit is shown as

z z12 21= 1=

z z11 12 1=

z11 1 2z12= + =


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z22 1 2z12= + =

z6 @ 21

1

2= > H

Now T16 @ zz

zz

z zz1

21

11

21

21 21

22

T

=

R

T

SSSS

V

X

WWWW

zT z z z z 11 22 12 21=

( ) ( ) ( )( )2 2 1 1 3= =

[ ]T1 /

/

/

/

2 1

1 1

3 1

2 1

2

1

3

2= => >H H

For the cascades connection

T

6 @ [ ] [ ]T T1 2=

2

1

3

2

2

1

3

2= > >H H

S

7

4

12

7

= > H

MCQ 1.23 If the output terminal ( )2 2 l is short circuited then input impedance of the

network is

(A) 1.71 (B) 1.6

(C) 1.5 (D) 0.33

SOL 1.23 Option (A) is correct.The ABCDparameters are defined as

V1 AV BI 2 2= ...(i)

I1 CV DI 2 2= ...(ii)

If output terminal is short circuited

V2 0=

Substituting V 02 = into equation (i) & (ii)

V1 BI2=

I1 DI2=

Input impedance,

ZI

Vin

1

1= DB=

1.71712

= =

Common data for Question 24 to Q. 25

Consider the circuit shown in figure below with a load impedance ZL .


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MCQ 1.24 The value of ZLwhich will absorb a maximum power, is

(A) 10 5j + (B) 10 5j +

(C) 8 4j1 + (D) 8 14j


We obtain the Thevenin impedance across the load terminal.

ZTh ( ) || ( )j j10 15 10= +

( )( )

j jj j

j10 15 1010 15 10

8 14 =+

+ =

For maximum power transfer

ZL 8 14Z jTh = = +)

MCQ 1.25 If ZLis taken to be purely resistive, then the value of load resistance which willabsorb a maximum power is

(A) 10 (B) 20

(C) 11.18 (D) 16.125


If ZLis purely resistor then for maximum power transfer

RL R X ZTh Th Th 2 2= + =

RL ( ) ( ) 16.1258 142 2 = + =

Answer Sheet

1. (A) 6. (A) 11. (A) 16. (B) 21. (B)

2. (C) 7. (C) 12. (C) 17. (A) 22. (C)

3. (A) 8. (C) 13. (B) 18. (D) 23. (A)

4. (C) 9. (D) 14. (A) 19. (D) 24. (C)

5. (B) 10. (A) 15. (A) 20. (D) 25. (D)


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