gate electrical engineering 2012

8/10/2019 GATE Electrical Engineering 2012

1/25

Q N o 1

5 C

arry On

e Ma r k

a

ch

1.

Th

e br idge meth

od commonly

used f or findin

g m utual indu

ctance is

A

) Heaviside C

ampbell bridge

B)

Schering bridg

e

C)

De

Sauty

bridge

D ) Wien bridg

e

Answ

er:- A)

2. A tw

o phase loa

d draws

the fo llowing ph

as e cur rent

s i

1

t )=I,. sin

rot -

cJt

,

t ) = Imc

os

ro

t

e s e currents

are balan ced

if i is equal to

A)

z

B) z

C) 1t 2 -

z)

Answe

r:- D)

Exp:-

I t) =

Im

sin r

ot -

cr

1

) = Imcos[90-

rot

-

r

1

] = Im

cos rot

- 4\ - 90)

lx t)

=

Imcos

rot- l

2

)

Angl

e difference b

etween two cu

rren ts should

be

- 18

0

or

) 180 for bala

need

-

cjll

+ l2 9

0

=

-180 cjll =

90 + cjl2

3 . The slip

of

an

induction

motor no rma l

ly does not de

pendon

A) Roto

r speed

B)

Synch

ronou s speed

C) S

haft to rque

Answ er:-

D )

Exp:-

N N

slip=

s

r

Ns

D ) Co

re-loss comp

onent

7 So d

epends on N

5

synchronous

speed)

7

So depends

on Nr ro tor speed

)

7

f

torque increases Nr decreas es

7

I t

will no

t dependent

on co re loss

4 .

A period

ic vo ltage wa

veform obser

ved

on an

osc

illoscope acro

ss a load is s

hown.

A permanen

t magnet mov

ing coil PMM

C) met

er

con

nected across

the same loa

d

rea

d s

A) 4V

B) S

V

v t)

c)

8V

D ) 10V

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2/25

Exp:-

PMMC

will read a

verage value

V

=

Area under cur

ve

avg Time pe

riod

= [ [

~ x 1 0 x 1 0 J-[5x2

]+[8x5J]x10

-

3

=

4

V

2

0 x 10-

3

5.

The bus ad

mit tance m atr

ix of a three-b

us three-l ines

ystem is

y

= -

150l

5 10 - 1

3

I f

e

ach transmiss

ion line betwe

en the two b

uses is repres

ented by

an

e

quivalent

Jt

-

network, th

e magni tude

of the shunt s

usceptance

of the line con

necting bus 1

and 2

is

A) 4

B) 2

C) 1

Answer:

- B)

Exp:-y

11

=

Y s

1n1

2 +y

12

+y

13

+

Yns > Ys

1n12

+

Ysml3 =

2

2

2

2 2

Ysnl3

Ysn 3 = 2

2 2

P

0

= P

1

+ P

2

-

P

L > 40 = 20 + P

2

-

2 > P

2

= 22

M

W

P

1

=

20 ; r

2

= 22

D) 0

6.

I f x [ n

J

(1

3)tl - (1

2

f

u [ n , then th

e region of

con vergence RO

C) of i ts

z-

tr ansform

in

th eZ-plane wi ll

be

A

) . .

1

0k)

ib)

= 100 I+

99ib

+ ib);

-1

0000ib = 1001

+ 100x 1

ib = 1001 + 10000i

b

- 20

000 ib = 1001

::::> =

-

100

i

=[ I

]

2000

0 200

V

=10

0 [ I + +

J

100

I

+

10 0

=

5 1

10

.

In the ci rcu it

shown below ,

the current

th

rough t he in d

uctor is

B) -

1

. A

1 + ]

C)

j A

D ) 0 A

Answer : -

C )

E x

p:-

11.

1 1

IL = 1l Q

=

A

1+j1 1+j1

10

1 2

Given

f z)

.

If C is a

counterc loc kw

ise path In th e

z-plane such

tha

t

Z + 1 Z+3

1

lz + 11= 1

, thev

alueof 1 t j ~

c f z ) is

A) -2

B)

-1

C) 1

D ) 2

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5/25

Exp:-

~ ~ f z ) d z

= ~ [ ~ -

d z - ~ - z dz

]

27tl c

2m c z

+

1

c z + 3

-or------ - - - - - - v - - - - - -

It

lz

z = - 1 is sing

ulari ty

inc

a

nd

z=-3is not

inc

By

cauchys integ

ral formula ~

= - d z =

0

cZ+3

1

:

1

1

=

~

dz

=

1;

- 1

2

=

1

c Z +

1

12

. Two inde

penden t rand

om variables

X and Y ar

e unifo rmly d

istributed in

the

inte

rval [ -1

,

1]

.

T

he probability

that

m ax[X, Y

]

is

less than 1/2

is

(A ) 3/

4 (B

)

9/16

C ) 1/4

D)

2/3

Ans

wer: - (B)

Exp:-

Uniform dist

ribut ion X, Yon

J ; f x ) = f y )

~

P

m a x x , y ) : o ;

~ ) = P X = ~ , - 1 : o

; Y : : ; ~ }

X ~ , Y = ~ )

1

>

1

2

1 3

3 9

=

J

dxJ dy

x

_ 1 2

-1

2

4 4

16

13.

For

the circuit shown

in

the figure, the voltage and current expressions are

v ( t )= E

1

sin

(rot)+ E

3

sin (3

rot)

and

i t) =

sin rot -


6/25

E

xp:- G iv e

n, x="'-1.; xx

= ( ) . J - 1 = i

W

e know that e

8

=cos

8+ is

in e

=>

e' = c o s ~

i s i n ~ = I

2

2

: (i)' = )' =

15. Th

e typical ra

tio

of

la tch ing curre

nt to hold in

g

c

urrent in a

20A thyrist

or is

A) 5.0

B) 2 .

0 C)

1.0

D) 0.5

Answ

er:- B )

16. A

half-control

led single -pha

se bridge

rec

t i f ier is supp

ly ing an R-L

load.

t

is

opera

ted at a

fi ring angle o an

d

the

load curre

nt

Is

cont in

uous. The fra

ct ion of

cycle that the

f reewhee l

ing d iode

conducts is

A)

_ .

2

Answer

:-

D )

C)

2 t

D)

o

t

17.

The seque

nce compon

ents

of

the

f

ault

c

urrent a re

as fol lows

:

pos

t

iv

e=

j1.5pu,I

negative = -j0 .5pu

,

zero

= - j1

pu. The typeo

f faul

t

in thes

ys tem is

A)

LG

Answe

r:- C)

Exp:

- 1

1

= 1

2

+

1

0

So LLG fa

ult

B)

LL

C) LLG

D) LLL

18. T

he f igure sh

ows a two-gener

ator sys

t m supply ing

a load of

P

0

= 4 0 MW,

co

nnected at bu

s 2

Bus 1

G,

v

Po

= 40 MW

Th

e fuel cost

of

generators G1

and G2

are:

C

1

(PG

1

= 1

0,000

RsI MWh and C

2

PG

2

) = 1250

0Rs I MWh and

the

loss in the

l ine

Is

P

1

oss(

pu)

= 0.

5

~ l p u )

whe

re

the

loss co

eff ic ient is sp

ecif ied

In

pu

on a

10

0 MV

A

base. T

he most e

conomic pow

er generation

sche

dule in MW Is

A) PG =

20, P

G = 22

B) P

Gl

= 22, PG

= 20

(C

) P

Gl =

20, PG2

= 20

swer:- A

)

D)

PG

=

0, PG = 4

0

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7/25

Exp:- A

l = -2

A

l

=

10 00

0

;

'),2 = 125 00

4 = 1

>

_ _ > 10 000

=

12500

1

_ oPL 1

P

1

1 P

1

a

P

1 10

000 p > 25 0

0 [

] u

1250

0

1

1250

0 5 p

1

PL =

s x 100 = 20MW

PL = 0.5 ~ r

=

~

p.u = x 100 =

2MW

P

0

=

P

1

+ P

2

PL

> 40= 20 + P

2

2 P

2

=

MW

19. Con

sider the given

circuit

A

8

In

th isc

ircuit , the ra c

e around

A) Doe

s

no

t occur

B) Occursw

hen CLK = 0

C) Oc

curs when

CLK

= 1 and A=

B = 1

D )O

ccurs when

CLK = 1 and A=

B = 0

Answ

er:- A )

Exp:- . A

1

Cllc

Qnext

= A CLK Q

= A

.CLK Q

Qnext =

A

CLK + Q

f

CLK = 1 andA and B = 1

th en

Qn = }

_ext

_ No rac

e a round

Qnext - 1

f

CLK = 1

and A = B = 0

~ ~ N o

race a

round

Q

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20. The

output

Y

of

a

2-bit

comparator is logic 1 whenever the 2-

bit

in

put A is greater


8/25

than the 2-bit input B. The number of combinations for which the output is logic

1, is

A) 4

Answer:- B)

B) 6

E x p : ~ :

: :o}

A>B

if

A

11

+[A

1

B

1

[A

00

]

- 1 0

1 0 0 0

[I]@]@][I]

[I] [I]@]@]

[I][I]@][I]

@][I]@]@]

[I] [I] [I]@]

C) 8

D) 10

21.

The i-v characteristics

of

the diode in the circuit given below are

l

v 0 7

i = A

v

o.7v

OA v

- -=

dt

dt t

t2

C) X = -

2

IF

=

f t ~

= e

109

t

=

t;

so lu ti on is x (IF) =

J

IF}

tdt

t

D)

X

2

f

1

xt =

J. tdt =

>

xt =

2

+

c; Given

that x (1} = 0. S

=> 0. S =

2

+ c

=> c = 0

.

. The req

uired so lut ion

isxt=f

. =>

x

=

.

2

2

Q No. 26

5 carry

T

wo Marks Ea

ch

2

6. A 220V

1S

kW

1000 rpm shunt

mo

tor

with a rm a

ture resistanc

e of 0.2S.Q h

as a

rated line

current

of

68

A and a ra te

d field curren

t of 2.2

A.

Th

e change in f

ield

flux r

equired

to

obt

a in a speed

of

1600

rpm w h

i le drawing a

l ine current

of S2.8 A

a

nd a field cur

rent of

1.8

Ais

(A)

18

.18 in cre

ase

(C)36.36

increas

e

A

nswer : - (D)

(B ) 18

.18 decre

ase

(D)

36.36 decreas

e www.examrace.com

N1

Ebl


10/25

Exp : -

X

N 2

Eb2

Ra =

0.25; Ia = 6-2.2 =

65.8A; Ia

2

=

52

.8-1.8

= 51A

1000 = [2

20- x

0 . 2 5 J x ~

1600

220-

51

X 0 25

2

=

0.6364


11/25

I=

136x0.45 = 12.24A

5

P

=VI

cos e =

136x

12.24x 0.45

=

50 W

Y n12 =1 ::::> y .

2

= 2

2

s1n

29. For the system shown below, S

01

and S

2

are complex power demands at bus 1

and bus

2

respectively. I f IV

2

1

= 1pu, the V R rating of the capacitor Q

2

connected at

bus

is

Bus

1

Bus 2

V,

=1/Q

pu

z

Z = j0.5pu

So, =

1

pu

So2 =

1

pu

A) 0 2

pu

Answer:- B)

B) 0 268 C) 0 312 D) 0 4 pu

Exp:-

v

1

= i[ Q v =

1l-3o

s .

~ ~ ; o ~ ~

I

s =

1p.u. s.

=

1p.u.

Line is loss ess

SG

1

= S

1

+ S

2

= 1 + 1 = 2p.u.

Power

transfer

from bus-1

to

bus-2 is 1p.u.

:.

1=

I V ~ :

1 s i n 9

- 9

= ~ ; s i n e

- 9

sin e

1

-9z)=0.5

9

1

9

2

=sin-

1

0.5=30; 9

1

=O{V

1

=11.Q};

9

2

=-30 ; V

2

=11-30

I = vl-

v2

= 1 i l l - =

1

o.288

12

2

j0.5

J

Current

S

2

= 11-30;

Current

in QG

2

=

~ 3 [ 1

j0.268J=

0.26a-120

VAR rating of capacitor = IV

2

11IQ

I

sin IV

2

11I

2

1 =

1x

0. 268 x sin +90) = 0. 268

30. The circuit shown is a

C R

+--11-- Vv\r- -

Input

1

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1


12/25

(

A ) Low pass

f

ilter

with f

3

dB

=

( ) ra

d

s

R

+ R

2

C

(B) Hi

gh pass

filter

with f

3

dB

= -

1

rad

s

R

1

C

(C) Low pass filter

with f

3

dB = -

1

rad s

R

1

C

(D ) High

pass

filter

w

ith f

3

dB = (

1

) rad

s

R

+

R

2

C

Answer : -

(B)

Exp:- Vo= R2

n

R 1

1 sc

1

31

. Let y[n]

denote the co

nvolution of

h[n] and g[

n], where h [ n

]= (112)"u[n

] and

g[n ]

Is a causal se q

uence. I f y(O ]

= 1 and y1]

= Y2 then g[1]

equals

(A) 0

(

B ) 112

(C) 1

(D

)

3 2

An

swer:-

(A )

(1)"

1

xp:-

h

[n] =

2

u(n);

y (o) =

1, y 1) =

2

s

ince y(n)=g(n)

*h(n)=

g(m)h(n-m)

g n}

g 0 }

g

1)

h

n)

'I

1

2

I

1 4

n

. n

0

0

1

2

h -n}

2

r

h --11)

2

4

1 / 4

n

2 1

0

1

0

mult ip ly

ing g(n) and h(-

n); y(O)=. xo+.

. x 0+

1 x g O)

4 2

1 1

y(

1) = -g

(O) + g(1) g(1 )

= 0

2 2

32.

The state tr a

nsit ion diagra

m

for

the logic

circuit sh own

Is

r

n

1

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13/25

Ans

wer:-

D)

Exp:

-

Prese

nt s ta te (Q) A Next

state

0

0

1

1 1

0

0

1 0

1 0 1

Sta te mach

ine

33. The

vo lt age gain

Avof the circuit

shown belo

w is

13 .7 Vo lts

(A) I

Avl=200

Answer:-

D)

Exp:-

VLIn

Inpu

t loop,

13.7- Ic+Ia)1

2k-100k I

)

- 0.7 = 0

>

I

6

=

9.91- A; Ic = j

3I

6

=

0.9

9mA; IE=

lmA

26mA

.

.

(lOOk

l l12k)

:

r.

= I = 2

6n, Z = j)r. = 2.6kn,

:

Av =

26

= 412

2

=

Z I

I ( lOOk)= 2

21 i l A

=A

=412

)

221

)

l i+412

vs

vZ

+

R,

221 + 10k

IA vsl

' 10

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34. I

f

VA-

V

8

= 6V, th en

Vc- V

0

is


14/25

(A)

-5V

A n

swe r :- (A )

5

V A

B) 2V

C) 3V

D) 6V

E xp :

- I = V

A;

Va

= = 3A; Sin c

e current

entering any

network is sam e

as leaving in

Vc - V

0

branch

also

i t

is I =

3A

R

10.

A

35. The m

ax im um va lue

of f

(x )

= x

3

9x

2

+ 24x + 5

in the

inter

v a

l [1, 6

]

is

A)

21

Answer

:-

C

)

B)

25

EXP:- G i

ven, f (x ) = x

3

- 9

x

2

+

24X 5

C)

41

D)

46

f '(x ) = 0

fo r s ta t ionary

v l

u e s ~ 3x

2

-18x + 24 = 0

> x=2,4

f (x) =

6x - 18; f

(2 )=12-18

0

Hen

ce f(x) ha s

maximum

va lue

at x=

2

: The m

aximum value is 2

3

-

9x

2

2

+ 24 x

2 + 5 = 25

B

ut

we

have

tofindthe m

aximum value in the

interval

[1,

6]

: f(6) = 6

3

-

9x6

2

+24x6+5=

41

36.

[

5

Given

that A =

2

(A )

1 5 A + 12I

A nsw er:- (B )

-

3

]

[

0

and

I=

0

~

the val

ue of A

3

Is

(B) 1 9 A +

3 1

(C)

17 A + 15I

(D) 17A +21

1

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E

xp: -

Given: A= [

- : ~

l


15/25

Chara

cterist ic equa

tion of A is IA

-D.I = 0

~

~ ~ . 1

= 0

( - 5 - A.)

-A.)

+ 6

= 0 SA

+

A

2

+ 6 = 0

~ A

2

= - S

A

- 6

and A

3

=

-SA.

2

- 6

A.

= -5

(-S A

.- 6) - 6A ( :

A

2

= - SA - 6)

~ A = 25A -6

A . + 3

0=19A.+30

Every

matrix

satisf ies i ts chara cter is t ic equati on. . A

3

= 1 9A +

301

37. A sing le

phase

10

kV

A

1

50 H

z transform

er

with

1 kV

pr

imary w

inding draws

0

.5

A

an

d 55

W,

atra

ted vol tage an

d frequency ,

on no load . A

second

transforme

r

has

a c

ore with a ll

its

l inear d im e

nsions .. /2 tim

es the corres

ponding dim e

nsions

of

the

first

t ra ns

fo rmer . The c

ore mater ia l a

nd lamina t io n

th ickness are

the same in

both tran

s fo rmers . The

prim

ary

w ind

ings o

f

both

the t rans fo rm e

rs have th e s

ame

num

ber of turns .

I f a rated vo lta

g e of 2 kV

at 50 Hz

Is

app

lied to the

primary

o

f

th e second tra

ns fo rmer , the

n the

no

load

curre

nt

and p

ower, respecti

ve ly , are

(A )

0.

7

A,

77.8

W

(B)

0.7 A, 155.6

W

(C)

1 110

W

An swe r:-

(B)

D) 1 A, 220

W

Exp: -

Coreloss o corevo

lume;Pc2= .

./2f; Pc

1

= .J2f x 55 = 155W

Core loss co

mponent

Ic2

=

.J2f

xicl

=

2 / 2 [

~ ~

=

0.155A

55

Ic

2

=

1 0 0 0

=0

.055A;

Magneti

z ing compone

nt, I+

1

= -

Ic

1

2

=

Jo

5

-

0.

055

2

= 0.

49 6

9A

R

11

1000

2000

Now

reluctance

= r:::,


16/25

m ult lp l ler set

ting

o

f 40 k.n,

the voltme

ter reads

(A) 371 V

Answ

er:-(D)

(B )

383 V

(C) 394 V

(D ) 4

06 V

Exp:- L

et resist anee of

vo l tmete r

be R k.n

352V

(

440 )

352

)

= R

20 + 440 ..

... (1);

V=35

2+ R

80 ..........

.. (2)

Solving, V=480; R= 220;

V

=

48

0

x 220

=

406V

40+

220

t

4

0. The input x t)

and output y{t)

ofa

system ar e r

elated as y t)

= x( t)cos(3

t)d t.

The system

is

(A) t ime- inva

riant and stab

le

C) t ime- invaria

nt and n ot sta

ble

Answer:- (B

)

t

Exp:- y

t) = X( t)c

os(3 t)d t

(

B) stable and

not t ime- inva

r iant

(D )

not t ime- Inva

riant and no t s

table

Since y t)

and

x t)

a

re related w it

h some functi

on of t ime,

so

they are not

t ime

in v

ariant.

Let

x

t)

be bounde

d

to

some f in i

te value k.

t

y

t) = Kco

s(3 t)d t

y t)

Is

a

lso bounded.

Thus System

is stable.

41.

The feedbac

k system show

n below oscil

lates

a

t 2 rad /

s when

(A) K = 2 a

nd

a=0.

75

(C) K

= 4 and a=0

.5

Answ er

:- (A)

(B) K = 3 an

d a =

0

. 7 5

(D) K=

2 and a = 0. 5

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Exp:- Characteristic e quatin is 1+G s)H s) = 0


17/25

k(s+1)

1 +

3

2

=0;

S

3

+S

2

a+s(k+

2)+(k+1)

= 0

s

-1- a

s + 2s +1

s

3

1

k+2

a k+

1

51

(k+2)a-

(k+1)

a

so

as

2

+

k + 1) = 0; s=

jro;s

2

= r

2

; r

= 2

-

aro

2

+ k+

1) =

0

;

aro

2

=k

+

1; 4a=k+

1

; From options

,

k=

2, a=0.75

42

. The Fouri

er transform

o

f

a signa I h(t

) is H jro) = 2

cos

ro

sin

2ro

)

I ro. The va lue

of

h

O)

is

A

) 1/4

Answer :-

C)

Exp:-

(B

) 1/2

sin

2ro

C) 1

D) 2

l

X=

(t)

2

2

t

h

(t) =

h. (t-

1 + h.

(t

1

. k

3 1 3

43 .

The state vari

able descriptio

n

of

an LTI sy

stem is given

by

where

y

Is the output and u is the in put. The system

Is

co ntro l lable for

A)

a

1

0, a

2

= 0,

a

3

0

B) a

1

= 0, a

2

= 0, a

3

0

C) a

1

= 0, a

2

= 0, a

3

= 0

D

) a

1

0, a

2

0, a

3

= 0

Answer : - D

)

E

xp. Q,

0

[

B AB

J A=[ ;} B=m;

ABo[

HA B

0

[ ]

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o 0

a

1

a

2

]

e

= 0


18/25

[

2

; I f rank

of e =3

=o

rder o

f matrix, th

en e is contr

ol lab le

1

a

1

O

a

2

t 0 th

en IQ

el

0

a

3

= 0

4

4. Assumin

g both the

v oltag e sourc

es are

in

ph

ase, the va lu

e of R for

wh ich

ma x im

um

po wer

is transferred

fr om c irc

uit

A

to

ci

rcui

t

B i

s

(A) 0.8.

Q

Answer:-

(A )

B) 1.4.Q

il

n

Carcun

(C) 2.Q

3

D )

2.8.Q

Exp:-

Po

wer t ransferr

ed f rom circuit

A to ci r

cuit A= VI= (_ _

2

)

(

6

+ lOR)=

42

+ ~

R+ R

+ 2 R+ 2

)

I

= 10 - 3 = _7

_ ; V =

I R =

+ ~

=

6

+lOR)

2+

R 2 + R

2+R 2+R

dP (R + 2)

2

(

70)- 42+7

0R)2 R+2)

=

=0

dR (R +

2 t

r.........................

- - - - ,

: .--. .

-

- - - - ~

J

,w

- + - - - - i - -

;

1o

70 R +2)

2

= 42

+

70

R)2 R +

2); SR+2) = 2

3 +

SR

SR + 10 = 6

+lOR; 4=5R;

R=O.S

.Q

45. Co

ns ider the dif

ferential equat

ion

+ 2 dy t)

+Y t)=o

t)

with

y t) l = 2 a

nd dyl = 0.

dt dt

t O

dt

toO

The num

erica l va lue of

is

(A )

-2

( B)

-1 C)

0

(D) 1

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Answer:- (D)

2


19/25

d y

(t) 2dy(t

) ) )

E

xp:-

2

+

+ y t

= t

dt

dt

Convert ing to

s - domain

1

s

2

y

(s) - sy (

0) -

y

(0)

+ 2 [

sy

(s)-

y (0)J y (

s)

=

1

[ s

2

+ 2

s

+ 1J (

s) + 2s

+

4

= 1

-

2s

y s ) - ~ - ~

-

(s

2

+ 2

s

+ 1)

Fin

d

in

verse lapalce

t ransform

Y (t ) =

[ -2e-t

-te-

1

]u(t

)

dy (t) = 2e

-t

+

te-t - e-t

d

t

dy(t)l

=

2-1=1

dt

t -0

4

6. The dir

ect ion of v

ector A is

radially out

ward f rom

the originr

with

IAI

= k

r where r

2

= x

2

+

y

2

+ z

2

and

k is constan

t. The va lue

of n for wh

ich

V'.A = 0 is

A) -2

Answer:- (A)

B) 2

- 1 a )

xp:-

We

k

now thatr V

'.

A

=

2

-

r

2

Ar

r ar

N

ow,V .A=

~

= _ _

(krrH

2

)

=

(n

+

2) r +

1

r

2

ar

r2

=

k (n+2)r +

1

C) 1

: For,

V

.

A=

1

=>

(n

+

2) = 0 => n =

-2

D ) 0

47. A fair c

oin is tossed t

i l l a head app

ears for the f

irst time. The

probabil i ty tha

t the

num

ber of require

d tosses is

oddr is

A)

1 3 B)

1 2 C)

23 D )

3 4

Answer:- (C

)

E

x p:- P (odd to

sses) = P (H)+

P(TTH) + P(TT

TTH) + .....

~ +

~ J + ~ J

+

.. . . . . = ~ [ 1

+ ~ J

+ ~ r

+ .. .

. J

= H

~ H ~ J +

[ l ~

J ~ =

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Common Data Questions

48

49


20/25

With

10V de conne

cted at port

A

in

the line

ar nonreciproc

al

tw

o-port n

etwork

s

hown below,

the following w

ere observed:

I

) 1n connect

ed at port B d

raws a curren

t

of

3 A

ii)

2.

5 n connec

ted at port Bd

raws a curren

t of 2 A

48. Fo

r the same n

etwork, with 6

V de connect

ed

at

port

A

1n connecte

d

at

port B

draws 7 3A. f

8 V de is

connected to por

t A, the op

en circuit volt

age

at

port B

is

A) 6V

B) 7V

C)

8V

D) 9V

Answer:-

B )

49. W

ith

1

0V de co n

nected at port

A,

the

curren

t drawn by

7

n

co nnected at po

rt

B

is

A) 3

/7 A B)

5/7 A C)

1 A

D)

9 7 A

Answer: - C)

Common

Data

Que

stions

5

51

In the 3-phas

e inverter circ

uit shown, th

e load is balan

ced and the g

ating scheme

Is

180 - co

nduction mode

. All the switc

hing devices a

re ideal

I

I

3 phase

inver t

r

i

JQa,

. . ;

so.

The rms value of load phase vo ltage is

A)

106.1 V B)

141.4 V

C

) 2

12.2

V

D)

282.8 V

Answer:- B)

Ex

p:-

RMS

value

of

line

volt

age=

V

=

~ s

V fi

fi

RM S

value of phas

e voltage = r ;

= V = x

00

=

141

.42V

v3

3 s 3

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51.

f

th

ed e bu s voltage

Vd

= 300 V, the power

con

sum ed b y 3-ph ase load is

A) 1. 5 kW

B) 2

.0 kW

C) 2 . 5 kW

D)

3.0

kW


21/25

An sw e

r:- D )

v

ph

2

141.4

2)

2

E

xp:- P = 3

= 3 x

= 3000W

R ph 20

Linked

Answer

Quest

ions

Q 52

to Q55 Carry

Two

Marks

Each

Statement

for

Link

ed

Answer Questions 5

2 53

I n t

he

c

ircu i t show n,

th

e th ree

voltmeter

re ad

ings are V

1

=

220V, V

2

= 1 2

2V,

V

3

=

136V

52. The po

wer facto r

of the load is

A)

0. 45 B) 0 .

5

C)

0 .

5

5

A n s

w e r:- A )

22 0

2

-122

2

-136

2

=0.4

5

2 x 122 x

136

D) 0 .60

53.

I f RL

=

5

.0

the appro

x i

ma

te power

consu m ption

in the load is

A ) 70

0 W

B)

750

W

C) 800

W D)

850W

Answer:-

B )

R

5

Exp : - c

as e

=

_L ;

0.4

5=-

> z =

11.11

z

z

54 .

V

3

136

.

I =

= =

12.

24A,

P =I

2

R = 12.2

4

2

x 5

=

750W

z

11

.11

Statement

for

Link

ed Answer

Questions 5

4 55

The

tr

an

s fe r function

of a co mp e nsa

tor is

giv

ena

s

Gs) = s+a

S+ b

e

s)

is a lead c

ompe ns

ato

r if

A) a =

1

b = 2

B) a= 3

, b = 2

C) a = -

3

b=

- 1 D) a =

3

b = 1

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Answer:- A)

E

xp: - = ta n -

1


22/25

fo r

phase lead

should

be

+ v

e

< b

both

option A) an

d C) satisf ier

but

o

ption C) wil l po

t polar and

zeroas

R

HS of s-plane

thus no

t

possible

Opt io

n

A

) is rig

ht

55. The

ph ase o

f

th

e above lead co

m pensator is

m a x im u m

at

A)

J

rad

s B) ./3

rad s

C) J

rad

s D)

11./3 rad s

A nswer : -

A)

E xp :- For a

lead compen

sator, a c 2, then C t

I f c t < C2,

then C2

f A t A2, th

en A2

Either A orB

would be hea

vier(Say A >B

At V

S A2

f A t

=

A2, then A 3

f A

t > A2, then A

t

65.

O

ne of

the

le

gacies of

t

he

Roman

legions

was discipl

ine. n the

legions,

m

ilitary la

w prevailed and

dis

cipline

was brust

al.

Discip

line o

n

the

battlefie

ld kept u

nits

ob

edient, intact

and

f ighting

ev

en w

hen th

e

odds

and condition

s

were

aga

inst the

m

Which one ofthe following s ta tements best sums up the meaning of

the above

pa

ssage?

(A) Throug

h regimenta ti

on was the m

ain reason for th

e efficienc

y

of

the Roma

n

le

gions even in

adverse circu

mstances.

(B) Th

e legions were

treated inhe

rita nee from their

se niors.

(C) Disc

ipline was the

a rmies in her

i tance from the

ir seniors.

(D) The

harsh discip lin

e

to

which

th

e

legions wer

e subjected t

o led

to

theo

dds

a

nd

co

nditions being

against them.

A

n sw e r: - ( A)

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gate electrical engineering 2012

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