gate electrical engineering 2012
TRANSCRIPT
-
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
www.examrace.com
-
8/10/2019 GATE Electrical Engineering 2012
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
www.examrace.com
-
8/10/2019 GATE Electrical Engineering 2012
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 -
-
8/10/2019 GATE Electrical Engineering 2012
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
www.examrace.com
-
8/10/2019 GATE Electrical Engineering 2012
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
www.examrace.com
20. The
output
Y
of
a
2-bit
comparator is logic 1 whenever the 2-
bit
in
put A is greater
-
8/10/2019 GATE Electrical Engineering 2012
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
-
8/10/2019 GATE Electrical Engineering 2012
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
-
8/10/2019 GATE Electrical Engineering 2012
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
www.examrace.com
1
-
8/10/2019 GATE Electrical Engineering 2012
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
www.examrace.com
-
8/10/2019 GATE Electrical Engineering 2012
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
www.examrace.com
34. I
f
VA-
V
8
= 6V, th en
Vc- V
0
is
-
8/10/2019 GATE Electrical Engineering 2012
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
www.examrace.com
E
xp: -
Given: A= [
- : ~
l
-
8/10/2019 GATE Electrical Engineering 2012
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:::,
-
8/10/2019 GATE Electrical Engineering 2012
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
www.examrace.com
Exp:- Characteristic e quatin is 1+G s)H s) = 0
-
8/10/2019 GATE Electrical Engineering 2012
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
[ ]
www.examrace.com
o 0
a
1
a
2
]
e
= 0
-
8/10/2019 GATE Electrical Engineering 2012
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
www.examrace.com
Answer:- (D)
2
-
8/10/2019 GATE Electrical Engineering 2012
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 ~ =
www.examrace.com
Common Data Questions
48
49
-
8/10/2019 GATE Electrical Engineering 2012
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
www.examrace.com
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
-
8/10/2019 GATE Electrical Engineering 2012
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
www.examrace.com
Answer:- A)
E
xp: - = ta n -
1
-
8/10/2019 GATE Electrical Engineering 2012
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)
www.examrace.com