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1
2
. In th
trans
a) –1
. The t
a) 0.5
s
e circuit s
sistor Q an
5
transfer fu
5 11
ss
++
Elec
shown bel
d an ideal
b) –0.7
nction 2
2
((
VV
b) 3ss
++
ctrical Eng
ow what
op‐amp ar
7
( )( )ss
of the c
62
++
gineering: E
is the out
re used?
c) +0.7
circuit sho
c) 21
ss
++
Exercise: 1
tput voltag
wn below
2
1
ge (Vout) in
d) +15
is
d) 12
ss
++
n Volts if a
a silicon
3
4
5
6
. Assum
unit s
a) u(t
. The i
a) 2
2t
c) (t −
. Whic
and s
a) A
b) Z
c) A
d) A
th
. Two
the o
a) pro
c) con
ming zero
step input
t)
mpulse res
( )u t
21) ( 12
u t−−
ch one of t
stable LTI s
All the pole
eros of the
All the pole
All the root
he jω axis.
systems w
overall imp
oduct of h
nvolution o
initial con
u(t) is
b) tu(t
sponse of
1)
the follow
system?
es of the sy
e system c
es must lie
ts of the c
with impuls
ulse respo
1(t) and h2
of h1(t) and
ndition, th
t)
a system is
b) t
d) t
ing statem
ystem mus
an lie anyw
within |s|
haracteris
e response
onse of the
(t)
d h2(t)
e response
c) 2
(2t u
s h(t) = t u
( 1) (2
t t u t−−
2 1 (2
t u t−−
ments is NO
t lie on the
where in th
= 1.
tic equatio
es h1(t) an
e cascaded
b) sum
d) subtr
e y(t) of t
)t
(t) . For an
1)−
1)
OT TRUE f
e left side o
he s‐plane
on must be
d h2(t) are
system is
of h1(t) an
raction of
he system
d) e‐t u(t
n input u(t –
or a contin
of the co a
.
e located
connected
given by
nd h2(t)
h2(t) from
m given bel
)
– 1), the o
nuous tim
axis.
on the left
d in cascad
h1(t)
low to a
utput is
e causal
t side of
de. Then
7. A source vs(t) = V cos100πt has an internal impedance of (4 + j3)Ω. If a purely
resistive load connected to this source has to extract the maximum power out of
the source, its value in Ω should be
a) 3 b) 4 c) 5 d) 7
8. A single‐phase load is supplied by a single‐phase voltage source. If the current
flowing from the load to the source is 10∠ ‐ 150° A and if the voltage at the load
terminals is 100∠60° V, then the
a) load absorbs real power and delivers reactive power.
b) load absorbs real power and absorbs reactive power.
c) load delivers real power and delivers reactive power.
d) load delivers real power and absorbs reactive power.
9. A single‐phase transformer has no‐load loss of 64 W, as obtained from an open‐
circuit test. When a short‐circuit test is performed on it with 90% of the rated
currents flowing in its both LV and HV windings, the measured loss is 81 W. The
transformer has maximum efficiency when operated at
a) 50.0% of the rated current.
b) 64.0% of the rated current.
c) 80.0% of the rated current.
d) 88.8% of the rated current.
10. The flux density at a point in space is given by B = 4xax + 2kyay + 8az Wb/m2. The
value of constant k must be equal to
a) ‐2 b) ‐0.5 c) +0.5 d) +2
1
1
1
1. A con
∞. Th
a) 0.3
2. The c
a) 4xy
b) 4a
c) (4x
d) 0
3. In th
the
a) inp
b) inp
c) inp
d) inp
ntinuous ra
hen P{X > 1
368
curl of the
yax + 6yz a
x + 6ay + 8a
xy + 4z2)ax
e feedbac
put impeda
put impeda
put impeda
put impeda
andom va
1} is
b) 0.5
gradient o
ay + 8zxaz
az
+ (2x2 + 6y
k network
ance incre
ance incre
ance decre
ance decre
riable X ha
of the scala
yz)ay + (3y2
k shown be
ases and o
ases and o
eases and o
eases and o
as a proba
c) 0.632
ar field def
2 + 8zx)az
elow, if th
output imp
output imp
output imp
output imp
bility dens
2
ined by V =
e feedbac
pedance de
pedance als
pedance al
pedance in
sity functio
d) 1.0
= 2x2 y +3y
k factor k
ecreases.
so increase
lso decreas
ncreases.
on f(x) = e‐
y2z +4z2x is
is increase
es.
ses.
‐x, 0< x <
s
ed, then
1
1
4. The i
infini
the v
a) 4.4
5. The B
The g
phase
a) 39
s
input impe
te. Assum
voltmeter i
46
Bode plot o
gain (20 log
e is negativ
9.8s
edance of
ing that th
n Volts is
b) 3.15
of a transfe
g |G(s)|) is
ve for all ω
b) 2
39.s
the perma
he diode sh
5
er function
s 32 dB an
ω. Then G(s
8
anent mag
hown in th
c) 2.23
n G (s) is sh
d –8 dB at
s) is
c) 32s
gnet movin
he figure b
hown in th
1 rad/s an
ng coil (PM
below is ide
d) 0
e figure be
nd 10 rad/s
d) 2
32s
MMC) volt
eal, the re
elow.
s respectiv
meter is
ading of
vely. The
1
1
1
1
2
6. A bul
other
by an
switc
a) an
c) an
7. For a
funda
a) 10
8. A ba
Acco
valid
a) 5
9. Cons
show
the e
a) k2
0. The a
a) A
lb in a stair
r one at th
ny one of t
ching of the
AND gate
XOR gate
a periodic
amental fr
0
and‐limited
rding to th
is
ider a de
wn below. I
elements o
angle d in t
Angle betw
rcase has t
he first floo
he switche
e bulb rese
signal v(t
requency in
b) 300
d signal w
he samplin
b) 12
lta connec
If all eleme
f the corre
b) k
the swing e
ween stator
two switch
or. The bul
es irrespec
embles
b) a
d) a
t) = 30 sin
n rad/s is
0
with a max
ng theorem
ction of r
ents of the
esponding
equation o
r voltage a
hes, one sw
lb can be t
ctive of the
an OR gate
a NAND gat
n100t + 1
c) 500
ximum fre
m, the sam
c) 15
esistors a
e delta con
star equiv
c) 1/k
of a synchr
nd current
witch being
turned ON
e state of t
te
10cos 300t
equency o
mpling fre
nd its equ
nnection ar
alent will b
onous gen
t.
g at the gro
and also c
he other s
t + 6 sin
d) 1500
of 5 kHz i
quency in
d) 20
uivalent st
re scaled b
be scaled b
d) k
nerator is t
ound floor
can be tur
witch. The
(500t + π
s to be s
kHz whic
tar conne
by a factor
by a factor
he
and the
ned OFF
e logic of
π/4), the
sampled.
ch is not
ction as
k, k > 0,
of
2
2
2
b) A
c) A
ro
d) A
sy
1. Leaka
a) Flu
b) Flu
c) Flu
d) Flu
2. Three
readi
readi
a) V
c)V =
3. Squa
Angular dis
Angular dis
otating axi
Angular di
ynchronou
age flux in
ux that leak
ux that link
ux that link
ux that link
e moving
ings V, V1
ings is
1 2
2 2V V
= +
V1V2
re root of
placement
splacemen
is.
splacemen
usly rotatin
an inducti
ks through
ks both sta
ks none of t
ks the stato
iron type
and V2, a
2
2
–i, where
t of the rot
nt of the
nt of an
ng axis.
ion motor
h the mach
tor and ro
the windin
or winding
voltmete
as indicate
b) V
d) V
1i = − , a
tor with re
stator mm
axis fixe
is
hine
tor windin
ngs
g or the rot
ers are co
ed. The c
V = V1 + V2
V = V2 – V1
re
espect to th
mf with r
d to the
ngs
tor winding
nnected a
orrect rela
he stator.
respect to
rotor w
g but not b
as shown
ation amo
o a synchr
ith respec
both
below. Vo
ong the vo
ronously
ct to a
oltmeter
oltmeter
a) i, ‐i
b) 3 3cos sin ,cos sin
4 4 4 4i iπ π π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞− + − +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
c) 3 3cos sin ,cos sin
4 4 4 4i iπ π π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
d) 3 3 3 3cos sin ,cos sin4 4 4 4
i iπ π π π⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ − − +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
24. Given a vector field F = y2xax – yzay – x2az, the line integral 1F d⋅∫ evaluated along
a segment on the x‐axis from x = 1 to x = 2 is
a) ‐2.33 b) 0 c) 2.33 d) 7
25. The equation 1
2
2 2 01 1 0
xx
− ⎡ ⎤⎡ ⎤ ⎡ ⎤=⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦⎣ ⎦
has
a) no solution b) only one solution 1
2
00
xx
⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥
⎣ ⎦⎣ ⎦
c) Non‐zero unique solution d) multiple solutions
26. A strain gauge forms one arm of the bridge shown in the figure below and has a
nominal resistance without any load as Rs = 300 Ω. Other bridge resistances are R1
= R2 = R3 = 300 Ω. The maximum permissible current through the strain gauge is 20
mA. During certain measurement when the bridge is excited by maximum
permissible voltage and the strain gauge resistance is increased by 1% over the
nominal value, the output voltage V0 in mV is
2
2
7. In the
main
rating
a) 12
c) 25
8. The o
conn
reduc
comp
e circuit sh
tain 5V ac
g of the Ze
5 and 125
0 and 125
open‐loop
ected in f
ce the tim
pared to th
hown belo
cross RL, t
ener diode
transfer f
feedback a
me constan
hat of the o
ow, the kne
the minim
in mW res
b) 1
d) 2
function o
as shown
nt of the
open‐loop
ee current
um value
spectively
125 and 25
250 and 25
of a dc mo
below, th
closed loo
system is
of the ide
of RL in Ω
are
50
50
otor is give
he approx
op system
eal Zener d
Ω and the
en as ((a
sV sω
imate valu
m by one
diode is 10
e minimum
) 10) 1 10s s
=+
ue of Ka t
hundred t
0 mA. To
m power
s. When
that will
times as
2
3
a) 1
9. In th
Thev
a) 10
0. Three
have
show
comb
capac
a) 2.8
he circuit s
enin’s equ
0∠90°
e capacito
breakdow
wn, the m
bination a
citance acr
8 and 36
b) 5
shown be
ivalent vo
c) 800
rs C1, C2, a
wn voltages
maximum s
nd the co
ross the te
b) 7
low, if the
ltage in Vo
∠0°
and C3 wh
s of 10V, 5
safe volta
orrespondi
erminals ar
and 119
c) 10
e source v
olts as seen
c) 800∠
ose values
5V, and 2V
age in Vo
ing total c
re respectiv
c) 2.8 a
voltage VS
n by the lo
∠90°
s are 10µF
V respectiv
olts that c
charge in
vely
and 32
d) 100
S = 100∠5
ad resistan
d) 100∠
F, 5µF, and
vely. For th
can be ap
µC stored
d) 7 and
53.13° V t
nce RL is
60°
d 2µF resp
he intercon
pplied acr
d in the e
80
hen the
ectively,
nnection
ross the
effective
3
3
3
1. A vol
meas
a) sin
c) (sin
2. The s
of 20
is us
arma
at the
a) 0.4
3. For a
per u
react
refer
tage 1000
sured acro
n ωt
nω t – |sin
separately
0 A and a r
ed to con
ature react
e rated spe
4
a power sy
unit. The
tance of –
ence node
sin ωt Vo
ss WX in V
nωt|) / 2
excited dc
ated arma
ntrol the a
tion, the du
eed and th
b) 0.5
ystem netw
voltage at
j3.5 per u
e, the curre
lts is applie
Volts is
b) (
d) 0
c motor in
ture voltag
armature
uty ratio o
he rated fie
work with
t node 3
nit is now
ent drawn
ed across Y
sinωt + |si
0 for all t
n the figure
ge of 150 V
voltage. If
of the chop
eld current
c) 0.6
n nodes, Z
is 1.3 ∠ –
added to
by the cap
YZ. Assum
inωt|)/2
e below h
V. An idea
f La = 0.1
pper to obt
t is
Z33 of its b
– 10° per
the netwo
pacitor per
ing ideal d
as a rated
l chopper
mH, Ra =
tain 50% o
d) 0.7
bus impeda
unit. If a
ork betwee
r unit is
diodes, the
armature
switching
= 1 Ω, ne
of the rated
ance matr
a capacitor
en node 3
e voltage
current
at 5 kHz
eglecting
d torque
ix is j0.5
r having
and the
a) 0.325 ∠ –100° b) 0.325 ∠ 80°
c) 0.371 ∠ –100° d) 0.433 ∠ 80°
34. A dielectric slab with 500 mm × 500 mm cross‐section is 0.4 m long. The slab is
subjected to a uniform electric field of E = 6ax + 8ay kV/mm. The relative
permittivity of the dielectric material is equal to 2. The value of constant ε0 is 8.85 ×
10‐12 F/m. The energy stored in the dielectric in Joules is
a) 8.85 × 10‐11 b) 8.85 × 10‐5 c) 88.5 d) 885
35. A matrix has eigen values –1 and –2. The corresponding eigenvectors are 1
1⎡ ⎤⎢ ⎥−⎣ ⎦
and
12
⎡ ⎤⎢ ⎥−⎣ ⎦
respectively. The matrix is
a) 1 11 2
⎡ ⎤⎢ ⎥− −⎣ ⎦
b) 1 22 4
⎡ ⎤⎢ ⎥− −⎣ ⎦
c) 1 0
0 2−⎡ ⎤
⎢ ⎥−⎣ ⎦ d)
0 12 3
⎡ ⎤⎢ ⎥− −⎣ ⎦
36. 2
2
44
z dzz
−+∫ Evaluated anticlockwise around the circle |z – i| = 2, where 1i = − , is
a) ‐4π b) 0 c) 2 + π d) 2 + 2i
37. The clock frequency applied to the digital circuit shown in the figure below is 1 kHz.
If the initial state of the output Q of the flip‐flop is ‘0’, then the frequency of the
output waveform Q in kHz is
3
3
a) 0.2
8. In the
and t
and Y
expre
a) XY
9. In the
25
e circuit sh
the diode d
Y are digit
ession for Z
Y
e circuit sh
b) 0.5
hown below
drops negl
tal signals
Z is
b) XY
hown below
w, Q1 has n
ligible volt
with 0 V
Y
w the op‐a
c) 1
negligible c
age across
as logic 0
c) XY
amps are id
collector‐t
s it under f
and Vcc a
deal. Then
d) 2
o‐emitter
forward bi
as logic 1,
d) XY
Vout in Vol
saturation
ias. If Vcc is
then the
ts is
n voltage
s +5 V, X
Boolean
4
a) 4
0. The s
this s
a) 5s
c) 2s
signal flow
system is
2
16 2
ss s
++ +
14 2
ss
++ +
b) 6
w graph for
r a system
b) s
d) 5
c) 8
m is given b
2
16 2
ss s
++ +
2
15 6 2s s+ +
below. The
2
d) 10
e transfer
function YU
( )( )
Y sU s
for
4
4
4
4
1. The i
3). Th
a) 0
2. Two
opera
serie
a) q1R
c) (q1
3. The f
which
to ge
appli
VWX2
a) 12
c) 10
4. Thyri
width
mpulse re
he value of
magnetica
ating frequ
s, their eff
R1 + q2R2
1R1 + q2R2)
following
h attenuat
et an open
ed across
/ VYZ2 are r
5/100 and
0/100 and
stor T in t
h 10 µs. It
sponse of
f the step r
b) 1
ally uncoup
uency. The
fective Q fa
/ (R1 + R2)
arrangeme
tes by a fac
n circuit v
YZ to get
respective
d 80/100
100/100
he figure b
t is given t
a continuo
response a
pled induct
eir respecti
actor at the
b) q
d) q1R2 + q
ent consis
ctor of 0.8
oltage VYZ
an open c
ly,
b) 1
d) 8
below is in
that L ⎛= ⎜⎝
ous time sy
at t = 2 is
c) 2
tive coils h
ive resistan
e same op
q1 / R1 + q2
q2R1
sts of an
. An ac vol
1 across Y
ircuit volta
100/100 an
80/100 and
nitially off
100 H⎞μ⎟π ⎠ a
ystem is g
have Q fac
nces are R1
erating fre
/ R2
ideal tran
ltage VWX1
YZ. Next, a
age VWX2 a
nd 80/100
d 80/100
and is trig
and 1C ⎛= ⎜
⎝
iven by h(t
d) 3
tors q1 and
1 and R2. W
equency is
sformer a
= 100V is
an ac volta
across WX.
ggered wit
100 H⎞μ⎟π ⎠. A
t) = δ(t – 1
d q2 at the
When conn
and an att
applied ac
age VYZ2 =
Then, VYZ
h a single
Assuming
1) + δ(t –
e chosen
nected in
tenuator
cross WX
100V is
1 / VWX1,
pulse of
latching
4
4
4
and h
zero,
a) 10
5. A 4‐p
sourc
negat
a) 10
6. A fun
point
a) 20
7. When
1 = 0
1.2 is
a) ‐0.
holding cu
T conduct
μs
pole induc
ce, is rota
tive seque
0
nction y =
t in this int
n the New
, the solut
s
.82
urrents of
ts for
b) 50 μ
ction moto
ating at 14
ence curren
b) 98
5x2 + 10x
terval, dydx
b) 25
wton‐Raphs
ion at the
b) 0.49
the thyrist
μs
or, supplie
440 rpm.
nt in the ro
is defined
is exactly
son metho
end of the
9
tor are bo
c) 100 μ
ed by a sl
The electr
otor is
c) 52
over an o
c) 30
d is applie
e first itera
c) 0.705
oth zero an
μs
ightly unb
rical frequ
open interv
ed to solve
ation with t
5
nd the init
d) 200 μ
balanced t
uency in H
d) 48
val x = (1,
d) 35
the equat
the initial g
d) 1.69
tial charge
s
hree‐phas
Hz of the
2). At leas
tion f(x) = x
guess valu
e on C is
e 50 Hz
induced
st at one
x3 + 2x –
e as x0 =
C
C
In
M
a
4
4
C
T
⎡⎢⎣
5
Common
Common D
n the figur
MOSFET Q
ssumed to
8. The a
a) 3/2
9. The P
a) 0.9
Common D
he state va
1
2
20
xx
−⎤ ⎡=⎥ ⎢
⎣⎦
0. The s
a) Co
b) No
Data Que
Data for Qu
re shown
is switche
o be ideal.
average so
2
PEAK‐TO‐P
96
Data for Qu
ariable for
1
2
01
xx
⎡ ⎤⎤⎢ ⎥⎥− ⎦ ⎣ ⎦
system is
ontrollable
ot controlla
estions
uestions 48
below, th
d at 250 k
urce curre
b) 5/3
EAK sourc
b) 0.14
uestions 50
mulation o
1,
1u⎡ ⎤
+ ⎢ ⎥⎣ ⎦
but not ob
able but ob
8 and 49:
e chopper
Hz, with a
ent in Amp
e current r
44
0 and 51:
of a system
1(0) 0,x =
bservable
bservable
r feeds a
duty ratio
s in steady
c) 5/2
ripple in Am
c) 0.192
m is given a
2 (0) 0 ax =
resistive l
o of 0.4. Al
y‐state is
mps is
2
as
[and 1y =
oad from
ll elements
d) 15/4
d) 0.288
] 1
2
0xx
⎡ ⎤⎢ ⎥⎣ ⎦
a battery
s of the cir
source.
rcuit are
5
L
S
In
v
re
5
c) Bo
d) Bo
1. The r
a) 12
−
c) 2te−
inked An
tatement
n the follo
oltage pha
eactances
2. The v
a) θ2
b) θ2
c) θ2
d) θ2
th control
oth not con
response y
212
te−−
t te−−
nswer Qu
for Linked
owing netw
ase angles
are equal
voltage pha
= ‐0.1, θ3 =
= 0, θ3 = ‐0
= 0.1, θ3 =
= 0.1, θ3 =
lable and o
ntrollable a
y(t) to a un
estions
d Answer Q
work, the
are very
to j1 Ω.
ase angles
= ‐0.2
0.1
0.1
= 0.2
observable
and not ob
it step inpu
b) 1
d) 1
Questions 5
voltage m
small, and
in rad at b
e
bservable
ut is
2112
te−− −
1 te−−
52 and 53:
magnitudes
d the line
buses 2 an
12
te−
:
at all bus
resistance
d 3 are
ses are eq
es are negl
qual to 1 p
ligible. All
p.u., the
the line
5
S
T
H
re
3
5
5
3. If the
respe
the s
a) ‐10
tatement
he Voltage
Hz, square‐
eference d
Ω, L = 9.5
4. In the
curre
a) Q1
5. Appr
(ZCS)
a) ZV
c) ZCS
e base im
ectively, th
lack bus is
0
for Linked
e Source In
‐wave ac o
direction o
5 mH.
e interval
ent is
1, Q2
opriate tra
) of the IGB
VS during tu
S during tu
pedance a
hen the re
b) 0
d Answer Q
nverter (V
output volt
f the outp
when V0 <
b) Q3,
ansition i.
BTs during
urn‐off
urn‐off
and the lin
al power i
Questions 5
SI) shown
tage (vo) a
ut current
< 0 and i0
Q4
e., Zero V
turn‐on/tu
ne‐to‐line
n MW del
c) 10
54 and 55:
in the figu
across an R
t io are indi
> 0 the pa
c) D1, D
Voltage Sw
urn‐off is
b) ZVS d
d) ZCS d
base volta
livered by
:
ure below
R‐L load. R
icated in t
air of devic
D2
witching (Z
during turn
during turn
age are 10
the gener
d) 20
is switche
Reference
he figure.
ces which
d) D3, D4
ZVS)/Zero
n‐on
n‐on
00 Ω and
rator conne
ed to prov
polarity of
It is given
conducts t
4
Current Sw
100 kV,
ected at
ide a 50
f Vo and
that R =
the load
witching
Key for Exercise: 1
Q.NO. Key 1. B 2. D 3. B 4. C 5. C 6. C 7. C 8. B 9. C 10. A 11. A 12. D 13. A 14. A 15. B 16. C 17. A 18. A 19. B 20. D 21. D 22. D 23. B 24. B 25. D 26. C 27. B
28. C 29. C 30. C 31. D 32. D 33. D 34. C 35. D 36. A 37. B 38. B 39. C 40. A 41. B 42. C 43. B 44. C 45. B 46. B 47. C 48. B 49. C 50. A 51. A 52. Marks to all 53. Marks to all 54. D 55. Marks to all
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