Yume.
Time:
1
30
Hours]
b
Q.1
Attempt
all
parts:
e
f
Q-2
ABES
Institute
of
Technology,
Ghaziabad
C
Subject
Code:
(BEE201)
Fundamentals
of
Electrleal
Enginssny
I
YEAR
SEC-7
to
12
(ALL
Branch)
CT-2
(EVEN
SEMESTER
2022-23)
Why
series
resonant
circuit
is
called
an
acceptor
circuit
and
parallel
resonant
circuit
a
rejector circuit?
SECTION-A
For
an
ac
circuit,
R=1002,
L=1H
and
C=luF.
Calculate
Resonant
Frequency
and
Q-factor.
What
do
you
mean
by
a
transformer?
How
the
emf
is
induced
in
a transformer?
Explain.
The
primary
winding
of
a
50
Hz
transformer
is
supplied
from
a
440V
source
has
200
turns.
Find
the
voltage
induced
in
secondary
if
it
has
50
turns.
Compare
a
single
phase
and
three
phase
ac
system.
Why
the
losses take
place
in
a transformer?
Attempt
any
TWO
parts
SECTION-B
A series resonant
circuit
has
an
impedance
of
500
2 at resonant
frequency
and
cut
off
frequencies
are
10kHz
and
100kHz.
Determine
()
Resonant
Frequency
(ii)
Value
of
R,
L and C
(iii)
Quality
Factor at Resonant Frequency and
(iv)
PF
of the circuit
Define
quality
factor
in
series
RLC
circuit.
Determine
the
half
power
frequencies
and
the
resonant
frequency
for
series
RLC
circuit.
Draw
the
phasor
diagram
and
circuit
diagram
fora three-phase
ac
circuit
in
case
of
(a)
STAR
(b)
DELTA.
Q3
Attempt
any
TWO
parts
Three
Impedances
of
52
each
are
supplied
from
a
400V,
50Hz
three
phase
ac
source.
Calculate
line
current
and
phase
current
when
the
load
is
connected
in
(a)
STAR
(b)
DELTA
SECTIOON-C
Derive
the
EMF
equation
of
the
transformer.
A
6600/440V
single
phase
transformer
has
a
primary
resistance
of
1402
and
a
secondary
resistance
of
0.25N.
Calculate
the
equivalent
resistances
referred
to
primary
winding
and
secondary
winding.
The
primary
resistance
of
a
440/110V
single
phase
transformer
is
0.28n
and
the
secondary
resistance
is
0.0182.,If
the
iron
loss
is
measured
to
be
160W
when
rated
voltage
is
applied,
find
the
kW
loading
to
give
maximum
efficiency
at
unity
power
factor.
Also,
derive
the
condition
of
maximum
eficiency.
A
1100/230V,
150KVA
single
phase
transformer
has
a
core
loss
of
1.4kW
and
a
full
load
copper
loss
of
1.6kW.
Determine
()Maximum
efficiency
(i)
Half
load
Efficiency
and
(iii)
Full
load
Eficiency.
The
power
factor
of
the
load
is
©.8
lagging.
(1/7)
(CO
2)
(CO-2)
(CO-3)
(CO-3)
(CO-3)
(CO-3)
(CO-3)
(7x2=14)
(Co-2)
(Co-2)
(C0-2)
(7x2=14)
(CO-3)
(CO-3)
(C0-3)
