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JEE Main Physics Class 12 Alternating Currents Part 1 Questions
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© examsnet.com
Question : 19
Total: 100
A sinusoidal voltage of peak value
250
V
is applied to a series
L
−
C
−
R
circuit, in which
R
=
8
Ω
,
L
=
24
m
H
and
C
=
60
µ
F
. The value of power dissipated at resonant condition is
x
k
W
. The value of
x
to the nearest integer is
[16 Mar 2021 Shift 1]
Your Answer:
Validate
Solution:
Given,
V
0
=
250
V
,
R
=
8
Ω
,
L
=
24
m
H
and
C
=
60
µ
F
We know that, at resonance power,
P
=
V
rms
2
R
⇒
P
=
(
250
/
√
2
)
2
8
[
∵
V
rms
=
V
0
√
2
=
250
√
2
V
]
=
(
250
)
2
∕
16
=
62500
16
=
3906.25
W
≈
4
k
W
x
=
4
.
Comparing with the given value in the question i.e.,
x
k
W
, the value of
x
=
4
.
© examsnet.com
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