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NEET 2018 Physics and Chemistry
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© examsnet.com
Question : 7
Total: 90
An inductor
20
m
H
, a capacitor
100
µ
F
and aresistor 50Ω are connected in series across asource of emf,
V
=
10
sin
314
t
. The power loss in the circuit is
[NEET 2018]
0.79 W
0.43 W
1.13 W
2.74 W
Validate
Solution:
👈: Video Solution
P
av
=
(
V
RMS
Z
)
2
R
Z
=
√
R
2
+
(
ω
L
−
1
ω
C
)
2
=
56
Ω
∴
P
av
=
(
10
(
√
2
)
56
)
2
×
50
=
0.79
w
© examsnet.com
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