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Byju's Answer
Standard XII
Physics
RLC Circuit
200 V AC sour...
Question
200
V
AC source is fed to series LCR circuit having
X
L
=
50
Ω
,
X
C
=
50
Ω
and
R
=
25
Ω
. Potential drop across the inductor is :
A
100
V
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B
200
V
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C
400
V
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D
10
V
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Solution
The correct option is
C
400
V
Here
V
r
m
s
=
200
V
,
X
L
=
50
Ω
,
X
C
=
50
Ω
,
R
=
25
Ω
Impedance of the circuit,
Z
=
√
R
2
+
(
X
L
−
X
C
)
2
=
√
25
2
+
(
50
−
50
)
2
=
25
Ω
Current in the circuit ,
I
r
m
s
=
V
r
m
s
Z
=
200
V
25
Ω
=
8
A
Voltage drop across the inductor is
V
L
=
I
r
m
s
X
L
=
8
A
×
50
Ω
=
400
V
Suggest Corrections
0
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