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JEE Mains 27-July-2022 Shift 1 Solved Paper
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© examsnet.com
Question : 54
Total: 90
To light, a
50
W
,
100
V
lamp is connected, in series with a capacitor of capacitance
50
π
√
x
µ
F
, with
200
V
,
50
Hz
AC
source. The value of
x
will be _________.
[27-Jul-2022-Shift-1]
Your Answer:
Validate
Solution:
P
=
V
2
R
⇒
R
=
V
2
P
(
V
R
)
(
V
C
)
R
=
100
×
10
2
50
=
R
=
200
Ω
V
R
2
+
V
C
2
=
V
2
(
100
)
2
+
V
C
2
=
(
200
)
2
i
=
100
200
=
1
2
;
V
2
=
40000
V
=
I
×
X
C
;
V
C
2
=
30000
V
C
=
100
√
3
X
C
=
200
√
3
200
√
3
=
1
ω
C
C
=
1
20
×
50
×
20
√
3
=
50
×
10
−
6
√
x
√
x
=
50
×
10
−
6
×
100
×
200
√
3
X
=
3
© examsnet.com
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