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Class 12 NEET Physics Electromagnetic Induction and Alternating Currents
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© examsnet.com
Question : 6
Total: 114
A
10
µ
F
capacitor is connected to a
210
V
,
50
Hz
source as shown in figure. The peak current in the circuit is nearly
(
π
=
3.14
)
:
[NEET 2024]
0.58
A
0.93
A
1.20
A
0.35
A
Validate
Solution:
👈: Video Solution
Capacitive Reactance
X
C
=
1
ω
C
=
1
2
π
f
C
=
1
2
×
3.14
×
50
×
10
×
10
−
6
=
1000
3.14
V
rms
=
210
V
i
rms
=
V
rms
X
C
=
210
X
C
Peak current
=
√
2
i
rms
=
√
2
×
210
1000
×
3.14
=
0.932
=
0.93
A
© examsnet.com
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