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JEE Main Physics Class 12 Alternating Currents Part 2 Questions
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© examsnet.com
Question : 11
Total: 75
An alternating emf
E
=
110
√
2
s
i
n
100
t
volt is applied to a capacitor of
2
µ
F
, the rms value of current in the circuit is ______
mA
.
[8 Apr 2024 Shift 2]
Your Answer:
Validate
Solution:
👈: Video Solution
C
=
2
µ
f
;
E
=
110
√
2
s
i
n
(
100
t
)
X
C
=
1
ω
c
=
1
100
×
2
×
10
6
=
10000
2
=
5000
Ω
begin
a
l
i
g
n
e
d
i
o
=
110
√
2
5000
i
ms
=
110
√
2
5000
√
2
=
110
5
mA
=
22
mA
end
a
l
i
g
n
e
d
© examsnet.com
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