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JEE Main 20 July 2021 Shift 1 Solved Paper
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© examsnet.com
Question : 5
Total: 90
AC voltage
V
(
t
)
=
20
sinωt of frequency
50
H
z
is applied to a parallel plate capacitor. The separation between the plates is
2
m
m
and the area is
1
m
2
.
The amplitude of the oscillating displacement current for the applied AC voltage is _______.
[
Take
ε
0
=
8.85
×
10
−
12
F
∕
m
]
[20 Jul 2021 Shift 1]
21.14
µ
A
83.37
µ
A
27.79
µ
A
55.58
µ
A
Validate
Solution:
From the given information,
C
=
ε
0
A
d
=
∈
0
×
1
2
×
10
−
3
F
∴
X
C
=
1
ω
C
=
2
×
10
−
3
2
×
50
π
×
∈
0
=
2
×
10
−
3
25
×
4
π
∈
0
Ω
∴
X
C
=
2
×
10
−
3
25
×
9
×
10
9
=
18
25
×
10
6
Ω
∴
i
0
=
V
0
X
C
=
20
×
25
18
×
10
−
6
A
=
27.47
µ
A
.
The value of amplitude of displacement current will be same as value of amplitude of conventional current.
Hence option 3.
© examsnet.com
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