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JEE Main 17 March 2021 Shift 2 Solved Paper
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© examsnet.com
Question : 30
Total: 90
Seawater at a frequency
f
=
9
×
10
2
H
z
, has permittivity
ε
=
80
ε
0
and resistivity
r
=
0.25
Ω
−
m
. Imagine a parallel plate capacitor is immersed in seawater and is driven by an alternating voltage source
V
(
t
)
=
V
0
sin
(
2
π
f
t
)
. Then, the conduction current density becomes
10
x
times the displacement current density after time
t
=
1
800
s
. The value of
x
is ..............
(
Take
,
1
4
π
ε
0
=
9
×
10
9
N
−
m
2
C
−
2
)
[17 Mar 2021 Shift 2]
Your Answer:
Validate
Solution:
As we know, conduction current,
I
c
=
[
v
0
sin
(
2
π
f
t
)
]
ρ
d
A
⇒
I
c
=
A
[
v
0
sin
(
2
π
f
t
)
]
ρ
d
..(i)
As we know the displacement current equation,
I
d
=
ε
0
ε
t
A
d
d
d
t
[
v
0
sin
(
2
π
f
t
)
]
I
d
=
ε
0
ε
t
A
d
v
0
(
2
π
f
)
[
cos
(
2
π
f
t
)
]
...(ii)
Divide Eq. (ii) by Eq. (i), we get
I
d
I
c
=
ε
0
ε
r
×
ρ
2
π
f
cot
(
2
π
f
t
)
Substituting the values in the above equation, we get
I
d
I
c
=
1
4
π
×
9
×
10
9
×
(
80
)
×
2
π
×
900
(
0.25
)
cot
(
2
π
(
900
)
1
800
)
I
d
I
c
=
1
10
6
Comparing with
10
x
, we get
x
=
6
.
© examsnet.com
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