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KVPY 2020-21 SB SX Exam Previous Papers
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© examsnet.com
Question : 29
Total: 120
An initially uncharged capacitor C is being charged by a battery of emf E through a resistance R upto the instant when the capacitor is charged to the potential E/2, the ratio of the work done by the battery to the heat dissipated by the resistor is given by :-
2 : 1
3 : 1
4 : 3
4 : 1
Validate
Solution:
i
=
E
R
e
−
t
∕
R
C
,
Q
=
C
E
(
1
−
e
−
t
∕
R
C
)
Capacitor is charged to
E
2
,
So
Q
=
C
E
2
∴
C
E
2
=
C
E
(
1
−
e
−
t
∕
R
C
)
1
2
=
e
−
t
∕
R
C
t
=
R
C
l
n
2
Work done by battery
=
(
Q
flown
)
(
Δ
V
)
=
(
C
E
2
)
(
E
)
=
C
E
2
2
Heat dissipated
=
R
C
l
n
2
∫
0
i
2
R
d
t
=
E
2
R
R
C
ln
2
∫
0
e
−
2
t
∕
R
C
.
d
t
=
3
4
(
C
E
2
2
)
Work done
Heat dissipated
=
C
E
2
∕
2
3
4
(
C
E
2
2
)
=
4
3
© examsnet.com
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