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AIEEE 2010 Solved Paper
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© examsnet.com
Question : 43
Total: 89
Let
C
be the capacitance of a capacitor discharging through a resistor
R
. Suppose
t
1
is the time taken for the energy stored in the capacitor to reduce to half its initial value and
t
2
is the time taken for the charge to reduce to one-fourth its initial value. Then the ratio
t
1
∕
t
2
will be
1
1
2
1
4
2
Validate
Solution:
Initial energy of capacitor,
E
1
=
q
1
2
2
C
Final energy of capacitor,
E
2
=
1
2
E
1
=
q
1
2
4
C
=
(
q
1
√
2
2
C
)
2
∴
t
1
=
time for the charge to reduce to
1
√
2
of its initial value
and
t
2
=
time for the charge to reduce to
1
4
of its initial value
We have,
q
2
=
q
1
e
−
t
∕
C
R
⇒
ln
(
q
2
q
1
)
=
−
t
C
R
∴
ln
(
1
√
2
)
=
−
t
1
C
R
.
.
.
(
1
)
and
ln
(
1
4
)
=
−
t
2
C
R
.
.
.
(
2
)
By (1) and
(
2
)
,
t
1
t
2
=
ln
(
1
√
2
)
ln
(
1
4
)
=
1
2
ln
(
1
2
)
2
ln
(
1
2
)
=
1
4
© examsnet.com
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