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Standard XII
Physics
Question
The charge flowing through a resistance
R
varies with time
t
as
Q
=
a
t
−
b
t
2
, where
a
and
b
are positive constants. The total heat produced in
R
is
b
3
R
6
a
True
False
A
False
B
True
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Solution
Verified by Toppr
Given:
Q
=
a
t
−
b
t
2
⇒
i
=
d
Q
d
t
=
a
−
2
b
t
Therefore, current will be zero at
t
=
a
2
b
We know that:
d
H
=
i
2
R
d
t
⇒
H
=
∫
a
/
2
b
0
(
a
−
2
b
t
)
2
R
d
t
⇒
H
=
[
(
a
−
2
b
(
a
2
b
)
)
3
R
−
3
×
2
b
]
−
[
(
a
−
2
b
(
0
)
)
3
R
−
3
×
2
b
]
⇒
H
=
a
3
R
6
b
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Q1
The charge flowing through a resistance
R
varies with time
t
as
Q
=
a
t
−
b
t
2
, where
a
and
b
are positive constants. The total heat produced in
R
is
b
3
R
6
a
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Q2
The charge through a conductor of resistance
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