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Question

An RC circuit consists of a resistance R = 5 MΩ and a capacitance C = 1.0 μF connected in series with a battery. In how much time will the potential diffrence across the capacitor become 8 times that across the resistor? (Given loge(3)=1.1 )
(IIT-JEE 2005)


A

5.5 s

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B

11 s

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C

44 s

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D

88 s

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Solution

The correct option is B

11 s


If the circuit is like this,

If Vc is the potential differnce across C, the power differences across R will be VVC.
We are asked when will be
8(VVc)=Vc
or, when is
8V=9Vc ...(1)
We know, for a capacitor
qc=q0(1et/RC)
qcC=q0C(1et/RC)
Vc=V(1et/RC
(since q0C=V)
substitution in (1) gives,
8V=9V(1et/RC)
89=1et/RC
et/RC=19
t/RC=loge9
t/RC=2loge3
t=2×1.1×106×5×106
= 11 s


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