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The capacitor of capacitance C in the ci...

The capacitor of capacitance C in the circuit shown in fully charged initially. Resistance is R.–

After the switch S is closed, the time taken to reduce the stored energy in the capacitor to half its initial value is

A

`(RC)/2`

B

RCln2

C

2RCln2

D

`(RCln2)/2`

Text Solution

Verified by Experts

The correct Answer is:
D
`Q = Q_(0)e^(-t//RC) rArr Q_(0)/sqrt(2) = Q_(0)e^(-t//RC) rArr 1/sqrt(2) = e^(-t/RC)`
`ln 1/sqrt(2) = -t/(RC) rArr t/(RC) = lnsqrt(2) rArr t = (RCln2)/2`
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