KVPY SX SB Exam 2012 Question Paper

Question : 22 of 120

Marks: +1, -0

The capacitor of capacitance C in the circuit shown is 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

Solution:

Discharging -

Q=Q_0e^{-t\/RC}

U'=U/2⇒Q_0^2/{2C}e^{-2t\/RC}

=Q_0^2/{2C}

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