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
KVPY 2012, Diffcult
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(d)
Stored energy in capacitor is
$U=\frac{q^2}{2 C}$
and charge of capacitor in an $R C$ circuit during discharging is
$q=q_{0 } e^{-t / R C}$
Combining both, we have
$U=\frac{q_0^2}{2 C} \cdot e^{-2 t / R C}=U_0 e^{-2 t / R C}$
Given, $\quad U=\frac{U_0}{2}$
So, $\quad \frac{U_0}{2}=U_{0} e^{\frac{-2 t}{R C}} \Rightarrow e^{\frac{-2 t}{R C}}=\frac{1}{2}$
$\Rightarrow \frac{-2 t}{R C}=\ln \frac{1}{2}$
$\Rightarrow t=\frac{1}{2} \cdot R C \ln 2$
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