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(D) The energy stored in a capacitor is given by $U = \frac{q^2}{2C}$.During the discharging of an $RC$ circuit,the charge on the capacitor at any tim

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$\frac{R C \ln 2}{2}$

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(D) The energy stored in a capacitor is given by $U = \frac{q^2}{2C}$.During the discharging of an $RC$ circuit,the charge on the capacitor at any time $t$ is given by $q = q_0 e^{-t / RC}$.Substituting this into the energy formula,we get $U = \frac{(q_0 e^{-t / RC})^2}{2C} = \frac{q_0^2}{2C} e^{-2t / RC} = U_0 e^{-2t / RC}$,where $U_0$ is the initial energy.We want the time $t$ when $U = \frac{U_0}{2}$.Setting the equations equal: $\frac{U_0}{2} = U_0 e^{-2t / RC}$.Dividing by $U_0$: $\frac{1}{2} = e^{-2t / RC}$.Taking the natural logarithm on both sides: $\ln(1/2) = -2t / RC$.Since $\ln(1/2) = -\ln 2$,we have $-\ln 2 = -2t / RC$.Solving for $t$: $t = \frac{RC \ln 2}{2}$.

The capacitor of capacitance $C$ in the circuit shown is fully charged initially. The 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

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