We apply Kirchhoff's Law around the circuit to find that

$\begin{align*}V_C = I R\end{align*}$
where $V_C$ is the voltage across the capacitor. We differentiate this equation with respect to time to obtain
$\begin{align*}\frac{d V_C}{dt} = \frac{d I}{dt} R\end{align*}$
But
$\begin{align*}I = C\frac{d V}{dt}\end{align*}$
where $I$ is the current through the capacitor. Actually, because of the sign conventions we have used, $I$ must be replaced by $-I$ :
$\begin{align*}-I = C\frac{d V}{dt}\end{align*}$
Therefore,
$\begin{align*}-I = \frac{d I}{dt} R C\end{align*}$
The solution to this equation, with the initial condition that $I(0) = V_0/R$ where $V_0$ is the initial voltage across the capacitor, is
$\begin{align*}I(t) = V_0/R \cdot e^{-t/(RC)}\end{align*}$
The voltage across the capacitor as a function of time is therefore
$\begin{align*}V(t) = V_0 \cdot e^{-t/(RC)} \label{eqn:1}\end{align*}$
The energy stored in the capacitor is given by
$\begin{align*}U = \frac{1}{2}C V^2\end{align*}$
Therefore, the half of the energy is dissipated at a time $t_1$ when
$\begin{align*}V(t_1) = V_0/\sqrt{2} \end{align*}$
By plugging this into equation (\ref{eqn:1}), we find that
$\begin{align*}V_0/\sqrt{2} = V_0 \cdot e^{-t_1/(RC)} \Rightarrow t_1 = \boxed{\frac{R C \ln 2}{2}}\end{align*}$
Therefore, answer (E) is correct.

Solution to 1992 Problem 11