Solution to 1992 Problem 11

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*}
\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*}
\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.

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