We apply Kirchhoff's Law around the circuit to find that
where is the voltage across the capacitor. We differentiate this equation with respect to time to obtain But where is the current through the capacitor. Actually, because of the sign conventions we have used, must be replaced by : Therefore, The solution to this equation, with the initial condition that where is the initial voltage across the capacitor, is The voltage across the capacitor as a function of time is therefore The energy stored in the capacitor is given by Therefore, the half of the energy is dissipated at a time when By plugging this into equation (\ref{eqn:1}), we find that Therefore, answer (E) is correct. |

Please send questions or comments to X@gmail.com where X = physgre.