Energy Stored by a Capacitor

The voltage across the whole circuit equals the external voltage: $\begin{aligned} U_0 &= U_C + U_R = \text{const} \quad (1)\\ U_C &= \frac{Q(t)}{C} \\ U_R &= R I(t) \end{aligned}$ Using the relation $I(t) = \frac{d Q(t)}{dt}$ for the current, equation (1) can be transformed into a differential equation. An additional time derivative results $\begin{aligned} \frac{d U_0}{dt} &= \frac{d U_C}{dt} + \frac{d U_R}{dt} = \frac{I(t)}{C} + R \frac{dI(t)}{dt} = 0 \\ \Rightarrow \quad \frac{dI(t)}{dt} &= -\frac{1}{RC} I(t) \end{aligned}$ Using the exponential approach $I(t) = C \cdot \exp(-a \cdot t)$ , the solution reads $\begin{aligned} I(t) &= \frac{U_0}{R} e^{-t/RC} \\ Q(t) &= C U_0 (1 - e^{-t/RC}) \end{aligned}$ which satisfies the initial condition $Q(0) = 0$ . Integrating the electrical power $P_i(t) = U_i(t) I(t)$ for both components results: $\begin{aligned} W_C &= \int_0^\infty P_C(t) dt = \int_0^\infty U_C(t) I(t) dt = \frac{U_0^2}{R} \int_0^\infty (1 - e^{-t/RC}) e^{-t/RC} dt = \frac{1}{2} C U_0^2 \\ W_R &= \int_0^\infty P_R(t) dt = \int_0^\infty U_R(t) I(t) dt = \frac{U_0^2}{R} \int_0^\infty e^{-2 t/RC} dt = \frac{1}{2} C U_0^2 \end{aligned}$ Therefore, $\eta = \frac{W_C}{W_R + W_C} = \frac{1}{2} = 50\,\text{\%}$ regardless of the resistance $R$ .