Solution

Clearly, if $\mathcal{E}\le V_0$, then there is no current in the circuit and hence no heat generated in the resistor. If $\mathcal{E}>V_0$, then the charge $q$ collected on the capacitor satisfies the differential equation

$$\dot qR+\frac{q}{C}=\mathcal{E}-V_0,$$

where the terms on the left are voltage drops across the resistor and the capacitor, respectively. Solving this equation subject to the condition that $q$ is initially zero, we get

$$q=C(\mathcal{E}-V_0)[1-e^{-t/(RC)}],$$

so the current in the circuit is

$$I=\dot q=\frac{\mathcal{E}-V_0}{R}\cdot e^{-t/(RC)}.$$

Thus, the heat generated in the resistor is

$$\int_0^\infty I^2R\,dt=\frac{(\mathcal{E}-V_0)^2}{R}\int_0^\infty e^{-2t/(RC)}=\frac{C(\mathcal{E}-V_0)^2}{2}.$$