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+### Statement
+
+$8.4.8.$ [Insert the problem statement]
+
+### Solution
+
+![For problem $8.4.8$ |1182x489, 31%](../../img/8.4.8/8.4.8.png)
+
+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
+
+\[\int_0^\infty I^2R\,dt=\frac{(\mathcal{E}-V_0)^2}{R}\int_0^\infty e^{-2t/(RC)}=\frac{C(\mathcal{E}-V_0)^2}{2}.\]
+
+#### Answer
+
+[Insert a concise answer or boxed result]