Capacitors - 5

Statement: The energy dissipated by the resistor is equal to the energy stored in the capacitor at steady state.

The energy in the capacitor at steady state, and therefore the energy dissipated by the resistor is $\boxed{ 9 \times 10^{-5} \text{ J}}$ $_\square$

---

Proof: Suppose the switch is closed at $t = 0$ . Then the current flowing in the circuit at time $t$ is $i = i_0 e^{-t/RC}$ . The total heat dissipated by the resistor is

$\begin{aligned} \int _{0} ^{\infty} i^{2} R \, dt &= \int _{0} ^{\infty} (i_0 e^{-t/RC})^{2} \times r \, dt\\ & = i_{0}^{2} R \int _{0} ^{\infty} e^{-2t/RC})^{2} \, dt \\ & = i_{0}^{2} R \times \frac{RC}{-2} \times \left[ e^{-2t/RC})^{2} \right]_{0} ^{\infty}\\ & = i_{0}^{2} R \times \frac{RC}{-2} \times [ 0 - 1] \\ & = i_{0}^{2} R \times \frac{RC}{2} \\ & = \frac{1}{2} C \times (i_{0} R)^{2} \\ & = \frac{1}{2} C \times V_{0}^{2} \end{aligned}$