Question

PARAGRAPH 1
Consider a simple RC circuit as shown in Figure 1 .
Process 1: In the circuit the switch $S$ is closed at $t =0$ and the capacitor is fully charged to voltage $V _{0}$ (i.e. charging continues for time $T \gg RC$ ). In the process some dissipation $\left( E _{ D }\right)$ occurs across the resistance $R$. The amount of energy finally stored in the fully charged capacitor is $E_{C}$.
Process 2: In a different process the voltage is first set to $\frac{ V _{0}}{3}$ and maintained for a charging time $T \gg RC$. Then the voltage is raised to $\frac{2 V _{0}}{3}$ without discharging the capacitor and again maintained for time $T \gg RC$. The process is repeated one more time by raising the voltage to $V _{0}$ and the capacitor is charged to the same final
voltage $V _{ o }$ as in Process 1 .
These two processes are depicted in Figure 2 .

In Process 2, total energy dissipated across the resistance $E_{D}$ is:

• A. $E_{D} = \frac{1}{2}CV_{0}^{2}$
• B. $E_{D} = 3\left(\frac{1}{2}CV_{0}^{2}\right)$
• C. $E_{D} = \frac{1}{3}\left(\frac{1}{2}CV_{0}^{2}\right)$
• D. $E_{D} = 3 \,CV_{0}^{2}$

Answer 1

Answer: (C)

If capacitor is charged from $V _{ i }$ to $V _{ f }$; heat dissipated is
$H = W _{\text {battery }}-\Delta U$
$= C \left( V _{ f }- V _{ i }\right) V _{ f }-\left[\frac{1}{2} C \left( V _{ f }^{2}- V _{ i }^{2}\right)\right]$
$=\frac{1}{2} C \left( V _{ f }- V _{ i }\right)^{2}$
Total heat dissipated
$=\frac{1}{2} C\left\{\left(\frac{V_{0}}{3}-0\right)^{2}+\left(\frac{2 V_{0}}{3}-\frac{V_{0}}{3}\right)^{2}+\left(V_{0}-\frac{2 V_{0}}{3}\right)^{2}\right\}$
$=\frac{1}{6} C V_{0}^{2}$