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 $\frac{2V_0}{3}$ 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. $E_D=\frac{1}{2}C{V}_0^2$
• B. $E_D=3\left(\frac{1}{2}C{V}_0^2\right)$
• C. $E_D=\frac{1}{3}\left(\frac{1}{2}C{V}_0^2\right)$
• D. $E_D=3C{V}_0^2$

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
$=\frac{1}{2}C\left\{{\left(\frac{V_0}{3}-0\right)}^2+{\left(\frac{2V_0}{3}-\frac{V_0}{3}\right)}^2+{\left(V_0-\frac{2V_0}{3}\right)}^2\right\}$
$=\frac{1}{6}C{V}_0^2$