Question

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 $V_e$ (i.e. charging continues for time $T>>RC$ ). In the process some dissipation ($E_D$) 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>>RC. Then the voltage is raised to $\frac{2V_0}{3}$ without discharging the capacitor and again maintained for a time $T>>RC$. The process is repeated one more time by raising the voltage to $V_0$ and the capacitor is charged to the same final voltage $V_0$ 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. $E_D=\frac{1}{3}\left(\frac{1}{2}{CV}_0^2\right)$
• B. $E_D=3\left(\frac{1}{2}{cv}_o^2\right)$
• C. $E_D=\frac{1}{2}{CV}_o^2$
• D. $E_D=3{CV}_0^2$

Answer 1

The correct option is A $E_D=\frac{1}{3}\left(\frac{1}{2}{CV}_0^2\right)$


$W_{\text{battery}}=\frac{v_0}{3}\left(\frac{{CV}_0}{3}\right)=\frac{{CV}_0^2}{9}$

Energy of capacitor =$\frac{{CV}_0^2}{18}$
Heat loss $E_1=\frac{{CV}_0^2}{9}-\frac{{CV}_0^2}{18}{E}_1=\frac{1}{18}{CV}_0^2$



$\begin{array}{cc}{U}_f& =\frac{q^2}{2C}\\ & =\frac{4}{18}{CV}_0^2\\ U_f& =\frac{2}{9}{CV}_0^2\\ & \\ \Delta U& =\frac{{CV}_0^2}{6}\end{array}$

$W_{\text{battery}}=\frac{2V_0}{3}\left(\frac{{CV}_0}{3}\right)=\frac{2}{9}{CV}_0^2\therefore \text{heat loss}=\frac{2}{9}{CV}_0^2-\frac{{CV}_0^2}{6}$

$=\frac{3}{54}{CV}_0^2$

$E_2=\frac{{CV}_0^2}{18}$




$\begin{array}{cc}\text{Work done by battery}=& V_0\left(\frac{1}{3}{CV}_0\right)\\ & =\frac{{CV}_0^2}{3}\\ \Delta U& =\frac{5}{18}{CV}_0^2\end{array}$

Total heat loss $E_3=\frac{{CV}_0^2}{3}-\frac{5}{18}{CV}_0^2$
$E_3=\frac{1}{18}{CV}_0^2$

$\therefore$ Total heat dissipated

$\begin{array}{cc}E& =E_1+E_2+E_3\\ =& \frac{1}{18}{CV}_0^2+\frac{1}{18}{CV}_0^2+\frac{1}{18}{CV}_0^2\\ =& \frac{1}{6}{CV}_0^2=\frac{1}{3}\left(\frac{1}{2}{CV}_0^2\right)\end{array}$