Question

In process $2$ , energy dissipated across the resistance ${E_D}$ is

\left( A \right){E_D} = 3\left( {\dfrac{1}{2}C{V_0}^2} \right) \\\ \left( B \right){E_D} = \dfrac{1}{2}C{V_0}^2 \\\ \left( C \right){E_D} = 3C{V_0}^2 \\\ \left( D \right){E_D} = \dfrac{1}{3}\left( {\dfrac{1}{2}C{V_0}^2} \right) \\\

Answer 1

Hint : In order to solve this question, we need to analyze the two processes as given in the question, then, the dissipation energy of the charged capacitor ${E_D}$ is calculated by subtracting the voltage change from the work done for charging the capacitor, the difference gives us the correct value of ${E_D}$ .
Formula used: The formula for the dissipation energy of a charged capacitor is:
${E_D} = {W_b} - \Delta V$
Where, work done is given by
${W_b} = \dfrac{{C{V_0}}}{3}\left( {\dfrac{{{V_0}}}{3} + \dfrac{{2{V_0}}}{3} + {V_0}} \right)$
And voltage change by
$\Delta V = \dfrac{1}{2}C{V_0}^2$

Complete Step By Step Answer:
Consider a simple RC circuit as shown in figure $1$
In the process $2$ ,as we can see in the figure $2$ the voltage is first decreased to one third of the voltage ${V_0}$ , i.e. $\dfrac{{{V_0}}}{3}$ and kept at that value for time $T > > RC$ , then the voltage is increased to twice this voltage , i.e. $\dfrac{{2{V_0}}}{3}$ and again kept at the same for time $T > > RC$ . Then, this process is repeated again and the voltage is increased to the value ${V_0}$ .
Now the dissipation energy for the capacitor is ${E_D}$ and its value is calculated as
{E_D} = {W_b} - \Delta V \\\ \Rightarrow {E_D} = \dfrac{{C{V_0}}}{3}\left( {\dfrac{{{V_0}}}{3} + \dfrac{{2{V_0}}}{3} + {V_0}} \right) - \dfrac{1}{2}C{V_0}^2 \\\ \Rightarrow {E_D} = \dfrac{{C{V_0}}}{3}\left( {2{V_0}} \right) - \dfrac{1}{2}C{V_0}^2 = \dfrac{{C{V_0}^2}}{6} \\\
This is the same energy value as given in the option $\left( D \right){E_D} = \dfrac{1}{3}\left( {\dfrac{1}{2}C{V_0}^2} \right)$ , hence, $\left( D \right)$ is correct. ${V_0}$

Note :
In the Process $1$ , as we can see in the circuit, the switch S is closed here at $t = 0$ and the capacitor has the full charge with the voltage. For the time, $T > > RC$ , the capacitor continues to charge . Across the resistance $R$ , the dissipation that occurs is ${E_D}$ . The energy stored in the fully charged capacitor is finally ${E_C}$ .