Physics - Capacitor Question with Solution | TestHub

For process (i)
Charge on capacitor$=\frac{C{V}_0 }{3}$
Energy stored in capacitor$=\frac{1}{2}C\frac{V_0^2}{9}=\frac{C{V}_0^2}{18}$
Work done by battery$=\frac{C{V}_0}{3}\times \frac{V}{3}=\frac{C{V}_0^2}{9}$
Heat loss$=\frac{C{V}_0^2}{9}-\frac{C{V}_0^2}{18}=\frac{C{V}_0^2}{18}$
For process (ii)
Charge on capacitor$=\frac{2C{V}_0}{3}$
Extra charge flow through battery$=\frac{C{V}_0}{3}$
Work done by battery:$\frac{C{V}_0}{3} .\frac{2V_0}{3}=\frac{2C{V}_0^2}{9}$
Final energy store in capacitor:$\frac{1}{2}C{\left(\frac{2V_0}{3}\right)}^2=\frac{4C{V}_0^2}{18}$
Energy store in process 2:$\frac{4C{V}_0^2}{18}-\frac{C{V}_0^2}{18}=\frac{3C{V}_0^2}{18}$
Heat loss in process (ii) = work done by battery in process (ii) – energy store in capacitor process (ii)
$=\frac{2C{V}_0^2}{9}-\frac{3C{V}_0^2}{18}=\frac{C{V}_0^2}{18}$
For process (iii)
Charge on capacitor$=C{V}_0$
Extra charge flow through battery:$C{V}_0-\frac{2C{V}_0}{3}=\frac{C{V}_0}{3}$
Work done by battery in this process:$\left(\frac{C{V}_0}{3}\right) \left(V_0\right)=\frac{C{V}_0^2}{3}$
Find energy store in capacitor:$\frac{1}{2}C{V}_0^2$
Energy stored in this process:$\frac{1}{2}C{V}_0^2-\frac{4C{V}_0^2}{18}=\frac{5C{V}_0^2}{18}$
Heat loss in process (iii):$\frac{C{V}_0^2}{3}-\frac{5C{V}_0^2}{18}=\frac{C{V}_0^2}{18}$
Now total heat loss$\left(E_D\right): \frac{C{V}_0^2}{18}+\frac{C{V}_0^2}{18}+\frac{C{V}_0^2}{18}=\frac{C{V}_0^2}{6}$
Final energy store in capacitor:$\frac{1}{2}C{V}_0^2$
So we can say that$E_D=\frac{1}{3}\left(\frac{1}{2}C{V}_0^2\right)$