Initially S is open capacitor A is charged up to potential difference  $\frac{V}{2}$ and capacitor B is uncharged. After closing the switch S the heat released in the circuit is  $\frac{C{V}^2}{2\times n}.$  Find n?

 

Let charge supplied by battery is q.
Apply KVL
 
$$\begin{gathered} \frac{\frac{CV}{2}+q}{C}+\frac{q}{C}=V \\ \frac{CV}{2}+2q=CV \end{gathered}$$

$2q=\frac{CV}{2};q=\frac{CV}{4}$

Energy supplied by battery  $=q\times V=\frac{C{V}^2}{4};$
Energy consumed by capacitors  $=U_{ƒ}-U_i$
 
$=\frac{9C{V}^2}{32}+\frac{C{V}^2}{32}-\frac{C{V}^2}{8}=\frac{3C{V}^2}{16}$

Heat loss$=\frac{C{V}^2}{4}-\frac{3C{V}^2}{16}=\frac{C{V}^2}{16}=\frac{C{V}^2}{2\times 8}$