In the circuit shown, $A \mathrm{and} B$ are equal resistances. When switch $S$ is closed, the capacitor charges from the cell of emf $\epsilon$ and reaches a steady state. 

Current through $A$ is the main current passing through the battery. So this current is more than the current passing through$B$. Hence during charging more heat is produced in $A$.

In steady state, 

$i_C=0$

and $i_A=i_B$

Hence heat is produced at the same rate in $A \mathrm{and} B$.

Further, in steady state

$V_C=V_B=\frac{\epsilon }{2}$

$\therefore U=\frac{1}{2}C{V_C}^2=\frac{1}{8}C{\epsilon }^2$