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

 

At any time , i = $i_B + \frac{dQ}{dt}...........\left(1\right)$

                    $\frac{Q}{C}= i_{B}R ............\left(2\right)$

  and            $E = iR + \frac{Q}{C}............\left(3\right)$

  Solving , Q = $\frac{EC}{2}( 1 - e^{-\frac{2t}{RC}}) . Therefore i_C =\frac{dQ}{dt}=\frac{E}{R}{e}^{-\frac{2t}{RC}}$

In steady state, i_C=0, i_A=i_B

Hence, heat is produced at the same rate in A and B. 

Further in steady state,

$$\begin{gathered} {\mathrm{V}}_{\mathrm{C}}={\mathrm{V}}_{\mathrm{B}}=\frac{\mathrm{\epsilon }}{2} \\ \therefore \mathrm{U}=\frac{1}{2}{\mathrm{CV}}_{\mathrm{C}}^2=\frac{1}{8}{\mathrm{C\epsilon }}^2 \end{gathered}$$