Find out the potential difference between the points x and y in figure

Given $\mathrm{E}=24\mathrm{V},\mathrm{R}=2\mathrm{\Omega },{\mathrm{R}}_1=4\mathrm{\Omega },{\mathrm{R}}_2=6\mathrm{\Omega },{\mathrm{C}}_1=(4/3)\mathrm{\mu F},{\mathrm{C}}_2=4\mathrm{\mu F}$.

As the capacitor offers a very high resistance to the current in the steady state, so the
current is prevented to pass through the capacitors. Now, the total resistance in
the circuit is
${\mathrm{R}}_{\mathrm{eq}}=\mathrm{R}+{\mathrm{R}}_1+{\mathrm{R}}_2=2+4+6=12\mathrm{\Omega }$
Hence, the net current in the circuit is
$\frac{\mathrm{E}}{{\mathrm{R}}_{\mathrm{eq}}}=\frac{24}{12}=2\mathrm{A}$
Therefore, the terminal potential difference across is
$\mathrm{AB}=24-2\times 2=20\mathrm{V}$
Note : As the capacitors and resistance are in per parallel the potential difference
of 20 V is available to both the capacitors and the resistors.
$\begin{array}{r}{\mathrm{V}}_{\mathrm{B}}-{\mathrm{V}}_{\mathrm{y}}=20\times \frac{6}{10}=12\mathrm{V}......\left(\mathrm{i}\right)\\ {\mathrm{V}}_{\mathrm{B}}-{\mathrm{V}}_{\mathrm{x}}=\frac{20\times 4/3}{4/3+4}=5\mathrm{V}......\left(\mathrm{ii}\right)\end{array}$
From Equations, (1) and (2), ${\mathrm{V}}_{\mathrm{x}}-{\mathrm{V}}_{\mathrm{y}}=12-5=7\mathrm{V}$