The potential difference between the point A and ,B in  the given circuit Fig will be 

Resistance of CBD part = $5+5+5=15\mathrm{\Omega }$
Resistance of CAD part =$5+5+5=15\mathrm{\Omega }$
The two pa s are connected in parallel. So, their equivalent resistance is given by
$\frac{1}{\mathrm{P}}=\frac{1}{15}+\frac{1}{15}=\frac{2}{15}\text{ or }\mathrm{R}=(15/2)$
Current through the cell
$\mathrm{i}=\frac{2}{(15/2)}=\frac{4}{15}\mathrm{amp}$
Current through the branch CAD or CBD= (2 / 15) amp
$\therefore {\mathrm{V}}_{\mathrm{AB}}=\frac{2}{15}\times 5+\frac{2}{15}\times 5-\frac{2}{15}\times 5=\frac{2}{3}\mathrm{V}$