In the figure, potential difference between A and B is:

Diode is in forward bias, so it will behave as conducting wire,  so

$\begin{array}{l}Two 10k\Omega are in parallel and in series with third 10k\Omega .\\ U\sin g voltage distribution, \\ V_{ab}=\frac{R_1}{R_1+R_2}\times E=\frac{5}{5+10}\times 30=10V\end{array}$