In the circuit the potential difference between A and B is

No current flows through wire AD and BC.

Hence, Potential difference across  AB will be equal to battery equivalent across CD

${\text{V}}_{\text{AB}}={\text{V}}_{\text{CD}}=\frac{\frac{{\text{E}}_{\text{1}}}{{\text{r}}_{\text{1}}}+\frac{{\text{E}}_{\text{2}}}{{\text{r}}_{\text{2}}}+\frac{{\text{E}}_{\text{3}}}{{\text{r}}_{\text{3}}}}{\frac{\text{1}}{{\text{r}}_{\text{1}}}+\frac{\text{1}}{{\text{r}}_{\text{2}}}+\frac{\text{1}}{{\text{r}}_{\text{3}}}}=\frac{\frac{1}{1}+\frac{2}{1}+\frac{3}{1}}{\frac{1}{1}+\frac{1}{1}+\frac{1}{1}} =\frac{6}{3}=2\text{V}$