Two batteries are joined as shown. The internal resistance of $6V$ battery is $2\Omega$ and $4V$ battery is $8\Omega$. The potential difference between the points $X$ and $Y$ is

$\text{ If given circuit can be redrawn as, }$

$E_{\text{net }}=\frac{E_1/{\pi }_7+E_2/r_2}{1/r+1/r_2}$

$=\frac{(6/2)+(4/8)}{(1/2)+(1/8)}$

$=\frac{28}{5}=5.6\text{ volt }$

Since, no current is taken from this battery.

$y_{XY}=E_{\text{net }}=5.6volt$