A circuit is as shown in the figure. Then, the current from A to B is:

Applying Kirchhoff second law for a closed loop CABDC, we get

$-10 \mathrm{I}_1-\left(\mathrm{I}_1-\mathrm{I}_2\right) 10+10=0$

$-20 \mathrm{I}_1+10 \mathrm{I}_2=-10$

$2 \mathrm{I}_1-\mathrm{I}_2=1$                    (i)

Again, applying Kirchhoff second law for a closed loop ABFEA, we get

$-\left(\mathrm{I}_1-\mathrm{I}_2\right) 10-5+15 \mathrm{I}_2=0$

$-10 \mathrm{I}_1+25 \mathrm{I}_2=5$

$2 \mathrm{I}_1-5 \mathrm{I}_2=-1$               (ii)

Solving (i) and (ii), we get

$\mathrm{I}_1=\frac{3}{4} \mathrm{~A}, \mathrm{I}_2=-\frac{1}{2} \mathrm{~A}$

The current flows from A to B is

$=-\mathrm{I}_1-\mathrm{I}_2=\frac{3}{4} \mathrm{~A}-\frac{1}{2} \mathrm{~A}=\frac{1}{4} \mathrm{~A}=0.25 \mathrm{~A}=250 \mathrm{~mA}$