Determine the current in branches AB, AC and BC of the network shown in figure.

For closed loop ADCA , $$10-4(\mathrm{I}_1-I_2)+2(\mathrm{I}_2+\mathrm{I}_3-\mathrm{I}_1)-\mathrm{I}_1=0$$ $$7I_1-6I_2-2I_3=10 \quad ------(i)$$ For closed loop ABCA , $$10-4I_2-2(\mathrm{I}_2+\mathrm{I}_3)-\mathrm{I}_1=0$$ $$I_1+6I_2+2I_3=10 \quad ------(ii)$$ For closed loop BCDED , $$5-2(\mathrm{I}_2+\mathrm{I}_3)-2(\mathrm{I}_2+\mathrm{I}_3-\mathrm{I}_1)=0$$ $$2I_1-4I_2-4I_3=-5 \quad -----(iii)$$ Current in branch AB = $\displaystyle I_2=\dfrac{5}{8}A$Current in branch AC = $\displaystyle I_1=2.5A$Current in branch BC = $\displaystyle I_2+I_3=2.5A$