Determine the current in branches $\displaystyle \mathrm{AB}, \mathrm{AC}$ 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 = $\displaystyle I_2+I_3=2.5A$