Let us first distribute the current in different branches. Now, equations for different loops using Kirchhoff’s II law, Loop 1 ΣE = ΣIR $10I_1+5I_g-5I_2$ = 0 or $2I_1+I_g-I_2$ = 0 ... (i) Loop 2 ΣE = ΣIR $5I_g + 10[I_2+I_g] - 5[I_1-I_g]$ = 0 $10I_2 + 20I_g-5I_1$ = 0 or $2I_2+4I_g-I_1$ = 0 ... (ii) Loop 3 $5I_2+10(I_2+I_g)+10I$ = 10 $15I_2+10I_g+10I$ = 10 or $3I_2+2I_g+2I$ = 2 ... (iii) Solving equations (i) and (ii) $2I_1+I_g-I_2$ + 2 $[-I_i+4I_g+2I_2]$ = 0 or $9I_g+3I_2$ = 0 or $I_2$ = - 3 $I_g$ ... (iv) In the loop ABCDA $10I_1+5[I_1-I_g]$ - 10 $[I_2+I_g]$ - $5I_2$ = 0 $15I_1-15I_2-15I_g$ = 0 or $I_1-I_2-I_g$ = 0 ... (v) Solving equations (ii) and (v) $2I_2+4I_g-I_1$ = 0 or $2(I_1-I_2-I_g=0)$ or $2I_g+I_1$ = 0 or $I_1$ = - 2 $I_g$ ... (vi) Now using the result of (iv) and (vi) in equation (iii) $3I_2+2I_g+2I$ = 2 - $3[3I_g] + 2I_g$ + 2I = 2 or 2I - $7I_g$ = 2 ... (vii) Using Kirchhoff’s law, I = $I_1+I_2$ ⇒ I = - $5I_g$ So, equation (vii) $2[-5I_g] - 7I_g$ = 2 or $-17I_g$ = 2 So, finally $I_g$ = - 2/17 A and I = $\frac{+10}{17}$ A. Also $I_1$ = $\frac{4}{17}$ A , $I_2$ = $\frac{6}{17}$ A Current in branch AB = $\frac{4}{17}$ A , Current in branch AD = $\frac{6}{17}$ A Current in branch BD = $\frac{-2}{17}$ A , Current in branch BC = $\frac{6}{17}$ A Current in branch DC = $\frac{4}{17}$ A