Determine the current in each branch of the network shown in Fig.

Current flowing through various branches of the circuit is represented in the given figure.

$I_1=$ Current flowing through the outer circuit

$I_2=$ Current flowing through branch AB

$I_3=$ Current flowing through branch AD

$I_2-I_4=$ Current flowing through branch BC

$I_3+I_4=$ Current flowing through branch CD

$I_4=$ Current flowing through branch BD

For the closed circuit ABDA, potential is zero i.e.,

$10I_2+5I_4-5I_3=0$

$2I_2+I_4-I_3=0$

$I_3=2I_2+I_4$  ...(1)

For the closed circuit BCDB, potential is zero i.e.,

$5(I_2-I_4)-10(I_3+I_4)-5/4=0$

$5I_2+5I_4-10I_3-10I_4-5I_4=0$

$5I_2-10I_3-20I_4=0$

$I_2=2I_3+4I_4$  ...(2)

For the closed circuit ABCFEA, potential is zero i.e.,

$-10+10(I_1)+10(I_2)+5(I_2-I_4)=0$

$10=15I_2+10I_1-5I_4$

$3I_2+2I_1-I_4=2$  ...(3)

From equations (1) and (2), we obtain

$I_3=2(2I_3+4I_4)+I_4$

$I_3=4I_3+8I_4+I_4$

$-3I_3=9I_4$

$-3I_4=+I_3$  ...(4)

Putting equation (4) in equation (1), we obtain

$I_3=2I_2+I_4$

$-4I_4=2I_2$

$I_2=-2I_4$  ...(5)

It is evident from the given figure that,

$I_1=I_3+I_2$  ...(6)

Putting equation (6) in equation (1), we obtain

$3I_2+2(I_3+I_2)-I_4=2$

$5I_2+2(I_3+I_2)-I_4=2$  ...(7)

Putting equations (4) and (5) in equation (7), we obtain

$5(-2I_4)+2(-3I_4)-I_4=2$

$-10I_4-6I_4-I_4=2$

$17I_4=-2$

$I_4=\frac{-2}{17}A$

Equation (4) reduces to

$I_3=-3(I_4)$

   $=-3\left(\frac{-2}{17}\right)=\frac{6}{17}A$

$I_2=-2(I_4)$

   $=-2\left(\frac{-2}{17}\right)=\frac{4}{17}A$

$I_2-I_4=\frac{4}{17}-\left(\frac{-2}{17}\right)=\frac{6}{17}A$

$I_3+I_4=\frac{6}{17}+\left(\frac{-2}{17}\right)=\frac{4}{17}A$

$I_1=I_3+I_2$

     $=\frac{6}{17}+\frac{4}{17}=\frac{10}{17}A$

Therefore, Current in branch $AB=\frac{4}{17}A$

In branch BC $=\frac{6}{17}A$

In branch CD $=\frac{-4}{17}A$

In branch AD $=\frac{6}{17}A$ 

In branch BD $=\left(\frac{-2}{17}\right)A$

Total current $==\frac{4}{17}+\frac{6}{17}+\frac{-4}{17}+\frac{6}{17}+\frac{-2}{17}=\frac{10}{17}A$