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

I₁ = Current flowing through the outer circuit

I₂ = Current flowing through branch AB

I₃ = Current flowing through branch AD

I₂ - I₄ = Current flowing through branch BC

I₃ + I₄ = Current flowing through branch CD

I₄ = Current flowing through branch BD

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

$10{\mathrm{I}}_2+5{\mathrm{I}}_4-5{\mathrm{I}}_3=0$
$2{\mathrm{I}}_2+{\mathrm{I}}_4-{\mathrm{I}}_3=0$
${\mathrm{I}}_3=2{\mathrm{I}}_2+{\mathrm{I}}_4 ...\left(1\right)$

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

$5({\mathrm{I}}_2-{\mathrm{I}}_4)-10({\mathrm{I}}_3+{\mathrm{I}}_4)-5{\mathrm{I}}_4=0$
$5{\mathrm{I}}_2+5{\mathrm{I}}_4-10{\mathrm{I}}_3-10{\mathrm{I}}_4-5{\mathrm{I}}_4=0$
$5{\mathrm{I}}_2-10{\mathrm{I}}_3-20{\mathrm{I}}_4=0$
${\mathrm{I}}_2=2{\mathrm{I}}_3+4{\mathrm{I}}_4 ...\left(2\right)$

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

$-10+10\left({\mathrm{I}}_1\right)+10\left({\mathrm{I}}_2\right)+5({\mathrm{I}}_2-{\mathrm{I}}_4)=0$
$10=15{\mathrm{I}}_2+10{\mathrm{I}}_1-5{\mathrm{I}}_4$
$3{\mathrm{I}}_2+2{\mathrm{I}}_1-{\mathrm{I}}_4=2 ...\left(3\right)$

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

${\mathrm{I}}_3=2(2{\mathrm{I}}_3+4{\mathrm{I}}_4)+{\mathrm{I}}_4$
${\mathrm{I}}_3=4{\mathrm{I}}_3+8{\mathrm{I}}_4+{\mathrm{I}}_4$
$-3{\mathrm{I}}_3=9{\mathrm{I}}_4$
$-3{\mathrm{I}}_4=+{\mathrm{I}}_3 ...\left(4\right)$

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

${\mathrm{I}}_3=2{\mathrm{I}}_2+{\mathrm{I}}_4$
$-4{\mathrm{I}}_4=2{\mathrm{I}}_2$
${\mathrm{I}}_2=-2{\mathrm{I}}_4 ...\left(5\right)$

It is evident from the figure that,

${\mathrm{I}}_1={\mathrm{I}}_3+{\mathrm{I}}_2 ...\left(6\right)$

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

$3{\mathrm{I}}_2+2({\mathrm{I}}_3+{\mathrm{I}}_2)-{\mathrm{I}}_4=2$
$5{\mathrm{I}}_2+2{\mathrm{I}}_3-{\mathrm{I}}_4=2 ...\left(7\right)$

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

$5(-2 {\mathrm{I}}_4)+2(-3 {\mathrm{I}}_4)-{\mathrm{I}}_4=2$
$-10{\mathrm{I}}_4-6{\mathrm{I}}_4-{\mathrm{I}}_4=2$
$17{\mathrm{I}}_4=-2$
${\mathrm{I}}_4=-\frac{2}{17}\mathrm{A}$

Equation (4) reduces to

${\mathrm{I}}_3=-3{\mathrm{I}}_4=-3(-\frac{2}{17})=\frac{6}{17}\mathrm{A}$
${\mathrm{I}}_2=-2{\mathrm{I}}_4=-2(-\frac{2}{17})=\frac{4}{17}\mathrm{A}$
${\mathrm{I}}_2-{\mathrm{I}}_4=\frac{4}{17}-(-\frac{2}{17})=\frac{6}{17}\mathrm{A}$
${\mathrm{I}}_3+{\mathrm{I}}_4=\frac{6}{17}+(-\frac{2}{17})=\frac{4}{17}\mathrm{A}$
${\mathrm{I}}_1={\mathrm{I}}_3+{\mathrm{I}}_2=\frac{6}{17}+\frac{4}{17}=\frac{10}{17}\mathrm{A}$

Therefore, the current in branch in AB$={\mathrm{I}}_2=\frac{4}{17}\mathrm{A}$

The current in branch in BC$={\mathrm{I}}_2-{\mathrm{I}}_4=\frac{6}{17}\mathrm{A}$

In branch CD$={\mathrm{I}}_3+{\mathrm{I}}_4=\frac{4}{17}\mathrm{A}$

In branch AD$={\mathrm{I}}_3=\frac{6}{17}\mathrm{A}$

In branch BD$={\mathrm{I}}_4=-\frac{2}{17}\mathrm{A}$

Total current$={\mathrm{I}}_1=\frac{10}{17}\mathrm{A}$