In the figure shown, the current in the $

Solution:

Using Kirchoff's loop law in loop $A B C D$

$-5 i_{2}-10\left(i_{1}+i_{2}\right)-2 i_{2}+20=0$

$\Rightarrow-10 i_{1}-17 i_{2}+20=0$     ..(1)

Using Kirchoff's loop law in loop $B E F C$

$\Rightarrow-10+4 i_{1}+10\left(i_{1}+i_{2}\right)=0$

$\Rightarrow 14 i_{1}+10 i_{2}+10=0$    ...(2)

Multiplying equation (i) by 10 , we have

$\left(10 i_{1}+17 i_{2}=20\right) \times 10$

$\Rightarrow 100 i_{1}-170 i_{2}=200$             ...(3)

Multiplying equation (ii) by 17 , we have

$\left(14 i_{1}+10 i_{2}=10\right) \times 17$

$\Rightarrow 238 i_{1}-170 i_{2}=170$                 ...(4)

On solving equations (iii) and (iv), we get

$-138 i_{1}=30 \Rightarrow i_{1}=-\frac{30}{138}=-0.217$

$i_{1}$ is negative it means current flows from positive to negative terminal.