In the figure shown, the current in the $10 \mathrm{~V}$ battery is close to
Correct Option: , 3
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.