Currents through different branches of the electric circuit is shown in figure , then the current through the  2 ohm resistance is 

Solution Applying Kirchhoff's first law (junction law) at junction $B$,

$i_1=i_2+i_3$

Applying Kirchhoff's second law in loop $1\left(ABEFA\right)$,

$-4i_1+4-2i_1+2=0$

Applying Kirchhoff's second law in loop $2\left(BCDEB\right)$

$-2i_3-6-4i_3-4=0$

Solving Eqs. (i), (ii) and (iii), we get

$i_1=1A$

$i_2=\frac{8}{3}A$

$i_3=-\frac{5}{3}A$

Ans.

Here, negative sign of $l_3$ implies that current $i_3$ is in opposite direction of what we have assumed.