In the circuit shown in figure potential difference between points A and B is $16V$  . The current passing through  $2\Omega$ resistance will be

$V_A-V_B=16V$

$\therefore 4i_1+2(i_1+i_2)-3+4i_1=16V\mathrm{..........}\left(1\right)$

Using Kirchhoff’s second law in the closed loop we have

$9-i_2-2(i_1+i_2)=0\mathrm{..............}\left(2\right)$

Solving Eqs. (1) and (2), we get

$i_1=1.5A$and    $i_2=2A$

Current through $2\Omega$ resistor

$=2+1.5=3.5A$