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

 

$V_A-V_B=16V$

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

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

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

Solving eqs (1) and (2) , we get $i_1=1.5A$ and  $i_2=2A$

Current through $2\Omega$ resistor  $=2+1.5=3.5A$