b
Apply $\mathrm{KVL}$ from $\mathrm{A}$ to $\mathrm{B}$
$\mathrm{V}_{\mathrm{A}}-4 \mathrm{i}-9-\mathrm{i}_{1}+3-4 \mathrm{i}-\mathrm{V}_{\mathrm{B}}=0$
$16-8 i-9-i_{1}+3=0\left(A s V_{A}-V_{B}=16 V\right)$
$8 i+i_{1}=10$ ...........$(i)$
$\mathrm{KVL}$ in loop
$-9-i_{1}+2\left(i-i_{1}\right)=0$
$2 i-3 i_{1}=9$ .........$(ii)$
solving equation $(i)$ and $(ii)$
$\mathrm{i}=1.5 \mathrm{\,A}$ and $\mathrm{i}_{1}=-2 \mathrm{\,A}$
Current through $2\, \Omega$ resistance will be $=i-i_{1}$
$=1.5-(-2)$
$=3.5 \mathrm{\,A}$
