\(\therefore 4 i _1+2\left( i _1+ i _2\right)-3+4 i _1=16\, V\)
Using Kirchhoff's second law in the closed loop we have
\(9-i_2-2\left(i_1+i_2\right)=0\)
Solving equations \((i)\) and \((ii)\), we get \(i _1=1.5 A ^{\text {and } i _2}=2 A\)
\(\therefore\) current through \(2 W\) resistor \(=2+1.5=3.5\,A\).