In the circuit shown in figure potential difference between points A and $B$ is $16\,V$. the current passing through $2 \Omega$ resistance will be $...........\,A$
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(b)
$\therefore 4 i _1+2\left( i _1+ i _2\right)-3+4 i _1=16\, V$
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 $=2+1.5=3.5\,A$.
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