In the circuit shown, the current in the $1\,\Omega$ resistor is
JEE MAIN 2015,AIPMT 1988, Medium
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From $KVL$
$-6+3 I_{1}+1\left(I_{i}-I_{2}\right)=0$
$6=3 \mathrm{I}_{1}+\mathrm{I}_{1}-\mathrm{I}_{2}$
$4 \mathrm{I}_{1}-\mathrm{I}_{2}=6$ .....$(1)$
$-9+2 I_{2}-\left(I_{1}-I_{2}\right)+3 I_{2}=0$
$-\mathrm{I}_{1}+6 \mathrm{I}_{2}=9$ ....$(2)$
On solving $( 1)$ and $( 2)$
$\mathrm{I}_{1}=0.13\, \mathrm{A}$
Direction $Q$ to $P,$ since $I_{1}>I_{2}$.
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