In the arrangement shown in figure, the current through $5\,\Omega$ resistor is ............. $A$
AIIMS 2015, Diffcult
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(a)
Using the Kirchhoff's voltage law
for loop $ABCDA , 2 i _1+5\left( i _1+ i _2\right)=12$ or $7 i _1+5 i _2=12 \ldots$ (1)
and for loop $EBCFE$, $2 i_2+5\left(i_1+i_2\right)=12$ or $5 i_1+7 i_2=12 \ldots(2)$
Now, (1) $\times 7,(2) \times 5 \Rightarrow 49 i_1+35 i_2=84 \ldots(3)$ and
$25 i_1+35 i_2=60 \ldots(4)$
(3) $-(4) \Rightarrow 2 i_1=24$ or $i_1=1 A$
Substituting the value of $i_1$ in (1) we get, $i_2=\frac{12-7}{5}=1 A$
Thus the current through $5 \Omega=i_1+i_2=1+1=2 A$
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