In the arrangement shown in the figure,the cu | vedclass

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(A) Using Kirchhoff's voltage law:For loop $ABCDA$,$2i_1 + 5(i_1 + i_2) = 12$,which simplifies to $7i_1 + 5i_2 = 12 \dots (1)$For loop $EBCFE$,$2i_2 +

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$2$

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(A) Using Kirchhoff's voltage law:For loop $ABCDA$,$2i_1 + 5(i_1 + i_2) = 12$,which simplifies to $7i_1 + 5i_2 = 12 \dots (1)$For loop $EBCFE$,$2i_2 + 5(i_1 + i_2) = 12$,which simplifies to $5i_1 + 7i_2 = 12 \dots (2)$Now,multiplying equation $(1)$ by $7$ and equation $(2)$ by $5$ gives:$49i_1 + 35i_2 = 84 \dots (3)$$25i_1 + 35i_2 = 60 \dots (4)$Subtracting equation $(4)$ from $(3)$ gives $24i_1 = 24$,so $i_1 = 1\,A$.Substituting $i_1 = 1\,A$ into equation $(1)$,we get $7(1) + 5i_2 = 12$,so $5i_2 = 5$,which means $i_2 = 1\,A$.The current through the $5\,\Omega$ resistor is $i_1 + i_2 = 1 + 1 = 2\,A$.

In the arrangement shown in the figure,the current through the $5\,\Omega$ resistor is ............. $A$.

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