Explanation

(d) [image:3_image193.emf]
Applying Kirchhoff’s second law for closed loop $AEFBA,$
We get
$- \left( I_{1} + I_{2} \right) \times 5 - I_{1} \times 2 + 2 = 0$ or $7I_{1} + 5I_{2} = 2$ …(i)
Again, applying Kirchhoff’s second law for a closed loop $DEFCD,$ we get
$- \left( I_{1} + I_{2} \right) \times 5 - I_{2} \times 2 + 2 = 0$
or $5I_{1} + 7I_{2} = 2$ …(ii)
Multiplying (i) by 5 and (ii) by 7, we get
$35I_{1} + 25I_{2} = 10$ …(iii)
$35I_{1} + 49I_{2} = 14$ …(iv)
Subtracting (iv) from (iii) we get,
$- 24{\ I}_{2} = - 4 \Rightarrow I_{2} = \frac{1}{6}A$
Substituting the value of $I_{2}$ in equation (i), we get
$7I_{1} = 2 - 5 \times \frac{1}{6} \Rightarrow 7I_{1} = \frac{7}{6} \Rightarrow I_{1} = \frac{1}{6}A$
The current through the $5\Omega$ is
$= I_{1} + I_{2} = \frac{1}{6}A + \frac{1}{6}A = \frac{1}{3}A$