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In the circuit shown, the current through the $5 \Omega$ resistor is
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The correct answer is:
$\frac{1}{3} \mathrm{~A}$
Applying Kirchhoff's second law for closed loop $A E F B A$, we get
$-\left(I_1+I_2\right) \times 5-I_1 \times 2+2=0$
Again, applying Kirchhoff's second law for a closed loop $D E F C D$, we get
$-\left(I_1+I_2\right) \times 5-I_2 \times 2+2=0$
On solving, we get $I_1=\frac{1}{6} \mathrm{~A}$ and $I_2=\frac{1}{6} \mathrm{~A}$ $I=I_1+I_2=\frac{1}{6}+\frac{1}{6}=\frac{1}{3} \mathrm{~A}$
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