Question

In the circuit shown, the current through the $5 \,\Omega$ resistor is

• A. $\frac{8}{3}\, A$
• B. $\frac{9}{13}\, A$
• C. $\frac{4}{13}\, A$
• D. $\frac{1}{3}\, A$

Answer 1

Answer: (D)

Applying Kirchhoff's second law for closed loop $AEFBA$, we get
$-\left(l_{1}+l_{2}\right) \times 5-l_{1} \times 2+2=0 $
or $7 l_{1}+5 l_{2}=2$
Again, applying Kirchhoff's second law for a closed loop $DEFCD$, we get
$-\left(l_{1}+l_{2}\right) \times 5-l_{2} \times 2+2=0$
or $ 5 l_{1}+7 l_{2}=2$
On solving, we get $I_{1}=\frac{1}{6} A$ and $I_{2}=\frac{1}{6} A$
$I=I_{1}+I_{2}=\frac{1}{6}+\frac{1}{6}=\frac{1}{3} A$