In the circuit shown, the current through the $5 \Omega$ resistor is
- $\frac{8}{3} \mathrm{~A}$
- $\frac{9}{13} \mathrm{~A}$
- $\frac{4}{13} \mathrm{~A}$
- $\frac{1}{3} \mathrm{~A}$
Solution
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 $I_2=\frac{1}{6} \mathrm{~A}$ $I=I_1+I_2=\frac{1}{6}+\frac{1}{6}=\frac{1}{3} \mathrm{~A}$
Asked in: NEET 2018