JEE Main · 2023 · 11 Apr 2023 · Shift 1 · Current Electricity

The two cells oppose, so the net emf driving the loop is $8-4=4$ V, and the internal resistances $0.5\Omega$ and $1\Omega$ add to the external resistors. The $3\Omega$ and $6\Omega$ resistors in parallel give $\frac{3\times 6}{3+6}=2\Omega$, so the total resistance is $0.5+1+4.5+2=8\Omega$ and the main current is $\frac{4}{8}=0.5$ A. This splits in the inverse ratio of the branch resistances, so the current in the $3\Omega$ resistor is $0.5\times \frac{6}{3+6}=\frac{1}{3}$ A. Comparing with $\frac{x}{3}$ gives $x=1$.