JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Current Electricity

Between $A$ and $B$ the network separates into independent branches that can each be reduced first. The upper branch has $1.5\Omega$ and $0.5\Omega$ in series, giving $2\Omega$, and the lower branch has $1\Omega$ and $1\Omega$ in series, also $2\Omega$. In the central branch the two resistors leaving $A$, namely $8\Omega$ and $2\Omega$, are in parallel ($1.6\Omega$) and join in series the two reaching $B$, namely $4\Omega$ and $6\Omega$, in parallel ($2.4\Omega$), totalling $4\Omega$; beside these run the direct $12\Omega$ and $6\Omega$ links. All five branches connect $A$ to $B$ in parallel, so $\frac{1}{R_{eq}}=\frac{1}{2}+\frac{1}{2}+\frac{1}{4}+\frac{1}{12}+\frac{1}{6}=\frac{3}{2}$, and the equivalent resistance between $A$ and $B$ is $\frac{2}{3}\Omega$.