JEE Main · 2023 · 10 Apr 2023 · Shift 1 · Current Electricity

Label the apex junction $P$ and the central junction $Q$. The branch $a\to P\to b$ is two $4\Omega$ resistors in series ($8\Omega$), and so is the branch $a\to Q\to b$ ($8\Omega$). The network is symmetric in these two arms, so $P$ and $Q$ stay at the same potential and the $4\Omega$ resistor joining them carries no current. The two $8\Omega$ arms then combine in parallel to $\frac{8\times 8}{16}=4\Omega$, which sits in parallel with the $16\Omega$ resistor directly across $a$ and $b$. Hence $R_{ab}=\frac{4\times 16}{4+16}=\frac{64}{20}=3.2\Omega$.