JEE Main 2023 — Current Electricity Question with Solution

The resistance in $ab$ is parallel to all other resistances. Since all the resistances other than $R_b$ is $4 \mathrm{\Omega }$ the potential is zero$\left(V_x=V_y\right)$ as the upper branch will become a case of balanced wheatstone bridge between $V_x\&V_y$. So, the simplified circuit looks like the one below.

Hence, the equivalent resistance is 

$$\begin{gathered} \frac{1}{R_{eq}}=\frac{1}{8}+\frac{1}{8}+\frac{1}{16}=\frac{5}{16} \\ \Rightarrow R_{eq}=\frac{16}{5} \mathrm{\Omega }=3.2 \mathrm{\Omega } \end{gathered}$$