JEE Main 2024 — Current Electricity Question with Solution

In the above diagram, $6\Omega$ is short circuit. Therefore, we can remove it.

In the left part, $2 \Omega$ and $2 \Omega$ are in parallel combination, which will give us $1 \Omega$ which again in turn is in series combination with the $2 \Omega$(adjacent to $3 \Omega$ resistor), therefore net contribution of this part will be $1 \Omega +2 \Omega =3 \Omega$. 

Now, all three $3 \Omega$ are in parallel across point A & B, therefore $R_{eq}=3\times \frac{1}{3}=1 \Omega$