Physics - Current Electricity Question with Solution | TestHub

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 $R_{eq}=3\times \frac{1}{3}=1\Omega$