JEE Main 31-Aug-2021 Shift 2 Solved Paper

In the given circuit, 5 Ω resistance is shorted. So, it can be discarded.
Now, we get a resolved circuit as shown below In parallel,
R1=

2×2

2+2

=1Ω In series,
R2=R1+2=1+2=3Ω Here, all the three resistances(3Ω) are parallel.
∴ Equivalent resistance across A and B,

1

Req

=

1

3

+

1

3

+

1

3

=1
⇒Req=1Ω