The equivalent resistance across P and Q in the given electric circuit will be

Options

A. `1 Omega`
B. `2 Omega`
C. `3 Omega`
D. `5 Omega`

Answer

A

Explanation

After simplifying the circuit, we get a balanced Wheatstone's bridge. This, the resistance `2Omega` is ineffective and `1Omega` and `1Omega` are in series and `1Omega` and `1Omega` in series.
`R_(s_(1))=1+1=2Omega` and
`R_(s_(2))=1+1=2` are in parallel
i.e. `R_(s_(1))` and `R_(s_(2))` are in parallel.
So, effective resistance will be
`R_(p)=(R_(s_(1))xxR_(s_(2)))/(R_(s_(1))+R_(s_(2)))=1Omega`