The equivalent resistance between the point `P` and `Q` in the network given here is equal to (given `r = (3)/(2) Omega)`

Options

A. `(1)/(2) Omega`
B. `1 Omega`
C. `(3)/(2) Omega`
D. `2 Omega`

Answer

B

Explanation

By seeing current distribution, `POQ` and `XOY` are not connected at `O`, hence make new diagram by placing `POQ` separately
`R_(eq) = (2r)/(3) = (2)/(3) xx(3)/(2) = 1 Omega`.