Two resistances wires, when joined in parallel, give an affective resistances of `6/5 Omega` When one of the resistances wire is disconnected the effective resistances charges to `2 Omega`. The resistance of the disconnected resistance wire is

Options

A. `4/5 Omega`
B. `2 Omega`
C. `6 Omega`
D. `3 Omega`

Explanation

It is given that `R_1=2 Omega` and resistance of parallel grouping of two resistances `R_p=6/5 Omega`
As per parallel arrangement `1/R_p=1/R_1+1/R_2`
`implies 1/R_2=1/R_p-1/R_1=5/6-1/2=(5-3)/6=2/6=1/3 implies R_2=3 Omega`