Two resistances are joined in parallel whose equivolent resistance is `3/5Omega`. One of the resistance wire is broken and the effective resistance becomes `3Omega`. The resistance (in ohms) of the wire that got broken was

Options

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

Answer

B

Explanation

`R_(P)=6/5 Omega, R_(1)=2 Omega` and `R_(2)=?`
`R_(P)=(R_(1)R_(2))/(R_(1)+R_(2))`
`6/5 = ( 2 R_(2))/(2+R_(2))`
`12+6R_(2)=10R_(2)`
`4 R_(2)=12`
`R_(2)=3Omega`