Two resistors when connected in parallel give the resultant of 2 ohm, but when connected in series the effective resistance becomes 9 ohm ? Calculate the value of each resistance.

Explanation

Effective resistance in parallel, `R_(P) = 2 Omega`
Effective resistance in series ,`R_(s) = 9 Omega`
`R_(s) = R_(1)+R_(2) : (1)/(R_(P)) = (1)/(R_(1)) +(1)/(R_(2))`
`R_(s) = R_(1)+R_(2) = 9`
`implies R_(2) = 9-R_(1)`
`(1)/(R_(1))+(1)/(R_(2)) = (1)/(R_(P)) = (1)/(2)`
`(1)/(R_(1)) + (1)/(9-R_(1)) = (1)/(2)`
`(9-R_(1)+R_(1))/(R_(1)(9-R_(1)) = (1)/(2)`
`implies (9)/(9R_(1) - R_(1)^(2)) = (1)/(2)`
`R_(1)^(2) - 9R_(1) +18 = 0`
`implies (R_(1)-3)(R_(1)-6) = 0`
`therefore R_(1) = 3Omega (or) 6 Omega`
When `R_(1) = 3Omega, R_(2) = 9-3 6 Omega`
When `R_(1) = 6Omega , R_(2) = 9-3 = 3Omega`