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

Resultant resistance of parallel combination `R_(P)=2Omega`
Resultant resistance of series combination `R_(S)=9Omega`
`(1)/(R_(P))=(1)/(R_(1))+(1)/(R_(2))`
`1/2=(1)/(R_(1))+(1)/(R_(2))`
`1/2=(R_(1)+R_(2))/(R_(1)R_(2))`
`2(R_(1)+R_(2))=R_(1)R_(2)` . . . (1)
Substitute equation (2) in equation (1)
`R_(S)=R_(1)+R_(2)`
`9=R_(1)+R_(2)`
`R_(1)=9-R_(2)` . . (2)
`2(9-R_(2)+R_(2))=(9-R_(2))R_(2)`
`18=9R_(2)-R_(2)^(2)`
`R_(2)^(2)-9R_(2)+18=0`
`(R_(2)-3)(R_(2)-6)=0`
`R_(2)=3,6`
(i) If `R_(2)=3,R_(1)=9-R_(2)=9-3`
`R_(1)=6`
(ii) If `R_(2)=6,R_(1)=9-R_(2)=9-6`
`R_(1)=3`