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

Explanation

We know that two resistances are in parallel and hence
`(1)/(R_(P))= (1)/(R_(1))+(1)/(R_(2)) implies R_(P)= (R_(1)R_(2))/(R_(1)+R_(2))`
Given, `R_(P) = 2Omega`
`implies 2= (R_(1)R_(2))/(R_(1)+R_(2))`
`2(R_(1)+R_(2))= R_(1)R_(2)` ..(1)
Now, same resistances are in series
`R_(s)= R_(1)+R_(2)`
Given `R_(s)= 9"Ohm"`
`9= R_(1)+R_(2)` ...(2)
From (1) and (2), we get
`R_(1)R_(2)= 18`
Again using (2), we have
`R_(2)= 9-R_(1)`
`R_(1)(9-R_(1))= 18`
`R_(1)^(2)- 9R_(1)+18=0`
`R_(1)^(2)- 6R_(1)-3R_(1)+18=0`
`implies (R_(1)-6)(R_(1)-3)= 0`
Either `R_(1)= 6` or `R_(1)= 3`
`R_(2)= 3Omega` or `R_(2)= 6Omega`
Thus two resistances are `3Omega` and `6Omega`