Two coils have a combined resistance of `9 Omega` when connected in series and `2 Omega` when connected in parallel. Find the resistance of each coil.

Explanation

Let `R_(1)` and `R_(2)` be the resistance of two coils
When coils are in series, then
`R_(s)=R_(1) +R_(2)=9`
When coils are in paralle, then
`R_(p) =(R_(1)R_(2))/(R_(1)+R_(2))= 2`
or `R_(1) R_(2)=2(R_(1) +R_(2))=2 xx 9 =18`
From (i), `R_(1) =9 -R_(2)`
Putting this value in (ii), we get `(9 -R_(2))R_(2)=18` or `9 R_(2) - R_(2)^(2) =18`
or `R_(2)^(2)-9R_(2) +18=0`
On solving, we get
`R_(2)=3Omega` or `6 Omega` and `R_(1)=6 Omega` or `3 Omega`