When two resistances connected in series and parallel their equivalent resistances are `15Omega` and `(56)/(15)Omega` respectively. Find the individual resistances.

Explanation

`R_(s)=R_(1)+R_(2)=15Omega " " …(1)`
`R_(p)=(R_(1)R_(2))/(R_(1)+R_(2))=(56)/(15)Omega" "…(2)`
From equation (1) substituting for `R_(1)+R_(2)` in equation (2)
`(R_(1)R_(2))/(15)=(56)/(15)Omega " " therefore R_(1)R_(2)=56`
`R_(2)=(56)/(15)Omega " "...(3)`
Substituting for `R_(2)` in equation (1) from equation (3)
`R_(1)+(56)/(15)=15` then, `(R_(1)^(2)+56)/(R_(1))=15`
`R_(1)^(2)+56=15R_(1) rArr R_(1)^(2)-15R_(1)+56=0`
The above equation can be solved using factorisation.
`R_(1)^(2)-8R_(1)-7R_(1)+56=0`
`R_(1)(R_(1)-8)-7(R_(1)-8)=0 rArr (R_(1)-8) (R_(1)-7)=0`
If `(R_(1)=8Omega)`, using in equation (1)
`8+R_(2)=15 rArr R_(2)=15-8=7Omega`
`R_(2)=7Omega` i.e, (when `R_(1)=8Omega, R_(2)=7Omega`)
If `(R_(1)=7Omega)`, substituting in equation (1)
`7+R_(2)=15 rArr R_(2)=8Omega`, i.e, (when `R_(1)=8Omega, R_(2)=7Omega`)