The total resistance of two resistors, when connected in series is `9Omega` and when connected in parallel, their total resistance ecomes `2Omega`. Find the value of each resistance.

Explanation

Let we have resistors of resistances `R_1 and R_2`. Total resistance given by them in series = `R_s`
Total resistance given by them in parallel `=R_P`
`R_s=18Omega` (given)
`R_P = 4Omega` (given)
`implies R_1 +R_2 =18 " "..(1)`
and `1/R_1+1/R_2=1/R_P`
`implies 1/R_1+1/R_2=1/4`
`implies (R_1+R_2)/(R_1R_2)=1/4`
`implies 18/(R_1R_2)=1/4" "` [Using eqn.(1)]
`implies R_1R_2 = 18 xx4 = 72`
We know that
`(R_1-R_2)^2=(R_1+R_2)^2-4R_1R_2`
`implies (R_1-R_2)^2=(18)^2-4(72)`
= 324 - 288
`(R_1-R_2)^2=36`
`implies R_1-R_2 =6 " "...(2)`
Adding eqns. (1) and (2)
`2R_1 = 24`
`implies R_1 =24/2= 12Omega`
Put value of `R_1` in eqn, (1)
`12+R_2=18`
`implies R_2 =18 - 12 = 6Omega`
` implies R_2 = 6Omega`