# Two resistors have a combined resistance of 38 when connected in series and a resistance of 6 when connected in parallel. What is the resistance of each resistor.

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It is given that two resistors when connected in series have a combined resistance of 38 ohm and when connected in parallel have a combined resistance of 6 ohm.

When two resistors R1 and R2 are connected in series the combined resistance is R = R1 + R2. On the other hand when two two resistors R1 and R2 are connected in parallel, the combined resistance R is related by `1/R = 1/(R1) + 1/(R2)` .

Let the resistance of the two resistors used in the problem be R1 and R2. `R1 + R2 = 38` and `1/6 = 1/(R1) + 1/(R2)`

Substitute `R1 = (38 - R2)` in `1/6 = 1/(R1) + 1/(R2)`

=> `1/6 = 1/(38 - R2) + 1/(R2)`

=> `R2*(38 - R2) = 6*R2 + 6*(38 - R2)`

=> `38*R2 - R2^2 = 6*R2 + 228 - 6*R2`

=> `R2^2 -38*R2 + 228 = 0`

The solution to this equation is `R2 = 19-sqrt 133` and `R2 = 19 + sqrt 133` .

R1 = `38 - (19 - sqrt 133)` = `19 + sqrt 133`

**The resistance of the two resistors is `19 - sqrt 133` and `19 + sqrt 133` .**