Question

Two resistors connected in series have an equivalent resistance of 690 Ohms. When they are connected in parallel, their equivalent resistance is 118 Ohms. Find the resistance of each resistor.

Answer 1

Step-by-step explanation:

Let
`R_1` and
`R_2` be the the resistances of the resistors. We are given that

`R_1 + R_2 = 690\:Ω\:\:\:\:\:\:\:(1)`

and

`(1)/(R_1) + (1)/(R_2) = (1)/(118\:Ω)\:\:\:\:\:(2)`

From Eqn(1), we can write

`R_2 = 690\:Ω - R_1`

and then plug this into Eqn(2):

`(1)/(R_1) + (1)/(690\:Ω - R_1) = (1)/(118\:Ω)`

or

`(690\:Ω)/((690\:Ω)R_1 - R_1^2)= (1)/(118\:Ω)`

`\Rightarrow R_1^2 - (690\:Ω)R_1 + (690\:Ω)(118\:Ω)= 0`

or

`R_1^2 - 690R_1 + 81420 = 0`

Using the quadratic formula, we find that the above equation has two roots:

`R_1 = 151.1\:Ω,\:\:538.9\:Ω`

This means that if you choose one root value for
`R_1`, the other root will be the value for
`R_2`.