Two resistors have resistances R1 and R2. When the resistors are connected in series to a 12.4-V battery, the current from the battery is 2.13 A. When the resistors are connected in parallel to the battery, the total current from the battery is 9.01 A. Determine R1 and R2. (Enter your answers from smallest to largest.)

Question

I have tried this, by finding the combined R and don't know where to go from there.

Thank you!

Answer 1

R1+R2=12.4/2.13

1/R1 +1/R2=conductnce=9.01/12.4

now solve for R1 in the first equation in terms of R2, then put that into the second equation, solve for R2.

A little bit of algebra fractions is involved.

Answer 2

Eq1: R1+R2 = E/I = 12.4/2.13 = 5.82 Ohms

(R1*R2)/(R1+R2) = 12.4/9.01 = 1.38 Ohms
Replace R1+R2 with 5.82 Ohms:
R1*R2/5.82 = 1.38
R1*R2 = 5.82*1.38 = 8.0
R1 = 8/R2

In Eq1, replace R1 with 8/R2:
8/R2 + R2 = 5.82
8 + R2^2 = 5.82R2
R2^2 - 5.82R2 + 8 = 0
Use Quadratic Formula.
R2 = 3.59 Ohms

In Eq1, replace R2 with 3.59 Ohms
R1 + 3.59 = 5.82
R1 = 2.23 Ohms