Two resistors when connected in series to a 120-V line use one-fourth the power that is used when they are connected in parallel. If one resistor is 4.8 kohms. what is the resistance of the other?

Answer 1

E = 120 V.
R1 = 4.8k
R2 = ?

R1+R2 = 4(R1*R2)/(R1+R2)
(R1+R2)^2 = 4(R1*R2)
R1^2+2R1*R2+R2^2 = 4R1*R2
23.04+9.6R2+R2^2 = 19.2R2
R2^2 - 9.6R2 + 23.04 = 0
Use Quadratic formula
R2 = 4.8k

Check:
Parallel Connection
P1 = E^2/2.4k = 120^2/2.4k = 6,000 mW = 6 W.

Series connection
P2 = 120^2/(4.8k+4.8k) = 1,500 mW = 1.5 W.

P2/P1 = 1.5/6 = 1/4

Answer 2

Your answer is 4.8 k omes .

If one resistor is 4.8, then so is the other.