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.

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?

Given,

E = 120 V
`"R"_1` = 4.8k
`"R"_2` = ?
`"R"_1` + `"R"_2` = `\frac{4("R"_1 \times "R"_2)}{"R"_1 + "R"_2}`
`("R"_1 + "R"_2)^2` = 4("R"_1 \times "R"_2)
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