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?

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  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

    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

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