How much current must be applied across a 60 Ω light bulb filament in order for it to consume 55 W of power? Unserious answers will be deleted, and reported. (Please show some work)


  1. Answer: current I = 0.96 Ampere


    Given that the

    Resistance R = 60 Ω 

    Power = 55 W

    Power is the product of current and voltage. That is

    P = IV …… (1)

    But voltage V = IR. From ohms law.

    Substitutes V in equation (1) power is now

    P = I^2R

    Substitute the above parameters into the formula to get current I

    55 = 60 × I^2

    Make I^2 the subject of formula

    I^2 = 55/60

    I^2 = 0.92

    I = sqr(0.92)

    I = 0.957 A

    Therefore, 0.96 A current must be applied.

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