The voltage and power ratings of a particular light bulb,which are its normal operating values,are 110 V and 60 W.Assume the resistance of the filament of the bulb is constant and is independent of operating conditions.If the light bulb is operated at a reduced voltage and the power drawn by the bulb is 36 W,what is the operating voltage of the bulb?
A) 66 V
B) 72 V
C) 78 V
D) 85 V
E) 90 V

Question

Quizplus · Accepted Answer

The power (P) of the bulb is given as P = V^2/R, where V is the voltage applied to the bulb and R is the resistance of the filament. Since R is assumed to be constant, we can write P1/P2 = (V1/ V2)^2, where P1 and V1 are the original power and voltage ratings, and P2 and V2 are the reduced power and voltage.

Substituting the given values, we get 60/36 = (110/V2)^2. Solving for V2, we get V2 = 85 V. Therefore, the operating voltage of the bulb when the power drawn is 36 W is 85 V. The answer is (D).

The Voltage and Power Ratings of a Particular Light Bulb,which