Power: 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?

Question

Quizplus · Accepted Answer

We can use the formula for power, P = V^2/R, where R is the resistance of the filament. Since the resistance is constant, we can set up the equation:

60 W = (110 V)^2/R

Solving for R, we get:

R = (110 V)^2/60 W = 201.67 Ω

When the bulb is operated at a reduced voltage of V, we can use the same equation with the new power rating of 36 W:

36 W = V^2/201.67 Ω

Solving for V, we get:

V = sqrt(36 W x 201.67 Ω) = 84.85 V

Therefore, the operating voltage of the bulb when the power drawn is 36 W is approximately 85 V, which corresponds to choice (D).

Power: the Voltage and Power Ratings of a Particular Light