A 40 watt fluorescent bulb is rated for use in a 120 volt line. During a brownout the actual voltage drops to 102 V. The actual power dissipated in the bulb under these conditions is
A) 0.283 W.
B) 28.9 W.
C) 30.0 W.
D) 47.1 W.

Question

Quizplus · Accepted Answer

The power dissipated in the bulb can be calculated using the formula $P = \frac{V^2}{R}$, where $P$ is the power, $V$ is the voltage, and $R$ is the resistance. Since the resistance of the bulb remains constant, we can set up a ratio of the power dissipated at the normal operating voltage to the power dissipated during the brownout: $\frac{P_1}{P_2} = \frac{V_1^2}{V_2^2}$, where $P_1$ is the original power (40 W), $V_1$ is the original voltage (120 V), $P_2$ is the unknown power, and $V_2$ is the reduced voltage (102 V). Solving for $P_2$, we find that $P_2 = P_1 \times \left(\frac{V_2}{V_1}\right)^2 = 40 \times \left(\frac{102}{120}\right)^2 \approx 28.9 W$.

A 40 Watt Fluorescent Bulb Is Rated for Use in a 120