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 with a current that is 50% of the current rating of the bulb,what is the actual power drawn by the bulb?
A) 10 W
B) 15 W
C) 20 W
D) 25 W
E) 30 W

Question

Quizplus · Accepted Answer

The power rating of the bulb at its normal operating condition is given as 60 W at 110 V. The power $P$ can be calculated using the formula $P = V^2 / R$, where $V$ is the voltage and $R$ is the resistance. When the current is halved, the power drawn by the bulb becomes a quarter of the original power because power $P$ is also given by $P = I^2R$, where $I$ is the current. Since the current is halved ($I/2$), the power becomes $(I/2)^2R = I^2R/4$, which is a quarter of the original power. Therefore, the actual power drawn by the bulb is $60W / 4 = 15W$.

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