Question Detail
Question
A light bulb has an input power consumption of 50 watts. The bulb was activated for 60 seconds and produced heat of 2400 joules. Find the efficiency of the light bulb.
Correct Answer
Option A is the correct answer.
Detailed Explanation
Total input energy = 50W×60s = 3000J. Since the 2400J of heat is wasted energy (not useful light output), efficiency = (3000-2400)/3000 = 20%.
Hint
Total input energy minus wasted heat gives useful output — remember heat here is the wasted energy, not the useful output.