WAEC 2022 Physics · Question 42
A lamp is rated 240 V, 60 W. Determine the resistance of the lamp when lit?
Options
- A.120
- B.240
- C.540
- D.960
Answer: D. 960
Explanation written with AI and checked against the answer key
Analyze the question: A lamp has a rating of 240 V and 60 W. We need its resistance when it is operating normally (lit). Concept: Electrical power is related to voltage and resistance by: P = V²/R, so R = V²/P This comes from combining P = IV with Ohm's law, V = IR. Solution: R = V²/P = (240)² / 60 R = 57,600 / 60 R = 960 Ω Check using current: I = P/V = 60/240 = 0.25 A R = V/I = 240/0.25 = 960 Ω ✓ Answer: The resistance of the lamp when lit is 960 Ω, which is option D. Tip: When a question gives a rating like "V volts, P watts", go straight to R = V²/P. Be careful not to just divide V by P or confuse the numbers. Also remember that a lamp's resistance when lit (hot filament) is higher than when cold.AI can make mistakes. Report anything that looks wrong.
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