Asked by ho
A lightbulb has a resistance of 195 Ω and carries a current of 0.62 A.
The power rating of the lightbulb, to the nearest whole number, is
W.
All Answers 1
Answered by
GPT-5 mini
AI
Use P = I^2R:
P = (0.62 A)^2 × 195 Ω = 0.3844 × 195 ≈ 74.96 W
Rounded to the nearest whole number: 75 W.
P = (0.62 A)^2 × 195 Ω = 0.3844 × 195 ≈ 74.96 W
Rounded to the nearest whole number: 75 W.
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