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Physics Help!!!!!!!!!!!

[From: ] [author: ] [Date: 11-09-06] [Hit: ]
296875 amp = .296875 C/sec x 120 = 35.......
Lightbulbs are rated by their brightness (power), assuming they are connected to a 120 V source of voltage. A lightbulb that is printed with a rating of 45 W is connected to a 95 V source. Assume the bulb's resistance is constant.
(a) Find the resistance of this lightbulb
(b) Find the power given off by this lightbulb when connected to the 95 V source
(c) Find how much charge passes through this lightbulb, connected to the 95 V source, in 120 seconds

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a) R = V²/P = 120²/45 = 320 Ω
b) P95 = 95²/320 = 28.2 W
c) i = V/R = 95/320 = .296875 amp = .296875 C/sec x 120 = 35.625 C in 120 sec
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