Nonideal situations of a lightbulb

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byeh2004

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Ok we know that the power of a light bulb would be

P= I (current of bulb) x V (voltage of bulb)

But that would be in an ideal situation. What sorts of factors would contribute to a real messurement of the power of a light bulb being less than the ideal power of the light bulb in this calculation?

Thanks!
 

mc4435

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Some of the electrical energy going through the lightbulb (as current) is dissipated as heat and light, so the energy going out of the bulb is less than what goes in. Because power is in units of Joules/second, if the joules decreases under "real" conditions, then the power is less than ideal conditions as well.
 

DrBowtie

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I'd say P=i^2R. Thats the power through a resistor and basically a non-ideal lightbulb has resisitance.
 
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