Engineer
Elite Member
- Oct 9, 1999
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Originally posted by: Spencer278
It isn't really that simple because your light bulb isn't purely resistive and your using AC current.
You would probably have a very small Inductance in there and likely, little or no capacitance (at least that is what I would imagine a "wound" filament (springy) would introduce (sorta like a minature choke). Also, the equation: Power = I^2*R = V*I would hold true (for the most part other than the small inductance factor) if the current and voltage were RMS numbers for AC, which is what the voltage is (120v).
Hell, if the inductance is small enough, it should be almost like a DC calculation, shouldn't it? (I forgot all that crap a little while ago!
