Quote:
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Originally Posted by Selectron
P = V * V / R
P = 120 * 120 / 484.0
P = 29.75 watts
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Agreed.
A lightbulb is simply a resistive load and we can assume (for a homework problem) that the resistance stays the same.
In reality it's a non ohmic device and will run cooler (and dimmer) on lower voltage so will have less resistance.
This will draw more current and power than predicted above, but that's far beyond the scope of a normal homework problem.