how much current does a 75W light bulb require in a 100V circuit?

Using Ohm's Law to solve this makes the false assumption that the bulb somehow or another always consumes 75 Watts no matter what voltage is applied to it. That's not the case.

Neither is an incandescent light a fixed resistance, otherwise they could be rated in ohms instead of watts, so the solution is more complex than using linear correlations like Ohm's Law. Even if you knew the resistance of the room temperature filament, it is non-Ohmic and Ohm's law does not apply for extrapolation purposes.

Empirically, the current is a logarithmic function of the voltage as follows:

(V2/V1)^.55 = I2/I1

75/130 = .58 = I1

==> I2 = .50 A

So your 130V, 75 watt light is now a 100 V, 50 watt light ==> 100 V x .5 A = 50 W.

(and we didn't expect it to remain a 75 Watt load)

@Marv

Your entire post is nothing more than generalizations and hand-waving which tells me that, to you, what you think is going on makes for a valid explanation. It does not.

For the specific reasons I stated, that 130v , 75W lamp does not draw 750 ma at 100v.

Do you think I just pulled the power of .55 basis out of thin air?

Get a variac and a bulb of any wattage rated at 120-130V and prove it to yourself if that's what you need. Until then, your lay understanding and generalized "explanation" is nothing more than vacuous rhetoric.

Yes, it is an interesting answer... it's also the correct answer.

Though it is true that Tungsten has a positive temperature coefficient so it's resistance increases with temperature, it is not "effectively zero" at room temperature.

Hit the books, kid. Start with looking up non-Ohmic resistances.

I just read this horribly incorrect advice on something as simple as biasing for an LED:

http://answers.yahoo.com/question/index;_ylt=Ai9m2...

You can fool some of the people some of the time...

Stick to what you know. You may be an "electronics wizard" compared to the pastry chef next door, but don't let that define you. By the way, the correct answer for one LED would be 100 ohms and 40mW, not 60 ohms and 600mW. If you want to know why and learn something, then post a question about it. (I'm betting you don't/won't)

Joule's Law is used for power, not Ohm's Law.

I detest posers.

Interesting answer Tom G.

In an incandescent bulb, (Edison Bulb), the filament resistance when cold is effectively zero. When the rated voltage is applied the filament heats up instantly emitting light. The filament resistance increases as it heats up. Sooner or later it hits a designed point and stabilizes. This all takes almost no time ( instant ). The filament gets to about 133 ohms resistance which works out to 75 watts at 100 volts and 750 milliamps. At that point all is stabilized and you get a steady light output. In actuality there are minute differences in some of these numbers and your bulb may not actually use these power amounts. But for all intents and purposes these numbers are correct.

I=P/V=75/100=.75amps.

P=I*E 75/100=.75 Amps

more data required, for ex: the current is itself needed to find the voltage and vice-versa. Ur question is actually like an attempt to find 2 unknown quantities with a single equation