Thats it. Well, except for the “?” that wouldn’t fit.
Nope. Except for the really, really tiny bit that leaks through the gas inside the bulb. Practically speaking, a burnt-out bulb is a zero power draw.
Ah. Excellent. Thanks.
Ok, I guess we’re done here, unless someone wants to refute the omnipetent QED 
They wouldn’t dare. Except Desmostylus. He’d say (correctly) that the leakage capacitances in the circuit would have a higher current draw than the leakage through the air. But I love saying that, because it sounds impressive.
Groovy.
(I have no idea what you just said.)
Well, I’d expand on the correct statement. When incandescent bulbs burn out they break the filament (opening the circuit).
In rare cases, it’s possible for the filament to land in such a way as to actually short circuit the lamp. However, a short circuit will draw enough current to blow whatever circuit protection device is around (fuse, breaker), so it’s usually obvious.
Fluorescent bulbs are occasionally referred to as ‘burnt out’ when they are flickering (though in some cases it’s not the bulb, but the fixture). It should be clear that there is a significant power drain in this case (even if the light is very dim).
Not only that, but if a bit of the filament should actually land like that, the current would likely vaporize it, too.
Q.E.D., you certainly got a charge out of this thread, didn’t you?
It’s actually the little appliance bulb from my lava lamp - it keeps burning out and then I forget to turn off the lamp. I’m just wondering if it’s hugely contributing to my massive electric bill - unfortunatly, I think I’ll just have to blame Ralph. 
Stupid Ralph.
What about other kinds of lights? I have a outdoor light fixture that is sodium vapor (I think, may be something similar) and takes a 300 watt bulb. Recently, I got annoyed at the loud buzzing noise (60 cycle hum?) that was coming out of the fixture. I thought taking out the bulb would end the hum, but it didn’t. Even with the light out of the base and no circuit completed, noise is still being made. I would think that the noise would require electricity to be generated. So do those kinds of lights use energy when there’s no bulb or a burnt out bulb?
Low-pressure sodium vapor lamps have a starter similar to fluorescent fixtures, and that’s where the hum comes from. Best to use a fluorescent substance which exists plentifuly in nature.
I see there’s no need for a comment on charging current in cabling. 
However, this bit:
Is a bit off. Fluorescence refers to a process of absorbing e-m radiation of one frequency, say UV, and re-emitting another frequency, usually visible.
What’s happening in a sodium vapour lamp isn’t fluorescence, it’s ionisation.
We can all agree that there is current in AC home wiring even when the circuit is off, albeit a slight amount. This is due to capacitance. But would this be recorded on the power meter? I was under the assumption that modern home power meters measure real power, not apparent power. But I could be wrong.
They measure V I cos (θ[sub]V[/sub] - θ[sub]I[/sub]), which is real power.
Even if the dielectric is lossless, there will still be a non-zero amount of real power lost, because the cables have a non-zero resistance.
Whether this is measurable using a home power meter is unlikely. If everything is off, there probably wouldn’t be enough torque to overcome the static friction of the meter’s bearings. If something else is turned on, the contribution of the charging current would most likely be swamped by the meter’s measurement error.
Desmostylus: Agree. It’s pretty much a non-issue, in that the meter would likely not be sensitive enough to measure the few uW that are being expended. I was really just curious about what the meter was measuring. I seem to recall that, at one time, they measured apparent power. Just wondering if this was still the case.
I also believe this is how white LEDs work. So would it be proper to call a white LED “fluorescent”?
AFAIK, they’ve always been the way they are now. The design is incredibly simple and cheap to make. It just has two electromagnets on opposite faces of the rotating disc. One generates a flux proportional to voltage, the other generates a flux proportional to current. The torque generated is propotional to VIcosθ. Simplest thing in the world. The disc also acts as a time integrator, to derive energy from instantaneous power.
Yes. The actual “LED” part is UV.
You know, I don’t even remember writing that. Damn beer. In any case I ought to have wrtiten ballast, not starter. The starter circuit only operates when turning the lamp on, to provide the initial voltage needed to start a current flowing.
I have no idea what I meant by my second sentence. :rolleyes:
As for the power meter issue (really an energy meter), I suspect at that sufficiently low power draw, there is insufficient torque to overcome the friction that exists in the mechanism. In digital meters a similar issue occurs due to resolution, since any voltages measured below the lowest discrete value will render digitally as zero…
Q.E.D., maybe ypu can answer this one:
I’ve been tolda TV set uses more electricity when on Stand-by then when turned off, completely.
I very docilely therefor always turn off the TV set completely, which irritates my BF to no end (as he just likes flopping on the couch, and flip on the TV withput moving at all).
My question: how does the TV set use electricity when on Stand-By? it can’t be just the little red light that show the TV is actually on Stand By, is it?
I once heard that when the TV is on SB, only the screen goes dark, but the TV is still receiving all the information, jus tnot showing it.
this true?
thanks!
I’ll let Q.E.D. answer the stand-by question.
But a bit more on the charging current thing. (I did the calcs.)
The common 2.5 mm[sup]2[/sup] PVC insulated twin-and-earth cable used here has an a.c. resistance of about .01 ohm/m, and a shunt capacitance of about 150 pF/m.
For simplicity sake, assume there’s 100 m of that cable in the house. Total series resistance = 1 ohm, total shunt capacitance = 15 nF.
Using our 240 V / 50 Hz, that gives a charging current of about 1 mA. Modelling as a pi network, half of that current flows through the series resistance. I[sup]2[/sup]R = 0.3 μW.