That’s the whole question, kids.
Assuming an incandescent bulb, No. A burnt bulb passes exactly as much current as an open switch: none. There will be a nearly immeasurable amount lost in the wire between the switch and the bulb, but I don’t think you’d see the difference in a year’s worth of utility bills.
Not even an immeasurable amount. Open is open. No current passes. A bulb burned out at the end of a long circuit uses the same amount of current as one burned out on a shorter circuit. Zero.
Tris
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Sorry to disagree with you, Trisky, but I believe there would still be some capacitive and inductive loss in a real world system, as opposed to a circuit diagram, since there is no such thing as a perfect conductor, as the potential of the wire cycles 50 or 60 times per second. The amount of power radiated would be tiny, probably in the microwatt range (and my electrical theory textbooks are in storage), but there would still be some loss. The loss would be greater at a higher frequency, but there will still be some at 50 or 60 Hertz.
Assuming a loss of 1 microwatt (and that is really a WAG), I get an annual cost at $0.10/KWHr of $0.876 (10[sup]-6[/sup]/10[sup]3[/sup]24365*0.1)
Sewalk- I think you’re confusing “voltage used as recorded at the meter” and “voltage/potential as measured with very sensitive instruments.”
The “leakage” of any given house wire running current can be detected at several feet with the correct equipment. Yes, that’s a miniscule “loss”, but the fact remains that, if you had a meter, a switch and a burned-out bulb wired up, there wouldn’t be a cent’s worth of recorded “usage” after even a year. (To say nothing of eighty cents’ worth, which yes, I know was a wild guess.)
Originally, I wrote my post using nanowatts as the WAG amount of loss, which did end up being well under a penny/year, but I hedged upward. This is about the fifth time in six months I could have used those books for some esoteric exercise. I may have to crawl up into Mom’s attic and get them out, now that I’m settled in a house. [sub]But good Lord, the heat! Maybe this winter I’ll get them and the rest of the stuff I really don’t need but kept anyway[/sub]
Of course, even a dollar per year is probably less than the accuracy of many utility meters. I was just trying to point out that the actual (not measured) loss is something greater than zero, even if infinitesimal.
I thought kilowatt-hour meters only measured real/average power (Watts), which comes from resistive circuit elements. Capacitance and inductance would produce reactive power (Vars), which wouldn’t show up on your electric bill, although it might be of concern to the power company.
Just a thought. I’m open to any clarification of the matter.
We’re talking a conventional light switch here Right?
In a dimmer switch you are using electricity as long as that switch is turned on.
You may be in a touch switch too.
Guess I’ll have to go to “www how stuff works” to find out for sure. It has been such a long time since I’ve fooled with that stuff.
There is a good explanation of touch switches over at how stuff works.
Nothing about dimmer switches , the kind with the rotating knob.
Upon further consideration they probably leave the third lead of the rheostat floating so there is no current drawn when the light bulb is burnt out.
Even ignorning the capacitance of air between two plates (conductive but open wires) the blob of air between the two is also a resistence, albeit a very large one. Current is flowing; you will never pay for that incredibly tiny amount.
Meters may or may not ignore apparent power. Remember, though that power is not used it still has to be there in the first place. Most likely it is paid for once. Once the power is used it isn’t returned, but it was still taken in the first place.
Depending on what is being powered (residential, factory, etc), the power company may put huge capacitors on your power lines to offset the inductive appliances’ apparent power consumption.
Whew! I had a feeling this might get too rich for my blood. Thank you all for your input. 