It's been a week since a light bulb in the bathroom died. I'm wondering if the bulb is dead, does it still consume electricity if the switch is in ON?
The light is completely dead, no light whatsoever.
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8Just a word of warning, even though the bulb might not be "consuming" electricity any longer, the circuit is still live! From a safety perspective, avoid trying to replace the bulb unless it's been switched off completely.– BobMay 5 '17 at 2:57
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Fun fact: some smart switches will not even power up unless a good bulb (load) is detected.– KrisMay 5 '17 at 18:45
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4"Does the light go out when you close the fridge door?" - am I the only one who thought that when seeing this question?! :)– MrWhiteMay 5 '17 at 21:23
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2A relevant anecdote: My family was very alarmed when we smelled burning electronics. It turns out that it was a dead LED bulb which someone left turned on (they didn't notice since there was no light). So, yeah, some do consume electricity even when they're dead.– AlohaMay 6 '17 at 11:51
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This Q should probably be more specific about what type of light: incandescent, LED, compact fluorescent, tube fluorescent... "Light bulb" sounds to me like a glass incandescent light bulb, while a LED is a light emitting diode that may or may not technically be in a "bulb" shaped case.– Xen2050May 7 '17 at 2:50
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It depends on the type of bulb.
Regular incandescents won't consume any electricity if the bulb is dead, since there's no continuous path for the current to take. It's just like an open switch.
With CFLs and LEDs, it depends on why the bulb burned out, but in general they will consume some amount of electricity even when burned out. Some CFLs may even consume up to 50% as much as a good bulb (older link, but a lot of burned out bulbs may be old). Newer bulbs may have circuits which eliminate most electricity usage on dead bulbs, as this answer from the electronics stack shows.
Smart bulbs have additional electronics, and so would consume even more electricity than an equivalent non-smart bulb, assuming of course that it's not the smart electronics that died.
The only way to be sure is to measure the usage, with a device like a Kill-a-Watt meter. You would need to install the bulb in a lamp or other fixture with a plug.
If it is a true incandescent light bulb: NO, apart from very very minor losses (through insulation imperfections and transmission line effects) due to the fact that a longer run of wiring is now live.
In addition, if it is an old school flourescent fixture: Very minor losses due to EMI filtering circuitry.
In addition, if it is a LED bulb using a capacitor-based passive power supply: Depends on how the LEDs itself failed. LEDs can fail in a way that they still generate heat (or even pose a short circuit, which would put all the energy into the current limiting circuitry) but no light.
In addition, if there is any active electronics inside (modern LEDs or CFLs), it depends on how these failed and/or react to failure of the actual lighting component - no general statement possible without knowing the exact circuitry.
Nope. Basic principles of electricity: electrical current doesn't flow through an open circuit (at least not at the voltages a residence sees). When a bulb burns out, the conductive path through the bulb is broken and the circuit becomes open - effectively an infinite load. Same as if a breaker were to open.
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7Assuming it's incandescent... I've seen crappy CFLs where the base stays warm after the bulb is dead.– TysonMay 4 '17 at 18:13
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1You're right. Not sure why I was only considering incandescents, especially considering they're obsolete.– Chris M.May 4 '17 at 18:21
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2@immobile My logic is that LEDs do very well what CFLs do in a very clunky and awkward way. The complex glasswork in particular is rather expensive, and the market will fall apart when subsidy contracts expire. Whereas incandescents have many applications not easily replaced, e.g. Oven lights. May 5 '17 at 1:36
Well we've learned one thing the word depends is the most commonly used verb in our industry.
Incandescent of course not.
Any lamp using a magnetic ballasts (oldstyle for any fluorescent or low and high pressure gas lamps) is an autotransformer and power will pass through it even though there is no load and like mmathis has said could be up to 50%.
Newer Electronic ballasts and drivers for LED's have the ability to sense whether or not there is a load and shut down. So if all lamps burn out, it will use some trace power but not enough where I would be concerned about usage.
answered May 4 '17 at 21:48
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1Oldschool FLs have autotransformers only in 110V countries. When ran on 220V, mains voltage is enough to have single-coil inductor simply connected in series as the ballast, so a burnt heater is a truly open circuit.– Agent_LMay 5 '17 at 10:34
There is one scenario I don't see mentioned: Christmas tree lights.
Between the eras of the old large bulb Christmas lights and the "modern" LED lights there was a period of small incandescent lights. Typically 10-30 small incandescent bulbs would be strung in series on a string (sometimes with several series strings physically assembled into a longer string).
Since the lifetime of these tiny bulbs was unpredictable, and since if any bulb in a series goes out the entire string goes out, a technique was developed to tolerate a few dead bulbs in a string.
Basically, at in the base of each bulb was a small glob of conductive material with carefully chosen characteristics. If you had a 10-bulb string of 12-volt bulbs, for 120v total, the conductive glob would only draw a small amount of current and not get very hot. But if the filament of a bulb burned out then almost the entire 120v would be applied across the conductive glob and it would carry 10 times the current and (if everything went as to plan) get hot enough to "melt" (change phases somehow). When it melted, it's resistance would drop to near (but not quite) zero, and the defective bulb would be effectively shorted out.
So a Christmas tree lamp of this style can be consuming a small amount of power when "burned out", whether the "glob" is "melted" or not.
