The irradiate is completely dead, no light whatsoever.
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It relies on the type of bulb.
Regular incandescents won"t consume any kind of electricity if the pear is dead, because there"s no continuous path for the existing to take. It"s just like an open switch.
With CFLs and LEDs, it relies on why the bulb burned out, yet in basic they will certainly consume part amount of electrical energy even when burned out. Part CFLs may also consume approximately 50% as lot as a an excellent bulb (older link, but a lot of of melted out bulbs might be old). More recent bulbs may have circuits which get rid of most electricity intake on dead bulbs, as this answer indigenous the electronic devices stack shows.
Smart bulbs have additional electronics, and also so would consume even much more electricity than an identical non-smart bulb, assuming of course that it"s no the smart electronics that died.
The only means to be sure is to measure up the usage, through a an equipment like a Kill-a-Watt meter. You would must install the pear in a lamp or various other fixture v a plug.
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answered may 4 "17 at 18:13
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If the is a true incandescent light bulb: NO, personal from an extremely very boy losses (through insulation imperfections and also transmission heat effects) due to the truth that a much longer run the wiring is now live.
In addition, if it is one old college flourescent fixture: an extremely minor losses due to EMI filtering circuitry.
In addition, if it is a LED bulb making use of a capacitor-based passive strength supply: counts on just how the LEDs chin failed. LEDs have the right to fail in a means that they still generate warmth (or even pose a quick circuit, which would put all the energy into the current limiting circuitry) but no light.
In addition, if there is any energetic electronics within (modern LEDs or CFLs), it relies on exactly how these failure and/or react to failure of the actual lighting ingredient - no basic statement possible without learning the exact circuitry.
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answered might 4 "17 at 20:11
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Nope. Simple principles of electricity: electrical present doesn"t flow through an open up circuit (at least not in ~ the voltages a residence sees). When a bulb burns out, the conductive path through the bulb is broken and the circuit becomes open - successfully an infinite load. Same as if a breaker were to open.
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answered might 4 "17 in ~ 18:11
kris M.Chris M.
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Well we"ve learned one point the word depends is the most frequently used verb in our industry.
Incandescent of food not.
Any lamp using a magnetic ballasts (oldstyle for any kind of fluorescent or low and also high push gas lamps) is an autotransformer and power will certainly pass with it even though over there is no load and also like mmathis has said could be as much as 50%.
Newer electronic ballasts and also drivers because that LED"s have actually the ability to feeling whether or not there is a load and also shut down. For this reason if every lamps burn out, the will usage some map power however not sufficient where I would be concerned about usage.
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answered may 4 "17 in ~ 21:48
Retired grasp ElectricianRetired master Electrician
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There is one scenario ns don"t view mentioned: Christmas tree lights.
Between the eras of the old big bulb Christmas lights and also the "modern" LED lights there to be a duration of little incandescent lights. Generally 10-30 little incandescent bulbs would certainly be strung in collection on a wire (sometimes through several series strings physically assembled into a much longer string).
Since the life time of these tiny bulbs to be unpredictable, and also since if any kind of bulb in a collection goes out the entire string goes out, a technique was developed to tolerate a couple of dead bulbs in a string.
Basically, at in the basic of each pear was a tiny glob the conductive material with closely chosen characteristics. If you had a 10-bulb string of 12-volt bulbs, because that 120v total, the conductive glob would certainly only attract a little amount of current and not get an extremely hot. But if the filament of a bulb burned out then virtually the whole 120v would be applied throughout the conductive glob and also it would lug 10 times the current and also (if everything went regarding plan) obtain hot enough to "melt" (change phases somehow). When it melted, it"s resistance would drop to near (but no quite) zero, and also the defective bulb would be effectively shorted out.
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So a Christmas tree lamp of this format can be spend a small amount that power as soon as "burned out", even if it is the "glob" is "melted" or not.