Do burnt out lightbulbs use any electricity?
February 10, 2007 12:36 AM Subscribe
Does an incandescent lightbulb continue to use electricity after it's burnt out?
I know the circuit is broken, but does the current just terminate there? Are there little electrons bouncing around inside with nothing to do?
I know the circuit is broken, but does the current just terminate there? Are there little electrons bouncing around inside with nothing to do?
Best answer: No.
You can prove this to yourself if you have access to a small electric motor. If you hook up that kind of motor to a lightbulb and spin it, you will generate current and make the lightbulb glow. You will also notice that it's quite hard to do this when the lightbulb is burning; you are doing work and expending energy. If you then break the filament in the bulb, you will instantly notice that the motor is much easier to turn... you are no longer doing the work to light the filament, and thus the crank is easier to turn.
When it's hooked up to the AC power, your broken bulb is just vibrating with potential... 60 times a second, each of the two sides of the circuit will peak with voltage while the other troughs, and vice versa. This is just potential, though, and causes essentially no load and no wasted energy. It's only when you complete the circuit that the electrons flow and emit energy as heat and light. That energy loss requires fresh input by the power company to replace, and is measured by your power meter.
If you are very patient, you can disconnect every device in your house except one working lightbulb, and go outside and look at your meter... you will see it moving slowly. If you then replace the working bulb with a broken one, and and go look at the meter again, the needle should have stopped moving completely.
posted by Malor at 1:01 AM on February 10, 2007
You can prove this to yourself if you have access to a small electric motor. If you hook up that kind of motor to a lightbulb and spin it, you will generate current and make the lightbulb glow. You will also notice that it's quite hard to do this when the lightbulb is burning; you are doing work and expending energy. If you then break the filament in the bulb, you will instantly notice that the motor is much easier to turn... you are no longer doing the work to light the filament, and thus the crank is easier to turn.
When it's hooked up to the AC power, your broken bulb is just vibrating with potential... 60 times a second, each of the two sides of the circuit will peak with voltage while the other troughs, and vice versa. This is just potential, though, and causes essentially no load and no wasted energy. It's only when you complete the circuit that the electrons flow and emit energy as heat and light. That energy loss requires fresh input by the power company to replace, and is measured by your power meter.
If you are very patient, you can disconnect every device in your house except one working lightbulb, and go outside and look at your meter... you will see it moving slowly. If you then replace the working bulb with a broken one, and and go look at the meter again, the needle should have stopped moving completely.
posted by Malor at 1:01 AM on February 10, 2007
Consider: The circuit is broken at the switch if it's off. The same logic applies. Breaking a circuit literally cuts the ability for electrons to bounce.
Are you perhaps tripped up by the concept of AC adapters still using power when their devices aren't plugged in?
Because those are stepping up or down voltages and use capacitors and other things I know not of, and so will continue performing their function and having the power seep out of the capacitors whether something's plugged in or not. Different for reasons people who are far more smart and intelligent than I in the world of electricity can explain better than I.
posted by disillusioned at 1:01 AM on February 10, 2007
Are you perhaps tripped up by the concept of AC adapters still using power when their devices aren't plugged in?
Because those are stepping up or down voltages and use capacitors and other things I know not of, and so will continue performing their function and having the power seep out of the capacitors whether something's plugged in or not. Different for reasons people who are far more smart and intelligent than I in the world of electricity can explain better than I.
posted by disillusioned at 1:01 AM on February 10, 2007
On preview: Malor is one of those people.
posted by disillusioned at 1:03 AM on February 10, 2007
posted by disillusioned at 1:03 AM on February 10, 2007
the question's answered ("burnt out" = broken/melted/evaporated filament = open circuit = no current so no power) so I'm gonna derail :)
AC adapters consume power when no load is attached for a couple of reasons, the primary one being magnetization current in the transformers used. That applies to switch-mode supplies as well since they too use transformers, just smaller high-frequency ones.
Yes, one can build a buck-mode switcher with only an inductor (no transformer) but it's not considered safe where the output may come into contact with humans.
posted by polyglot at 3:36 AM on February 10, 2007
AC adapters consume power when no load is attached for a couple of reasons, the primary one being magnetization current in the transformers used. That applies to switch-mode supplies as well since they too use transformers, just smaller high-frequency ones.
Yes, one can build a buck-mode switcher with only an inductor (no transformer) but it's not considered safe where the output may come into contact with humans.
posted by polyglot at 3:36 AM on February 10, 2007
If you then break the filament in the bulb, you will instantly notice that the motor is much easier to turn... you are no longer doing the work to light the filament, and thus the crank is easier to turn.
Is this some sort of special electric motor? I've only superficial electrical experience, but I've never heard or experienced a motor that's physically harder or easier to crank depending on the load. I'm kinda interested now!
posted by wackybrit at 4:48 AM on February 10, 2007
Is this some sort of special electric motor? I've only superficial electrical experience, but I've never heard or experienced a motor that's physically harder or easier to crank depending on the load. I'm kinda interested now!
posted by wackybrit at 4:48 AM on February 10, 2007
There is still a tiny, tiny, tiny resistive and capacitive element present across the open ends of the filaments.
A megohmmeter would probably show a super high resistance, and a few femtoFarads of capacitance would be there for sure...(if from nothing other than the internal interconnects).
In practical terms, the load is infinitely small. It is non-zero, but pretty damn close to zero.
posted by FauxScot at 6:00 AM on February 10, 2007
A megohmmeter would probably show a super high resistance, and a few femtoFarads of capacitance would be there for sure...(if from nothing other than the internal interconnects).
In practical terms, the load is infinitely small. It is non-zero, but pretty damn close to zero.
posted by FauxScot at 6:00 AM on February 10, 2007
Regenerative braking in hybrid electric vehicles works in sort of same way wackybrit - a circuit is closed that is next to the spinning wheel/axel, and an electromagnetic load is placed on the axel to slow or stop its spinning, even though there is no direct physical contact between the two. The resulting energy flows through the closed circuit. (basically its the reverse of the example cited above)
posted by SirOmega at 6:01 AM on February 10, 2007
posted by SirOmega at 6:01 AM on February 10, 2007
According to James Thurber's aunt, electricity does leak out of empty light sockets, so keep those bulbs in there!
(Sorry.)
posted by kmennie at 6:59 AM on February 10, 2007 [1 favorite]
(Sorry.)
posted by kmennie at 6:59 AM on February 10, 2007 [1 favorite]
I've never heard or experienced a motor that's physically harder or easier to crank depending on the load.
Yes, all motors react like this to load. How could they not?
posted by kindall at 7:17 AM on February 10, 2007
Yes, all motors react like this to load. How could they not?
posted by kindall at 7:17 AM on February 10, 2007
Incandescent Chistmas-tree bulbs have a shunt to conduct current past a burned out element, to solve that "one bulb goes out, they all go out" problem. As a result, burned out Christmas-tree bulbs are still "running" when fed current, they're just not using any of the current they carry to produce light. If the whole string of christmas lights was burned out, It would be a complete circuit of some resistance (low), and could conceivably blow a fuse.
So: burned-out is the same as an open-circuit, except if the bulbs are engineered to allow other bulbs with which they are running in series to remain lit.
posted by Crosius at 1:04 PM on February 10, 2007
So: burned-out is the same as an open-circuit, except if the bulbs are engineered to allow other bulbs with which they are running in series to remain lit.
posted by Crosius at 1:04 PM on February 10, 2007
wackybrit: if motors (generators really, since you're converting mechanical into electrical energy) didn't behave that way, they'd be violating conservation of energy. Have you ever tried riding a bicycle with a lighting dynamo attached?
Output voltage is mostly defined by rotational rate, output current is defined by the load. Output electrical power = current * voltage. Input mechanical power = torque * speed.
So for a given speed, torque increases with load current to maintain conservation of energy. Considered alternately, the load current in the motor's (generator's) windings is exerting a torque on the shaft that must be counteracted or it will stall.
posted by polyglot at 8:41 PM on February 10, 2007
Output voltage is mostly defined by rotational rate, output current is defined by the load. Output electrical power = current * voltage. Input mechanical power = torque * speed.
So for a given speed, torque increases with load current to maintain conservation of energy. Considered alternately, the load current in the motor's (generator's) windings is exerting a torque on the shaft that must be counteracted or it will stall.
posted by polyglot at 8:41 PM on February 10, 2007
Incandescent Chistmas-tree bulbs have a shunt to conduct current past a burned out element, to solve that "one bulb goes out, they all go out" problem. ... If the whole string of christmas lights was burned out, It would be a complete circuit of some resistance (low), and could conceivably blow a fuse.Not that this changes the answer to the original question, but in the case of this kind of christmas light, it's drawing the most current (i.e. most likely to blow a fuse) when all lights are working.
Think of that shunt as another little electric bulb that's in parallel with the regular bulb. If you connect two lamps in parallel, you get twice as much light as with just one lamp, so the pair must be drawing about twice as much current as a single bulb.
Two resistors (or lamps) in parallel have lower resistance than either one by itself, and connected to the same voltage the pair will draw more current than either will on its own.
(Conversely two resistors in series have a resistance that's the sum of their individual resistances, i.e. greater than either one; two lamps connected in series will deliver less total light than just one lamp.)
posted by phliar at 4:32 PM on February 13, 2007
This thread is closed to new comments.
posted by disillusioned at 12:59 AM on February 10, 2007