If the bulb is out, or empty, but the switch is on, are you still paying for power?

October 1, 2005 1:04 PM

If the bulb is out, or empty, but the switch is on, are you still paying for power?
posted by airguitar to science & nature (17 comments total)

No.
posted by delmoi at 1:06 PM on October 1, 2005

No electricity is being used, so no. The switch is actually off, only not at the wall switch but at the point where the tungsten broke.
posted by geoff. at 1:07 PM on October 1, 2005

Nope - you pay based on the power consumed (usually measured in Watts), which is directly proportional to current squared. If there's no bulb in the socket, or if the bulb has blown, it's an open circuit -> infinite resistance -> no current (since current follows the path of least resistance).

On preview, what geoff. said.
posted by muddgirl at 1:09 PM on October 1, 2005

No. A light bulb's sole duty is to convert electrical energy to light (and heat). It's basically no more than a wire.

Something like a phone charger or power adapter, on the other hand, does use power if it's not plugged in to the appliance; the AC-DC conversion generates heat. So don't leave your chargers plugged into the wall.
posted by holgate at 1:12 PM on October 1, 2005

directly proportional to current squared

ask.mefi rocks
posted by airguitar at 1:21 PM on October 1, 2005

Here's a related question. If I have a light on a dimmer, and I turn it down half way, do I use half as much electricity? Or is the half that is not used to produce light somehow thrown out in the dimmer?
posted by alms at 1:53 PM on October 1, 2005

alms - it depends on the switch. Old dimmer switches just used a variable resistor - the same amout of current flows through the whole circuit, but the resistor causes a voltage drop which decreases the light output of the bulb. Less light, same power consumption.

Newer dimmer switches simply turn off the flow of current at certain points in the AC cycle, so overall it does consume less power (though not linearly - a lighbulb turned down to half-brightness will not consume half as much power).
posted by muddgirl at 2:10 PM on October 1, 2005

My high school chemistry teacher told us that when she was little and the light socket was empty she used to guard the light switch to make sure no one turned it on. She was afraid that if someone turned it on electricity would come pouring out and electrocute everyone.
posted by leapingsheep at 2:11 PM on October 1, 2005

If it is an electronic dimmer control, the dimmer circuit itself is about 97% efficient, so at half setting, you'd be using just a little more than half the energy of full setting. The electronic dimmer appears to the AC current source as a switching load, turning "on" for some portion of each AC waveform for some portion of the wave form from 0 to 100%, based on your setting. The switching element, usually a device called a triac, isn't perfectly efficient, so there is a slight amount of energy dissipated as heat by the dimmer itself. Note that you cannot use these dimmers safely to control non-incandescent lights, or anything with a motor, since at anything other than 100% setting the wave form to the load is far from sinusodial. Loads on these kinds of dimmers are limited to 600 Watts of lighting load.

An older kind of dimmer control, still in use for heavy load applications in commercial, industrial and theatrical applications used a special kind of transformer called an autoformer to provide full cycle AC at varying voltages. Sometimes these controls were motorized, where the large autoformer element could be mounted in a concealed location, and a small control switch for the adjustment motor on the autoformer could be installed in a conventional light switch box on the wall. These kind of systems can control thousands of watts of lighting load, and can operate and even reverse appropriately designed motors. They are quite expensive but they last forever, and have generally decent efficiencies of about 90%.
posted by paulsc at 2:20 PM on October 1, 2005

nthdegx has informed be that power is directly proportional to current, not current squared (since the voltage is constant, not the resistance). What do I know?
posted by muddgirl at 2:33 PM on October 1, 2005

airguitar -- until you put an active load on a circuit (turn something on) no electrical energy will be transferred through the socket. People think that they need to turn sockets off at the wall to prevent power consumption, but this isn't the case.

"you pay based on the power consumed (usually measured in Watts), which is directly proportional to current squared...

ask.mefi rocks"

Weeeeeell...

P = IV
That is, electrical power = electrical current x potential difference. Potential difference (voltage) in this case is a constant.

What muddgirl said would have been the case if resistance was a constant and potential difference was not.

Therefore electrical power is directly proportional to the current, not the current squared. That is, if you double the wattage of your light bulb you double the current in the circuit.
posted by nthdegx at 2:45 PM on October 1, 2005

This is one of those places where the naive analogy of electrical circuits to water pipes breaks down (which seems to be the way a lot of people understand it).

No circuit equals no (net) electron movement equals no bill. Now, if you go out to your sprinkler system and break one of the pipes (which is a lot like breaking the light bulb filament if you analogize electricity and water flow), and turn your water on, you WILL be paying for all the water that flows into the ground.
posted by teece at 2:56 PM on October 1, 2005

Well, to make the water analogy hold up you could say something like "break one of the pipes in such a way that both ends of the break seal off and prevent the flow of water" but that's not really a great analogy any more.
posted by Rhomboid at 7:33 PM on October 1, 2005

What muddgirl said would have been the case if resistance was a constant and potential difference was not.
Unless you're discussing a 3-way bulb (which has 2 filaments), the resistance in a light bulb is constant (give or take heat effects).

It's true that P=VI. That is a derivative of Ohm's Law, which states that V=IR. If we replace the V in the first equation, we get P=IIR. From that, it's apparent that P varies as the square of the current. (If we rearrange the equation differently -- P=VV/R -- it's also apparent that P varies as the square of the voltage.) As current increases through a constant resistance, power increases in proportion to the square of the current. muddgirl is correct in the general case where no variables are constrained.

In a bounded case like a lightbulb in a standard lamp in a standard home, however, the voltage is constant. That's why the lightbulb makers can rate their bulbs as 60W, 75W, etc. -- they know that the voltage will be about 110V, so they put in a wire with about 200 ohms resistance to produce a 60W bulb. Since the voltage is constant, the only way to change the power consumed is by varying the resistance. If the resistance is halved, twice the current must be applied to keep the voltage constant. In this particular case where the voltage is constrained, nthdegx is correct.
posted by forrest at 8:08 PM on October 1, 2005

True enough, forrest. I hope I covered that with "in this case is a constant". What your post suggests to me, though, is that it is a rare exception for potential to be constant. In the case of mains electrcitiy, potential difference *is* constant. It's not like light bulbs are a special case.
posted by nthdegx at 3:19 AM on October 2, 2005

Actually, the converse is true -- most devices that represent loads will have a constant resistance/impedance.
posted by forrest at 10:26 AM on October 3, 2005

k...
posted by nthdegx at 3:20 PM on October 3, 2005