I mean a lightbulb or heating filament. In school, they said they use tungsten because it has a high melting point and high electrical resistance. To increase the resistance further, the filament is thin and double-coiled. Ok that doesn’t make sense. If high resistance is so good, why don’t they make it out of wood? Oh, it needs a high melting point as well? How about glass?
Wouldn’t a LOW resistance increase current, and hence heat produced? If copper’s melting point were increased to above tungsten’s, would copper make a good filament?
You also need an incandescent material that glows below its burning or melting point. Glass glows when hot, but it also becomes gooey. Red hot glass can be blown into shape, red hot steel still needs to be hammered.
All materials are incandescent, and all glow the same amount at the same temperature, so that’s not an issue. If the melting point (or gooifying point) is high enough, the rest will fall into place.
Like Goldilocks, the resistance needs to be just right.
Too low, and the filament would have to be enormously long to provide enough resistance to prevent huge currents from flowing. Too high, and not enough current flows to create heating.
So, Tungsten works. So does Carbon, but it’s impractically brittle (although that’s what the first incandescent lamp used). Copper is both too conductive and has too low a melting point.
Resistance heating is determined by the formula V^2 / R. If you want to create a given amount of heat (say, 100W), then for a 100V supply, you would need a resistance of 100Ω. That would cause 1A to flow in the filament. If you replaced that 100Ω filament with a 1Ω one (say, made of copper), the lamp would pull 100A, and would trip the breaker (and it would dissipate 10,000W before it vaporized).
This is another question, but - is this true? The last time I asked (http://boards.straightdope.com/sdmb/showthread.php?p=16115336), someone said some materials put out more visible light and some put more infrared, which is why limestone and lantern mantles are used in their applications.
There’s another effect that I suspect would happen - energy dissipated is proportional to voltage drop, which is proportional to resistance. Meaning most of the energy is dumped at the high resistance points in a circuit. With a normal bulb and wires, that’s at the bulb. But if I were to use a “low resistance bulb”, with similar resistance to the wiring, the energy would be dumped throughout the circuit?
Actually the resistance of the filament is surprisingly low. I don’t remember offhand, but something like 10 ohms.
The low resistance causes a big initial current, but as the fillament heats up it’s resistance increases and the current draw drops.
In a way the current is self regulating. In some applications incadescent lamps are used as a constant current source. When you get past a certain voltage, any voltage increase wont result in higher current
Beowolf is correct … “[Blackbody] radiation has a specific spectrum and intensity that depends only on the temperature of the body.” (Black-body radiation - Wikipedia). Whether iron, tungsten or hydrogen gas … all glow exactly the same for a given temperature, barring any other circumstance.
A black body is an idealization, a body that absorbs 100% of incident light at all wavelengths. Real objects have an emissivity, the ratio of the emission to what it would be for an ideal blackbody at the same temperature and wavelength. An ideal filament would have an emissivity in the visible near 100% and zero emissivity outside the visible. Tungsten is far from this ideal, but its other properties, in particular its very high melting temperature, make it an excellent material for incandescent filaments.
You can buy the old carbon filament bulbs (made in China)-they run about $12 each. for those who like the old orange glow…for their Victorian homes.
You can get the same result with regular bulbs and a dimmer switch.
If a bulb had similar resistance to the wiring (when the bulb is on), the entire circuit (wiring + connections + bulb) would have very low resistance. If we assume the 120/240 VAC entering your home has an extremely low source impedance, this would create very high current in the circuit. Which means the circuit breaker would trip.
The black body calculation is only an approximation, but for the vast majority of materials and visible-light temperatures, it’s a pretty good approximation.
Yes, as Chronos mentioned, the blackbody approximation is good for many applications, but fails, for example, for gases, which have very discrete spectral lines. Tungsten’s emissivity is in the neighborhood of 50% for temperatures that peak in the visible. Using the emissivity at the peak gives a pretty good approximation.
It would, but I think if (for example) resistance was 1/10 of the original, the bulb wouldn’t produce 10x as much heat. It may produce 6x and the other 4x would be released in the wiring.
The black body calculation is very precise for a black body, such as an entirely enclosed cavity in which radiation gets to bounce off of surfaces any number of times before interacting with a surface.
For real physical surfaces, the dimensionless coupling coefficient between surface and radiation also matters. This coupling coefficient, the emissivity, is generally a function of wavelength, but can be averaged over the relevant wavelengths for convenience. If it’s a flat function, the object is called a graybody.
Emissivities have to be between 0 and 1 noninclusive, for real surfaces. For shiny metals they can be pretty small, for example 0.02 for gold mirrors.
Since radiation is proportional to emissivity, incandescence DEFINITELY depends on the material (unless we are talking about an entirely enclosed cavity). And, you can get more efficient radiation of visible light if you reduce or eliminate radiation outside the visible wavelength range. This is the point of Welsbach mantles.
Blackbody radiation depends upon the hot body being, well, black. More to the point, at the wavelengths you are concerned with, it needs to adsorb 100% of any incident radiation - ie appears perfectly black. Ironically, for an approximation that is so useful and widely applicable, a close to perfect black body radiator is rather difficult to create. ETA - Napier above ninja’ed me with a description of how you do make one.
It is fairly easy to work out what is going to happen. First up, assume, as Crafter_Man says, that the feed power supply has a very low - so say zero impedance. Zero is more than accurate enough. Then you simply need to look at the resistances. Say you have 2.5mm[sup]2[/sup] copper wire. This gets you about 0.013 ohms per metre. Say 10 metres from the switchboard, is 20 metres there and back, to a total of 0.26 ohms. Say your lamp has a nominal resistance of 100 ohms. So the total resistance is 100.26 ohms, and thus 99.74% of the power is dissipated in the lamp. Drop the lamp to 10 ohms and you have a total resistance of 10.26 ohms, and 97.4% of the power is dissipated in the lamp, and 2.6% in the cable. This is why cables get warm, and why building codes get very enthusiastic about correct sizing of cable runs.
Slightly complicating matters, that 2.6% of the power lost in the cable run also transltaes to a lower voltage appearing across the lamp - and thus the lamp may not dissipate as much power. Power is proportional to the square of the voltage so the lamp will only be dissipating 95% of the power you expect. As noted above, incandescent lamps tend to be somewhat self regulating, but other items (a cooking range for instance) would deliver the lower power.
Thanks for the clear reply. It seems to be different to what several other posters are saying. I wonder how to reconcile this?
So I guess the requirements so far are:
High temperature strength
Just right resistivity
Positive thermal coefficient of resistance (resistance increases with temperature)
High emissivity in visible region and low elsewhere
Ease of making into wires, etc.
You know, it’s funny this ended up being about blackbody radiation. I was considering starting another thread asking if it was possible to heat a body hotter than a source using only radiation, but I thought I knew the answer. Now I’m not so sure.
Here’s an article about changing the structure of a Tungsten filament in order to make it a more efficient radiator.
But really - Incandescent lamps are dead.
R.I.P