for example a 1mm gap with a strand of copper wire across it with 12v at one end of the gap and 0v at the other end
and unlimited current available (but only so much can flow)
and the wire being so incredibly thin that it wont even heat up or even try to draw any current at all?
is it possible for the wire to be thin enough to do that?
and if made slightly thicker and thicker at what point will it be unable to shed away heat and burn out with thicker and thicker wire?
how thin would the wire have to be to draw almost no current?
a few molecules thick? a few hundred molecules thick? a few thousand molecules thick? a few million or billion molecules thick?
and unlimited current available (but only so much can flow)
and the wire being so incredibly thin that it wont even heat up or even try to draw any current at all?
is it possible for the wire to be thin enough to do that?
and if made slightly thicker and thicker at what point will it be unable to shed away heat and burn out with thicker and thicker wire?
how thin would the wire have to be to draw almost no current?
a few molecules thick? a few hundred molecules thick? a few thousand molecules thick? a few million or billion molecules thick?
Use the formula for resistance, R = (r x l / a), where r is the material's resistivity,
l is the length, and a is the cross section area. For copper r = 1.68 x10E-8 ohm-meters.
With 12V and (say) 1uA current, the resistance is 12Mohms. Then 12M = (1.68x10-8 x 1x10E-3m / a).
And so, a = 1.47 x 10E-18 sqmeter, or 1.47 x 10E-12 sqmm cross section.
That's a pretty small wire, about a millionth of a mm across, or around 10 atoms.
Ohm's law probably doesn't work at this scale, though.
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Can you have a strand of wire so thin it wont burn out?
Not if you let ME make a few attempts. The first thing I'm likely to try, is to connect 440VAC (20 amp circuit breaker) to the strand of wire. I have a feeling it will burn out. Unless of course the (resistance per unit length) times the (length) is a truly gigantic number of ohms. That'd be a really, really long wire.
Not if you let ME make a few attempts. The first thing I'm likely to try, is to connect 440VAC (20 amp circuit breaker) to the strand of wire. I have a feeling it will burn out. Unless of course the (resistance per unit length) times the (length) is a truly gigantic number of ohms. That'd be a really, really long wire.
Dang thats a thin wire..
What about 3 inches of wire?
is there a point where ohms law doesnt work anymore because of how physically THIN the wire is where it literally can't carry enough current to burn itself up?
it seems the thinner the wire. the shorter it can be and not burn up.
What about 3 inches of wire?
is there a point where ohms law doesnt work anymore because of how physically THIN the wire is where it literally can't carry enough current to burn itself up?
it seems the thinner the wire. the shorter it can be and not burn up.
Don´t waste time on realflow´s *weekly* impossible questions.
Just trying to stir the pond for fun or out of boredom, never a *real* question which may lead to useful answers.
jm2c
Just trying to stir the pond for fun or out of boredom, never a *real* question which may lead to useful answers.
jm2c
We call that a light bulb!
The vacuum has been replaced with an inert gas....its cheaper and makes light bulbs safer when they break.
You can rephrase it as follows:
Let's say I have a strand of wire that 'just' burns out. If I make the diameter smaller, will it stop burning out? I think the answer lies in whether the heat capacity decreases slower than the diameter. In other words, if the diameter is halved, is the heat capacity also halved, or more, or less? If the diameter goes down faster than the heat capacity, it will stop burning. If the heat capacity goes down faster than the diameter, it will always burn.
I am not a physicist but my gut feeling says it will continue to burn when made smaller. But I may be wrong. Has happened before ;-)
Jan
Let's say I have a strand of wire that 'just' burns out. If I make the diameter smaller, will it stop burning out? I think the answer lies in whether the heat capacity decreases slower than the diameter. In other words, if the diameter is halved, is the heat capacity also halved, or more, or less? If the diameter goes down faster than the heat capacity, it will stop burning. If the heat capacity goes down faster than the diameter, it will always burn.
I am not a physicist but my gut feeling says it will continue to burn when made smaller. But I may be wrong. Has happened before ;-)
Jan
Yes. You can calculate the R if you know the diameter and length. You also know the mass of the copper. From the R and the V you can determine the dissipation. Then you can calculate how much the temp rises for that mass with that dissipation. I assume the 'burning' will mean it has to reach a certain temperature so you now know whether it is above or below.
The crux is whether the function that describes the delta-T as function of diameter is linear or quadratic. If quadratic it will have a minimum (or maximum depending on what you are looking for) and therefor a diameter where the burning starts or stops.
Any math wizzards here?
Jan
More thinking. I could write a sort of meta-function:
delta-T = Dissipation/Mass.
Mass=f(L, dia).
Dissipation is V*I but I=V/R and R= f(L, dia). So dissipation= V^2/f(L, dia).
That makes delta-T = V^2/[f(L, dia)]^2. Looks like delta_T (and thus 'burn point') is a quadratic function of the variable 'diameter'. A parabola. Does it have a maximum or a minimum?
Jan
delta-T = Dissipation/Mass.
Mass=f(L, dia).
Dissipation is V*I but I=V/R and R= f(L, dia). So dissipation= V^2/f(L, dia).
That makes delta-T = V^2/[f(L, dia)]^2. Looks like delta_T (and thus 'burn point') is a quadratic function of the variable 'diameter'. A parabola. Does it have a maximum or a minimum?
Jan
Other things being equal, heat dissipation ability varies as something like wire surface area which varies like cross-sectional radius.
Other things being equal, power dissipated varies like resistance (if constant current) or 1/resistance (if constant voltage). Resistance varies like 1/(cross-sectional area) which is 1/(radius^2).
So for constant current increasing the radius reduces the heat developed and increases the heat dissipation ability. The converse for smaller radius. So constant current means that there is a minimum radius below which the wire melts/burns.
For constant voltage the power dissipated goes like radius^2, and the heat dissipation ability goes like radius. So bigger radius gets hotter. Smaller radius is cooler, so if it is thin enough then it can't get very hot.
Other things being equal, power dissipated varies like resistance (if constant current) or 1/resistance (if constant voltage). Resistance varies like 1/(cross-sectional area) which is 1/(radius^2).
So for constant current increasing the radius reduces the heat developed and increases the heat dissipation ability. The converse for smaller radius. So constant current means that there is a minimum radius below which the wire melts/burns.
For constant voltage the power dissipated goes like radius^2, and the heat dissipation ability goes like radius. So bigger radius gets hotter. Smaller radius is cooler, so if it is thin enough then it can't get very hot.
I think I see your logic here, of course if the wire exists it must draw some current, but taken to the extreme of smallness till it almost doesn't exist it would very nearly be an open circuit, having a very high resistance and draw almost no current. It would be interesting if someone here could do the maths to figure out at what size it would survive.........I think they are nearly there 😉
if its a conductor, then enough V, backed by huge I capacity...
https://en.wikipedia.org/wiki/Exploding_wire_method
https://en.wikipedia.org/wiki/Exploding_wire_method
If a conductor is made small enough, it will eventually fail due to long term electro migration, even if heat dissipation is not an issue. It is like a long term fusing action, caused by the crystal structure being rearranged. The critical current densities are several tens of thousands of amps per square cm. It sounds like a lot until you think about dimensions involved in integrated circuits.
Well it'll take a lottt longer with the super thin wire than if you used a thicker wire that was just the right thickness to where it'd "pop" right away
many orders of magnitude longer i think.
many orders of magnitude longer i think.
Pretty sure I can fix you up with a wire so thick you can't burn it out. At least not with any common home electric service.
I dont mean using a wire thick enough that it wont burn out.
I'm talking just some amount thicker to where it'll burn out from a reasonable amount of current
the thinner the wire is. the more resistance it has. so the thinner it is. the less current will flow. once it's thin enough. it wont draw enough current to burn itself out and it becomes a resistor of whatever value it happens to be.
I'm talking just some amount thicker to where it'll burn out from a reasonable amount of current
the thinner the wire is. the more resistance it has. so the thinner it is. the less current will flow. once it's thin enough. it wont draw enough current to burn itself out and it becomes a resistor of whatever value it happens to be.
That's a joke son (said Foghorn Leghorn). Actually, if the math holds (which I doubt) there's a crossover at about 175 AWG where the fusing current exceeds the current at 1mm and 12 VDC. I doubt you'll be able to find anybody to draw 175 AWG, though there'd certainly be a lot of it on a spool. Google the Preece equation.
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