Book wrong? Ohms law, Graph, wire, slope change question

[insightful]

...the answer is that with an insulated filament of tungsten at a high temperature the gradient could go negative.

At what temperature does your analysis say the gradient goes negative, i.e., dI/dV=0? I would posit that it is much higher than the melting point.

 

[haruspex]

At what temperature does your analysis say the gradient goes negative, i.e., dI/dV=0? I would posit that it is much higher than the melting point.

If you look at the result I quote in post #25 (which you can check yourself) it is not the wire's temperature that's critical but the ambient temperature, i.e. the temperature on the far side of the insulator. So it should be possible to observe the effect by adequate cooling. Of course, the wire's temperature does need to be in the range where the 1.2 power law applies.

 

[insightful]

If you look at the result I quote in post #25 (which you can check yourself) it is not the wire's temperature that's critical but the ambient temperature, i.e. the temperature on the far side of the insulator. So it should be possible to observe the effect by adequate cooling. Of course, the wire's temperature does need to be in the range where the 1.2 power law applies.

Insulated, but with adequate cooling? Then why insulated? I apologize, but this makes absolutely no sense to me. I'll await a calculated amps vs. volts curve with a negative dI/dV.

 

[haruspex]

Insulated, but with adequate cooling? Then why insulated? I apologize, but this makes absolutely no sense to me. I'll await a calculated amps vs. volts curve with a negative dI/dV.

The question is about plotting current against voltage in an environment where higher power results in higher temperature, hence higher resistance. To make the question meaningful you have to presume that at each voltage the system is allowed to reach steady state. The resulting temperature will depend on the relationship between temperature and heat dissipation. For a wire in the open, the primary dissipation will be by radiation, giving a fourth power relationship. If insulated it becomes a linear relationship. But in order for there to be a steady state you still need the outside of the insulation to be in a constant temperature environment.

I have a correction to my last post - it does depend on the actual temperature The criterion I ended up with can be rewritten in the more useful form $\theta > \theta_0\frac{\mu}{\mu-1}$. In the case of Tunsgten that gives approximately $\theta >6\theta_0$. Since Tungsten melts around 3700K, it should be quite possible to demonstrate the effect with the outside of the insulator at room temperature.

 

[insightful]

Since Tungsten melts around 3700K, it should be quite possible to demonstrate the effect with the outside of the insulator at room temperature.

Calculating steady-state heat transfer from a heated wire through insulation should be straightforward then. Only a few points would be needed to demonstrate a falling amperage with rising voltage. I believe it would be a futile exercise, however.

 

[haruspex]

Calculating steady-state heat transfer from a heated wire through insulation should be straightforward then. Only a few points would be needed to demonstrate a falling amperage with rising voltage. I believe it would be a futile exercise, however.

Why futile? It would be interesting because it is such a surprising result.
I believe an analogous condition can arise in fluid flow, though maybe to do with turbulence in that case.

 

[insightful]

Why futile? It would be interesting because it is such a surprising result.

Futile as in "that could not succeed" (Webster's New World Dictionary). I await your calculations.

 

[itfitmewelltoo]

This is a lightbulb. It is used in speakers to limit the current when the power amp is cranking out sound.. As the voltage is increased, and current increases, the lightbulb gets hotter and brighter. The resistance is increasing and it creates an increasing resistance that helps limit current going to the speakers. Note that in this application, voltage is the independent variable.

Oh, read back a bit in the thread and see a reference to tungsten. What is the material used in incandescent light bulb filaments? I forget.

 

[haruspex]

Futile as in "that could not succeed" (Webster's New World Dictionary). I await your calculations.

What calculations do you need from me? I posted a set of starting equations based on a model. Do you see anything wrong with those equations? I then posted a criterion for negative slope that can be deduced from those equations. Do you need to see how I did that? Can you not go through the same steps? The final result appears to show that for Tungsten a negative slope could be observed. Do you see a problem with that conclusion?
Yes, I know what futile means (pointless), so I did not know whether you were saying you did not think the effect could ever be demonstrated in practice or that it was uninteresting even if it could. Anyway, the key point of my posts on this thread is that everyone (me included) was quick to say it was quite impossible as though there were some fundamental reason, but it turns out there isn't.

 

[haruspex]

This is a lightbulb.

There is no mention of a lightbulb in the OP. Do you have further information about the source of the question?
Even if the original question did state a lightbulb, my argument still stands: that there is no really obvious reason why the slope could not go negative. It wouldn't happen for a lightbulb, but only for the subtler reason that the power dissipation of a lightbulb rises faster than linearly with temperature.
(I haven't checked whether a putative material for which the resistance rises faster than the fourth power of absolute temperature might even demonstrate a negative slope in the radiative model.)

Edit: of course, even if it is a lightbulb, the question arises as to how the heat gets away from it. In an enclosed space it would still be a conductive model, not a radiative one.