Resistance: Dissipated power in collision model

AI Thread Summary
Resistance in metals is explained by electron collisions with stationary atomic cores, where more collisions typically suggest higher resistance. However, power dissipation is influenced by both resistance and current, with the relationship defined by P=U^2/R and P=I^2R. While increased collisions can lead to higher resistance, they also correlate with increased electron flow, which can enhance power dissipation. Lowering resistance at a constant voltage increases current, leading to greater power dissipation despite fewer collisions. The discussion highlights the complexity of these interactions, emphasizing that resistance is not solely determined by collision frequency but also by material structure and electron flow dynamics.
greypilgrim
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Hi.

A simple model explains resistance in metals with collisions of the electrons with the stationary atomic cores. So I assume more collisions result in a higher resistance?

But for the dissipated power we have ##P=U^2/R## , which is large for small resistance. I have difficulties combining those concepts. Shouldn't more collisions result in more energy being transferred to the lattice and hence more energy being dissipated?
 
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For a fixed voltage, a larger resistance means a smaller current. If the resistance increases indefinitely, the current goes to zero and thus there is less power dissipated because there is less "flow" being opposed. On the other hand, the smaller the resistance the larger the current for a fixed voltage. So even though there is less resistance, more power is dissipated because there is more current being opposed.

The power dissipation depends linearly on the resistance, but quadratically on the current.
 
greypilgrim said:
So I assume more collisions result in a higher resistance?

In the context of this simple model, no. More obstructions will cause higher resistance. If no current is flowing, there are no collisions, yet the obstructions remain.

Shouldn't more collisions result in more energy being transferred to the lattice and hence more energy being dissipated?
In the context of this simple model, yes.

What causes more collisions, in a given material, is increased electron flow. The other power equation, ##P=I^2R##, should make this clear.

Regarding ##P=U^2/R##, yes, when you lower the resistance while keeping ##U## constant, power will increase. This does not give you a sense of the current though. Ohm's law: ##I=U/R## should make it clear that for a fixed ##U##, reducing ##R## increases ##I##.
 
greypilgrim said:
A simple model explains resistance in metals with collisions of the electrons with the stationary atomic cores. So I assume more collisions result in a higher resistance?

But for the dissipated power we have ##P=U^2/R## , which is large for small resistance. I have difficulties combining those concepts. Shouldn't more collisions result in more energy being transferred to the lattice and hence more energy being dissipated?
First, I don't like the Drude model or most other "mixed" quantum/classical models.

That said, the problem is the assumption in your first paragraph that more collisions means more resistance, which is not correct. Consider two resistors of identical material, identical length, and with identical applied voltage. The only difference is that one has twice the cross sectional area of the other. The one with the greater area will have lower resistance, but more collisions.
 
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lewando said:
More obstructions will cause higher resistance.
But what is the mechanics behind those "obstructions" if not collisions or scattering?
 
I think this model is indeed based on collisions/scattering. What I meant by "more obstructions" would be variations of the lattice structure which would cause a reduction in ways for an electron to travel unimpeded (or less impeded).
 
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