Why does increasing resistance in a conductor decrease power dissipation?

[aaronll]

I read that resistivity is related to the mean free path of electrons inside the material and on density of charged within it.

The electric field does work to flow electrons through a resistor, and the energy is converted into heat due to collision with lattice ion, right? ( in free space they accelerate, but in a conductor they have a drift velocity due to collision, like the limit velocity when a body fall through the air (althoug it is in gravity field constantly)?)

When I have a potential difference through a resistor, with a given resistance R, there is a current that flow, due to Ohm's law, that is V/R, I imagine that when I increase the resistance, electrons collide more, maybe return back, interact with other electrons and they are slow down, so less current.. is correct?
But when I increase the resistance,with the SAME V, and then increase the collision, why don't increase the energy transfer and so the power?

I mean If i increase resistance the current decrease due to more collision but these more collision doesn't increase power spent? But I know power decrease due to V^2/R... why?

 

[BvU]

Hello @aaronll ,

$\qquad$ !​

You might want to look at the Drude model for conductivity.

$\$

 

[aaronll]

Hello @aaronll ,

$\qquad$ !​

You might want to look at the Drude model for conductivity.

$\$

Thanks for your regards...
Yes, I was referring to the Drude model that I studied, but if I look at a resistor like lattice ions + electrons that bouncing (like a pinball machine I envision) I don't understand why If I increase the number of lattice ( increase resistance I think ) the power delivered is less, with the same V.
I read because current is less, but is less because there are more collision, due to Drude model.
So maybe moreover collision doesn't means more energy trasfer to lattice ions, bu why?

 

[BvU]

But when I increase the resistance,with the SAME V, and then increase the collision

I don't understand what you are saying here; it appears to me you think you can increase things independently ?

 

[aaronll]

I don't understand what you are saying here; it appears to me you think you can increase things independently ?

I can increase the resistance, with the same potential difference across it I think...
So the current decrease, but the reason I found in many text is "because there are more collision with ion lattice" but this don't bring to more power dissipated? Or collision are elastic with no energy transfer?

 

[DaveE]

When there are more collisions, the charges have a shorter mean free path so they don't accumulate as much energy before they collide again. So, more collisions, but less energy in the colliding charges.

 

[aaronll]

When there are more collisions, the charges have a shorter mean free path so they don't accumulate as much energy before they collide again. So, more collisions, but less energy in the colliding charges.

Ok, and so less energy transfer to ion lattice, right? So less "heat"

 

[vanhees71]

Indeed, the power is $P=I U=R I^2=U^2/R$. This is the energy per unit time dissipated to heat in the resistor. The resistance gets larger the smaller the mean-free path of the conduction electrons in the metal is, and to get through two wires of the same length the conduction electrons suffers more collisions and thus dissipates more heat if you have the same currents in both wires (resistors).