# How can heat dissipation from a resistor have such a simple relationship with resistance?

I understand that when current runs through a resistor in a circuit, Joule heating occurs. I've read that you can think of the energy conversion as happening due to some of the electrons of the current being blocked on their way through the resistor. When you have a collision of course there is energy transfer.

The formula for the rate of dissipation is $I^2R$

$$\text{Resistance} = \frac{\text{material resistivity} \times \text{length}}{\text{sectional area}}$$

Say we have 2 wires, both with length 5 but one with resistivity 2 and area 2, and the other with resistivity 1 and area 1. Both have resistance 5. The latter has lower resistivity, but since it also has less area in which the electrons can travel, the resistance remains the same.

If we hook both up to identical 5 V batteries, 1 A will flow in both cases. The heat dissipation will be $1^2 \times 5 = 5$ in both cases. But surely the higher resistivity of the first wire should imply more of the energy-transferring collisions per unit area, which combined with the higher area should result in a higher rate of dissipation?

• One possible answer just occurred to me. Maybe I was wrong to suggest a wire whose resistance is kept low by being slender with less paths, will have less collisions. Maybe the lower resistivity of said wire means electrons will fly more readily down the paths that do exist, which would serve to factor out the lower resistivity? – Justin Poirier Mar 9 '18 at 0:01