Why, microscopically, is the voltage drop across a resistor affected by the current through it?

Question

If voltage is a potential difference, describing how much energy will be lost per unit charge as charge carriers pass through the resistor, why would the rate at which charge flows (i.e. current, $=dq/dt$) affect the amount of energy lost per unit charge?

I suppose I am asking why a quasi-"extensive" property of the system (current) would affect a quasi-"intensive" property of the system (voltage drop across the resistor).

Apologies if the answer is obvious; I was having trouble understanding why the rate of charge flow would factor into voltage drop whilst trying to rationalize Ohm's law for myself.

Answer 1

First a small correction: voltage is a difference in potential between two points, in this case the difference in potential between the to ends of a resistor.

I suppose your question comes from the interpretation of $U_R=R.I$ with $U_R$ the voltagedrop over the resistor.

Now, the direction of your causal interpretation is wrong. Assuming an Ohmic resistor (constant, current-independent resistance), R is given and the current going through is dependent on the voltage, not the other way around (the way you say it, is like looking at a motorway and saying 'look at all the traffic, that's why the road is so big', while of course the road is so big and that's why all those cars are there.)