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

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.

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