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.


1 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.)

  • $\begingroup$ Ah, I see your point. So the current through the resistor is a result of the potential difference that must exist across the resistor. But why would charge flowing through at a different rate affect the amount of energy lost per unit charge? $\endgroup$
    – GnomeSort
    Jun 23, 2015 at 20:51
  • $\begingroup$ Now you are reasoning the same way again. The voltage over the resistor is there because of a battery or something like that, and depending on the voltage provided by that battery, the current is different. $\endgroup$
    – Dries
    Jun 23, 2015 at 20:59

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