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Suppose that we have a closed circuit that consists of a single wire. If we said that the wire has resistance $R_1$, then the electrons would travel through the wire and the potential difference across two random points of the wire would be given by Ohm's Law $V=I(\sigma l/A)$ (where $\sigma$ is resistivity, $l$ the length that separates both points and $A$ the cross-sectional area). However, if we had the same circuit but substituting a part of the wire with a resistor of higher resistance $R_2$, we would get that the potential difference between the terminal points of the resistor is different from the difference at those points when there wasn't any resistor. I have read that these could be explained due to different charge densities inside the wire that affect the potential, but this brings one doubt. Wouldn't different densities across the wire cause that the potential difference between the battery points depends on the resistors you put? As a difference in charge densities will generate a potential that can be added to the initial one.

Edit: Maybe my question could be reformulated in the following way. If we had that there is no resistor nor wire with any resistance and we had a vacuum inside the wire we would have that the potential at every point will be given by gauss Law for example. My question is how putting resistors, changes the potential at every point? As putting one wire or another both will end up adding V, but my question is more related to how this potential is defined at every point.

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    $\begingroup$ In a circuit with an ideal battery that delivers unlimited current no matter what kind of resistors you put between its points the potential difference is always the same. In reality a battery has internal resistance that limits the current it can deliver. This also leads to a decrease in the external potential difference when you decrease the external resistor. Just use Ohms law for external and internal resistors who are in series. $\endgroup$
    – Kurt G.
    Commented Mar 27, 2022 at 16:49
  • $\begingroup$ Okay, thank you. And what about the charge densities? $\endgroup$ Commented Mar 27, 2022 at 16:59
  • $\begingroup$ They follow from Ohms law once you calculated the current. $\endgroup$
    – Kurt G.
    Commented Mar 27, 2022 at 17:00
  • $\begingroup$ I understand what you say but I don't think it answers my main question. How can a resistor change the potential at a random point of a wire? You can imagine just a wire with higher resistivity if it is necessary to answer the question. Since potential depends on the charge inside the terminals of the battery. Assuming an ideal battery. $\endgroup$ Commented Mar 27, 2022 at 17:11
  • $\begingroup$ OK. Assume ideal battery. Total constant potential difference $U$. Only the wire with certain resistance per wire length. Can you calculate from this the potential difference between two points on the wire? Now put in a resistor in the middle of the wire. Then apply Ohms law to three resitors: half of the wire, resistor that we put in, other half of the wire. Three potential differences that add up to $U$. $\endgroup$
    – Kurt G.
    Commented Mar 27, 2022 at 17:15

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Usually you have

  1. a given potential difference that is the cause of a current in a circuit,
  2. a resistor with a given resistance (dictated by resistivity - a material property - and sizing).

They together determine the current in the circuit.

Your idea of changing the resistance to change the potential difference does assume an equal current, which is usually not what happens.

Check out the Drude model. The explanation for Ohm's law in it is that the electrons that make up the electric current are slowed down by kicking against the material atoms. The electrons are accelerated by the external potential difference (which equals a force). Think of the potential difference calculated for the resistor as the force balancing that external force according to Newton's actio est reactio (to obtain a steady current, an equilibrium between forces accelerating the electrons and those slowing them down).

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