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As far as I know electric potential is a characteristic of a point in space and the difference in electric potential (voltage) is the difference between the electric potential of two points.

I was also taught that the electric potential of a point with respect to a charge depends on the charge and the distance from the charge. Also the electric potential of a point due to a system of charges is the sum of the electric potentials caused by every single charge of the system.

Now, in an electric circuit I assume that the Voltage of a battery is caused by a spot where there are more negative charges (negative terminal) and a spot where there are more positive charges (positive terminal). Since the electric potential as discussed above depends just on the system of charges and distances, I expected that the voltage would change just in base of the distance between a terminal, but Ohm's Law states that the voltage changes also because of the characteristic of the conductor (Resistance).

I can think of this drop as the potential energy transformed into heat because of the collisions with the conductor particles, but I find this contrasting with the definition of voltage I was given.

Can anyone help me to solve this doubt?

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  • $\begingroup$ It depends on the circuit. Typically, batteries produce a (roughly) fix voltage difference on their output, thus your initial assumption doesn't stay in the most simplest, most general case. Maybe you could insert a circuit picture, to make your question more clear. $\endgroup$
    – peterh
    Aug 8, 2017 at 22:25
  • $\begingroup$ Could you please include paragraphs in future posts, people will stick with your question longer? ( and uppercase I for "I think" etc)? Thanks $\endgroup$
    – user163104
    Aug 8, 2017 at 22:28
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    $\begingroup$ I try to understand it as a model connecting $J=\rho v$ to the force $F=\rho E + \rho v \times B$, one could complicate the model by adding convection or diffusion I think. $\endgroup$
    – Emil
    Aug 8, 2017 at 22:56
  • $\begingroup$ @Emil Thanks, it is what I was searching for. Correct me if I'm wrong. If we have a resistor with an higher potential to the right and lenght L, the work required to move a negative charge q from right to left considering just the magnitude is $W = E * q * L + F * L$. $F$ is a force that models the collisions of the electrons with the resistor particles, so it depends on the charge q and the material of the resistor. In Ohm's Law, the drop (gain if you consider the electrons' path) of voltage($W/q$) is modelled using the $R$ term. Is this correct? $\endgroup$
    – Giorgio Buttiglieri
    Aug 9, 2017 at 9:59
  • $\begingroup$ @Giorgio Buttiglieri: Sorry, I do not have that much insight. $\endgroup$
    – Emil
    Aug 9, 2017 at 10:02

2 Answers 2

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but Ohm's Law states that the voltage changes also because of the characteristic of the conductor (Resistance).

Actually, Ohm's law does not state that. Ohm's law is just a relationship between - in this case - 3 parameters: Resistance $R$, current $I$ and voltage $V$.

$$V=RI$$

But while it is the relationship between them, it does not state which ones are dependent and will change when another one changes. That depends on the situation.

In the case of a battery connected to a simple circuit, we know that the voltage is constant. Regardless of the resistance in the circuit, the voltage is always, say, 5 V across the two battery terminals. This is true when not connected (open-circuit, effectively infinite resistance), when connected (some specific value of resistance) and when short-circuited (effectively zero resistance).

The voltage is constant and doesn't depend on the resistance that happens to be in the way. So, when looking at Ohm's law, what is then changing, if not $V$? Mathematically, you are right that something else must change when $R$ changes. But it doesn't have to be $V$ - it can also be $I$. And it is.

Think of $V$ as the "pressure" that "pushes" on charges.

  • In the open-circuit case, they are being pushed with 5 V, but they still can't move because there is no conducting path - infinite $R$ but zero $I$. Ohm's law obeyed.
  • In the connected case, they are being pushed still with 5 V, and they flow with whichever current $I$ that fulfills Ohm's law. $R$ limits and alters $I$, not $V$.
  • In the short-circuit case, they are still being pushed with 5 V and this time nothing stops them. They speed up and up and up. At any flow speed (current), the 5 V pushes them faster to gain higher speed, which continues forever (or until the heat generated melts the wire). $R$ is zero and $I$ infinite - Ohm's law obeyed.

In general, be careful when reading a mathematical formula. It only shows a relationship between parameters - it doesn't show which of them that are dependent and which that are fixed. That depends on the particular situation.

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It appears that OP is confused about the voltage powered by a battery and the electric potential (voltage) due to a system of charges which depends just on charges and distances.

The voltage developed across a cell's terminals depends on the energy release of the chemical reactions of its electrodes and electrolyte$^1$.

Another important concept is conductivity, wires and conectors are ideally modeled as perfect conductors. With this wires you can make terminals to conect an object (resistor, diode, led, etc). So if you change the distance between this terminals the electric potential between both points does not change, this electric potential is caused by the negative charges in one terminal and a the abscence of negative charge (absence of electrons) in the other terminal. This charges are free to move in the wires and metal, remember that they are perfect conductors. When you change the distance between terminals, this free charges move and the electric potential is always the same.

Now if you conect an object to those terminals and it is an ohmic device, then the Ohm's law states $$I=\frac{V}{R}$$ If you change the resistance then what will change is the current through the conductor: $$I_1=\frac{V}{R_1} \ \ \ \ I_2=\frac{V}{R_2}$$

because you are not changing the batery.

  1. https://en.wikipedia.org/wiki/Battery_(electricity)
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