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Series circuit with 3 resistors and 3 voltmetersIn my physics class, we have begun talking about potential difference, charge, resistance etc. While watching online videos for extra help on some clarification, I often hear that as electrons move through a resistor they lose some of their electric potential energy, creating a difference in voltage (v1 greater than v2). I noticed than in series when multiple resistors have been connected the resistors with higher resistance (ohms) have a greater voltage.

Does this voltage mean how much electric potential energy was lost, or something completely different? I know that potential difference refers to how much work a coulomb of charge can do when travelling from high to low potential, but how does this principle relate to before and after an encounter with one or more resistors in series?

Essentially, I don't really understand how potential difference functions across multiple resistors in series, and why a voltmeter produces the numbers that they do in a logical manner (not mathematically).

I have included an example picture to further show my problem:

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  • $\begingroup$ In a series circuit, the resistors with higher resistance have a higher voltage DROP. Your volt meter is actually measuring $\Delta V$ rather than $V$. $\endgroup$ Commented Jun 5, 2020 at 15:01

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Does this voltage mean how much electric potential energy was lost,

  • Each point in a circuit corresponds to some amount of potential energy $U$. This energy is a measure of how strongly a charge is being repelled from or attracted to that point (the energy represents the effect of the electric forces).
  • On a per-charge basis, we usually simply call it a potential $V$. So, the electric potential associated with a point is the potential energy per charge, $$V=U/q$$
  • A charge will only move, if it is more strongly repelled from one point than from another. Meaning, if there is a potential difference. Two points at equal potential show equal amounts of repulsion/attraction, and thus the charge will have no tendency to move from one to the other. The difference in potential is such a crucial idea that it has been given its own name: voltage, $$\Delta V=V_2-V_1=\frac{U_2}{q}-\frac{U_1}{q}$$

Oftentimes voltage is not symbolised with $\Delta V$ but simply with $V$, just as the potential is. Still, always remember that voltage is nothing more than the difference in potential.

A voltage drop across a resistor is another way of saying that there is a different potential on either side (a better term would have been "potential drop", since "voltage" already implies a drop. But anyways...). We can think of the potential as the "pressure" which squeezes electrons through this tough resistor. As pressure in a hose when you bend it in the middle. Surely, it takes more "pressure", if the resistor is larger. So, yes, the voltage across a resistor does show how much potential energy per charge that was lost/used/spent/absorbed as the charged passed through the resistor.

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