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Say we have a simple circuit with three resistors in series, each of which have the same resistance. We also have a battery with a given constant potential difference hooked up in series with these resistors.

According to Ohms law and Kirchhoff's law, the current passing through each of the resistors will be

  1. Determined by the sum of the resistances of the resistors (V/(Total R) = I)

  2. Constant, as there is only a potential difference drop for each resistor in series.

So the amount of charge passing through each resistor is always constant, just like water flowing through pipes.

But the "pressure" on these electrons decreases- meaning that the amount of energy each electron has is reduced, part of their energy converted into energy for the element that the resistor represents to use (electrons doing work.)

So logically, this should mean that the speed of each electron is reduced, and that extra energy they once had in kinetic energy is the energy used on the element.

So if this is the case, how is it possible that Current stays constant? Yes, by the definition Q/t current is constant because the number of electrons passing through a point is the same, but by the more involved definition I=nqAV, current is reduced because the drift velocity must be reduced, or else no work could be performed.

Where is my intuition wrong?

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It’s better not to think about individual electrons in macroscopic circuits, so you can avoid confusions like this. But since you asked:

The electrons don’t move the entire distance in one sweeping move. That’s your intuition, but it’s not what happens.

Rather, electrons in resistors move a tiny distance, pick up a bit of energy & speed from the electric field, then scatter and dump all that into the atomic lattice as heat. They do this over and over and over again Their average speed in each little step is about the same. The steps at the top of the resistor chain happen the same as those at the bottom hence have the same speed.

To an electron, gaining a picovolt and then losing it at the top, high potential end of the resistor chain looks just like gaining and losing a picovolt at the bottom: speed up a little, thump to a stop, repeat. The absolute potential (voltage) at which that happens doesn’t matter.

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  • $\begingroup$ I think my question is about the macro-view of a basic circuit system: If there is a higher potential difference at one point in the circuit, the force present on the electrons is greater than the force applied at a different point in the circuit at which potential difference is lower, meaning they should carry more energy and move faster or slower, and current is dependent on that. Is the argument that "on average" all electrons move at the same speed throughout a circuit with one type of conductor? $\endgroup$ Aug 4 '19 at 21:39
  • $\begingroup$ “Higher potential difference” than what? All an electron is affected by is the electric field where it is. The potential has no effect on electron movement, only the potential difference right around where it is, which is the E field. $\endgroup$ Aug 4 '19 at 22:46
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"extra energy they once had in kinetic energy"

That may be why you're confused. The electrons don't start off with lots of kinetic energy which they gradually lose. They start off, having passed through the battery, with lots of electrical potential energy. If they had a 'clear run', in other words if they didn't keep colliding with the vibrating ions in the metal, they would get faster and faster, in other words they would steadily lose electrical potential energy and gain kinetic energy. But because they do keep bumping into ions they pick up kinetic energy and very soon lose it. The process happens over and over again, so within a few nanoseconds of completing the circuit they reach a constant mean 'drift' speed.

There is quite a good analogy with very small raindrops falling from a cloud. These start off with plenty of gravitational potential energy, but rather than getting faster and faster as they fall, they quickly reach a constant terminal velocity (owing to collisions with air molecules).

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  • $\begingroup$ Thank you- that analogy at the end is quite good. So to conclude, yes, they do gain kinetic energy that is quickly lost, but the forces and conversion of energy for the electrons that happens due to the potential gradient isn't black and white at different points in the circuit- the load of resistance all plays into the way electrons move through the circuit as a whole, and the drift velocity is the mean velocity that is a result of contributions from all the resistors. Which is why the current equation uses Drift velocity. $\endgroup$ Aug 4 '19 at 21:44
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current is reduced because the drift velocity must be reduced, or else no work could be performed.

Electrical Work can be performed without reducing drift velocity (and thus without reducing current). Consider the analogy of pushing a box on a floor with friction. You can exert a constant force to push the box at constant velocity if your force equals the opposing kinetic friction force. If the force you apply is F and you push the box a distance d the work you do is Fd. But the energy you put into the box is dissipated as friction heat.

It's similar with current and resistance. Work is done in moving charge at constant drift velocity through a resistor. The energy is dissipated at resistance heating.

Hope this helps.

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