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The electric field created by a battery through a section of wire in a series circuit is constant. This means that the force on each electron as it travels through the wire is constant.

According to Newtonian mechanics, the total work done on an electron by the electric field as it moves from position $A$ to position $B$ is the change of that electron's kinetic energy of between position $A$ and position $B$.

We have 2 resistors - Resistor $A$ and Resistor $B$.

$R_A>R_B$

The resistor $A$ is at a distance of $0.2 m$ from the negative pole of the battery and $0.8 m$ from the positive pole. The resistor $B$ is $0.5 m$ from the negative pole and $0.5 m$ from the positive pole.

Then the kinetic energy gained by the electron as it travels between the two resistors must be greater than the kinetic energy gained by the electron as it travels from the negative pole to resistor $A$, since the distance from the negative pole to $A$ is less than the distance between the two resistors.

So the kinetic energy loss at the first case must be less than the kinetic energy loss in the second case.

How is this possible?

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  • $\begingroup$ Electrons do not move in the wire like a sphere would move in space. They are continuously colliding. See drift velocity $\endgroup$ – Eagle Mar 3 at 19:09
  • $\begingroup$ In quantum mechanics, the concepts are explained by considering the Fermi velocity of the electron. $\endgroup$ – KV18 Mar 3 at 19:30
  • $\begingroup$ @Natasha had the right idea. I tried elaborating below. Hope it helps! $\endgroup$ – Joshua Ronis Mar 3 at 19:35
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Short answer:

You're making an incorrect assumption: electrons don't gain kinetic energy as they travel through a section of wire - not from the negative pole to $A$, and not from $A$ to $B$. As they travel through any section of wire, they gain some kinetic energy due to the electric field, but then they smash with an atom and lose it all, then they gain some again until they smash with another atom and lose it all again. And this repeats over and over Overall, however, their average kinetic energy (and thus average velocities) stay constant through any section of wire. You can prove this to yourself by seeing that the current is constant throughout a single series section of wire.

Longer answer:

First of all, look at your first two lines (with a few edits just to make them more grammatically pleasing, at least compared to the way in which they were first written):

The electric field through a closed series circuit is constant. This means that the force on each electron moving through the circuit is constant. According to Newtonian mechanics, the total work done to the electrons as they move from position A to position B is the difference in the electrons' kinetic energy at position $B$ and position $A$.

I'm gonna stop you there. Think about this: if what you're saying is true, and there indeed WAS a change in the kinetic energy of the electrons between "position $A$ and position $B$" in the circuit, that would mean electrons would be moving faster at position $B$ than at position $A$ (assuming they're moving from $A \rightarrow B$), since they gained kinetic energy.

Is that what truly happens?

The current throughout a circuit is constant, so the electrons CAN'T have more kinetic energy at position $B$ than at $A$. A change in the velocity of the electrons would imply a change in the current through the circuit. Since the current is constant throughout a series circuit, we know the electrons can't have sped up - they can't have gained Kinetic Energy.

The work done on the electrons by the electric field doesn't increase the kinetic energy of the electrons - it simply counters the work done on the electrons by the atoms within resistors, atoms that the electrons are constantly smashing into. The work done by the resistors on the electrons cancels out the work done on the electrons by the battery, causing the electrons to maintain a constant velocity through the circuit.

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  • $\begingroup$ I know what drift velocity is. Drift velocity is the average velocity of one electron through the circuit . With no resistance applied to the circuit,the path an electron follows is the same with how the shape of the wire. $\endgroup$ – Max Destiny Mar 3 at 20:14
  • $\begingroup$ In case of superconductor there is no work done by collisions with atoms on an electron so the electrons are accelerated. $\endgroup$ – Max Destiny Mar 3 at 20:16
  • $\begingroup$ Usually quantum mechanic is used for electrons inside the atom where the electrons move randomly within specific energy levels. $\endgroup$ – Max Destiny Mar 3 at 20:18
  • $\begingroup$ If electrons dont move then why is there the meaning of current and drift velocity? $\endgroup$ – Max Destiny Mar 3 at 20:51
  • $\begingroup$ Electrons DO move, but their kinetic energy isn't increasing as they move through a resistor. They're only gaining kinetic energy between collisions with atoms in the resistors - then they lose it all after they collide, and gain kinetic energy (speed up) again before colliding with the next atom. $\endgroup$ – Joshua Ronis Mar 3 at 21:03
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The electric field inside that a battery creates is constant so this means that the force on each is electron is constant.

The electric field in a battery is not constant. As charges are moved around by the electrochemical potential and by the circuit, the field may be larger or smaller.

According to Newtonian mechanics the total work of the force from position A to position B is the change of kinetic energy of between position A and position B.

The total work of all forces, not any single force. As an example, we can use forces on an elevator cable to lift the cab from the ground to the roof. But the kinetic energy doesn't increase because of gravity is pulling down at the same time. Similarly, the battery can provide a force to move charges forward, but forces from resistors may prevent the kinetic energy from increasing.

At steady-state, all losses in a circuit are of potential energy, not kinetic energy.

Please do not explain this by using the Ohm law. Resistance is caused by collisions of free electrons with atoms . Kinetic energy is lost not potential energy.

Kinetic energy is only lost if nothing else is replacing the energy. We can set up a block that slides down a ramp at a constant speed. In this case, energy is being lost as friction turns it into heat. But no kinetic energy is lost. Instead the loss due to friction is exactly balanced by the gain from gravity. So we say that potential energy is lost (as the block falls), not kinetic energy.

The same thing happens in the circuit. Within the resistor, energy is lost to collisions. But the electrons gain the same amount as they fall to a lower energy level in the electric field. So the velocity of the electrons remains constant.

And how is potential energy lost since the electric field remains the same?

The electric field defines the gradient of the potential. Any charge that is moved by an electric field is necessarily losing electric potential energy. Just like a mass moved by the gravitational field is losing gravitational potential energy.

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  • $\begingroup$ Please do not explain this by using the Ohm law. Resistance is caused by collisions of free electrons with atoms . Kinetic energy is lost not potential energy. $\endgroup$ – Max Destiny Mar 4 at 8:56
  • $\begingroup$ And how is potential energy lost since the electric field remains the same? $\endgroup$ – Max Destiny Mar 4 at 9:01
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There is no flow of Electrons. The energy is transferred from one valance electron to the adjacent atoms valance electron, and so on.

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  • $\begingroup$ There can be a flow of electrons depending on the model you consider. It's not about what model is more accurate, but about how a simple model really works. $\endgroup$ – FGSUZ Mar 3 at 19:33
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There are a few common misconceptions that need to be addressed, and I will give you a QM explanation:

Electrons that move in a conductor are not free. They are loosely bound to the valence shells of the atoms.

Now these electrons, as they gain kinetic energy, from an external field, move from one atom's valence shell to the other atom. This is called drift velocity and is very slow. How can then electricity move almost at light speed?

It is because electrons in the conductor are so densely packed, they are shoulder to shoulder and as the first one moves, it will move all the other ones. It is basically the energy of the external field, that transforms into the kinetic energy of the first electron, and that kinetic energy gets passed by as the electrons move.

Electrons move in empty space inbetween the atoms' valence shells. As they move inbetween the atoms in empty space, they do move close to light speed. Why is the drift velocity then slow? It is because of the interaction that the electrons have with the atom, that takes time.

AS the electron gains kinetic energy from the external field, it gets excited, and if the energy gained is at a certain level, the electron gets kicked off the valence shell, and moves to another atom's valence shell. If there is more energy coming from the external field, the electron can get excited again, and get kicked off again and move to another atom again. But even with one move, it is because the electrons are so densely packed, that the first electron moving pushes the other electrons and so the speed of electricity will be close to light speed.

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  • $\begingroup$ Extremely bad, wrong answer. The electron-electron interaction is usually negligible. The E field only affects less than one free electron per billion in average and a not so wrong model can indeed consider the outer electrons as free, forming a Fermi gas. $\endgroup$ – thermomagnetic condensed boson Mar 4 at 10:23

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