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The Drude Model helped me quite a lot to visualize how current could flow in a circuit. However, there is still a point that I cannot grasp in the explanation given by some people when they talk about energy. When considering a round trip of one electron along a circuit, it is usually said that :

The electron gains energy when going through the battery and loses that energy along the wire as it collides with the lattice of the conductor.

My problem with this reasoning is that it is implying that the electron has more energy when it leaves the battery than anywhere else in the circuit. However, according to the Drude model, when the electrons collide with the lattice, they indeed lose kinetic energy but they regain kinetic energy as soon as they start accelerating thanks to electric fields produced along the wire. This causes electrons to have an average drift speed. In other words, their kinetic energy is constant on average, which is at first glance, in contradiction with the reasoning given above. What am I missing here?

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    $\begingroup$ The energy is in the electric field, not in the electrons. All the electrons in the wire are doing is to "move" an electric potential from one place to another. The conductors act as a boundary condition to Maxwell's equations and that's how they allow us to shape the electromagnetic field, but the internal mechanism doesn't actually transport the energy, it merely converts some of it into heat. $\endgroup$
    – CuriousOne
    Apr 6, 2016 at 16:04

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The electron gains potential energy in the battery; this is transformed to kinetic energy, which from time to time gets dissipated in the inelastic collisions with the lattice.

It's quite analogous to a (semi-elastic) ball jumping down some stairs (and then, in the battery, taking the elevator to get up again). This might be more intuitive.

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The battery sets up an electric field in the external circuit which all the mobile electrons feel in all the external circuit. This means that there is a force on all the mobile electrons in the external circuit.
So these mobile electrons are accelerated by the electric field and gain kinetic energy from the electric field which is maintained by the battery.
The accelerated electron then interact with the vibrating lattice ions and on average there is a transfer of energy from the mobile electrons to the lattice ions is the ions vibrate more.
After the interaction with the lattice ions the mobile electrons are accelerated again thus gaining kinetic energy, interacting with the lattice ions etc.
So on top of the random thermal motion of the mobile electrons there is a drift velocity away from the negative terminal of the battery and towards the positive terminal of the battery as they gain kinetic energy from the electric field and then lose some to the lattice ions.
Within the battery the chemical process moves electrons from the positive terminal to the negative terminal thus giving the electron electric potential energy from the chemical energy of the reacting chemicals in the battery.

The key point is that outside the battery all the mobile electrons drift round the circuit from the negative terminal to the positive terminal of the battery. They continuously lose electric potential energy, gain kinetic energy and then lose the kinetic energy on colliding with the lattice ions which vibrate more - the conductor's temperature is increased which is called "ohmic heating".

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  • $\begingroup$ I totally agree with that. My next question is : Can you say that this following explanation is right? : "The electron gains energy when going through the battery and loses that energy along the wire as it collides with the lattice of the conductor." I feel like this last one is wrong or at least badly formulated because it can lead to different interpretations. It doesn't tell that kinetic energy is regained and it assumes that potential energy is only gained in the battery although the electric field is uniformely distributed along the circuit, right? $\endgroup$
    – Dory
    Apr 6, 2016 at 12:33
  • $\begingroup$ Electrons whilst passing through the battery do gain electric potential energy and electrons in the external circuit do lose electric potential energy however they do not convert their electric potential energy into the kinetic energy all in one go it is a continuous process all around the circuit. Think of many masses sliding down a slope with the frictional force such that they move down the slope at constant velocity. Gravitational potential energy is continuously being converted into heat. Then at the bottom of the slope the masses are lifted up and allowed to slide down the slope again. $\endgroup$
    – Farcher
    Apr 6, 2016 at 12:41
  • $\begingroup$ Better still the slope has alternate friction free stripes and friction stripes so arranged that the average velocity of the masses between adjacent pairs of friction free and friction stripes down the slope is constant. $\endgroup$
    – Farcher
    Apr 6, 2016 at 12:55
  • $\begingroup$ Your last analogy is just perfect :D . I totally understand it know, thanks :) $\endgroup$
    – Dory
    Apr 6, 2016 at 13:19
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1) Battery constantly moves electrons from the positive terminal to the negative terminal creating non-zero positive and negative charges at the respective terminals. These net non-zero charges create electric field inside the conductor (E=V/d).

2) This electric field inside the conductor accelerates electrons which from time to time collide with the lattice loosing their energy.

3) So, all in all, chemical energy of battery is transformed, first into kinetic energy of electrons, and secondly into heat.

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