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In the following video clip at 2:10,

http://www.youtube.com/v/YslOUw5oueQ ,

Professor Walter Lewin talks about a misconception people have that the energy going through a wire to a resistor is in the form of kinetic energy of electrons. He proves this cannot be so as follows. The current density is J = I/A = Vne where V is the drift velocity (or average velocity), n is the number of electrons per volume, and e is the charge of an electron. A is a cross-sectional area of the wire.

We can make A as large as we want (keeping the current constant), and therefore V will have to become very small, and the electrons will have very little kinetic energy. Yet the resistor (say a light bulb) dissipates the same amount of power P=(I^2)*R. Therefore, it must be that the form of energy is not the kinetic energy of electrons.

My first question is, if we make A larger why does it have to be that V goes down? Perhaps n goes down - we increased the volume (by increasing the cross-sectional area), so there should be fewer electrons per volume?

My second question, my main question is, if the energy is not the kinetic energy of the electrons, what does in fact bring energy to the resistor and how does it heat up?

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my main question is, if the energy is not the kinetic energy of the electrons, what does in fact bring energy to the resistor and how does it heat up?

We assume steady state operation.

The drift velocity of the electrons entering the resistor must equal the drift velocity of the electrons leaving the resistor. This follows from the fact that the current into the resistor equals the current out of the resistor.

However, the electrons leaving the resistor have less potential energy than those entering the resistor. This follows from the fact that there is an electric field through the structure of the resistor and, thus, there is a potential difference between the ends of the resistor.

The electric field within the resistor structure accelerates (does work on) the electrons increasing their kinetic energy however, this energy is quickly given up to the structure of the resistor via collisions; the resistor gets hotter.

On a more fundamental level, the energy flows from the battery to the resistor through the space around the conductors via the electromagnetic field. See, for example, William Beaty's description of energy flow in a simple circuit here.

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Yes, it is the electron kinetic energy gained between collisions with the lattice atoms which is transferred to the lattice atoms who heats the lattice. After collision the electron changes it direction essentially, so the drift velocity is small whereas the instant velocity is high.

The drift velocity has nothing to do with the temperature. In absence of $V$ there is no drift velocity at all, but the resistor can be hot. The external field moves the electrons and cannot move the lattice, so it is the electron subsystem, which receives energy from the external field.

If you make the area $A$ large, then the resistance will go down.

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  • $\begingroup$ But, once again, Doesn't Walter Lewin clearly state that the energy is not in the form of kinetic energy of electrons. thanks. $\endgroup$ – user35687 Oct 30 '14 at 21:15
  • $\begingroup$ I did not watch the video, but his explanation operates with the drift velocity, not with the true kinetic energy. $\endgroup$ – Vladimir Kalitvianski Oct 31 '14 at 6:53

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