# Reconciling the drift velocity with Joule heating at high and low voltage

I'm trying to reconcile the particle behaviour in an electric field on the micro level with the conductor and the current properties on the macro level.

As far I understand - with resistance being constant an increase in voltage causes the power to increase and current flow to decrease. Also, an increase in voltage causes an increase in the drift velocity of the electrons.

The Joule heating happens because the drift velocity and electron-ion collisions cause the part of the energy of the electric field to be converted into the kinetic energy of the ions, and thus into thermal energy of the conductor.

To reduce these heat losses we increase the voltage applied to the conductor using transformers on the entry to a power line, and on the exit from it, like shown in this simple experiment.

On the macro level, the transformer increases the voltage applied to the conductor, and that allows using a thin wire that would overheat and melt when the same power would be drawn through it under lower voltage.

But surely on the micro level the increase in voltage will increase the electrons' drift velocity and should increase the average number of collisions with ions and the total amount of energy transferred through those collisions, not decrease it. And given that the same conductor is used on low and high voltage - the total amount of electrons moving through the conductor would not change with the change of voltage, only their speed of movement.

So why does the Joule heating decreases when voltage goes up when it seems to have to increase?

increase in voltage causes the power to increase and current flow to decrease. Also, an increase in voltage causes an increase in the drift velocity of the electrons.

You seem to be talking about power transmission. In the rationale for using the high voltage there, the voltage that is getting increased isn't the one between two fixed points on a single line. It is the voltage between two different lines that is getting increased.

By increasing this voltage between two different lines, the current $$I$$ flowing through the lines is decreased (because consumers drawing the same power cause smaller current to flow) and because losses are proportional to $$I^2$$, the losses are decreased. At the same time, since current is decreased, voltage between two fixed points on single line also gets decreased.

So by increasing the voltage between the lines, we decrease voltage between the ends of a single line (which would be zero ideally, but is not in practice and we want to minimize it).

• Erm, not sure I follow. As far as I understand - this is a single circuit, from the power source, through transformers and one thin wire, to the consumer, from the consumer, back through transformers and another thin wire and back to the power source. So when you say "between the lines" - what parts of the circuit you mean? Btw, the "power source" here is a mains outlet and there is a whole other world behind it, but I assume that's not material for this question. Tell me if I'm wrong here. Commented May 27, 2021 at 7:15
• Also, I'm looking to understand what happens on the particle level within the thin wire under those different conditions and how that connects to the macro level. Commented May 27, 2021 at 7:15
• It is not one electrical circuit, there are at least two (and sometimes even more): there is the high voltage circuit (the transmission), and there is the low voltage circuit. The latter is galvanically isolated from the former. The word cluster "voltage between two different lines" in my answer means difference of potential between two overhead lines that transport power over long distance. Commented May 28, 2021 at 13:54
• In the second example from Practical Engineering, the high voltage is between the two thin lines connecting the transformers. The voltage between ends of a single line is very low, in fact lower than it would be in the first example without increasing voltage. So increasing voltage there makes both current in the line and voltage between its lends lower. This means the drift velocity in the thin high voltage lines is lower. Commented May 28, 2021 at 14:01

The answer and discussions above don't really go beyond school physics course territory, and that does not provide the answer to my question at all.

I finally did find data that helped me formulate the answer I'm happy with in this youtube video. My explanation fits all the data I have. Here it is.

The energy is not transferred at all by the electrons moving around, the energy flow is created when there are both electric and magnetic fields present between two points, and it is the energy flow that we really mean when we talk about electricity. This is what's stipulated in that video that is different and contrary to the school physics explanations of electricity.

The energy flow needs both electric and magnetic fields to be present at the same time. The electric field by itself is not sufficient. E.g. if you take a well-charged AA battery - there is an electric field constantly present between its cathode and anode, but without a magnetic field to accompany it - the energy is not flowing, so the battery does not discharge and no work is done.

The magnetic field is created by electric charges moving within an electric field. The charges need to be moving to create a magnetic field, so once we create conditions where charges can move - we allow for energy to start flowing. I imagine that "charges" here can really be any kind of objects with a positive or negative electric charge, but the most common charged objects we use are electrons in a material that allows them to freely move - metal conductors.

When we apply an electric field to the conductor - the electrons in this conductor start moving. They start moving - and this creates a magnetic field around the conductor, and this magnetic field together with the electric field that was there already allows for power to flow from one point to another.

So in fact - the moving electrons themselves are not transferring any power whatsoever, the only reason we need them moving is to create a magnetic field. Moving electrons are a means to an end, but not a vehicle for power transfer in itself.

Btw, this also nicely explains the alternating current - if electrons are not really transferring the power and they only have to move somehow - it doesn't matter in what direction they are actually moving, and they can just jiggle around back-n-forth in an AC circuit. They still move, the field is created, and power is flowing.

Now, as for the original question - why increasing voltage with constant power decreases current and decreases the side effects of the current on the conductor?

I'm making a further leap here to accommodate for that. Extending on the idea that the two fields are necessary and sufficient for transferring power - I think the fields don't have to be of the same intensity to achieve that.

Generally, the stronger the fields - the more power will be transferred. The voltage is directly measuring the intensity of the electrical field between two points. The current is measuring the amount and speed of moving charges, so it indirectly measures the magnetic field intensity.

Movement of the charges creates negative side-effects we want to minimise (energy loss through heat and conductors melting). So we have to minimise the movement of charges and thus - the magnetic field. We need some magnetic field still, but don't want much of it.

And to compensate for that - we at the same time increase the electric field. The huge electric field and small magnetic field will transfer the same amount of power overall as the average-size magnetic and electric fields together. But creating a small magnetic field will create fewer unwanted side effects.

This is why a cable at high voltage will heat up less than at low voltage. There is no relation to drift velocity at all in here. We will increase the voltage and decrease current by placing resistance on the way, and that will transfer all the power we need and not melt the cable.