What carries the energy in an electric current?

[Jano L.]

Does the voltage drop because the moving electrons collide with other particles in the conductor, and transfer some of their kinetic energy to them (manifested as heat)? This would make sense to me, but evidently the energy of an electric current is not the kinetic energy of the electrons.

The potential drops down along the wire because the potential is determined by the battery or other source of voltage that the wire is connected to. Transport of kinetic energy by the electrons is negligible. The work done (the heat evolved in the wire) by the electric current far away from the energy source (power plant) is done at the expense of energy of electromagnetic field at this place; most of this energy comes there from outside the wire (The Poynting vector points towards the wire slightly in the direction of the current).

 

[alva]

Particle accelerators, cathode ray tubes, etc.

I think the OP is talking about normal/simple circuits.

In a wire that is carrying a constant current "I" there is an energy stored in the kinetic energy of the moving electrons. But this current does not change, nor the kinetic energy, so you does not have to take it into account into any formula.

But what happens in the transient time when you switch on the circuit ? Where does the energy needed to speed up the electrons come from ? There is an amount of energy stored in the magnetic field of the current: 1/2*I(exp 2)*L. But this energy can be recovered when you switch off the circuit.

What formula takes into account the very small kinetic energy of the electrons ?

Please, no water analogies.

 

[Dale]

I think the OP is talking about normal/simple circuits.

Agreed, which is why I specifically responded about normal circuits until you directly asked about the exceptions.

But what happens in the transient time when you switch on the circuit ? Where does the energy needed to speed up the electrons come from ?

The energy comes from the battery, or generator, or whatever source is supplying energy to the circuit.

What formula takes into account the very small kinetic energy of the electrons ?

The standard KE formula 1/2 mv² still applies. Just since m and v are so small the KE is also small.

 

[alva]

What formula takes into account the very small kinetic energy of the electrons ?

I meant "electrical formula" such as V = I* R, P = V * I, W = 1/2*L*I²

The standard KE formula 1/2 mv² still applies. Just since m and v are so small the KE is also small.

When you switch on a circuit with a battery and an inductance ALL the energy that the battery supplies is stored in the magnetic field, not in the kinetic energy of the electrons.

Does the inductance L of the circuit include the kinetic energy of the electrons ? But L only depends on the geometry of the wires.

 

[sophiecentaur]

Does the voltage drop because the moving electrons collide with other particles in the conductor, and transfer some of their kinetic energy to them (manifested as heat)? This would make sense to me, but evidently the energy of an electric current is not the kinetic energy of the electrons.

Electricity will never "make sense" when you try to explain it in terms of anything concrete and mechanical. Why do you think it could? Charge is just another quantity which is not part of the simple, mechanical world. This whole thread is full of comments from people who seem to image in there is no need to treat electricity as anything different or special. Some things in life are not just familiar or intuitive and they never will be. There is no short cut here. If there were, then many (clever / brilliant) people would not have gone down the road that they have done.

 

[Dale]

Does the inductance L of the circuit include the kinetic energy of the electrons ?

That is an excellent idea. I hadn't thought of that before, but the KE would be proportional to I^2 and the inductance energy is also proportional to I^2, so you could model the KE as a very small addition to the inductance.

When you switch on a circuit with a battery and an inductance ALL the energy that the battery supplies is stored in the magnetic field, not in the kinetic energy of the electrons.

... But L only depends on the geometry of the wires.

The "effective inductance" including the KE would be imperceptibly larger than the "true inductance".

 

[Naty1]

I'm with sophiecentaur here [post#35]. focus on the electricity.

What carries energy, electric potential, in an electric current are the individual electrons.


Each moves rather slowly and is imparted a potential difference from the [rapidly moving] electric field propagation. Each electron carries the electric potential required to break it free and allow drift.

These individual electrons are what move along in lock step in a typical conductor...and resistor...loosely bound electrons in a conductor, more tightly bound electrons in a resistor...'More tightly bound' means a stronger local bond [more Coulomb attraction] to an atom ...so it takes more potential [electrical energy] to dislodge them.

So as you move a voltmeter further and further apart over the length of a resistor [or conductor] to measure voltage you encounter more and more displaced electrons, each with a tiny electric potential difference, and they each add up. And of course some of the energy is dissipated as heat, so called I²R losses.

 

[alva]

There is no short cut here. If there were, then many (clever / brilliant) people would not have gone down the road that they have done.

A very scientific argument.

 

[alva]

The question is still unanswered. Which electrical formula takes into account the kinetic energy of the electrons ?

 

[Dale]

None. But $E=1/2 \; LI^2$ could be adapted to take it into account.

 

[Naty1]

KE = 1/2mv

²

KE is utterly inconsequential in an electrical circuit...lost somewhere beyond ten decimal places...or maybe 8 or 12...you can calculate the driftspeed ' v 'from here:

http://en.wikipedia.org/wiki/Electron_current#Drift_speed


and look up th mass of an electron...and figure out how many are in your circuit.

 

[Dale]

KE is utterly inconsequential in an electrical circuit...lost somewhere beyond ten decimal places...or maybe 8 or 12

Agreed. It might be a good exercise for alva to calculate the "effective inductance" for a straight copper wire and see how small the KE contribution is.

 

[sophiecentaur]

A very scientific argument.

So you don't subscribe to the existence of cleverer minds than your own? That's a risky attitude. You can only afford to disagree with established Science when you actually understand it sufficiently.

Wouldn't the scientific approach be to follow the evidence?

 

[OmCheeto]

I'm still trying to figure out what "how to carry energy" means.

Can someone translate the OP's question into an equation, please.

 

[sophiecentaur]

I'm still trying to figure out what "how to carry energy" means.

Can someone translate the OP's question into an equation, please.

An excellent point, there.
But things are even worse than that; some perfectly satisfactory answers that have used Maths have largely been ignored and 'convincing' arm-waving answers have been demanded instead.

 

[vanhees71]

It's an interesting question, that imho can't be answered with hand-waving arguments but only by a careful analysis (more careful than in some textbooks!) of the solution of the Maxwell equations (for the simplifying case of time-independent fields and sources).

A very nice paper on the subject, solving the problem for a toroidal conductor (with an infinitely thin battery :-))

J. A. Hernandes and A. K. T. Assis. Electric potential for a resistive
toroidal conductor carrying a steady azimuthal current. Phys. Rev. E. 68, 046611 (2003).

 

[Naty1]

I'm thinking that if an analogy for insights is required, maybe there are better ones than water...
Solid state material electronics [semiconductors] sure had a load of theoretical current formulas when I studied it many years ago...that is likely one source for analogies.

[I still rememeber the author ,Mohammed Ghausi; google: oh my gosh, he is STILL associated with UC Davis!...and some of his books from the 1960's are listed on Amazon! ]

Here is the first analogy I looked at:
Photovoltaic effect
http://en.wikipedia.org/wiki/Photovoltaic_effect

When the sunlight or any other light is incident upon a material surface, the electrons present in the valence band absorb energy and, being excited, jump to the conduction band and become free. These highly excited, non-thermal electrons diffuse, and some reach a junction where they are accelerated into a different material by a built-in potential (Galvani potential). This generates an electromotive force, and thus some of the light energy is converted into electric energy. ...

Besides the direct excitation of free electrons, a photovoltaic effect can also arise simply due to the heating caused by absorption of the light. The heating leads to an increase in temperature, which is accompanied by temperature gradients. These thermal gradients in turn may generate a voltage through the Seebeck effect. Whether direct excitation or thermal effects dominate the photovoltaic effect will depend on many material parameters.

 

[Naty1]

Oh, I missed an obvious choice for insights into energy and circuits:
superconductivity. especially perhaps high temp superconductivity.

That surely offers some interesting insights into the E and B fields in conductors.

 

[scoobmx]

The electric and magnetic fields carry the energy, but the movement of electrons is influenced by these fields. I have always thought of the kinetic energy of the electrons as mechanical energy, that become heat when they scatter on impurities (the phenomenon of resistance).

 

[DrZoidberg]

Imagine you put a table inside a big solenoid. On that table you place a magnet and put a heavy object on top to make sure the magnet can only move slowly.
Then you turn on the solenoid so that the magnet moves slowly over the table at a !constant! speed.
Because of friction, the table and the magnet will become warmer. You could claim that the kinetic energy of the magnet was converted to heat. But you could also say that the energy of the field was converted to heat in a relatively direct manner.

 

[scoobmx]

Imagine you put a table inside a big solenoid. On that table you place a magnet and put a heavy object on top to make sure the magnet can only move slowly.
Then you turn on the solenoid so that the magnet moves slowly over the table at a !constant! speed.
Because of friction, the table and the magnet will become warmer. You could claim that the kinetic energy of the magnet was converted to heat. But you could also say that the energy of the field was converted to heat in a relatively direct manner.

Yes but the question was what carries the energy in current. The part lost to resistance is no longer part of the current.

 

[DrZoidberg]

I guess that means the part lost in the device you are running is also no longer part of the current?
You don't lose current in a resistor. The current stays the same, what you loose is potential.

 

[sophiecentaur]

The electric and magnetic fields carry the energy, but the movement of electrons is influenced by these fields. I have always thought of the kinetic energy of the electrons as mechanical energy, that become heat when they scatter on impurities (the phenomenon of resistance).

Have you not read all the preceding posts that tell you the amount of KE carried by electrons is as near zero as makes no difference? Resistive wire can get pretty hot. Where would the energy come from if it relied on electron KE?

You could prove this for yourself if you work out the actual mass of all the conduction electrons in a wire and then multiply it by v²/2 , with mean v = 1mm/s.

 

[scoobmx]

Have you not read all the preceding posts that tell you the amount of KE carried by electrons is as near zero as makes no difference? Resistive wire can get pretty hot. Where would the energy come from if it relied on electron KE?

You could prove this for yourself if you work out the actual mass of all the conduction electrons in a wire and then multiply it by v²/2 , with mean v = 1mm/s.

That 1 mm / s is the drift velocity, not the actual velocity. Even at 0 drift velocity electrons have kinetic energy that approximately follows the equipartition theorem (hence 'electron gas').

If truly electrons carried the energy in the circuit then when you flip a light switch you would have to wait for the 1 mm / s electrons to crawl to the light bulb. But we all know it's nearly instantaneous. It's the electromagnetic field that conveys the energy.

 

[sophiecentaur]

That 1 mm / s is the drift velocity, not the actual velocity. Even at 0 drift velocity electrons have kinetic energy that approximately follows the equipartition theorem (hence 'electron gas').

If truly electrons carried the energy in the circuit then when you flip a light switch you would have to wait for the 1 mm / s electrons to crawl to the light bulb. But we all know it's nearly instantaneous. It's the electromagnetic field that conveys the energy.

I thought we'd been there before / knew that. That post of mine that you are quoting is a (several times) repeated reply which tried to knock the electron KE idea on the head. It was just a suggestion to try it out for himself.