Energy of an electron in an electric circuit

[Trollfaz]

When an electron flows through a circuit say 250V, does it mean the electron possesses 250eV at any time or does it give out a total of 250eV

 

[DaveE]

When an electron flows through a circuit say 250V, does it mean the electron possesses 250eV at any time or does it give out a total of 250eV

"give out a total of 250eV" ?
I'm a bit confused about potential energy vs. kinetic energy here. Also what is your reference for potential energy. So, let's compare the electron that leaves the - side of the battery to one entering the + terminal. The electron at the - terminal does have 250eV of potential energy wrt the + terminal. That is the potential difference between those locations. If It could jump "through" (across?) that potential without any interactions, then when it arrived at the + terminal it would have 250eV of kinetic energy and no potential energy left. But you said it "flows through a circuit", there is undoubtedly resistance in that circuit, or if you prefer interactions with other charged particles. That will slow the electron down and make heat, it will lose its energy as it flows through the resistive circuit. In that case it arrives with essentially no kinetic energy with all of the original energy left behind in the resistor as heat. Halfway through the resistor the potential energy will be one half of the original PE. It the other case, where it moves through a vacuum, at the halfway point it will have 50% PE and 50% KE, 125eV each.

 

[Dale]

When an electron flows through a circuit say 250V, does it mean the electron possesses 250eV at any time or does it give out a total of 250eV

Yes, to both. But it is best to think of the energy flow as flowing through the fields, not the charge carriers.

 

[tech99]

If the current in a circuit is everywhere the same, the electron must travel at constant speed through the circuit. Its KE does not then alter as it passes through a resistor.

 

[anorlunda]

If the current in a circuit is everywhere the same, the electron must travel at constant speed through the circuit. Its KE does not then alter as it passes through a resistor.

That's correct. But it is also useful to point out that electron volts (ev) is often used as a unit of velocity for electrons.

But I think the bigger and unfortunately common misunderstanding of the OP is that electrons are like capsules of energy and become "used up" in the circuit.

Nor is it helpful to mention that energy is carried in the fields outside of the wire. As I see it, the OP question has to do with energy consumption, not transmission. With a resistive load, the energy is dissipated in the resistor, and it is the resistor that gets warm first, not the surrounding air.

@Trollfaz, electric power is calculated as P=V*I. It takes both a voltage drop across the load and a current through it to deliver power. Current alone won't do it.

[By the way, if you think that P=I²R means that V doesn't matter, the equation is better written as P=V*I=(I*R)*I .]

Electrons are neither consumed nor depleted in a circuit. Kinetic energy of electrons contributes almost zero of the energy delivered.

 

[hutchphd]

Also how do you know it is the "same" electron.??
I'm with @Dale...the energy is in the field(s) as we have recently discussed endlessly here.

 

[anorlunda]

Also how do you know it is the "same" electron.??
I'm with @Dale...the energy is in the field(s) as we have recently discussed endlessly here.

Our disagreement is pedagogy, not physics.

The OP used the word circuit. The rules of CA circuit analysis, assume no magnetic fields and no electric fields. Therefore, it muddies up the waters for a beginning to introduce field effects into a circuits discussion.

[It is possible to contrive devices to emulate L and C in a circuit without being physical inductors or capacitors. We still use the same CA tools to analyze the circuit nevertheless. CA is a mathematical tool, not a physics simulation.]

If we do consider fields, the tools are Maxwell's Equations, not CA. Beginners almost never start with fields because the math is more difficult. And when beginners do begin with fields, they typically analyze isolated charged particles, or individual components like an inductor or capacitor, not complete circuits.

A wire transmits electricity (via external fields, i.e. Poynting vector), but it also dissipates energy in I²R losses, which are internal to the wire. It is customary to simplify by assuming R=0 so there is no dissipation portion of the energy.

With a load resistor being fed by a battery in a circuit, it is customary to neglect energy transmitted past the resistor, and assume that 100% of the energy is dissipated inside the resistor.

Opposite, conventions 180 degrees apart for the wires and the resistors in the circuit. These conventions (assumptions) are seldom stated explicitly, especially when trying to answer a simple question simply.

So in a circuit with both wires and resistors, it becomes exceedingly complex to be physically correct in all details. That's exactly why CA is so popular; it is a simplification.

All those nuances are far to much for a beginner who doesn't understand the basics yes. So I am opposed to introducing advanced concepts until the student learns the basics. Pedagogy, not physics.

 

[Dale]

The OP used the word circuit. The rules of CA circuit analysis, assume no magnetic fields and no electric fields.

Well, electrons aren’t part of circuit theory, so the question itself isn’t clearly only a circuit-analysis question.

 

[hutchphd]

All those nuances are far to much for a beginner who doesn't understand the basics yes. So I am opposed to introducing advanced concepts until the student learns the basics. Pedagogy, not physics.

Your point is certainly well taken, but I have heard similar arguments for the "water pipe" analog.
One should be very circumspect here: I know I had to unlearn a lot of wrong physics in this realm because of simple wrong teaching. There is much bad information promulgated, and some of it is truly detrimental. I am reminded of the snafu with Walter Lewin and Kirchhoff Laws.

 

[vanhees71]

Our disagreement is pedagogy, not physics.

It's obviously physics too!

The OP used the word circuit. The rules of CA circuit analysis, assume no magnetic fields and no electric fields. Therefore, it muddies up the waters for a beginning to introduce field effects into a circuits discussion.

This is of course a very superficial point of view. It's true after the laws for circuit theory are derived, you don't care about fields anymore and it's simply used to calculate the circuit, but for a physicist the point is not so much to derive a bunch of recipies to calculate circuits but to understand how these rules are derived from the fundamental principles, i.e., Maxwell's equations, and there the electromagnetic field and the charge-current distributions are the fundamental entities you have to work with.

Indeed, what happens in a metallic conductor is, in the most simple semiclassical point of view, that you have a (degenerate!) gas of conduction electrons in a background of a posively charged lattice of ions (in the most simple "jellium model" you can smear it as a homogeneous charge background). When an external field is applied the conduction electrons move under the influence of the Lorentz force (including electric and magnetic fields, where usually in the non-relativistic limit, which applies here very well, given the fact that in house-hold circuits the electrons crawl with a speed of the order of 1mm/s). The conduction electrons are scattering with the lattice (due to its thermal motion and/or other "defects"), and the friction coefficient is closely related to the electrical conductivity. This makes up the classical Drude model of conduction in a metal modified by Sommerfeld to take into account degeneracy.

To answer the question, how the energy is transported from the source ("battery") along the circuit you need the fields, and as the long debates in these forums show, it's surprising that indeed the fields transport the energy. There's almost no energy in the electrons (remember the drift velocity is on the order of 1mm/s only!).

[It is possible to contrive devices to emulate L and C in a circuit without being physical inductors or capacitors. We still use the same CA tools to analyze the circuit nevertheless. CA is a mathematical tool, not a physics simulation.]

If we do consider fields, the tools are Maxwell's Equations, not CA. Beginners almost never start with fields because the math is more difficult. And when beginners do begin with fields, they typically analyze isolated charged particles, or individual components like an inductor or capacitor, not complete circuits.

A wire transmits electricity (via external fields, i.e. Poynting vector), but it also dissipates energy in I²R losses, which are internal to the wire. It is customary to simplify by assuming R=0 so there is no dissipation portion of the energy.

This is no simplification but a big trouble. You cannot describe a superconductor by simply setting $R=0$, and that's the only physical situation where you can say with some right that you have "zero resistance".

With a load resistor being fed by a battery in a circuit, it is customary to neglect energy transmitted past the resistor, and assume that 100% of the energy is dissipated inside the resistor.

Which of course is a good approximation as long as electromagnetic radiation can be neglected, which is the case when the quasistationary approximations are applicable, i.e., when the typical wavelength of the em. waves, $c/f$, are large compared to the typical extension of the circuit.

Opposite, conventions 180 degrees apart for the wires and the resistors in the circuit. These conventions (assumptions) are seldom stated explicitly, especially when trying to answer a simple question simply.

So in a circuit with both wires and resistors, it becomes exceedingly complex to be physically correct in all details. That's exactly why CA is so popular; it is a simplification.

All those nuances are far to much for a beginner who doesn't understand the basics yes. So I am opposed to introducing advanced concepts until the student learns the basics. Pedagogy, not physics.

Well yes, but with only using the end result of the analysis, i.e., Krichhoff's rules, you have no chance to understand the basics ever!

 

[jbriggs444]

If the current in a circuit is everywhere the same, the electron must travel at constant speed through the circuit. Its KE does not then alter as it passes through a resistor.

This is not correct.

The current in a circuit is a rate of charge flow. It is not the velocity of the elemental charges. Just like in rivers or pipes, one can have fast and narrow stretches and slow and wide stretches while maintaining the same flow rate through each.

The net velocity of a charge carrier in a typical circuit is very low. Far lower than its thermal velocity. So you have this electron (for instance) that is jiggling and bouncing back and forth randomly at 10,000 meters per second or so while its "drift velocity" averages perhaps some few millimeters per second.

If the resister is uniform in geometry and electron mobility is constant throughout then yes, the drift velocity should be constant from one end of the resistor to the other.

The mental image that I sometimes use for electric flow is that of an incredibly incompressible fluid that is pushed by insanely high pressures so that one can transmit meaningful amounts of energy with very low flow rates. Hydrostatic drive on steroids. The kinetic energy of the flow does not enter in.

 

[alan123hk]

When an electron flows through a circuit say 250V, does it mean the electron possesses 250eV at any time or does it give out a total of 250eV

From what I can tell, Electronvolt seem to be generally used to describe electrons accelerated in a vacuum, and less often to describe the flow of electrons in circuits. If we insist on applying the concept of Electronvolt to circuits, of course each electron gains 250eV of energy as it flows from positive to negative.

https://en.wikipedia.org/wiki/Electronvolt

 

[DaveE]

each electron gains 250eV of energy as it flows from positive to negative.

gains and also probably loses, assuming there are things to bump into.

 

[alan123hk]

gains and also probably loses, assuming there are things to bump into

Yes, as it is a closed loop circuit, electrons can be considered to gain or lose 250eV of energy depending on the point of view of different observers

 

[jbriggs444]

From what I can tell, Electronvolt seem to be generally used to describe electrons accelerated in a vacuum, and less often to describe the flow of electrons in circuits. If we insist on applying the concept of Electronvolt to circuits, of course each electron gains 250eV of energy as it flows from positive to negative.

https://en.wikipedia.org/wiki/Electronvolt

Suppose that you have a switch connected to the negative terminal of your power supply and ten meters of wire running from the switch to an incandescent light bulb. The other side of the light bulb has another ten meters of wire running back to the positive terminal of the power supply.

The switch is open for now. In effect, the "negative" wire is only connected on the light bulb end. It floats at the positive terminal's voltage.

Using the positive terminal of the power supply as our voltage reference, all of the electrons in the wire on the positive side are at 0 eV. All of the electrons in the wire on the negative side are also at 0 eV.

Close the switch.

Before any of the electrons have moved appreciably, the electrons already in the wire on the negative side are suddenly at 250 eV potential. It was not because a new batch came out of the power supply with 250 eV potential and filled up the wire, replacing those old tired 0 eV electrons. Instead, it was because the field in the wire changed to have a 250 V potential.This does not contradict what you are saying. But it does suggest that the model of electrons coming out of the negative terminal with potential energy and then losing it as they pass through the circuit misses something.

 

[alan123hk]

This does not contradict what you are saying. But it does suggest that the model of electrons coming out of the negative terminal with potential energy and then losing it as they pass through the circuit misses something.

Or try looking at it from another angle. In the system in the image below, the electrons apparently gain 3eV of energy when passing through the battery and lose 3eV when passing through the light bulb.

 

[David Lewis]

D

If the current in a circuit is everywhere the same, the electron must travel at constant speed through the circuit. Its KE does not then alter as it passes through a resistor.

Current by itself doesn't tell us anything about the speed of the electrons. If we hold current constant, then average drift velocity is inversely proportional to the product of charge carrier density and cross-sectional area. Either or both of these values may be different for the resistor than for the wires to which it is connected.

 

[alan123hk]

Current by itself doesn't tell us anything about the speed of the electrons. If we hold current constant, then average drift velocity is inversely proportional to the product of charge carrier density and cross-sectional area. Either or both of these values may be different for the resistor than for the wires to which it is connected.

Different materials have different carrier densities, but if for the same material and constant cross-sectional area, then a constant current should represent a constant electron drift velocity.

 

[David Lewis]

Absolutely, but sometimes the resistor is made out of carbon, and the wires are made out of copper, for example.

 

[vanhees71]

D

Current by itself doesn't tell us anything about the speed of the electrons. If we hold current constant, then average drift velocity is inversely proportional to the product of charge carrier density and cross-sectional area. Either or both of these values may be different for the resistor than for the wires to which it is connected.

The current density is related to the velocity of the charge carriers by $\vec{j}=\rho \vec{v}$, where $\rho$ is the charge density and $\vec{v}$ the velocity field of the charged matter. This is the most fundamental (and thus also most simple) local description of moving electric charges. Charge conservation in local form is described by the continuity equation,
$$\partial_t \rho + \vec{\nabla} \cdot \vec{j}=0.$$
A current is a global concept. It is defined as the flux of charges through a surface $A$,
$$I=\int_A \mathrm{d}^2 \vec{f} \cdot \vec{j},$$
it's the amount of charge per unit time streaming through the surface. The sign is determined by both the direction of $\vec{j}$ and the choice of the surface-normal vectors, i.e., the orientation of the surface.

 

[alan123hk]

ρ is the charge density and v→ the velocity field of the charged matter

I wish to ask a conceptual question.

I used to think that $\vec{v}$ should be the velocity of charged matter, in general circuits it is the drift velocity electrons. So what is the difference between the velocity field of charged matter and the velocity of charged matter? Are they equivalent? Can I simply call it the velocity of charged matter?

Thanks

 

[vanhees71]

The velocity/flow field refers to the Eulerian description of fluid dynamics; $\vec{v}(t,\vec{x})$ is the velocity of the fluid cell at time, $t$, that is located at the position $\vec{x}$. That's why the "material time derivative" of a field is $\mathrm{D}_t=\partial_t + \vec{v} \cdot \vec{\nabla}$.

 

[alan123hk]

Thank you for the explanation.