# What type of energy is transfered in a circuit?

1. Feb 10, 2010

### p.tryon

In an electrical circuit the electrons kinetic energy is transfered to other forms of energy at the components. True or false? If it is false what type of energy do the electrons transfer?
Thanks

2. Feb 10, 2010

### Staff: Mentor

False. Electrons transfer electrical energy. Electrical energy has nothing to do with kinetic energy in electrons. Electromagnetism is one of the four fundamental forces and electrical energy is due to moving charges with that force.

3. Feb 10, 2010

### p.tryon

Thanks for that ;-)

4. Feb 10, 2010

### bjacoby

So by your definition of fundamental forces, no such thing as "kinetic energy" exists!

And that implies that the energy expressed when large masses smash into each other is simply electromagnetic transfers and radiation. Well, that is obviously true, but then how does one explain the term "kinetic energy"? Just perhaps, electromagnetic energy just might in some way be related to the effects that have been defined in some circles as "kinetic energy"?
That is rather a different answer than "has nothing to do with kinetic energy" isn't it?

But I take it what you are really saying is that energy in an electrical circuit is transferred in ways different from the simplistic "kinetic" model of ball bearings going down a hose. I'd agree with that.

5. Feb 10, 2010

### Staff: Mentor

No, russ is correct. The point is that electrons are so light and their drift velocity is so small that there is never a significant amount of energy in the form of kinetic energy. It is not that kinetic energy does not exist, simply that it is not relevant for electrical circuits.

Here is an example I worked out a while back in a similar thread:
https://www.physicsforums.com/showpost.php?p=1668323&postcount=27

Last edited: Feb 10, 2010
6. Feb 10, 2010

### Staff: Mentor

Good point - Kinetic energy isn't really related to one of the fundamental forces, so that part of the sentence is somewhat misleading.

7. Feb 11, 2010

### Repainted

I'm not quite sure how this works. Isn't the reason that electrons possess low average KE because they've transferred most of their KE to other forms in the components?

For example, in the absence of components(hence no or negligible resistance), an electron traveling through a circuit with a 6V potential across it would gain a KE of 6 eV, which is a quite a large amount of KE for an electron. Hence its maximum drift speed would be 1.03 x 10^6 m/s, and assuming it starts at rest and is uniformly accelerated through the circuit, its average drift speed would be half of that, 5.13 x 10^5 m/s. This agrees with the fact that a circuit with near zero resistance would have almost infinite current flowing through it, since I = nAve and all.

Because of this I concluded that the electrons lose their KE within the components of a circuit converting it to heat and light or whatever the components do, causing them to move at a much lower average drift speed.

Is this reasoning flawed?

8. Feb 11, 2010

### Born2bwire

I would say yes because outside of DC circuits, the energy in a circuit is transferred via electromagnetic waves. If we have an AC signal, then the charges have a zero net displacement. In addition, the propagation of the signal (the group velocity, not just the phase velocity), is much much higher than the drift velocities of the charges. As such, the energy cannot be transferred via kinetic energy. The mechanism for drawing out energy from the circuit in elements like a resistor do rely on the kinetic energy of electrons as they rely on the Drude model to transfer the energy from the charges to heat. However, that is not the mechanism for all devices (like an antenna) nor is it the mechanism for how the energy is transferred between components.

9. Feb 11, 2010

### Repainted

So what happens to the KE that the electrons are supposed to gain? The E-field produced by the battery exerts a force on the electrons, causing them to accelerate and hence gain KE.

Even in an AC circuit, while the electrons have no net displacement, the direction of the force due to the AC source is changing as well, hence it always(I think?) points in the same direction as the displacement at anytime?

10. Feb 11, 2010

### Born2bwire

In an AC circuit, you are not driving the charges. Instead, you excite an electromagnetic wave that you input into the circuit. The electromagnetic wave induces currents in the circuit's waveguide. These are the currents that we measure, but they are not directly created by the source but rather by the electromagnetic signals. When the wave reaches a component, it sets up a voltage potential that induces the movement of carriers in the component. In addition, most components will also allow the wave to migrate through the component (IC chip).

The kinetic energy of the electrons is given up by their collisions with the conductor's lattice. These phonons transfer the charges' energy into heat in what is known as the Drude model. Still, this is unsatisfactory for how energy is extracted for all but the most basic circuit elements (resistors). This does not explain how energy is stored in a capacitor by building up a potential difference, how antennas work by accelerating charges to excite electromagnetic waves, or how signals are propagated between circuit components when we are not discussing true DC circuits.

11. Feb 11, 2010

### Repainted

I am confused by the AC part. Isn't an AC produced by a coil of wire placed within a uniform magnetic field and rotated to change the magnetic flux through it, inducing a potential? Or is this a simplified explanation of how an AC generator works that doesn't talk about EM waves and such?

Also, if this Drude model is inaccurate at describing how energy is transferred to most components, then how is energy transferred from the moving electrons to the components? Or is it something that varies for every component?

12. Feb 11, 2010

### Phrak

Potential Energy. Specifically electric potential energy, eV.

13. Feb 11, 2010

### Staff: Mentor

It is flawed. Let me give you a completely mechanical example about energy that will hopefully help you understand about energy.

Let's say that we have 2 masses attached by a light rope with a pulley arranged to pull the second mass up as we move the first mass to the left (see http://farside.ph.utexas.edu/teaching/301/lectures/node48.html" [Broken]). Now, suppose we want to lift the second mass to the top of the table, but can only directly do work on the first mass. We have two basic options:

1) we can start with a little slack in the rope, suddenly do work on the first mass, accelerating it very quickly to a high KE, let it snap the line taut, and pull the second mass up by exchanging its KE for gravitational PE.
2) we can start with no slack in the rope, do work on the first mass, which will be transferred directly to the second mass via the rope with no significant amount of KE needed to effect the transfer.

Remember, work is force times distance, so as long as you can exert a force on something you can do work on it, regardless of KE. True, you can use KE to do work, but it is not necessary, and true if you don't have some other interaction then when you do work on something it will gain KE. But neither of those statements imply that you must use KE to do work nor that the first mass had a significant amount of KE in option 2) at any time.

So, in the case of electrical circuits, the rope is equivalent to the electromagnetic forces, the first mass is equivalent to the electrons, and the second mass is equivalent to whatever the electrons do work on. As I showed in my previous calculation there is never any significant amount of KE in the electrons, the electromagnetic "rope" is always taut and the energy which the power source supplies to the electrons is transferred directly to the circuit without gaining and losing KE.

Last edited by a moderator: May 4, 2017
14. Feb 11, 2010

### sophiecentaur

I think the AC factor is a red herring. The (mean) KE of the electrons is constant DC and varies in AC but it is negligible in both cases.

But the notion that the energy is transferred in the form of electronic KE is like saying that the energy transmitted through a bicycle chain is due to the KE of the links in the chain. That KE would only contribute (briefly) a bit of drive if you took your feet off the pedals!
(edit: this is more ore less what dalespam is saying - I think I was writing whilst his post went up)
In the case of a stream of electrons, the KE gained during acceleration would be transferred, on impact with the collector. If you apply Kirchhoff 2 to the two situations - the PD across a copper supply wire is very small- the majority of the PD is dropped across the load. In the case of an electron beam, the majority PD is across the acceleration gap and a small bit at the surface of the metal in the collector. In the first case, the energy transferred to the load is in the form of electrical. In the second case, the energy transferred to the collector is KE.

15. Feb 11, 2010

### Repainted

Yeah I sort of meant something like that. Not exactly colliding and stuff. More like an electron moving against another force while still under the force of the E-field. Like that the block on the table, if there was no load attached to it, it would have gained more KE, but work has to be done to gain GPE of the load, it doesn't end up with as much KE as it should.

So negative work is done on the electron while it is moving through the component, it does positive work on the component causing heat or light to be given off or whatever the component does while positive work is still being done on it by the E-field due to the battery... Is this more accurate?

16. Feb 11, 2010

### Born2bwire

How the original electromagnetic wave is generated is immaterial. For low frequency signals we can be rather basic with the generation of the wave. A simple generator works by physically rotating magnets about stationary wires. The changing magnetic field automatically creates an electromagnetic wave and the wires act as the waveguides to propagate the wave in the desired direction. At normal AC frequencies associated with residential electricity, the wavelength is so long that no real thought needs to be given in terms of designing a waveguide. It is not until we start going to higher frequencies like RF that we need to pay closer attention to waveguide design and start thinking about other ways of generating the signal. But the physics never change.

The Drude model is the classical model for normal ohmic losses. Whenever we lose energy due to the resistance of the device, it is typically the Drude model at work. However, outside of a resistor, we are not interested in converting the electrical energy into heat. As such, the mechanisms for extracting power vary from component to component. For example, a transistor works by first setting up a voltage difference between the desired terminals. This voltage difference then induces the movement of carriers within the transistor. Energy is spent to move these charges so that we build up voltages on the outputs, this can be lost due to phonon collisions akin to the Drude model but we can also emit electromagnetic radiation like with an LED. An inductor stores energy by inducing a magnetic field or locally constrained electromagnetic waves. And as I mentioned before, an antenna can work by providing a way to move the electromgantice wave from a guided mode to a propagating mode. The current side to this theory is that the accelerating currents in the antenna, say like in a simple dipole antenna, will gnerate the electromagnetic waves. It is the acceleration that matters, not their velocity or collision rate (which would affect the losses predicted by the Drude model). We can also extract power from the fields by coupling with another inductor to induce a current in the other inductor, like what we do in a simple transformer.

Energy can be extracted by the electromagnetic fields that are created by the induced currents, from the electromagnetic fields of the signals itself, or from quantum mechanical behavior of the charges themselves (dropping down from an excited state to a lower state and thus emitting a photon). Since heat itself is very rarely a desirable output from a circuit outside of a heating element, most desirable work is done via different mechanisms.

Last edited: Feb 11, 2010
17. Feb 11, 2010

### Repainted

Okay I think this sort of clears it up. Thanks a lot.

So to sum it up, the collisions in the Drude model are simply useful in portraying power output due to resistance, but for everyday uses like my computer for instance, it depends on how the individual circuitry is designed to extract power for the power source. Am I right to say this?

Oh and are these 'collisions' in the Drude model really collisions? or as DaleSpam showed with an example, the E-field causing the electrons to simply doing work against another agent continuously and not some haphazard movement through the circuitry?

18. Feb 12, 2010

### Born2bwire

More specifically the Drude model is how we classically explain resistance and drift velocity with the energy given up by the electrons to the lattice dissipated as heat. Yes, the physics behind how a component dissipates energy varies by component to component. All of them will have some amount of ohmic losses via the Drude model but generally we are uninterested in generating heat.

I do not feel that this is in conflict with what DaleSpam has been stating. To start, the force acting on the charges in the circuit are the electromagnetic fields. It is true that if an electron came in close contact to another charge, the forces from their fields would cause attraction or repulsion. Here the latter could be thought of as a collision that is mediated by the electromagnetic force (to be truthful, every collision that we normally talk about is mediated by this force). However, that is not to say that the reason why charges flow in a circuit is because they bump into each other. Instead, we set up an electric field throughout the circuit that acts on all of these charges. This is a static electric field in a DC circuit and an electromagnetic wave in an AC circuit. This field is the taut rope that DaleSpam references.

As these charges move along inside the lattice of the conductor, they will collide with the lattice and cause phonons; vibrations in the lattice. These phonons dissipate their energy in the form of heat. These collisions, governed by the statistical rate of their occurance, will dictate the overall drift velocity and resistance of the bulk conductor. So, as these charges move in response to the applied fields, they will give up kinetic energy to the lattice via these collisions as dictated by the Drude model. The transference of course requires that we take energy out of the applied fields to reaccelerate the charges. The amount of energy given up in a good conductor is very very low, as calculated by DaleSpam previously and as evidenced by the fact that we normally treat a good conductor as having zero resistance. Finally, on the whole we treat this as a statistical bulk behavior of the charges. There are so many charges that are moving and the bulk is so large that what seems like a very random phenomenon takes on a statistically consistent behavior at macroscopic levels. So that is why if any of us take draw out a piece of copper wire of similar dimensions and use similar sources we will end up with the same resistances and currents.

19. Feb 12, 2010

### Repainted

Oh I see. Its because fundamentally there are only 4 forces, so am I right to say that when we talk of 'collisions' taking place, they are simply some electromagnetic forces between electrons and the particles in the material which affect the electrons motion? Sorry for reiterating what you stated, but I'm doing so in my own terms to ensure I got it right.

One last question, for energy losses not related to heat loss due to resistance, can we still use Ohm's law and such to calculate current and power? Or can we still use Ohm's law, but the resistance here is not the same 'resistance' in terms of the Drude model, and is simply a measure of that component ability to produce power given a certain Voltage across it?

20. Feb 12, 2010

### Born2bwire

Yes, there are collisions going on, however they are not so much with actual particles but with the entire atomic lattice that makes up the conductor's bulk. The odds that the charge hits the particle directly is very low. However, the influence is extended due to some of the local fields of an atom or molecule. In the case of the lattice, the atoms or molecules that make up the constituents of the bulk will have bonds between them. These bonds are modified electron orbitals of the atoms. A very crude way of saying it would be that electrons are shared between the atoms in the lattice. This creates additional fields that the charges can "collide" with. In addition, there are the standard Van der Waals-London forces which are the typical bulk forces between molecules.

Suffice to say at least, that the conductor bulk presents a region of varying electrodynamic potentials that are a bit more complicated than a bunch of point source charges. As the charge moves throughout the bulk, it is influenced by these potentials, causing it to deflect slightly in some cases or to reflect completely backwards in other cases. In the end we just characterize the macroscopic bulk behavior.

We can still characterize an effective resistance to describe the power draw of various components even if they are not dissipating energy exactly like an ohmic loss. For example, antennas will often be described by a conductor loss resistance and a radiation loss resistance. The conductor loss is the power lost due to the ohmic heating that we have been discussing. The radiation resistance is the effective power that is radiated away by the antenna. As long as the power dissipation behaves ohmic, that is it follows the V=IR relationship, we can effectively model it as a resistor. However, this resistance maybe frequency dependent, like with an antenna, or it may be only valid for a range of voltages. For example, your skin is what we call a non-ohmic resistor. For dry, unbroken skin with low voltages the resistance is very high. But eventually you can apply a voltage high enough that the skin breaks down and the resistance drops very dramatically. Another example would be a diode where it has a very high resistance below the turn-on voltage and very low resistance above it. In these cases we often simply model the different situations with different circuit models. So a diode that is in the turn off is modeled as an open circuit or a very large resistor and a diode that is in the turn on is modeled as a resistor and a voltage source in series.

But bringing this all back to the question of the OP. Are these collisions, that is the transferrence of the kinetic energy of the charges to the circuit, the means by which the energy is transported to components. The answer is no, the fields that are applied under which these charges react are the means for propagating power throughout a circuit. In the context of how power is used in a component, that is dependent upon the component itself.