What type of energy is transfered in a circuit?

by p.tryon
Tags: circuit, energy, transfered, type
 P: 51 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
 Mentor P: 22,302 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.
 P: 51 Thanks for that ;-)
P: 132
What type of energy is transfered in a circuit?

 Quote by russ_watters 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.
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.
Mentor
P: 17,330
 Quote by bjacoby 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?
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:
http://www.physicsforums.com/showpos...3&postcount=27
Mentor
P: 22,302
 Quote by bjacoby So by your definition of fundamental forces, no such thing as "kinetic energy" exists!
Good point - Kinetic energy isn't really related to one of the fundamental forces, so that part of the sentence is somewhat misleading.
 P: 35 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?
PF Gold
P: 1,777
 Quote by 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?
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.
 P: 35 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?
 Sci Advisor PF Gold P: 1,777 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.
 P: 35 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?
P: 4,512
 Quote by 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
Potential Energy. Specifically electric potential energy, eV.
Mentor
P: 17,330
 Quote by Repainted Isn't the reason that electrons possess low average KE because they've transferred most of their KE to other forms in the components? ... Is this reasoning flawed?
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 Fig 30). 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.
 Sci Advisor Thanks PF Gold P: 12,186 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.
 P: 35 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?
PF Gold
P: 1,777
 Quote by 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?
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.
 P: 35 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?