[Physics] Is the same amount of electricity used for ceiling and table fans on different speeds

electricityspeed

We use ceiling and table fans in home which are can be set to low or high speeds using regulators. I want to ask that does it uses or consumes same amount of electricity at different speeds? Here, "same amount of electricity" means exactly I want to know that, my electricity bill will be different for different speeds or will it be same. I know that, when it has less speed, less electricity flows through it, but my friend told me that, when it is at less speed, the remaining electricity get wasted at regulator, so total electricity consumed by fan + regulator is same, so same electricity bill will be generated regardless on which speed we use it. So is my fried correct or wrong?

image of ceiling fan with regulator

Best Answer

I have checked the power consumption for various speed of fan.

It consumes $80 W$ at full speed $45 W$ at second speed and subsequently reduced at lower speed.

This I have measured from the Digital Electrical Meter.

Particle exchange force model is not causal

There is a flaw in your thinking, in that you are formulating the electromagnetic interaction in terms of photon emission and absorption and at the same time telling a story forward in time. These two ideas are both ok separately, but not together.

The particle emission/absorption picture is not a causal picture--- it requires that the particles go back and forth in time--- so you can't use causal language, like an electron emits a photon which kicks an electron etc. That's part of the story, but another part of the story is: an electron emits a photons which had already kicked an electron earlier, which emitted a different photon earlier than the first, etc, etc. If you go to a unitary Hamitonian causal picture, you renounce the idea that the field is due to particle emission and absorption (I only said unitary for a technical reason: it is conceivable somebody could make a nonunitary Hamiltonian formulation with unphysical photon polarizations which contain the Coulomb force, but then these unphysical photons would only be intermediate states, since the physical photons are not responsible for the Coulomb interaction anyway).

The acausality in the Feynman description is not a problem with consistency, because there are causal formulations of QED, one of which is Dirac's. Here the electrostatic repulsion is not due to photon exchange, but is instantaneous action-at-a-distance, while photons travel with only the physical transverse polarization. In Feynman's particle push picture, the electrostatic interaction is due to unphysically polarized photons travelling much faster than the speed of light, and these photons just are not present in Dirac's equivalent formulation.

Anyway, the best way to understand electron motion is using the classical electric and magnetic fields produced by the electrons.

It's not the electrons pushing

The electrons in a wire are not pushed by other electrons. They are pushed by the external voltage applied to the wire. The voltage is a real thing, it is a material field, it has a source somewhere at the power plant, and the power plant transmits the power through electric and magnetic fields, not by electron pushes.

The electron repulsion in a metal is strongly screened, meaning that an electron travelling along at a certain speed will not repel an electron 100 atomic radii away. In many cases, it will even attract that electron due to weak phonon exchange (this weak attraction gives superconductivity, and essentially all ordinary metals become superconducting at some low enough temperature).

You can completely neglect the interelectron repulsion for the problem of conduction, and just ask about external fields rearranging charges in the wire.

Fermi surface, not wire surface

The only electrons that carry current are those near the Fermi surface. The Fermi surface is in momentum space, it is not a surface in physical space. The electrons which carry the current are distributed everywhere throughtout the wire. But they all have nearly the same momentum magnitude (if the Fermi surface is spherical, which I will assume without comment in the remainder).

The behavior of a Fermi gas is neither like a particle nor a wave. It is not a wave, because the occupation number is 0 or 1, so that there is no coherent superposition of a large number of particles in the same state, but it is also not like a particle, because the particle is not allowed to have momentum states lower than the Fermi momentum, by Pauli exclusion. The particle is traveling through a fluid of identical particles that jam up all the states with momentum smaller than the Fermi momentum.

This strange new thing (new in the 1930s, at least), is the Fermi quasiparticle. It is the excitation of a cold quantum gas, and to picture it in some reasonable terms, you have to think of a single particle which is always required to move at faster than a certain speed, it cannot slow down below this speed, because all these states are already occupied, but its can vary its direction. It has an energy which is proportional to the difference in speed from the lower bound. This lower bound on the speed in the Fermi velocity, which in metals is the velocity of an electron with wavelength a few angstroms, which is about the orbital velocity in the Bohr model, or a few thousand meters per second.

The Fermi liquid model of dense metals is the correct model, and it supersedes all previous models. The speed of the current carrying electrons is this few thousand meters per second, but at longer distances, there are impurities and phonons which scatter the electrons, and this can reduce the propagation to a diffusion process. The electronic diffusion doesn't have a speed, because distance in diffusion is not propotional to time. So the only reasonable answer to the question "what is the speed of an electron in a metal?" is the Fermi velocity, although one must emphasize that an injected electron will not travel a macroscopic distance at this speed in a metal with impurities.

1.Presumably at this level, electrons are acting more like waves and less like particles, but is there any classical component in the picture, ie are electrons coming in imparting other electrons with kinetic energy through repulsion, or does it not work that way?

In order to using a time-ordered causal langauge (this does that, then this does that), you need electric and magnetic fields, not photons. The electrons are not what is coming into the wire to make it conduct, the thing that is coming in is an electric field.

When you switch on a light, you touch a high voltage metal to a neutral metal, instantly raising the voltage, and making an electric field along the metal. This field accelerates the electrons near the Fermi surface (not on the wire surface, those near the fermi momentum) to travel faster in the direction of (minus) the electric field E. It can only accelerate those electrons which can be sped up into new states, so it only speeds up electrons which are already running around at the Fermi velocity. These electrons keep moving until they build up enough charge on the surface of the metal to cancel out the electric field, and to bend the electric field direction to follow the wire wherever the wire curves. This causal propagation is Field-Electrons-Field, and the only electrons which serve to shunt the field are those which are building up charges on the surface of the wire (and the protons on the surface which also redirect the field where there needs to be positive charge)

When you apply a constant voltage, the electrons come to a steady state where they are carrying the current from the negative voltage to the positive voltage, making the voltage drops line up in space along the direction of the wire, no matter what the shape, and bouncing off impurities and phonons to dissipate the energy they get from the field into phonons (heat). The local electric field drives their motion, not their mutual repulsion. In that sense, it is not like water in a pipe. It is more like a collection of independent ball bearings pushed by a magnet, except that the ball bearings shunt the magnetic field to go along the direction of their motion.

2.If electrons momentarily have energy, then pass it on by a photon, what determines when that photon is emitted, and what frequency it will be? I assume that electrons in this cloud are not limited by any kind of exclusion principal, and that any frequencies are possible?

The electrons in the cloud are not only limited by exclusion, they are dominated by exclusion, this is the Fermi gas. It is not the electrons pushing other electrons, it is the field pushing the electrons. The photon particle-exchange picture is irrelevant to this, but if you insist on using it, then the photons are coming out of the wall socket, having followed the high-voltage wires from the power-plant in a back-and-forth zig-zag in time, and a negligible fraction of the photons are emitted by the conduction electrons, since all those photons are absorbed into phonons by the metal within a screening length.

The photons coming from the wall are bounced around by surface charges on the wire (static electrons and protons) so that they bounce around to follow the path of the wire in steady state.

3. Why should a photon emitted by an electron be in the direction of travel? Conservation of momentum tells me that if an electron is moving, the photon should be emitted in that direction, slowing the electron, but could an electron emit a photon in the opposite direction? If it did, I assume it would somehow have had to absorb energy from elsewhere? That sounds possible by analogy with quantum tunneling.

Photons are emitted in all directions, and back in time. It just is not useful to think of Feynman picture when you want to think causally.

4. What is the mechanism by which electrons propagating increase the temperature of the material? Are they transmitting energy to the electrons in the valence shell, which tug at the nucleus, do some photons hit the nuclei directly, or is there some other way?

So far, I have been treating the electrons as a gas of free particles. But you might be upset--- there are lots nuclei around! How can you treat them as a gas? Don't they bounce off the nuclei?

The reason you can do this is that a quantum mechanical particle which is confined to a lattice, which has amplitudes to hop to neighboring points behaves exactly the same as a free particle obeying the Schrodinger equation (at least at long distances). It does not dissipate at all, it just travels along obeying a discrete version of the Schrodinger equation with a different mass, determined by the hopping amplitudes.

In Solid State physics, this type of picture called the "tight binding model", but it is really more universal than this. In any potential, the electrons make bands, and the bands fill up to the Fermi surface. But the picture is not different from a free gas of particles, except for losing rotational symmetry.

If the lattice were perfect, this picture would be exact, and the metal would not have any dissipation losses at all. But at finite temperature there are phonons, defects, and a thermal skin of electrons already excited at a little more energy than the Fermi surface. The phonons, defects, and thermal electrons can scatter the conducting electrons inelastically, and this is the mechanism of energy loss. The electrons can also emit phonons spontaneously, if their energy is far enough above the Fermi surface so that they are no longer stable. All of these effects tend to vanish at zero temperature (with the exception of defects, which can be frozen in, but then the defects become elastic). But at cold enough temperatures, you don't go to zero conduxctivity smoothly. Instead, you tend to have a phase transition to a superconducting state.

5. Presumably, electricity travels slower than light, because there is some time in each exchange, and some time when electrons are moving at sublight speeds before emitting a photon. By how much is this slower than light, and what is the speed of each interaction?

This is again confusing Feynman description with a causal description. But I did this experiment as an undergraduate, and along a good coaxial cable, the speed was 2/3 the speed of light. I assume that if you use an ordinary wire in a coil on the floor, its going to be significantly slower, perhaps only 1% of the speed of light, because it requires more finnagling of surface charges for the wire to set up the field to follow it curves.

[Physics] Force required to drive car

In a perfect vacuum, on a frictionless road, you could just turn off the engine and the car would keep moving, never slowing down. However, in the real world, there are several effects that exert a force on a moving car, slowing it down, such as:

rolling drag between the tires and the road surface,
fluid drag from the air that the car moves through, and
various friction losses between moving parts in the car itself, which, unless compensated by engine power, cause the wheels to slow down and exert a torque on the roadbed slowing down the car.

To keep a car moving at a constant speed, the engine needs to exert enough force to balance all these forces.

The important thing to realize here is that, at high speeds, the main force slowing down the car is actually the fluid drag, which grows roughly in proportion to the square of the speed. Thus, to double the speed, the engine needs to exert four times as much force. (At least, that approximately holds at normal highway speeds. Things get even more interesting when you approach the speed of sound and wave drag starts to play a role.)

Because of this, minimizing the drag coefficient is a critical feature of high-speed automobile design, and is why essentially all modern cars (but especially high-speed models) feature streamlined shapes designed to minimize aerodynamic drag.

Also, as Sachin Shekhar notes in his answer, the rolling drag for pneumatic wheels is also somewhat speed dependent, mainly because the wheels are flexible, and thus deform as they rotate, losing energy as heat. These losses also increase with velocity, meaning that, even in vacuum, maintaining a higher velocity still needs more power. In principle, you could minimize these losses by making both the wheels and the road surface as hard and inflexible as possible — say, by making both out of steel, as is done for trains, which also minimize aerodynamic drag by their long and narrow shape. That's one reason why trains can travel at considerably higher speeds than would be practical for a car to maintain.

(Of course, to reduce drag even further, you could put wings on the car and have it fly high above the road, eliminating rolling drag entirely and reducing fluid drag significantly due to the lower air pressure in the upper atmosphere. Or, even better, go even higher up where the air is even thinner.)