What carries the energy in an electric current?

[Doug1943]

Electric current is often analogized to water under pressure in pipes.

When water is moving through a pipe, the energy of the system is the potential energy of whatever is giving the water its pressure -- a column of water, say -- and the kinetic energy of the water molecules as they move. Increase either their velocity (by increasing the pressure), or their total mass-per-unit-time passing a given point (by substituting a larger pipe for a smaller one, for example), and you increase the total energy they are carrying.

Now, water pressure is usually analogized to voltage, and the amount of water flowing per second, to electric current.

Is the energy in an electric current -- let us say, a direct current, to start with -- the kinetic energy of the electrons? The water-current-to-electric-current seems straightforward: water molecules = electrons. Litres per second is coulombs per second.

But what about the voltage? Does higher voltage (more energy) manifest itself in faster electrons? Is the energy of an electric current just kinetic energy, the way it is for a water current?

Since they are actually not moving that fast (the electrons, as opposed to the wave that passes down the electrons in the conductor), how can this be so?

 

[CWatters]

The water analogy isn't perfect. Normally the KE of the water is ignored. A load causes a pressure drop which is analogous to a voltage drop.

 

[mikeph]

Think of the kinetic energy as being entirely (and instantaneously) transferred to the conductor as resistive heating. The electrons do not "speed up" as they travel down a voltage gradient, the coulomb interactions occur so quickly (when compared to the drift velocity) that you can think of them as permanently traveling through the conductor at their terminal velocity.

 

[Dale]

Electric current is often analogized to water under pressure in pipes.

When water is moving through a pipe, the energy of the system is the potential energy of whatever is giving the water its pressure -- a column of water, say -- and the kinetic energy of the water molecules as they move. Increase either their velocity (by increasing the pressure), or their total mass-per-unit-time passing a given point (by substituting a larger pipe for a smaller one, for example), and you increase the total energy they are carrying.

In an inviscid fluid flow there are basically three forms of energy: 1/2 mv², PV, and mgh.

Now, water pressure is usually analogized to voltage, and the amount of water flowing per second, to electric current.

Yes, so in such an analogy PV->VQ, this is the primary form of energy in an electric circuit, voltage times charge. If you are interested in the rate of energy transfer then it is VI, voltage times current. This is the only significant form of energy in an electric circuit.

Is the energy of an electric current just kinetic energy, the way it is for a water current?

The KE of the electrons is essentially completely irrelevant for practical circuits. Their mass is so low and their speeds are so low that the amount of KE they contain is truly negligible. It is a worthwhile exercise to calculate it.

Similarly, the gravitational PE is essentially negligible since the mass of the electrons is so low.

 

[DrZoidberg]

The energy is transferred through an electric field. That field pushes all the electrons in the circuit.
The field induces surface charges which then in turn produce a field themselves. In that way the electric field propagates along the wire.

http://www.astrophysik.uni-kiel.de/~hhaertel/PUB/voltage_IRL.pdf
http://www.phy-astr.gsu.edu/cymbalyuk/Lecture16.pdf

 

[Doug1943]

My thanks to everyone who has taken the trouble to reply.

With respect to the argument just above, I believe the 'electron speeds' being referred to by the Hyperphysics link are speeds of the random motions of the electrons, not their motion down the conductor, which is very slow. Thus neither carry significant energy from one end of the conductor to the other.

Is it then correct to say that an increase in voltage -- i.e. an increase in force on each electron, pushing it 'harder' down the conductor, does NOT result in speeding it up? (Or is it perhaps incorrect to say that an increase in voltage increases the force on each electron?)

 

[Dale]

It does result in speeding it up. From 0 mm/s on average to a mm/s on average. It is just that the increased KE is such an irrelevantly small amount of energy that you can neglect it in ordinary circuits.

 

[Doug1943]

Okay. Then does the increase in voltage from, say, 1 volt to 100,000 volts, result in a 100 000-fold increase in the drift speed of the electrons?

 

[Dale]

Yes.

 

[russ_watters]

By the way, the water analogy itself works better if you ignore kinetic energy: for a typical hydroelectric dam, the vasst majority of the energy recovered is pressure energy. Kinetic energy is insignificant there just like it is with electricity - indeed, it is mostly lost anyway..

 

[alva]

It does result in speeding it up. From 0 mm/s on average to a mm/s on average. It is just that the increased KE is such an irrelevantly small amount of energy that you can neglect it in ordinary circuits.

In what type of circuits you can not neglect the kinetic energy of the electrons ?

 

[Dale]

Particle accelerators, cathode ray tubes, etc.

 

[Doug1943]

Dale Spam: thank you for the reply. So, is it correct to say that the electrons in a 450 kilovolt current are moving a hundred thousand times as fast as those in a 4.5 volt current?

Russ_watters: I am not sure what is meant by 'pressure energy'.

 

[Dale]

Dale Spam: thank you for the reply. So, is it correct to say that the electrons in a 450 kilovolt current are moving a hundred thousand times as fast as those in a 4.5 volt current?

Current isn't measured in volts, but other than that yes.

 

[Nugatory]

Current isn't measured in volts, but other than that yes.

Assuming you have the same number of electrons in both cases, of course.

 

[russ_watters]

Russ_watters: I am not sure what is meant by 'pressure energy'.

It isn't an easy concept to visualize, so I'll explain it backwards, starting with why kinetic energy isn't used:

If water flows into a turbine, it has to flow out otherwise it will collect somewhere. As a result, if you want a flow velocity (and therefore kinetic energy) much higher on the inlet than the outlet you need a much smaller pipe on the inlet. But high velocity flows create a lot of friction, so it is better to use a lower flow rate and a higher pressure to drive the turbine. Thus, the kinetic energy of water flowing in and out of most turbines is pretty constant, but there is a very large pressure drop across the turbine.

 

[Doug1943]

Russ_watters: Thank you for that very clear example.

It's easy to see that in a closed channel, of course there will be same number of particles per second moving on each side of a turbine, for the reason you stated. But some energy has been passed on to the turbine (which from the fluid's point of view is, I guess, just another type of 'friction').

That energy comes not from a drop in the velocity of the particles but from ... a drop in pressure. My problem is trying to visualize what is going on when fluid pressure changes: that's easy enough in a gas, which is compressible and whose particles exert their force by hitting the sides of their container, but liquids are, for practical purposes, incompressible. So how is the pressure 'stored'?

If I seal a sample of gas at a given pressure into a container, it will retain that pressure, even if the pressure outside that container changes, provided their temperature and volume does not change. But is this true of a liquid?

I'm sorry if this sounds like a stupid question.

I think my problem with trying to understand electrical 'pressure' stems from not having a clear grasp of fluid, or at least liquid, pressure, which in turn may stem from an inadequate understanding of 'force'.

 

[DrZoidberg]

There is no such thing as an incompressible substance.
http://www.engineeringtoolbox.com/fluid-density-temperature-pressure-d_309.html

If I seal a sample of gas at a given pressure into a container, it will retain that pressure, even if the pressure outside that container changes, provided their temperature and volume does not change. But is this true of a liquid?

That depends on the compressibility of the container vs the compressibility of the liquid. Water or even air inside a rubber container (balloon) will indeed change it's pressure if the outside pressure changes. Water inside a thick walled steel container however will stay at approximately the same pressure. In a thin walled steel container it should change it's pressure somewhat but not as much as the outside of the container.

 

[CWatters]

If energy was carried in the KE of the electrons then wouldn't you have to apply Newtons laws to batteries and wires (for example at bends).

 

[Doug1943]

Dr Zoidberg: as I visualize liquid pressure, it is caused directly by the weight of the liquid above it, pressing down on it (as opposed to gas pressure). So, presumably, if there were a 'swimming pool' in free fall (say, in orbit), there would not be any more pressure at the 'bottom' of a 'deep end' of the pool than anywhere else. (In contrast to gas pressure, which we can certainly have inside a container in free fall. I assume the pressure of a gas in a sealed container is not due to the 'weight' of anything, even if it was originally caused by weight.)

If a liquid is not (very) compressible, why would the pressure inside a sealable container stay at its original level, once the container was sealed and the weight of the liquid above it no longer had effect?

If I filled a (very strong) container with water at 10 000 meters below sea level, and then sealed it, would it retain its very great pressure when it was returned to the surface, assuming there was no air inside the container, nor any dissolved gases?

I understand that at that depth, it would be compressed by about one or two percent. Would this compression be the cause of its pressure?

Thank you for taking the time to answer these (perhaps naive) questions.

 

[russ_watters]

Russ_watters: Thank you for that very clear example.

It's easy to see that in a closed channel, of course there will be same number of particles per second moving on each side of a turbine, for the reason you stated. But some energy has been passed on to the turbine (which from the fluid's point of view is, I guess, just another type of 'friction').

That energy comes not from a drop in the velocity of the particles but from ... a drop in pressure. My problem is trying to visualize what is going on when fluid pressure changes: that's easy enough in a gas, which is compressible and whose particles exert their force by hitting the sides of their container, but liquids are, for practical purposes, incompressible. So how is the pressure 'stored'?

If I seal a sample of gas at a given pressure into a container, it will retain that pressure, even if the pressure outside that container changes, provided their temperature and volume does not change. But is this true of a liquid?

I'm sorry if this sounds like a stupid question.

I think my problem with trying to understand electrical 'pressure' stems from not having a clear grasp of fluid, or at least liquid, pressure, which in turn may stem from an inadequate understanding of 'force'.

"Pressure energy" is not due to compression it is just due to pressure, so it doesn't matter if you are using a liquid or a gas. For a gas at a small pressure drop (for example, airflow through a centrifugal fan), compressibility is generally ignored. This energy doesn't exist in a static situation (so if you isolate part of the system with valves you'll get nothing): it is the energy of a fluid driven to move by the pressure.

 

[DrZoidberg]

If a liquid is not (very) compressible, why would the pressure inside a sealable container stay at its original level, once the container was sealed and the weight of the liquid above it no longer had effect?

Again, it's down to the compressibility of the liquid vs that of the container.
If the container walls are rigid enough, the liquid will stay compressed and that compression is the reason for the pressure.
The force that compressed the liquid/gas can have different sources. It could e.g. be due to the weight of the liquid/gas above it. Air pressure at sea level for example is about 100,000 N/m^2. That pressure comes from the weight of the air above it. You could say we are living at the bottom of an ocean of air.
If you pump air into a container the pressure comes from the force that the walls of the container exert on the air because air molecules are constantly bumping into the walls resulting in a force between the molecules and the walls. The more the air is compressed, the more molecules will bump into the walls each second.

 

[jartsa]

Let's say I open the faucet for one seconds time, and let's say the water tower is 10 miles away.

The kinetic energy of the water did ot come from the water tower. Not enough time for that.

After ten minutes I open the faucet again for one second. The kinetic energy of the water comes from the same place as in the first case, obviously. During the 10 minutes time the energy content of that place were the energy comes from, was replenished by the water tower.

 

[Jano L.]

That energy comes not from a drop in the velocity of the particles but from ... a drop in pressure. My problem is trying to visualize what is going on when fluid pressure changes: that's easy enough in a gas, which is compressible and whose particles exert their force by hitting the sides of their container, but liquids are, for practical purposes, incompressible. So how is the pressure 'stored'?

When liquid is compressed by external forces, it shrinks a bit. The molecules get little bit closer and hence act on each other with greater force, hence the greater pressure.

The difference between liquid and gas is that in adiabatic compressions, liquid's energy (in the original sense of the word, like ability to do work) depends only weakly on its pressure, while for ideal gas, it depends on it strongly. This is because increasing pressure to high values in liquids requires only low amount of work (due to their small compressibility), while increasing pressure to similarly high value requires large amount of work, since large change of volume of the gas is necessary.

 

[russ_watters]

Again, it's down to the compressibility of the liquid vs that of the container.

Again (and for JanoL): compressibility plays no role here. It is a completely separate term in Bernoulli's equation that is left off in the situations we are discussing.

The basic form of Bernoulli's equation can be found here:
http://www.princeton.edu/~asmits/Bicycle_web/Bernoulli.html

The three terms are: Pressure energy, kinetic energy and gravitational potential energy.

 

[sophiecentaur]

"Pressure energy" is not due to compression it is just due to pressure, so it doesn't matter if you are using a liquid or a gas. For a gas at a small pressure drop (for example, airflow through a centrifugal fan), compressibility is generally ignored. This energy doesn't exist in a static situation (so if you isolate part of the system with valves you'll get nothing): it is the energy of a fluid driven to move by the pressure.

I think I see his confusion. Energy (work) needs a force times a distance. We spend all our time telling people that the table is doing no work on the book it supports. Your point is that the volume displaced one end of the pipe is the same as the volume displaced the other end. So you could say that work (Pressure times volume change) is done to (at one end) and by (at the other end). This also allows for the two quantities to be slightly different due to energy transfer in the pipe due to friction etc..

This reveals an interesting point. In an effort to make electricity 'easier', people try for the water analogy but, in fact, exactly the same kind of 'abstract' idea has to be used to explain energy transfer by water flow as is needed to explain electrical energy transfer. This totally supports my general reservations about using analogies. They can be a snare and a delusion, if you re not very very careful - and very few people are that careful.

 

[DrZoidberg]

Again (and for JanoL): compressibility plays no role here. It is a completely separate term in Bernoulli's equation that is left off in the situations we are discussing.

Did you look at the question that I was answering? I was talking about a container that is sealed at the bottom of the ocean and then lifted to the surface. Bernoulli's equation is not applicable in that situation.

 

[Doug1943]

I have read contemporary and historical descriptions of bringing up apparatus from great depths, and of how the great pressures that may exist inside containers that have had a small leak, can result in powerful 'explosions' if they are opened quickly. There is a description of such an incident when the original 'bathysphere' was being tested eighty years ago, where the bolts of the hatch which was being removed were blown forcefully across the deck of the ship, followed by a column of water.

But my assumption is that when this occurs, the effect is due mainly to air that was originally in the apparatus and which was compressed, and only a little of the effect is from the water expanding, since it has only been compressed by one or two percent of its volume.

But clearly there is a lot of potential energy stored in the compressed liquid, so perhaps that one or two percent expansion does contribute a major part to the dramatic effects reported. There doesn't seem to be an easily-discovered consensus on this.

 

[rcgldr]

When liquid is compressed by external forces, it shrinks a bit. The molecules get little bit closer and hence act on each other with greater force, hence the greater pressure.

Assuming temperature doesn't change, then the speed of molecules between collisions remains the same, but the number of collisions per second increases, resulting in more impulses per second, increasing the pressure. If the temperature increases, then the speed of the molecules between collisions also increases, and the kinetic energy of the molecules is propotional to the temperature in Kelvin.

I'm not sure this is a good analogy for energy in an electrical current. The energy at some point in a circuit over some period of time is equal to the voltage at that point in the circuit times the current (coulombs) that flowed over that period of time. As the voltage drops across components in a circuit, the energy (per unit time) also drops.

 

[Doug1943]

As the voltage drops across components in a circuit, the energy (per unit time) also drops.

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.

 

[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.