# What carries the energy in an electric current?

1. Dec 23, 2013

### 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?

Last edited: Dec 23, 2013
2. Dec 23, 2013

### 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.

3. Dec 23, 2013

### 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 travelling through the conductor at their terminal velocity.

4. Dec 23, 2013

### Staff: Mentor

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

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.

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.

5. Dec 23, 2013

### 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 [Broken]
http://www.phy-astr.gsu.edu/cymbalyuk/Lecture16.pdf [Broken]

Last edited by a moderator: May 6, 2017
6. Dec 23, 2013

### 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?)

7. Dec 23, 2013

### Staff: Mentor

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.

8. Dec 23, 2013

### 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?

9. Dec 23, 2013

### Staff: Mentor

Yes.

10. Dec 23, 2013

### Staff: Mentor

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..

11. Dec 23, 2013

### alva

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

12. Dec 23, 2013

### Staff: Mentor

Particle accelerators, cathode ray tubes, etc.

13. Dec 23, 2013

### 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'.

14. Dec 23, 2013

### Staff: Mentor

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

15. Dec 23, 2013

### Staff: Mentor

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

16. Dec 23, 2013

### Staff: Mentor

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.

17. Dec 24, 2013

### 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'.

Last edited: Dec 24, 2013
18. Dec 24, 2013

### DrZoidberg

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

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.

19. Dec 24, 2013

### 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).

20. Dec 24, 2013

### 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.