# High voltage, where is the power?

1. Apr 8, 2013

### ealbers

If I transmit 1000VAC at 1 amp, I can use a small 14 gauge wire with no issues, transferring 1000watts with little line loss.

If I transmit 1000Amps at 1 volt, I better use a large cable to move all those electrons, and will still have a large loss.

Ok, good with that....sort of.

Question, with the 1000Volts AC situation, how do so few electrons move so much power???, theres 1/1000th the number of electrons (1amp vs 1000 amps), yet they move the same amount of power (watts)??

Where is this power 'stored', and it seems awful efficient. It can\t be in the magnetic field, thats a function of the number of electrons, which is fewer!

Then the next logical question is, how few electrons can I use, could I push ALL the power onto one tiny little electron? Force the voltage up and amperage down to a limit where num electrons==1??

2. Apr 8, 2013

### mrspeedybob

power = amps x volts.

It is true that in theory you could raise the voltage arbitrarily high and use an arbitrarily low current, and therefore an arbitrarily small conductor. The catch is that the higher the voltage the more insulation you need to keep it contained. I'm not sure how practical it would be to wrap a .01 mm stand of copper in a meter of insulation.

3. Apr 8, 2013

### Staff: Mentor

Though not an exact analogy, the water flow analogy works well here: voltage is like pressure and amperage like flow rate, so if you double the height of a hydro dam you can produce the same power with half the flow.

4. Apr 8, 2013

### DrZoidberg

That water analogy doesn't work perfectly though. If you raise the pressure in a pipe, all the water will experience a stronger force i.e. every water molecule will collide with other molecules more frequently. But in a thin wire carrying 1000V the electrons won't necessarily experience a greater force than in a thick wire carrying 1V. If you reduce the cross sectional area of a wire by a factor of 1000 it's resistance will be 1000 times larger. If you additionally also decrease the current 1000 times you end up with the same voltage dropping over the resistance of the wire. Same voltage and same wire length means the electric field strength inside the wire will be the same. So the electrons experience the exact same force. They also move at the same speed.
The solution to this is that the energy is not transferred by the electrons that are moving in the wire. It's the electric field that transmits the energy to the load at the other end of the wire. And that is 1000 times stronger at 1000V.

5. Apr 8, 2013

### Crazymechanic

Nicely explained Zoidberg , simple and true :)

6. Apr 8, 2013

### Staff: Mentor

As I stated and by definition, of course.
Yes.
I'm not an electrical engineer, but that doesn't sound right to me: voltage is electromotive force, so how can you not have a greater force if you have a greater voltage? That seems like a tautology to me.
Agreed.
It kinda sounds to me like you are mixing together voltage and voltage drop as well as pressure and pressure drop. The voltage drop along the wire isn't the same as the voltage carried by the wire, nor is the pressure in a pipe the same as the pressure drop along the pipe. If the voltage is equal, the electromotive force is equal even if the voltage drop along the wire is lower.
I'm not seeing a problem needing a solution, nor a difference between water and electricity here.

Where the water vs electricity analogy does fail though is in the math: in electricity, voltage drop is directly proportional to charge flow rate whereas in water, pressure drop is a square function of water flow rate.

7. Apr 8, 2013

### DrZoidberg

Maybe I should clarify this some more.
The voltage carried by the wire is the voltage between the wire and some reference point (e.g. ground). However what determines the speed of the electrons in the wire is the current density and the properties of the metal.
Take a look the generalized form of Ohm's law.
http://en.wikipedia.org/wiki/Ohm's_law
J=σE
That equation means that the electric field inside the wire and therefore the force that the electrons experience depends only on the current density and the resistivity. It does not depend on the voltage relative to ground.
And the current density in turn depends only on the voltage that drops across the wire, it's resistance and it's cross sectional area.

8. Apr 8, 2013

### marcusl

No, that is incorrect. It is electric field and properties of the metal.

9. Apr 8, 2013

### DrZoidberg

Did you read the rest of my post?

10. Apr 8, 2013

### AlephZero

Without getting into EM fields and the movement of electrons, you can understand this just with Ohm's law. To keep it simple, let's ignore inductors and capacitors, and just think about resistance

If you have 1000V and 1A, the total resistance in the circuit must be 1000 ohms. The total power dissipated is 1000 W.

If your 14 gauge connecting wire has a small resistance compared with 1000 ohms, nearly all the power will be dissipated in the 1000 ohm load, not in the low resistance of the wire.

In the second example, you have 1V and 1000A. The total power dissipated is 1000 W as before, but this time the total resistance of the circuit is 0.001 ohms, which is 1 million times smaller than the the first example. Therefore, the resistance of the connecting wire also has to be 1 million times smaller, otherwise most of the power will be dissipated in the wire not in the load.

11. Apr 9, 2013

### Staff: Mentor

What happens if your wire is charged to a very high voltage, but there is no current? Electrons will still spontaneously jump off of the wire because of the voltage. The existence of an electric field does not require the electrons to be moving. It does not require any current.

You said yourself that 1000 times the voltage means 1000 times the electric field. Now you're saying the electric field doesn't depend on the voltage!
That's fine, but it is missing the point: Electric power lines don't obey V=IR when you look at the line voltage because the wires aren't the biggest, much less the only "R" in the circuit.

It still doesn't appear to me that what you are describing is an area where the analogy breaks down.

Last edited: Apr 9, 2013
12. Apr 9, 2013

### ealbers

Wow, the answers are great, I've heard the power is in the E field, obviously its not the magnetic field since thats dependant upon the number of electrons and in our simple example there only being 1 lonely electron B must be pretty small.

So the E field is where the 'power' is stored when V is high and I is low.

How come I never hear about big E fields being around transmission lines? I mean, you can use a simple compass to 'see' the B field, but the E field seems well contained, is it fully 'in' the wire?

It seems a very very efficient field as well! I mean, you can basically transfer as much power as you want almost free (I said almost folks)....

Given a hypothetical 'super' insulator, you could transfer all the power a whole city needs using a single electron, in a tiny diameter wire, now THATS cool...and curious...

Are there 'visible' manifestations in that case of the E field 'leakage', I mean is there a way to detect this E field from outside the wire as its transmitting the power? I suppose the detection would act as a drain on the power.

13. Apr 9, 2013

### ealbers

OH, one other thing, does relativistic effects come into play as the E field grows? In our simple example of 1 electron carrying huge power, would its 'apparent' Mass increase? Does it effect ones ability to detect the spin of said electron vs another entangled electron, each carrying a huge power in their respective E fields....i.e. does their effective radius increase when doing the calculations for wave function collapse?

14. Apr 9, 2013

### sophiecentaur

Instead of diverging more and more, it might more fruitful for the thread to sort out one thing at a time, I think.
The fact is that, for a given thickness of wire, carrying the same power, transmitting at a higher voltage involves a lower.voltage drop over the length of the wire. This is because the higher operating voltage requires less current.

15. Apr 9, 2013

### marcusl

Yes I did and your statements

are incorrect--backwards, actually. The current is induced by the E field, and not the other way around. To use your words, the force that the electrons experience depends on the E field. The electrons set in motion by E constitute the current density.

16. Apr 9, 2013

### DrZoidberg

Maybe I shouldn't have said "depend". But the field and the force on the electrons can certainly by derived from the current density. However that feels like splitting hairs. I mean is it incorrect to say e.g. that the voltage that drops over a resistor depends on the current?
First off, the equation that I mentioned - J=σE - is always true in materials that obey Ohm's law. So field strength and current density always depend on each other. If you have a perfect insulator the current density is 0 but the law is not violated. Inside a metal wire you can't have an E field without a current, and you can't have a current without a field.

I said the field INSIDE the wire doesn't depend on the voltage between the wire and ground which is not the same as the voltage across the wire. If you have 1V dropping over the wire and 1000V between the wire and ground, the current density and the field strength inside the wire will depend on that 1V that is dropping over the wire but not on the 1000V.

If you only have one single electron in your wire, that wire is not going to behave like a normal metal wire. You can not transmit an electric field along a wire with only one electron. You need a large number of them to do that.

There are strong E fields between an hv transmission line and ground. In the case of AC, you can measure them with a simple multimeter with one lead connected to ground and the other to some form of antenna. (But please don't do any dangerous experiments)

Not with a single electron.

You could measure the field that exists outside the wire with a field meter but it's usually different from the one inside the wire.
If you want to determine the field on the inside, all you have to do is measure the voltage that drops across the wire and divide it by the wire length. More important though is the field inside the load that's being powered. Because that is the one that transmits the power into the load.

Last edited: Apr 9, 2013
17. Apr 9, 2013