Does voltage exist across an unloaded generator?

[MotoMike]

In a recent discussion the output of a single-phase permanent magnet alternator was the topic. The alternator has a 6-pole rotor and 8 coils in series wound on 4 cores. Output is taken from the opposite ends of the series wound stator.

Another fellow made the statement that with the alternator spinning; the two stator outputs did not have voltage across them until connected to a load. He explained that it is only “potential” voltage and not true voltage. Though I suspect we may be into semantics, it did not seem right to me. Further discussion has not changed his mind. He disputed my contention that voltage is the potential that causes current to flow. I said that the rotating magnetic field passing through the conductors of the coil induced voltage across the windings of the stator and that this voltage was present weather the stator was loaded or not. I think his reasoning is based on Ohms law where E=IxR and since with no path for current, zero x R =0. But I am not sure why he feels this way. He suggested a physics professor could clear it up for me. Sadly I don’t know a physics professor.

My training (20 years Navy Sonar Tech) and 15 years as an amateur radio enthusiast leads me to believe that there are other ways to define voltage.

My thought was that (Please excuse my ignorance here as it becomes obvious) we get voltage by separating electrons from protons. In this case by passing magnetic lines of flux through the conductors in the stator. So we have a surplus of electrons moved to one stator output terminal and a deficit on the other. That it took work to separate them and they hold the potential to do work as they are naturally drawn towards one another. (I understand that with an alternator that condition will change during the next half cycle but for simplicity sake am considering a snap shot.) This difference of potential or charge is what I have come to know as voltage. And that it exists even in the absence of a current path for the electrons to follow.

I appreciate any help on this you can give.

Thanks,
Mike

 

[phinds]

I think his confusion (you are right, by the way) is that yes, E=IR but in the case you describe, you have R = infinity and I = 0 and that that equation does not apply. It only applies to a closed circuit. You have an open circuit.

 

[KingNothing]

In reality the resistance is finite, though incredibly high, and the current is finite, though incredibly low.

 

[phinds]

In reality the resistance is finite, though incredibly high, and the current is finite, though incredibly low.

Yes, I was simplifying

 

[MotoMike]

Phinds and KN Thanks very kindly.

the finite but very large resistance argument may allow me to actually use ohms law to show him what I'm getting at. if I use a hypothetical of 10billion ohms and .000000001 amps, I can show that there is 10 volts across the output of the alternator.

I seem to recall from a very long time ago that there are formulas that are used to determine the voltage across a capacitor that did not have current flow or resistance as part of the equation. one for inductors as well, but I think that would only confuse the issue at this point. Do you know of formulas that would do this and support the conclusion that voltage exists across a source in and unloaded circuit?

Or in the alternate, an example or explanation that would serve to convince him?

Kind regards,
Mike

 

[Jiggy-Ninja]

Even with infinite resistance and 0 current, it still works out fine.

When doing limits in calculus, it's possible to get results that basically evaluate to infinity * 0, and have a finite product.

Though that might be a little too high level for him.

 

[phinds]

I seem to recall from a very long time ago that there are formulas that are used to determine the voltage across a capacitor that did not have current flow or resistance as part of the equation.

capacitors do not have to have a current flow to have a potential (voltage) across them. That's what they are FOR --- to store electrons as a potential. As soon as the current starts to flow, the voltage starts going down (unless it's in a circuit where it gets recharged, and of course in an AC circuit things are more complicated)

Or in the alternate, an example or explanation that would serve to convince him?

trying to convince people who adamantly refuse to be convinced by facts is a fool's errend.
[/QUOTE]

 

[MotoMike]

Even with infinite resistance and 0 current, it still works out fine.

When doing limits in calculus, it's possible to get results that basically evaluate to infinity * 0, and have a finite product.

Though that might be a little too high level for him.

Thanks JN
Might be a little too high level for me as well.

Mike

 

[MotoMike]

capacitors do not have to have a current flow to have a potential (voltage) across them. That's what they are FOR --- to store electrons as a potential. As soon as the current starts to flow, the voltage starts going down (unless it's in a circuit where it gets recharged, and of course in an AC circuit things are more complicated)

trying to convince people who adamantly refuse to be convinced by facts is a fool's errend.

[/QUOTE]

Phinds
Sounds like good advice. thanks

Mike

 

[KingNothing]

As a thought experiment, think of a rectangle with sides x and 1/x. What is the area? Of course, the area is 1. What about the area as x approaches infinity? The area is still 1.

 

[Antiphon]

The fellow was right but with the wrong generator.

A permanent magnet generator will have a voltage on it.

If instead of permanent magnets there is a field coil, then there will be no voltage because there is no current in the field coil. With a load attached and the shaft rotating, the tiniest residual magnetic field in the core will be enough to force a small current which is then enough to bootstrap the generator to a full current and voltage.

 

[MotoMike]

The fellow was right but with the wrong generator.

A permanent magnet generator will have a voltage on it.

If instead of permanent magnets there is a field coil, then there will be no voltage because there is no current in the field coil. With a load attached and the shaft rotating, the tiniest residual magnetic field in the core will be enough to force a small current which is then enough to bootstrap the generator to a full current and voltage.

Thanks Antitiphon

He was not confused on the construction of the alternator in question. Just has long held ideas that he is reluctant to revisit.

regards,
Mike

 

[phinds]

Thanks Antitiphon

He was not confused on the construction of the alternator in question. Just has long held ideas that he is reluctant to revisit.

regards,
Mike

To restate what I said, trying to convince someone with facts when they don't want to change their mind is an exercise in what in the military is called "pissing up a rope".

 

[Bob S]

A coil intercepting the changing magnetic field lines of a spinning permanent magnet rotor produces an output voltage, based on the Faraday induction equation:
$$V=∮_CE⋅dℓ=−∂t∫_SB⋅ndS$$
Bob S

 

[MotoMike]

A coil intercepting the changing magnetic field lines of a spinning permanent magnet rotor produces an output voltage, based on the Faraday induction equation:
$$V=∮_CE⋅dℓ=−∂t∫_SB⋅ndS$$
Bob S

Bob S

Thanks for that. I won't be using it to convince my friend but will dig out my old texts and see if I can do a little study to make that work. Does that formula make provisions for pole numbers?

As Phinds has mentioned, I've tendered a last effort which if it fails will just be the end of it. I can see that this forum will come in handy.

I also wanted to make sure my understanding was substantially correct. I don't work with this anymore but have it as a hobby of sorts.

Kind regards,
Mike

 

[Bob S]

Bob S

Thanks for that. I won't be using it to convince my friend but will dig out my old texts and see if I can do a little study to make that work. Does that formula make provisions for pole numbers?

The formula I posted above is the per-turn induced voltage. For N turns on M in-phase poles, the total open-circuit output voltage is
$$V=\sum_M N∮_CE⋅dℓ=−\sum_MN\cdot∂t∫_SB⋅ndS$$
assuming all the turns are in series.

Bob S

 

[DragonPetter]

The alternator will have a back EMF (voltage) proportional to the velocity it is moving. This is the motor constant and it is load independent and comes from the voltage generated by the coils crossing the magnetic flux. Obviously, as the alternator is spinning faster, more flux is crossing the coils, and so the voltage is increased even if the load is the same or not present. Current is only flowing to keep the motor moving, which would be enough torque to overcome the internal friction if there is any, so the actual power being dissipated by an unloaded alternator should be very low even with these finite voltage and currents.

I think your friend doesn't understand what voltage is. Voltage is a measure of potential energy, and describes the strength of an electric field, it is not defined by Ohm's law. You can have an electric field without current flow, and this constitutes a storage of energy.. basically the alternator has stored energy in its coils.