AC generator, current and ground

[Dormetar]

Hi,

Let's say we have a basic AC generator, where a single wire is rotated within two magnets:

Due to the electromotive force, a voltage is generated so that Va - Vb = 120V.
Let's now ground Vb side directly to earth, while leaving Va still open:

We can say that Vb = Vc = 0V and Va alternates between -/+120V compared to Vb.

Now I have the following questions:
1) In the first figure, even though there is no load connected, there is alternating current between the ends of the wires due to potential differences, yes?
2) In the second figure, does the current continuously flow to/from the ground as the right side of the coil tries to "remain" 0V?
I.e., during half of the rotation the voltage generated let's say is higher than 0V, meaning current will flow to the ground and the second half lower than 0V, meaning current will flow from the ground.
3) If the answer to previous question, is yes, can we also deduce that adding some low resistance load L (e.g., 1 Ω) between Vb and Vc would have current running through it?

Thank you for your time.

 

[Baluncore]

Welcome to PF.

1) In the first figure, even though there is no load connected, there is alternating current between the ends of the wires due to potential differences, yes?

There is an alternating voltage, but no current will flow.
For any current to flow, you must close the circuit.

2) In the second figure, does the current continuously flow to/from the ground as the right side of the coil tries to "remain" 0V?

No. There is only one ground connection, so no return circuit. The potential of the alternator circuit will be held near ground, but no current will actually flow to or from ground unless you add another ground connection somewhere.

3) If the answer to previous question, is yes, can we also deduce that adding some low resistance load L (e.g., 1 Ω) between Vb and Vc would have current running through it?

No. There is no current flowing without a return path to make a circuit.

 

[Dormetar]

There is an alternating voltage, but no current will flow.
For any current to flow, you must close the circuit.

I understand your answer(s) from a circuit theory point of view, but theoretically, shouldn't there be a current, even if it's negligible? I mean, how antennas would work if open circuits wouldn't have current?

 

[Baluncore]

I understand your answer(s) from a circuit theory point of view, but theoretically, shouldn't there be a current, even if it's negligible?

I can only analyse the circuit you present. If you want to add a capacitor to ground, then there would be a reactive current flowing in the ground connection.

If you show a resistive load, there will be a circuit current flowing that will generate heat in the load.

If you add inductance between the circuit connections, there will be a magnetic field generated and radiated, by the circuit current that flows through the inductors.

 

[Rive]

but theoretically, shouldn't there be a current, even if it's negligible?

It is negligible. That's why it was not mentioned.

Any circuit can be discussed at different levels. You can have just a wire, but if that wire is on a PCB and fed with high frequency then suddenly you have to deal with a bunch of capacitances and inductances at every cross and turn...

It's better to describe the question carefully (often that makes up half of the answer).

Your initial question might be discussed as a simple circuit with a reference point (ground): it can be taken one with a protective ground: with that last part it might be about noise filtering/EMI or antennas...

It's a bad/inadequate setup/description for discussing the last two, though. And missing some important details to be discussed in the second context. So you got an answer based on the most common understanding.

If you provide an extended context we might be able to help.

 

[Dormetar]

Thank you both for your answers.

I agree that the drawn circuit isn't the best and I mainly added it to try to help/visualize the question(s) at hand. It seems it adds more confusion than clarity, so let's try to approach this purely from a theoretical perspective first and move on from there.

So I understand that there will be some current in the open circuit.
To calculate it, we'd need more information about the setup, but let us set it aside for now as it's hopefully not that important for the thought experiment.

If we agree that there's some current, could we say that when literally connecting one end of the AC generator to a grounding rod, there will also be current alternating between the generator and the rod? My thought process here is that between the grounding rod and the area where emf induces voltage, there will be a potential difference, which in effect causes the current to flow.

 

[Rive]

Feels like you are fighting one of those useless opinion wars of the internet somewhere ... ?

You may take that pole of the generator as a stray capacitance or antenna, but then you need to add that to the schematics to convey that it's not an open circuit.

If it's not there, it means it is some practical issue/question and that negligible load could/should be taken as negligible, and the pole as an open circuit.

If it's there then it's OK to calculate the resulting load (current) through the ground connection, but unless that 'open circuit' (which is at that point no longer a real open circuit) includes some county long transmission lines the resulting load will be ... well: negligible, by any practical sense.

 

[Dormetar]

Feels like you are fighting one of those useless opinion wars on the internet somewhere ... ?

Incorrect, not sure why would one state that. I'm trying to learn.

resulting load will be ... well: negligible, by any practical sense.

Why is that so? Yet again reiterating my thought process - if there's a potential difference and earth has capacitance to receive and give charge, why is there only negligible current?
To my understanding during one cycle of the rotation of the coil, current would flow to the open end of the wire. At the end of the wire, charge accumulates and simultaneously additional charge is pulled from the ground. During the next cycle, the process reverses and the current will flow to the ground while the accumulated charge is depleted from the open end.
If the earth can supply and receive large amounts of charge, how come the current in this case is only negligible?

 

[Baluncore]

To my understanding during one cycle of the rotation of the coil, current would flow to the open end of the wire. At the end of the wire, charge accumulates and simultaneously additional charge is pulled from the ground.

You have no choice but to add a stray capacitance, with a realistic value, to the schematic. That will close the circuit, and make the ground current computable. If the symbol and value is missing, the loop current cannot be evaluated.

You or I could add any number of imaginary stray components, without telling the other their values, or where we imagine, they might be connected. To analyse and discuss a circuit, requires you define that circuit, or we have no "common ground" for any discussion.

 

[Rive]

Singe wire capacitance can be calculated. Around a few pF/m, as I recall.
The current through that can be calculated too...
Might be a nice homework
Gives a perspective about what negligible is about.

 

[Baluncore]

Singe wire capacitance can be calculated.

That assumes a conductive ground as a boundary condition, at an infinite distance. It holds only for electrostatic DC, not for AC, because the transmission line connections, or EM propagation, must be of infinite length and so have infinite delay time.

 

[Rive]

Right. Need to make it more constrained... Then a few cm of parallel conductors, maybe.

 

[Baluncore]

Then a few cm of parallel conductors, maybe.

I would guess, 2 pF of stray capacitance to ground.

Now that we have a reactance, we need to specify the frequency and the voltage of the alternator output.

 

[sophiecentaur]

I would guess, 2 pF of stray capacitance to ground.

That would be mighty short piece of cable. With an assumed 2pF how would you manage to hang any measuring gear on it and expect a proper answer. I think we're heading nowhere with this. Any device will have parasitics so, without some specifics of the putative AC generator we can't give an answer.

 

[hutchphd]

Incorrect, not sure why would one state that. I'm trying to learn.

The issue is that any schematic diagram is an approximate representation of the world and (some bad) EE's forget this fact and promulgate nonsense far too often. (Ask Prof Walter Lewin. )
If one wishes an exact answer, one starts with Maxwell's Equations and solves for the current densities and the fields. There will be current and charge densities and E and M Fields all over the place but only a small subset are relevant. And the calculation will take a lifetime (or two).
So when you ask "are there currents" you must first define which version of reality you are adopting. (This is the important part of the answer to a very good question.)

 

[Dormetar]

So when you ask "are there currents" you must first define which version of reality you are adopting. (This is the important part of the answer to a very good question.)

My goal is more or less to understand intuitively why there is only negligible current in the described scenario.
I'm not sure entirely what mental model ("reality") to use to wrap my head around it, as there are plenty and they are easy to intermix.

But so far to my basic understanding (please correct any steps if wrong):

1. There is a potential difference introduced by the generator to the wire when compared to the ground rod that's connected to earth.
2. Earth is considered to be a large capacitor that can receive and return a lot of electrons.
3. If there wouldn't be a "path" for the charge carriers, i.e. electrons to move, there wouldn't be current.
   But given the wire is connected to earth, it's able to absorb and return a lot of electrons as stated in 2)
4. If our wire (let's say copper) would run out of free electrons, there would be no current.
   But there are plenty - copper wire has 8.4 × 10^22 free electrons per cm3 that could theoretically flow to the ground, given enough time (drift velocity is usually really tiny) and a large enough potential difference.
5. Due to 1. and 3., there has to be some current present.
6. Earth has resistance which will hinder the current.
   

Alas, the current will be insignificant.
And yet when the other loose end would be also grounded, then suddenly the current will be significant.
What am I missing?

 

[Baluncore]

Earth is considered to be a large capacitor that can receive and return a lot of electrons.

Not that I know of.
If you connect to ground, it can be a reference voltage.

 

[hutchphd]

Not that I know of.

A conducting ball on an insulated stand has a capacitance. So does the earth. It is, in the scheme of earth objects, quite large

 

[Baluncore]

It is, in the scheme of earth objects, quite large

OK. You run another wire across to infinity and connect that to the other side of the Earth capacitor. If it is not part of the CIRCUIT, it is not relevant to the analysis. The capacitance of the Earth to infinity, is irrelevant when you think of it as being connected to a local alternator, with a stray open circuit wire.

 

[hutchphd]

I thought it might be confusing to the OP not to mention it. I would also recommend to him https://en.wikipedia.org/wiki/Coefficients_of_potential
Otherwise I find the discussion incomplete and a bit perplexing

 

[Baluncore]

And yet when the other loose end would be also grounded, then suddenly the current will be significant.
What am I missing?

You are missing a complete circuit diagram. You need to show a load on the alternator, and any hypothetical stray capacitance or inductance that you wish to discuss. The value of the load can be unspecified, but it needs to be shown.

"... the other loose end would be also grounded, ..." suggests you will short circuit the alternator through ground. The "other loose end" needs to be clearly shown on the circuit diagram, not just posted in your imagination.

Why will you not post a full circuit diagram?

 

[sophiecentaur]

1. Earth is considered to be a large capacitor that can receive and return a lot of electrons.

This totally irrelevant. An 'Earth Return' always involves a (low) resistive connection to Earth at both ends. The approximately 700μF is a red herring because it can only be regarded as part of a series connection to another planet.

I seriously suggest that @Dormetar should do a bit more basic reading about all this stuff.

 

[Dormetar]

You are missing a complete circuit diagram. You need to show a load on the alternator, and any hypothetical stray capacitance or inductance that you wish to discuss. The value of the load can be unspecified, but it needs to be shown.

Well I guess it would be something like this:

Not sure what Earth capacitance would be, infinity?. Resistance assumed to be 5 ohms.
Wire inductance values got from an online calculator (d 2mm, L 1m).

Why will you not post a full circuit diagram?

Because I know that according to circuit theory the current is considered to be near zero.
And I want to build intuition why is that so for AC (maybe I'm in a wrong sub-forum?)
So, yet again, not a circuit diagram, but a picture/concept that maybe you guys could use to help me build my understanding upon, if you have the time:

Please let me know if somewhere I'm making a mistake or missing some crucial part of information:

As the alternator rotates, we're going to have a position where it's parallel to the magnetic field so maximum emf is generated. The charge distribution around the wire spreads near the speed of light. Electrons will slowly start to move - the wire of the generator has billions of free electrons, which now slowly drift to the right.
We consider the movement of charge, or in particularly in a wire, the movement of electrons to be current.
If the wire wouldn't be connected to the earth on the right side, the movement would end rather quickly - the electrons will not "pile" up endlessly, after all, they oppose each other.
But in the picture the wire is firmly connected to earth. And here my understanding breaks down.
I've read from multiple places that earth can receive and give a lot of charge, is this incorrect?
If earth can't receive huge amounts of charge and can be considered to be either a huge resistor or really small capacitor, I think we can end this discussion.
But if not, why doesn't a lot of current flow into it? What opposes it?
The wire has plenty of electrons to generate a ton of current.

 

[sophiecentaur]

How can any current flow here unless you include a finite capacitance between the top left wire end and some Earth? The diagram is incomplete, as a proper schematic. Also, what does the 1μF capacitor represent? What is the 'fat triangle', bottom right supposed to represent and what model are you using to represent the Earth connection?

 

[Baluncore]

If earth can't receive huge amounts of charge and can be considered to be either a huge resistor or really small capacitor, I think we can end this discussion.
But if not, why doesn't a lot of current flow into it? What opposes it?
The wire has plenty of electrons to generate a ton of current.

Every conductor must be terminated somewhere, with a terminal component. If not connected, then there is still a small stray capacitance to nearby conductive objects that are connected to the chassis or building structure. At 60 Hz, those stray currents are insignificant, and they alternate.

The chassis or building is effectively connected to the Earth at one point. Without a second connection, including a voltage or current source, to make a return circuit, there can be no circulating current through the conductive connection to the Earth.

A single connection to Earth will carry only electrostatic discharge currents from the atmosphere. That discharge may include a lightning strike, which may give you the impression "that earth can receive and give a lot of charge", but that is a different electrical circuit, and is not part of the grounded alternator circuit.

The two separate circuits simply share one conductor.

The Earth to cloud capacitor is charged by movement of the atmosphere, or by the "fair weather current". When a breakdown voltage is reached, the lightning strike discharges the atmospheric capacitor to ground.

 

[Tom.G]

In your diagram, the wire going to the left will VERY rapidly run out of free electrons to supply any current. Perhaps others here can come up with some actual number or electrons.

In the real physical world, that left wire will have some small capacitance to Ground, the exact value of that capacitance depends on the diameter of the wire and its distance from the Ground.

So to make an actual Circuit you need something that is connected to that left wire AND to the Earth or Ground on the right.

By the way, I asked Google to define 'circuit' and got this:
1 a roughly circular line, route, or movement that starts and finishes at the same place

View attachment 1692859099196.webp

Cheers,
Tom

 

[sophiecentaur]

1 a roughly circular line, route, or movement that starts and finishes at the same place

A bit like a lot of PF threads, then.