Electron flow in 240V circuits

[mbrooke]

That is just so wrong. That is as bad as saying current wants to get to Earth. Sorry I don't have time to elaborate at the moment but you should re-think that. Suffice it to say you could take bolt-cutters to "C" and it would not matter.

While phase angle has been omitted and more finite details have been left out, in a very simplified manner it does help when someone in grasping it for the first time. I do not agree with the half and half part, but I do agree that any phase is trying to get back on the others phases/neutral.

 

[mivey]

Is it so wrong that it does a disservice to an electrician working in the field? I don't pretend to understand the intimate intricacies of exactly why electricity behaves the way it does, but I've always considered this to be an acceptable shortcut to visualizing what's going on. I'm sorry you find it detrimental to the OP. Pray, what simple illustration do you have to offer?

The current flows back to the source. The circuit has to have a closed loop for this to work.

To illustrate without a picture, "C" is not in the loop. You could eliminate it entirely and it would not make any difference to the A-N and B-N loads.

Current is trying to return to the source, not some other unconnected path. Just as it is wrong to say current wants to get to Earth when Earth is not in the circuit, it is also wrong to say current wants to get to "C" when "C" is not in the circuit.

 

[mivey]

With the hole wire size question, I guess it comes down to how many atoms are in the copper wire with valence electrons that can be excited out of orbit. So can a #10 AWG wire carry 200 amps? The answer to that questions would probably be "Yes" but the wire and insulation would be burnt to pieces. Maybe a #22 AWG could not carry 200 amps because there are not enough valence electrons to excite out of orbit. Am I thinking along the right lines here?

You are not exciting them out of orbit. The "sea" of electrons is already there awaiting use.

A #22 can carry 200 amps, just not very long.

 

[mivey]

I do agree that any phase is trying to get back on the others phases/neutral.

You are wrong. See my response above.

Current tries to return to the source using the circuit path. It will not return through some phase that is not part of the circuit.

 

[mbrooke]

You are wrong. See my response above.

Current tries to return to the source using the circuit path. It will not return through some phase that is not part of the circuit.

And that circuit path can be another phase.

 

[iwire]

The current flows back to the source. The circuit has to have a closed loop for this to work.

To illustrate without a picture, "C" is not in the loop. You could eliminate it entirely and it would not make any difference to the A-N and B-N loads.

Current is trying to return to the source, not some other unconnected path. Just as it is wrong to say current wants to get to Earth when Earth is not in the circuit, it is also wrong to say current wants to get to "C" when "C" is not in the circuit.

Maybe I am missing something but it seems to me you are really splitting hairs here.

I have a working circuit connected to A-B, I do something dumb and put myself between either A or B & C and I am going to get whacked as current from A or B tries to get back to the source on C.

 

[mivey]

Maybe I am missing something but it seems to me you are really splitting hairs here.

I have a working circuit connected to A-B, I do something dumb and put myself between either A or B & C and I am going to get whacked as current from A or B tries to get back to the source on C.

Yes, you are missing something. George said with line to neutral loads that the electrons are trying to get back to the other phase. The other phase is not involved otherwise single phase supplies would never work. Listen Neo, there is no C.

When you get between A & C you completed a circuit from A to C. The B phase does not get involved. There is no B. The current for the load from A to B does not flow through you over to C; no way, no how.

When you get between B & C you completed a circuit from B to C. The A phase does not get involved. There is no A. The current for the load from A to B does not flow through you over to C; no way, no how.

This is not splitting hairs. It is a fundamental understanding of how circuits work. You know better than that but you are not thinking straight. You don't size a C phase conductor based on loads between A&B or those to neutral. C has nothing to do with it. Come on now, stop it.

 

[mivey]

And that circuit path can be another phase.

What the heck?!!! A load from A to N will NEVER have a path through C.

 

[mivey]

Back to the OP to emphasize:

Think of it even with line-neutral loads, that the electrons are seeking a path to the other phases. With single phase, the current leaving L1 is seeking L2, and the only path there is through the windings of the transformer between N and L2.

The load from L1 to N will not have current hitting the second winding at N and travel through that over to L2. L2 does not get involved. The current circulates through the first winding over to L1. There is no L2 in the circuit and it plays NO role.

That is why if you have a 10A load from A-N, and a 10A load from B-N, then 10A will travel on the neutral; that 10A is seeking C, and the shortest path there is to N, then on to C through the transformer windings.

That 10 amps in the neutral is seeking A & B. There is no C in the circuit and it plays NO role.

 

[iwire]

Yes, you are missing something. George said with line to neutral loads that the electrons are trying to get back to the other phase. The other phase is not involved otherwise single phase supplies would never work. Listen Neo, there is no C.

When you get between A & C you completed a circuit from A to C. The B phase does not get involved. There is no B. The current for the load from A to B does not flow through you over to C; no way, no how.

When you get between B & C you completed a circuit from B to C. The A phase does not get involved. There is no A. The current for the load from A to B does not flow through you over to C; no way, no how.

This is not splitting hairs. It is a fundamental understanding of how circuits work. You know better than that but you are not thinking straight. You don't size a C phase conductor based on loads between A&B or those to neutral. C has nothing to do with it. Come on now, stop it.

You are splitting hairs.

If have a load connected A to N and A faults to C you will have current between a and c. That will be a new path but it is path back to the source

 

[mbrooke]

What the heck?!!! A load from A to N will NEVER have a path through C.

But A and N will always have potential to C. Otherwise I could use 30 volt rated insulation on a 277/480 volt system.

 

[mivey]

You are splitting hairs.

If have a load connected A to N and A faults to C you will have current between a and c. That will be a new path but it is path back to the source

You are creating a straw-man. It has nothing to do with George's proposed scenario.

If the circuit includes C, whether by normal load or by a fault as you propose, then C is of course included.

This is not at all what George had in his OP. Completely different. Read it again.

Think about a 3-wire open-wye service with A, B, N. C is not and will not be involved. The current on N will NEVER try to get to C back at the windings.

 

[mivey]

But A and N will always have potential to C. Otherwise I could use 30 volt rated insulation on a 277/480 volt system.

A & N can have potential to all kinds of stuff. Without a circuit path to that stuff, the charges flowing in A & N will NEVER try to flow to the other stuff.

The 10 amp neutral current on George's circuit CANNOT flow to C. It will never happen. It is mathematically and physically impossible.

Just because C exists in the world does not mean something is going to use it. Just because we have three-phase available does not mean current to a single-phase load will try to eventually travel through all three phases. That is what George proposed and it is completely and utterly wrong. Fundamentally and any other way you want to look at it.

 

[George Stolz]

To simplify this: why is the following true yet my statement is false? Here are the currents measured at the panel, and the loads are all single phase L-N loads:

10A ... 10B ... 0C ... 10N
10A ... 10B ... 10C ... 0N

I am saying that if you are starting with a three phase source, current will opt for the phase instead of the neutral. I describe this as I have for simplification. Why is it wildly inaccurate? Obviously, if you start with a single phase source, there are no Bs or Cs, so I'm not understanding your objection.

Edit: if you are taking issue with this because if you wire some kind of open transformer source, current will behave in a certain way because you're using A and B and there is no C, I would question the relevance to 99% of the systems an electrician will see and wire in the field.

 

[iwire]

Just because C exists in the world does not mean something is going to use it.

But it can 'use it' and often does.

 

[iwire]

Y
If the circuit includes C, whether by normal load or by a fault as you propose, then C is of course included.

Bingo

This is not at all what George had in his OP. Completely different. Read it again.

I disagree.

 

[mivey]

To simplify this: why is the following true yet my statement is false? Here are the currents measured at the panel, and the loads are all single phase L-N loads:

10A ... 10B ... 0C ... 10N
10A ... 10B ... 10C ... 0N

Because you stated:

...With single phase, the current leaving L1 is seeking L2, and the only path there is through the windings of the transformer between N and L2.
...
That is why if you have a 10A load from A-N, and a 10A load from B-N, then 10A will travel on the neutral; that 10A is seeking C, and the shortest path there is to N, then on to C through the transformer windings.

The A & B currents seek A & B. Once the current reaches the winding through N, they will NEVER try to get to C. They have no interest in reaching C through the winding. They want to get back to their source. There is no "on to C" through the transformer winding.

I am saying that if you are starting with a three phase source, current will opt for the phase instead of the neutral.

Only for a complementary load on the other phase. If the load on the other phase is not complementary, the current travels on the neutral instead, even if the other phase is present.

I describe this as I have for simplification. Why is it wildly inaccurate? Obviously, if you start with a single phase source, there are no Bs or Cs, so I'm not understanding your objection.

It is your assumption that C will be used and that A & B currents will seek C through the winding once they reach the winding through N. The OP scenario had a load on A & B with neutral current. The premise was that the neutral current wanted to get to C. That is simply not the way it works. The only way C gets involved is if there is a complementary load on C, otherwise C is not part of the in the circuit path for the A & B currents.



Think of truck hauls as an illustration: If truck A makes a haul, think of the delivery route as phase A, and the back-haul route as the neutral. Then consider that the only way truck B gets involved with A's route is if B makes a delivery and can share in the back-haul. With no complementary load on B, A is on its own.

By complementary, consider that a pick-up truck and semi truck would not fully share back-hauls. The pick-up may carry some load, but just because the pick-up route exists does not mean it will fully share in the back-haul. For the circuit, unbalanced load current will flow on the neutral even if all three loads have the same amps. Only if they are balanced loads will the neutral have no current.

Now let trucks A & B be doing their thing and sharing back-haul routes (the original premise). Truck C does not get involved in the A & B hauls unless C has a complementary load route.

Current returns to the source. A currents want to get back to A, B to B, and C to C and that is just the way it is. Unless there is an equivalent load on C that will allow it to share in the back-haul, the A & B currents will not try to get to C through the winding via N as was stated in the OP.

C is not involved unless it has a load of its own. AND IT HAS TO BE A COMPLIMENTARY LOAD OR IT WILL NOT SHARE IN THE BACK-HAUL. If the loads are not matched, then the shared current burden starts to divert away from C.

The OP premise was that the 10 amp loads on A & B resulted in a 10 amp neutral current and that current wanted to get to C by going back on N and through the winding to C. That is completely wrong. Those currents have no desire to get to C as they only want to get back to their sources A&B. C is not in the circuit.

If you want C in the circuit, then put a load on C that equals A & B loads and then C will share the back-haul.



Now Iwire proposed that just because he dumby "got between" C and A or "got between" C and B that he has now created a path and A & B current will flow through him to reach C. His load will differ from the loads connected to A & B so the A & B resultant neutral current will not disappear and seek C through him.

First, it is not seeking C, it is seeking A & B. Second, only if his load complements the A & B loads will he & C share in the current path back-haul to A & B. Thirdly, even if he shares some back-haul through C, he probably won't fully share in the back-haul so the residual A & B current will get back to A & B through the N path, not C. That residual current has no interest in finding C through the windings of the transformer between N and L2. Once it reaches the winding at the N point, it will only go through the A-N winding to A and B-N winding to B. It will NEVER go through the C-N winding to reach C.

Edit: if you are taking issue with this because if you wire some kind of open transformer source, current will behave in a certain way because you're using A and B and there is no C, I would question the relevance to 99% of the systems an electrician will see and wire in the field.

Just because C is there does not mean C will be used. Without a complementary load on C, it is out of the picture.

 

[jaggedben]

What the heck?!!! A load from A to N will NEVER have a path through C.

If you lose the neutral on an MWBC with C-N loads turned on, then sure it will. It may fry the load but it sure as heck will have a path.

 

[ggunn]

In an AC circuit the amount of electrons that wiggle back and forth is the amount of current in that circuit rather than the speed at which they move, correct?

It's both. In terms of current, a few electrons moving fast is the same as a lot of electrons moving slowly. Think of it like cars on a freeway. When traffic is low the cars are moving fast and when traffic is heavy they are moving slowly, but the number of cars that pass a given point in a given time interval may be the same.

 

[mivey]

If you lose the neutral on an MWBC with C-N loads turned on, then sure it will. It may fry the load but it sure as heck will have a path.

That is not the OP scenario. It was A & B loads only with neutral current if you will recall with a magical trip to C.

If we want to come up with scenarios for C current, just put in a balanced load. No need for some abnormal loss of neutral thang. Then with a complementary load on C, A & B currents will run down the C-N winding to N and back out the other windings to A & B source points (at that part of the cycle's current direction).

But again, once A and B currents reach the N point at the winding, they will seek the source A or B, not run up through the C-N winding to try to get to C.