Electron flow in 240V circuits

mbrooke

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We know, of course, that you mean current (including phase) since energy generally leaves the winding permanently rather than going out one end and back in the other.
True, but if I connect 20kw across A-B, don't the A-B wingdings "share" the energy evenly?
 

mivey

Senior Member
I should have noted: The current that A-phase needs to supply its half of the load is the same magnitude and phase angle as the current that B-phase needs to supply its half of the load.

The lowest energy solution for the A-phase current leaving terminal A is the path through the load, through B terminal, through the N terminal, and back to A. The B-phase load is thus totally complementary to the A-phase load.
 

iwire

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mivey, you seem totally invested in how the load is connected when it seems to me it is the source that matters.

You point out it takes a closed circuit to make current flow, yet no one here has said otherwise.

I am entirely lost and perplexed at what you are trying to point out that was so egregious in George's post.
 

mivey

Senior Member
mivey, you seem totally invested in how the load is connected when it seems to me it is the source that matters.
Without the load the source sources nothing. The load determines the needed current characteristics.


I am entirely lost and perplexed at what you are trying to point out that was so egregious in George's post.
His analogy stated current circulating in a circuit loop on phase A ( A->load->N->A) wanted C terminal as a destination. His premise was that A current wanted to reach C so bad that it would make a turn at the winding's N terminal and go backwards through the C-N winding to try to reach the C terminal.

A current does not want C as a destination. Neither does circuit source current want Earth as a destination.

A current can take a path through C to get back to A but C is not the destination.

Circuit current can take a path through Earth to get back to the circuit source but Earth is not the destination.

Understand?
 
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GoldDigger

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True, but if I connect 20kw across A-B, don't the A-B wingdings "share" the energy evenly?
Sure, but that is still not energy going out one wire and back in through the other. :)

Since the two windings are 120 degreesout of phase there is, however, reactive "power" going back and forth . But I don't think that is what you were referring to.
 
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iwire

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Without the load the source sources nothing. The load determines the needed current characteristics.
I disagree with that and the fact you feel differently does not make me wrong.


His analogy stated current circulating in a circuit loop on phase A ( A->load->N->A) wanted C terminal as a destination. His premise was that A current wanted to reach C so bad that it would make a turn at the winding's N terminal and go backwards through the C-N winding to try to reach the C terminal.

A current does not want C as a destination. Neither does circuit source current want Earth as a destination.

A current can take a path through C to get back to A but C is not the destination.

Circuit current can take a path through Earth to get back to the circuit source but Earth is not the destination.

Understand?
I say you did not understand George's post.
 

mivey

Senior Member
I disagree with that and the fact you feel differently does not make me wrong.
Well then that's another thing you are wrong about.:p

I thought some pics & examples for you would clarify and actually started last night but was too tired to finish. If I get back to town next weekend I may finish. I wanted nice clean graphics but may just settle for hand-drawn.

I say you did not understand George's post.
I'll invoke Charlie's Rule. Plus his follow-ups did not indicate he meant anything different in his OP.

Kudos for sticking with your peeps though.
 

George Stolz

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Mivey, the shortest path back to the same phase is never leaving the terminal in the transformer. Voltage from "A" to "A" in any circuit is going to be zero, meaning there is nothing to cause current to flow.

You can attack me personally for being an idiot if you like, that is a teaching style I haven't encountered in quite a while and try not to employ personally. I am just a simple electrician speaking to other simple electricians in most cases. I don't much appreciate being called a liar, either: I told you that if you can clearly illustrate why my analogy actively does a disservice in visualizing current flow, I would remove it from my teaching. You then claimed that I must be so invested in it that I can't hear you and would not remove it. No, I respect your experience and your knowledge, and I am listening with rapt attention. You are not conveying the problem very clearly, and frankly, being damned rude about it. I had forgotten about this since the last time I checked in, that's how completely enraptured by this topic I am. Trust me, if you can make your case I will abide by my agreement and cease teaching the method. I do preface that portion of the class with making it clear that I'm not an EE, and this is a cheat to visualize current flow - not that it was fact, for the record.
 

mivey

Senior Member
Mivey, the shortest path back to the same phase is never leaving the terminal in the transformer. Voltage from "A" to "A" in any circuit is going to be zero, meaning there is nothing to cause current to flow.
KCL, George. The voltage across the A-N winding causes current to flow from A, through the load, to N through some path, and back to A. Current leaving A and returning to A does not require a voltage across A. Current leaving one part of terminal A MUST return on a different part of terminal A because of KCL. Think of a node: no voltage across, but all current leaving the node must also enter the node.

If you want to return to a source, then the A-N winding can be used: Leave from A terminal & return to N terminal. Even if the current for an A-N connected load goes through C, it MUST return to the N terminal of the A-phase source.

You can attack me personally for being an idiot if you like, that is a teaching style I haven't encountered in quite a while and try not to employ personally.
Never called you an idiot. Don't think you are an idiot, quite the contrary in fact. I did, however, call you wrong.

I am just a simple electrician speaking to other simple electricians in most cases. I don't much appreciate being called a liar, either: I told you that if you can clearly illustrate why my analogy actively does a disservice in visualizing current flow, I would remove it from my teaching. You then claimed that I must be so invested in it that I can't hear you and would not remove it.
Did not call you a liar either. Nor do I think you are a liar.

Just trying to point out why it might be hard for you to absorb changes to your analogy. It is not all about you the individual. Just something I have observed over many years. People in general do not like change and often cling to what they are familiar with. Nothing so unusual about that characteristic. I do that as well. You may or may not have to watch that but I have no way of knowing.

No, I respect your experience and your knowledge, and I am listening with rapt attention.
Thank you. Then recognize that it is hard to convey in a thread what could be done much quicker in person with a sketch pad and some back and forth. I am trying to find a way to make it clear.

You are not conveying the problem very clearly, and frankly, being damned rude about it.
Not my intent. I thought your comment about my illustration being a "novel" was quite rude but I did not get up in arms about it. I assure you I did not type that illustration for my benefit but only to try and help. Had I known it would have been dismissed so I would certainly have typed something shorter.

I had forgotten about this since the last time I checked in, that's how completely enraptured by this topic I am. Trust me, if you can make your case I will abide by my agreement and cease teaching the method. I do preface that portion of the class with making it clear that I'm not an EE, and this is a cheat to visualize current flow - not that it was fact, for the record.
Your analogy needs changing, not abandoning.
 

mivey

Senior Member
Ok George. I re-read some of my prior posts. I could have been more patient and worded them better. Sorry.
 

mbrooke

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Sure, but that is still not energy going out one wire and back in through the other. :)

Since the two windings are 120 degreesout of phase there is, however, reactive "power" going back and forth . But I don't think that is what you were referring to.
I was not, but you are correct. If the connection is delta there is also "circulating" current, but I understand what you are saying. :)
 
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