Why is residential wiring known as single phase?

[jim dungar]

Finegood.
So, in the circuit in post #1004, are I_a and I_b in phase?
That's just yes or no.

Wish it were that easy.

I stated this previously, a good 1/2 hour before you asked this specific question, so our posts may have crossed.:

I know that an AC waveform has a portion above the 0V point, which we often refer to as the positive half cycle, as well as a portion below, which we call negative. However when performing circuit analysis we look at a 'snapshot in time' RMS value when assigning a direction, and not the individual portions of the AC waveform(s).

The load sees (2) different currents flowing in the same direction 'from the top of the load towards the bottom of the load'.
Because current Ia and Ib are not flowing at the same time, the phase relationship between them is not readily apparent from your graphics.

 

[rattus]

Haven't I been including a neutral when I write: Van+Vnb=Van-Vbn, where Vbn=-Vnb?


So then, including a neutral.
One group says: single phase systems are made up of (2) out-of-phase voltages connected in series with a neutral point.
One group says: single phase systems are made up of (2) in-phase voltages connected in series with a neutral point.

Or supply your own summations.

Yes, the summations are OK, but your original statements did not include the neutral. My problem is calling it a series connection because a third wire is actually included. I know what you mean though.

 

[rbalex]

My diagram very clearly indicates that their conduction periods do not start at the same point.

I_a starts at π/4. I_b starts at 5π/4.
Neither is at 0?.

Surely, as a professional engineer, you can understand that.

As a Professional Engineer, I?ll give you one last chance to open your eyes before I?m forced to conclude you are so intractably entangled in the leads of the ?Oscilloscope Jungle? that you can?t see the phase for the traces.

Set your circuit up just as before but open circuit branch I_a. If you can?t see that the I_b trace has the identical period of its driving voltage, you best call Mr. Ohm and tell him his law doesn?t apply when you monkey with the circuit with clips and rectifiers.

 

[rattus]

As a Professional Engineer, I?ll give you one last chance to open your eyes before I?m forced to conclude you are so intractably entangled in the leads of the ?Oscilloscope Jungle? that you can?t see the phase for the traces.

Set your circuit up just as before but open circuit branch I_a. If you can?t see that the I_b trace has the identical period of its driving voltage, you best call Mr. Ohm and tell him his law doesn?t apply when you monkey with the circuit with clips and rectifiers.

Now I would say that Ia and Ib do have the same period, but what does that prove? It is irrelevant.

It should be obvious to anyone who understands a full wave rectifier circuit that the voltages that drive Ia and Ib are out of phase by PI. That is the crux of the matter is it not?

Strictly speaking we can't assign a phase to non sinusoids, but we can easily see that Ia and Ib are separated by T/2, or if you insist, PI radians.

Bob, why don't you knock of the questions, and answer some instead.

 

[mivey]

Where have I ever said that certain voltages do not exist.

Many places. You often call one just a math manipulation of the "one real" voltage in existence. But just as an example, let's just look further in your post:

So does it matter if we say the source is -V43 when it is really V34.

"when it is really V34" is saying the voltage across the terminals is really only V34. That means that a voltage across the terminals can't really be V43. If V43 really can't be a voltage across the terminals then you are saying V43 really does not exist because we are just looking at V34 the wrong way. Ergo, you are saying that V43 does not exist except as a mathematical manipulation of a voltage that is really only V34.

Now do you see how what you are saying reads?

In both cases I mathematically swapped V34 with V43 and got identical answers, but the two actual physical connections provide different results. So does it matter if we say the source is -V43 when it is really V34.

Ok. So we see what you have done by using the transformer in two different configurations. Now how about addressing the transformer I presented that is connected in only one configuration?

For my generator example, we have the 0? displaced, series combined voltages Van + Vnb = Vab at the single-phase transformer. We also have the two 180? displaced, differentially combined voltages Van - Vbn = Vab at the exact same terminals. Both voltage designations are based on their real physical make-up.

It is not that a voltage between terminals is "really only V34" but that there also "really exists" a voltage between the terminals of V43.



Another demonstration that both really physically exist is given in my open-wye example where I show that both series voltages X1->X2+X3->X4 and X2->X1+X6->X5 are present. That clearly shows that both X1->X2 and X2->X1 really are there as the transformer is connected. They produce real physical results in both directions.

It is not that one or the other of "X2->X1" or "X1->X2" is "really only the other one mathematically manipulated" but that both really are voltages.

I know that an AC waveform has a portion above the 0V point, which we often refer to as the positive half cycle, as well as a portion below, which we call negative. However when performing circuit analysis we look at a 'snapshot in time' RMS value when assigning a direction, and not the individual portions of the AC waveform(s).

Perform the circuit analysis however you want, but don't somehow think that a voltage is "really only" something as used in a series additive direction. It is also true that a voltage can "really be" something when we have a potential difference as well.

Because current Ia and Ib are not flowing at the same time, the phase relationship between them is not readily apparent from your graphics.

Then consider the circuit halves separately. If we looked at the top half, we would define the positive direction and time based on the relationship of the voltage and the direction of current leaving the voltage terminal. The same for the bottom half. The two halves are taking positive pulses at different points in time. Bringing them together does not change that.

 

[rbalex]

Now I would say that Ia and Ib do have the same period, but what does that prove? It is irrelevant.

It should be obvious to anyone who understands a full wave rectifier circuit that the voltages that drive Ia and Ib are out of phase by PI. That is the crux of the matter is it not?

Strictly speaking we can't assign a phase to non sinusoids, but we can easily see that Ia and Ib are separated by T/2, or if you insist, PI radians.

Bob, why don't you knock of the questions, and answer some instead.

Ah ? rattushopper, it is indeed relevant although, as you typically do, you have misidentified the relevant periods. But rather than put words in his mouth, I will permit Bes to explain why.

As for answering questions, you have been on the ?OK to ignore with a clear conscience list? since failing to respond to Post 1176.

 

[mivey]

As for answering questions, you have been on the ?OK to ignore with a clear conscience list? since failing to respond to Post 1176.

Oh, puh-leeze. Answered more than once.

Post #1176

Let me get this straight - are you saying that the voltage functions of the phasors can't be written as equivalent functions using either (wt) or (wt + 180)?

Do you really want to commit to the statement,"Don't count on identities either. They are useful, but there is nothing hard about them."?

The phasor has a length given by the maximum value of the function and an angle given by the phase angle value from 0? to 360?. A phasor does not have a negative length that cancels out part of the phase angle.

 

[rbalex]

Oh, puh-leeze. Answered more than once.

Post #1176


The phasor has a length given by the maximum value of the function and an angle given by the phase angle value from 0? to 360?. A phasor does not have a negative length that cancels out part of the phase angle.

THANKS - if you buy into rattus' drivel, I can put you on the list too.

 

[mivey]

THANKS - if you buy into rattus' drivel, I can put you on the list too.

The drivel that says Trig identities are not hard? Maybe they are for you, but for the rest of us: not so much.

Or are you referring to the drivel that says you can't just ignore negative signs because you want to use fake math terms?

Yeah, put me on your fake list right along side the list of the fake math terms you made up.

 

[jim dungar]

Many places. You often call one just a math manipulation of the "one real" voltage in existence. But just as an example, let's just look further in your post:

"when it is really V34" is saying the voltage across the terminals is really only V34. That means that a voltage across the terminals can't really be V43. If V43 really can't be a voltage across the terminals then you are saying V43 really does not exist because we are just looking at V34 the wrong way. Ergo, you are saying that V43 does not exist except as a mathematical manipulation of a voltage that is really only V34.

Ergo nothing.

Wow, do you have a hard time following a line of reasoning.
And it appears you have not read any of my posts where I call V43 an actual voltage.

Here let me lay it out again.
Is the connection actually X1-X2 joined to X3-X4? or is it X1-X2 joined to X4-X3?
If it is X2 to X3, then the actual voltages are V12 joined to V34 with the actual result of V14.
If it is X2 to X4, then the actual voltages are V12 joined to V43 with the actual result of V13.

Where does this say something exists and ergo something else does not?
Does not even mention any angles, much less in-phase versus out-of-phase?
In fact, all it does is reference voltages to transformer winding terminals.

 

[rattus]

Oh, puh-leeze. Answered more than once.

Post #1176]

The (fixed or static) phasors in question can be written as

120Vrms(cos(0) + jsin(0))
AND
120Vrms(cos(PI) + jsin(PI)).

(wt does not appear in the arguments of static phasors)

The phasor diagram looks like this:

V2<----------0----------->V1

They are irrefutably out of phase by PI radians. No question about it. For sure.
Really. Scout's honor. Cross my heart. No way out of it. Phasors Rule!

 

[mivey]

Wow, do you have a hard time following a line of reasoning.

Right back at you.

Why do you insist on changing the transformer configuration when that is not what I am talking about?

And it appears you have not read any of my posts where I call V43 an actual voltage.

Maybe that gets dimmed when you say that the output voltage of the winding is really only V34 if the transformer is connected series additive.

Here let me lay it out again.
Is the connection actually X1-X2 joined to X3-X4? or is it X1-X2 joined to X4-X3?

What do you mean by "joined"? X1-X2 is in series with X3-X4 to yield the larger voltage from X1 to X4.

If it is X2 to X3, then the actual voltages are V12 joined to V34 with the actual result of V14.

I have no problem with that except to note that you have only demonstrated series additive.

If it is X2 to X4, then the actual voltages are V12 joined to V43 with the actual result of V13.

I have not connected X2 to X4. I am not re-configuring my transformer. Why can't you follow that in my line of reasoning?

Where does this say something exists and ergo something else does not?
Does not even mention any angles, much less in-phase versus out-of-phase?
In fact, all it does is reference voltages to transformer winding terminals.

Let me try again:

Do you not see that in my SINGLE CONFIGURATION EXAMPLE that I have both the series voltage given by X1->X2+X3->X4 and the series voltage given by X2->X1+X6->X5 ? Can you not follow that reasoning that shows that both X1->X2 and X2->X1 exist in the SINGLE CONFIGURATION. Again, I have not connected X2 to X4. X2 stays connected to X3.

 

[Rick Christopherson]

Many places. You often call one just a math manipulation of the "one real" voltage in existence. .......

I'm the one calling you out on the math manipulation, but you, Rattus, and Besoeker are the ones dodging the discussion.

• Mivey dodged the question via deflection and tried to move down some unrelated path without addressing the real topic.
• Rattus blew it off by claiming that we're only dealing with "Ideal" systems, and therefore only mathematical...all while denying that this is a mathematical discussion.
• Besoeker simply remained silent on the discussion altogether until it blew over.

The three of you keep making your absolute assertions, but when you get cornered by them, you run away. When I ask you guys to confront these assertions, you either run away or deflect the discussion to something else.

 

[Besoeker]

Set your circuit up just as before but open circuit branch I_a. If you can?t see that the I_b trace has the identical period of its driving voltage,

I didn't dispute that V_b has the same period as I_b.
My simple point is that I_a and I_b are not in phase.
I'm sure you know that.

 

[rbalex]

The drivel that says Trig identities are not hard? Maybe they are for you, but for the rest of us: not so much.

Or are you referring to the drivel that says you can't just ignore negative signs because you want to use fake math terms?

Yeah, put me on your fake list right along side the list of the fake math terms you made up.

Again, as typical for you, you didn't read in context or for comprehension, you just ran off at the mouth. You had no idea what he was saying. Go all the way back and read his relevant posts that led up to 1176.

Do you really want to commit to buying into his position?

 

[Besoeker]

I'm the one calling you out on the math manipulation, but you, Rattus, and Besoeker are the ones dodging the discussion.
The three of you keep making your absolute assertions, but when you get cornered by them, you run away. When I ask you guys to confront these assertions, you either run away or deflect the discussion to something else.

I absolutely assert that I_a and I_b are not in phase.
The diagram in post #1004 clearly shows that to be the case.
My experience designing and manufacturing power electronics for forty years required that to be the case.
And here's a physical arrangement that depends on that being the case:

No discussion dodged on my part.

 

[rattus]

I'm the one calling you out on the math manipulation, but you, Rattus, and Besoeker are the ones dodging the discussion.

• Mivey dodged the question via deflection and tried to move down some unrelated path without addressing the real topic.
• Rattus blew it off by claiming that we're only dealing with "Ideal" systems, and therefore only mathematical...all while denying that this is a mathematical discussion.
• Besoeker simply remained silent on the discussion altogether until it blew over.

The three of you keep making your absolute assertions, but when you get cornered by them, you run away. When I ask you guys to confront these assertions, you either run away or deflect the discussion to something else.

From the omniscient one who brought us the "Charge Reservoir", aka as a helicopter.

 

[Besoeker]

Wish it were that easy.

It is. If only you'd let it be so.

Because current Ia and Ib are not flowing at the same time,

Then they are flowing at different times!
Which means they are not in phase.
So different phases. We're getting there.

the phase relationship between them is not readily apparent from your graphics.

Well, one starts at π/4, the other at 5π/4 so it's not too difficult to work out that the phase difference is π.
π is pi. It doesn't display terribly well here.

 

[Rick Christopherson]

I absolutely assert that I_a and I_b are not in phase.
The diagram in post #1004 clearly shows that to be the case.

No it doesn't, because according to Rattus, you cannot have phase angles on anything other than pure mathematical sinusoids. :lol:

That asinine comment not withstanding, your diagram does not dispel any difference between real and mathematical transformations. you are using a mathematical transformation to prove your mathematical transformation. That is a circular argument.

I am not arguing your mathematical transformation. What I am stating is that you cannot present your argument in the real world without a mathematical transformation.

You have created a physical inversion that is mathematically equated it to a phase shift, but are stating your phase shift as though it were physical.

[/QUOTE]

 

[rbalex]

I didn't dispute that V_b has the same period as I_b.
My simple point is that I_a and I_b are not in phase.
I'm sure you know that.

Thanks, then you recognize that each and every relevant voltage or current function of your circuit has the same period P and the fraction t/P of each and every relevant voltage and current function must be identical. (Making the wild assumption they all started at the same time.) Since polarity is not relevant to phase, every relevant voltage and current has the same phase. That means there is only one phase in your circuit; i.e., it is single-phase.

The currents I_a and I_b must have the same phase by definition, but they aren’t synchronized – which is why your circuit works.