Why is residential wiring known as single phase?

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rbalex

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Alright, then let phi0 be zero. Now the phases of the V1 and V2 are:

phi1 = (wt)

phi2 = (wt +/- 180)

They are clearly not equal. The angular positions of the two waves are not the same. We have a phase displacement or phase shift. They are out of phase. You have shown V2 is the negative of V1 but that does not prove your point.

Maybe there is something I missed. Tell me if I am wrong.
Those equations definitely aren't equal; but when used as the arguments for reduced periodic functions, such as Sine or Cosine, the phases [wt] remain the same.
 

mivey

Senior Member
I have reduced a periodic function - properly.
No problem with the reduction. I had no problem when I corrected your original reduction either. The corrected math is fine.

Either you don't understand that or you simply refuse to accept mathematically valid substitutions because you don't like them.
My likes or dislikes have nothing to do with it. But for the record, I like math. What you do not understand is that you have changed the phase constant. What you are left with is a phase constant for a different function. Whether you realize it or not, you have shifted the original function. Whenever you get the time, review your diff-eq texts or physics texts dealing with oscillating systems and hopefully you will understand what you have done. If not, I'll be glad to help.

Until then, I suspect you will continue to miss the fact that you have nothing to substantiate your claims. You somehow think the reduction did not change the function. The physics and math say otherwise.
 

mivey

Senior Member
Those equations definitely aren't equal; but when used as the arguments for reduced periodic functions, such as Sine or Cosine, the phases [wt] remain the same.
So now you are calling "wt" alone the phase? The phase of a function also includes the phase constant. The phase constant is found when solving the differential equation for the iniial conditions. That is what the source you keep referencing is telling you but you keep overlooking.
 

rbalex

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... You somehow think the reduction did not change the function. The physics and math say otherwise.
Yes it changes the function - what you don't understand or accept is that it doesn't change the characteristic phase of the function. And differential equations has nothing to do with it any more than any of the other rabbits you have been chasing.
 

mivey

Senior Member
Yes it changes the function - what you don't understand or accept is that it doesn't change the characteristic phase of the function.
Something you have completely made up.

And differential equations has nothing to do with it any more than any of the other rabbits you have been chasing.
Then I'll just leave you to it and let you keep living the dream.
 

gar

Senior Member
Location
Ann Arbor, Michigan
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EE
120213-1042 EST

T.M.:

If we consider a generic scope where we directly connect to the deflection plates, then since these are insulated from each other we can apply any voltages that we want up to breakdown voltage or deflection sensitivity limits.

If we consider a typical scope with single ended inputs for each channel (no isolation and no differential amplifiers at the input), then both channels have an electrically common terminal for one side of the input, and this also connects to the AC power EGC. In slang terms this is called ground whether or not the scope EGC is floated. Thus, one can not put one input ground lead on neutral, and the other ground terminal on a hot wire. Big sparks.

The two of you were not on the same page in your discussion. This last sentence is all slang and may not be fully understood by an Indian, a Scotchman, and me. While I have English, German, Welsh, Norwegian, Swedish, French, and American Indian ancestry in my background I am not native to any of these. I just discovered the French part by looking into a US census of Indians in the early 1800s. I am just a US great lakes person. So the details of my language are different than yours. This whole thread is basically a definition or language problem.

.
 

rbalex

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Professional Electrical Engineer
So now you are calling "wt" alone the phase? The phase of a function also includes the phase constant. The phase constant is found when solving the differential equation for the iniial conditions. That is what the source you keep referencing is telling you but you keep overlooking.
One more rabbit - you apparently didn't review the post I was responding to.
 

rattus

Senior Member
Those equations definitely aren't equal; but when used as the arguments for reduced periodic functions, such as Sine or Cosine, the phases [wt] remain the same.

But the phase of the waveform seen on L2 is not wt; it is (wt +/- 180). No need to reduce anything.
 

rbalex

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Professional Electrical Engineer
This is getting to be like saying, "1 Ξ sin2θ + cos2θ" and the next guy saying, “Well it sure as hell doesn’t look like ‘1’ to me.”

Edit Add: No one is obligated to agree with my definition of phase. But my math is consistent with the authoritative definition I cited. And it is consistent with why we call conventional single, two and three-phase systems what we do.
 
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Besoeker

Senior Member
Location
UK
120213-0830 EST

Besoeker, T.M., and others:

In post 501 I discussed scopes and Lissajous figures. To eliminate the ground problem I described the signals to the deflection plates. This eliminates the common connection problem. But this can also be solved by using differential inputs to the scope channels.

Gar, a couple of points if I may.
Your post #501 referred to CRT scopes and signals applied to the deflection plates.
I'm old enough to have used CRT scopes but have not done so in years. Can you still buy them?
I also don't recall having direct access to the X and Y deflection plates.
But that's probably academic given that modern and even not so modern scopes are mostly, if not all, solid state these days.

Differential probes, I agree, are another solution. But we are looking at somewhat specialist kit for that. The pictures I displayed earlier were from a Tektronix TDS210 and Gossen 100:1 2kV probes.
It's a pretty good arrangement for what I do.
The Tektronix differential probes that come closest in terms of bandwidth to suit the scope and voltage capability would cost 50% more than the scope itself.
Per probe.
 

Besoeker

Senior Member
Location
UK
gar
I request your comments for above with respect to your post #714

TM, here's what we have that you have been trying to tell me how to connect:

120-0-120Scoped01.jpg


Two signal leads each with a ground.
What do you want me to connect to each of the transformer terminals?
 

pfalcon

Senior Member
Location
Indiana
FINALLY. Forget all the rest Mivey.

Of course it is single-phase.
BINGO!!

But it is capable of delivering two phases displaced by 180?.
Yes it can. But it requires additional circuitry to do so. Otherwise you're just shorting the two ends of a coil with a periodic 240V difference.

Which is where you and I had arrived on the prior thread. If there is a COMPELLING NEED to know it's single-phase then yes that's exactly what it is. Otherwise for all practical purposes you can treat it as two opposing phases because that's exactly how it will behave.

At peak: The peak readings are -120, 0, +120 at 90o and +120, 0, -120 at -90o for the 240V waveform. This is a simple voltage gradient from one end to the other. The neutral is a simple voltage division point centered on the coil, which, because of the convenient but not required convention of assigning the neutral a value of 0V gives those readings. If you are wanting to utilize the +120V from the alternating 90o and -90o you can do so with additional circuitry. Can we say rectifier circuit?
 

gar

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Location
Ann Arbor, Michigan
Occupation
EE
120113-1401 EST

Besoeker:

I may have two, but at least one scope with direct access to the deflection plates. But I have never made direct connection. My two newer scopes, Tektronix, have no easy connection to the plates.

Then I have a Dataq data acquisition device that has 16 inputs or 8 differential inputs.

The one scope that I have that definitely has direct plate access is one I built from a WWII surplus 5BP1 CRT tube. Made it in two days, could see sine waves on it, not very wide bandwidth, just audio. I believe my Heathkit scope has terminals on the back for direct plate connection. Much of my older equipment does not work correctly or not at all. A Heathkit audio VTVM, and audio oscillator are still working. Whereas a slightly newer HP oscillator and several VTVMs are not working. Some time I need to fix them.

.
 

pfalcon

Senior Member
Location
Indiana
I seem to read that some think the two 180? displaced phases do not exist because they are not really displaced. That is what I take issue with. I contend that both voltage output options exist and are usable as such and that the displacement really does exist.

With respect of one end of the circuit to the other end of the circuit, just as with a teeter-totter, the two ends are effectively 180o out of phase, can be used in that manner, and for all practical purposes may be considered in that manner. Only when being .... retentive is it necessary to acknowledge that at no time are they really displaced but are rather two ends of a single whole piece.
 

Besoeker

Senior Member
Location
UK
120113-1401 EST

Besoeker:

I may have two, but at least one scope with direct access to the deflection plates. But I have never made direct connection. My two newer scopes, Tektronix, have no easy connection to the plates.

Then I have a Dataq data acquisition device that has 16 inputs or 8 differential inputs.

The one scope that I have that definitely has direct plate access is one I built from a WWII surplus 5BP1 CRT tube. Made it in two days, could see sine waves on it, not very wide bandwidth, just audio. I believe my Heathkit scope has terminals on the back for direct plate connection. Much of my older equipment does not work correctly or not at all. A Heathkit audio VTVM, and audio oscillator are still working. Whereas a slightly newer HP oscillator and several VTVMs are not working. Some time I need to fix them.

.

Gar

Thank you for your response.
I do remember WWII surplus kit being on the market. It was to mail order catalogues what ebay is today for those searching a bargain.
A friend of mine made fishing rods out of tank ariels. He made one for me but fishing just wasn't my thing.
 

pfalcon

Senior Member
Location
Indiana
120213-1440 EST

What is the period of a sine wave? Pi or 2*Pi? (180 or 360)

.

The proverbially sine wave starts at a value of 0 at 0; rises to +peak at 90 (pi/2); drops to 0 at 180 (pi); drops further to -peak at 270 (pi*3/2) and returns to 0 at 360 (2*pi).

So when you eat your pi you still have pi :D
 
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