# Why can any two phases be connected together?

1. Jan 17, 2010

### foo

The way I understood it and that made sense when connecting two phases together was that one of the phases was in the exact opposite range(polarity). So one phase would allow current to enter the circuit and then the other phase would allow the current back out of the circuit, back to the source. So it appeared to me that only certain combinations of 2 phases would work.

Well, then I read that ANY two phases could be connected together which totally blows that theory up.

How come a phase that's only 180' will add up voltage the same as the phase that's a whole 360' in the negative range?

I guess any phase to neutral voltage times the square root of 3 will equal the two phases. But why is this true for any phase combination?

How does two phases complete a circuit for current to travel?
I thought that the 180' would bring in because current is traveling in one direction(positive scale) and then the -180' would take it back as current would then be traveling in the opposite direction(negative scale)?

2. Jan 17, 2010

### Snoogans

When you say 180', do you mean 180 degrees in terms of phase. In a standard 3 phase sytem the phases are 120 degrees out of phase with each other.

The problem is usually best visualised using vector addition.

3. Jan 17, 2010

### foo

oops, yes 120

So when phase 1 is lets say at 120, and it's working in conjunction with a phase that's near zero, how is current able to return?

4. Jan 17, 2010

### gnurf

In the linear circuits that you refer to, luckily (!), the http://en.wikipedia.org/wiki/Superposition_principle" [Broken] is at play. From the wiki:

I'm going to leave this question for someone who will give you a better answer, but what you have stumbled upon is a fundamental physical concept, and an invaluable tool in electronics and elsewhere.

Last edited by a moderator: May 4, 2017
5. Jan 17, 2010

### Snoogans

Sorry, I said addition earlier, it should be subtraction.

It's best to look at one of the phases as being a reference, say 0 volts, then the other phase is the vector difference of the 2 phases. (yes this is superposition)

Lets say,

100V 0$$\phi$$ and 100V 120$$\phi$$

[Isert maths here] <--- I'm at work

And you get the resultant vector as 100$$\sqrt{3}$$ -30$$\phi$$, this is the Voltage between the 2 phases. It has the same frequency, just and different magnetude and phase.

As for where the current goes, that is hard to visualise. I may try to explain later.

6. Jan 17, 2010

### gnurf

The situation is not one where you have three childish currents fighting and pulling each other this way and that. You have three components, but there's only one spoon. Yes, spoon as in current.

7. Jan 17, 2010

### foo

100V 0 and 100V 120

See, that's not how I thought it was suppose to work. I thought it was like this.
One phase is 100V at 120 and another phase is -100V at 120.
That makes sense; I see one side as being the positive and the other as being the negative. So initially I thought, yes we can use two phases but only when those two phases are fully inverse of each other. This appears to me as being the way that current can be carried back and forth throughout the cycles. One phase is coming in positive while another phase is going out negative.

But instead I am now finding out that we can also have 0V at 120(one of the phase wires) and 100V at 120(the second phase wire) or -100V at 120 and 0V at 120, which doesn't explain how current travels back to the source.

EDIT, wups; I'm not trying to use 120. I'm trying to use the full 360 degree range.
The top wave will be at 180'.
The bottom wave is at -180'.

Last edited: Jan 17, 2010
8. Jan 17, 2010

### Staff: Mentor

What the heck are you guys talking about?

9. Jan 17, 2010

### dlgoff

I didn't know what to say!

10. Jan 17, 2010

### Snoogans

For simpicity we don't refer to the sinusoidal aspect of the system because we are assuming a contstant frequency. The voltage on each phase is described by the following functions:

P1 = V sin(ft + 0)
P2 = V sin(ft + 120)
P3 = V sin(ft - 120)

Where V is the peak Voltage, f is the frequency and t is time. If you plot these functions against time you get the diagram you showed earlier.

This is something that you may need someone to show you in person.

11. Jan 17, 2010

### foo

I'm sorry, I've gotten confused and really goofed up what I'm trying to say.

In the picture, we have a wave that is at peak for one phase. It's at 180. The three waves are displaced 120 degrees that's why they are at different heights at any given instant.
The one phase is at it's positive range meaning that voltage is positive.
There is another phase that is at a negative range so voltage is negative.

What I thought this showed was that the one phase that's positive is kind of like our hot and the phase that's negative is kind of like our return for current since it's going in the opposite direction. This was how I believe current was forced to travel through a circuit.

But when a phase that is near zero is being introduced;
that one phase can be at it's positive and another can be near zero in height - put together doesn't explain how current can follow a complete path back to the source. Voltage is pushing in, but no voltage is pulling out of the other phase; for example two phases connected to a transformer.

12. Jan 17, 2010

### Averagesupernova

I think there are some misconceptions going on here.

13. Jan 17, 2010

### foo

I'll start from scratch, blank slate.

AC

2 phases coming in.

How does current move when these two phases are connected?

14. Jan 17, 2010

### Averagesupernova

Any pair of wires with a voltage between them with a suitably low source impedance is able to source current. Your drawing doesn't really mean much to me. Incidentally, are we talking about 3-phase delta, 3-phase wye, single phase with 2 hots and a neutral, all of the above?

15. Jan 17, 2010

### foo

Lets consider 3 wire, 3 phase delta.

16. Jan 17, 2010

### gnurf

I can only speak for myself, but I was not talking about three phase electric power as I now realize the OP was. I guess the superposition principle still holds, but maybe the spoon thing was pushing my luck a little too far... (ahem)

17. Jan 17, 2010

### dlgoff

Last edited by a moderator: Apr 24, 2017
18. Jan 17, 2010

### Averagesupernova

I like to explain things in terms that makes it easy to visualize. 3-phase delta is a pretty easy one to answer. There are three transformer windings (secondaries) that are hooked in a series. Drawn out they appear as a triangle, hence the reason we call it delta. Each 'phase' comes off of a node from two windings. So, grab any two phases and you can see they are directly across a transformer winding. I don't see how you could not see that you can source power from any two phases. Maybe I missed the point?

19. Jan 17, 2010

### Phrak

I think a phase diagram is the only thing that can return sanity to this thread.

20. Jan 18, 2010

### sophiecentaur

foo - your red and blue picture is 'unconventional' and I think it shows that you are confused about what is happening. Arrows, such as you have drawn are usually taken to mean current flow. In that diagram, there would be no net current flow because it doesn't have anywhere to go - the two currents are in opposite directions.

Two wires with potentials which are alternating and in anti-phase will be in the same situation as if one wire is at Earth and the other is at twice that potential - the difference in potential is the same in each case.
If they are supplied with enough current to maintain them at these potentials then you can get power out of the arrangement. "Back at the supply" you could have a transformer winding which is connected to the two wires. If you connect the centre turn of this winding to Earth, the two outputs will be 180 degrees out of phase ' about Earth'.
You could also achieve this with two transformers, one for each wire and with its other output connected to Earth. The outputs could be chosen to he in phase or in anti phase- depending on which way round you connect the wires.
In a three phase supply your generator will produce three outputs, each of which is 120 degrees out of phase to the others. The generator often have three windings, connected as a 'star' or "Y" with one end of each of its windings at the centre and the other ends will have the three phases. This centre point may be connected to Earth, keeping the three phases nicely symmetrical about 0V. The 'return path' for currents flowing through a load connected across two of the phases will be via two of the windings - that was one of the original questions.

You really need to browse through Wikkers (and all the rest) to get a better idea of what's going on.

21. Jan 18, 2010

### zgozvrm

First of all, you would never want to connect 2 different phases together. I think what you mean to say is “connect a load to 2 different phases.”

I know, I know … it sounds like a minor detail. But then again, if someone owes me $1000.00 dollars but misplaces the decimal point and only writes me a check for$10.00, I’m going to be a little upset.

The point is, precision and accuracy are important (not only in calculations, but in what you say as well).

22. Jan 18, 2010

### zgozvrm

I think you're under the misconception that, at any given time, there is a positive voltage on one leg and a negative voltage on another leg. That is not true. Take 120V, single phase for example: You have a hot wire and a neutral wire. The neutral wire is always at 0 volts, whereas the hot wire oscillates between positive 120V and -120V (60 times per second). So there is never a time where you have 120V on one leg and -120V on the other (or +10V on one and -10 on the other, etc.) since one is always at zero volts.

23. Jan 18, 2010

### zgozvrm

Two different phases of a standard three-phase system are never in phase with each other. Each phase is at the same frequency (60 Hz in the U.S.), so they rise and fall at the same rate and one never "catches up" with another. Therefore, you can't have one leg (a "phase wire" as you call it) at 0V, 120 degrees and another leg at 100V, 120 degrees.

I'm not sure, but what you may be trying to say is something like this:

"I am now finding out that when one leg is at 120 degrees in its cycle, it is at 0V and, at the same time, another leg is at 100V."

But this isn't even possible, because when one phase is at 0V, the other 2 phases are at approximately 103.9V and -103.9V, respectively. Similarly, when one phase is at 100V, the other 2 phases are at approximately 7.4V and -107.4V respectively.

24. Jan 18, 2010

### zgozvrm

You don't need to have one line positive and one line negative in order to get current to flow. All you need is a difference in potential.

For example, in a DC circuit, you could supply a load with 2 different positive voltages, say 10V and 8V, you would have a potential difference of 2V. Therefore the circuit would be the same as if it were supplied by a single 2V battery where the positive side of the 2V battery would be pointing in the same direction around the circuit as the 10V battery did. (see attachment)

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25. Jan 18, 2010

### zgozvrm

Also, if you have 2 equal voltages from different phases, you still have a potential difference, and therefore you will have current flow. For instance, say you have one leg measuring 100V at 0 degrees, and another leg measuring 100V at 120 degrees, there is a potential difference of 100V at 60 degrees. This can be shown using vector addition (see attachment).

Note that the only time the voltages between any 2 phases of a 3-phase system coincide is at plus or minus 60V.

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