Why is residential wiring known as single phase?
- Thread starter TimWA
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Three-phase generators are actually quite common......
T
T.M.Haja Sahib
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The only way to prove which one is correct either 120-0-120 or 0-120-240v is to measure the currents in two outdoor conductors and the neutral for equal loads.
Well-equipped,you could easily do it...........
Why on earth would you want do that when it's a matter of a quite simple calculation?
As shown in post #64.
If, for example, the two loads are resistive and the currents balanced in each winding, the current in the neutral will be zero.
kwired
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about 20% of the way to 600 posts
T
T.M.Haja Sahib
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How it can be?If there are two opposite phases,the current in the neutral is double.
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Why?
It will be entirely in line with expectations.
about 20% of the way to 600 posts
Amazing, innit?
Especially since the discussion is being conducted mainly by two furriners........
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- Connecticut
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- Engineer
No it won't be double, precisely because the load currents are "opposite."
The current in the neutral will be In=Ia+Ib.
If Ia=10A at an angle of zero, and Ib=10A at an angle of 180 (180 is "opposite" of zero), then
In=Ia+Ib=10<0 + 10<180 = 0
T
T.M.Haja Sahib
Guest
Take into account the sine of 180 is -1 and the two current vectors are in opposite direction.So the net current through neutral should be difference of one positive vector and one negative vector.The result is double.
Sorry. That's just plain wrong.
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- Connecticut
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- Engineer
Huh?:?
The sine of 180 is zero, not -1.
But the cosine of 180 is -1, and the cosine of zero is 1.
That's why the current in the neutral will be zero.
In=Ia+Ib=10<0 + 10<180 = (1)*(10) + (-1)*(10) = 0.
T
T.M.Haja Sahib
Guest
Sorry I blundered.
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- Wisconsin
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- PE (Retired) - Power Systems
The voltages are opposite so they add, the currents are opposite so they subtract.:?
So far we have proved that Vnb=-Vbn, pretty cutting edge stuff.
How about solving the equations involved in typical center tapped circuits where the neutral is not common to all of the loads. The one I have asked about previously is a standard 3-wire stove, one load (say 1A) connected L1-N and a resistive load (say 30A) connected L1-L2. Then add a third load (say 10A) from L2-N. Are all of your voltages and currents 'in phase' with each other? Have values changed simply because a neutral point exists?
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- Connecticut
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- Engineer
:? When did I say anything like that? By the way "opposite" was T.M's word, not mine. That's why I put it in quotes.
I said the load on the neutral would be the sum of the current Ia & Ib: In=Ia+Ib.
Assuming all your loads are resistive: IL1=31<0A, IL2=40<180A and In=9<180A
Yes, I think they have. Without the neutral point, it would be difficult to connect the loads from line to neutral. If we left the loads from your example and disconnected the circuit at the neutral point, the currents would change to IL1=31.82<0, IL2=31.82<180 and In=0.
- Location
- Wisconsin
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- PE (Retired) - Power Systems
The major discussion has been that the voltages are out of phase and therefore combine to a larger number. Now you are saying that the currents are out of phase and are combining to a smaller number.
I would like you to show the current and voltages for situation #1 (2 loads) including the source voltages.
Then show the currents and voltages for situation #2 (3 loads).
Did any of the 'math' need to be modified when the source neutral point became common to multiple loads?
Joethemechanic
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K8MHZ
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Joethemechanic
Senior Member
- Location
- Hazleton Pa
- Occupation
- Electro-Mechanical Technician. Industrial machinery
Oh, look! There is the square root of 2!
lol, how did I miss that. BTW, here in Philly we did two phase the correct way, we used a 5 wire service. None of that messing around with that silly square root of two and the neutral that wasn't really a neutral that those places that had 3 wire 2 phase had to do.
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