Neutral current in a single phase installation confusion

[Calebp43]

In a 3 phase system the neutral carries current that is unbalanced. Okay.
So in a single phase domestic household the neutral carries the same current as the live otherwise the RCD will trip.
So does that mean if we have a perfectly balanced 3 phase system then if I go to one of the phases and test the neutrals going out to one of the lights for example, there would be no current?

 

[Electro-tech]

 

[Rockingit]

..... which is why traditionally you could use a 1/2 sized Neutral conductor for your Line size, and now also the same reason why the DNO's keep having cables explode under pavements because of the imbalance of modern systems due to SM PSU devices all over the land!

 

[timhoward]

So does that mean if we have a perfectly balanced 3 phase system then if I go to one of the phases and test the neutrals going out to one of the lights for example, there would be no current?

Excuse the appalling artwork, but are you asking whether you will see a difference if you clamp point A (pink) as the N returns from one phase or point B where the N from loads on all phases is combined?

 

[davesparks]

So does that mean if we have a perfectly balanced 3 phase system then if I go to one of the phases and test the neutrals going out to one of the lights for example, there would be no current?

If you are asking whether you will see current in the neutral of a single phase circuit taken from a 3 phase supply yes you will.

 

[cireland]

Only if it's a 3P light......etc.

Stupid rule.

 

[brianmoooore]

Test at A, and you'll measure a current. Test at B and you won't.

 

[timhoward]

I can't say I've had cause to use it much, but if you ever need to know the amount of N current in a 3 phase system current my notes say:
it's the square root of ( L1² + L2² + L3² - L1L2 - L1L3 - L2L3)

where L1 is L1 current etc.

 

[Rockingit]

I can't say I've had cause to use it much, but if you ever need to know the amount of N current in a 3 phase system current my notes say:
it's the square root of ( L1² + L2² + L3² - L1L2 - L1L3 - L2L3)

where L1 is L1 current etc.

OK - so some maths in action on this one…. A while back I was explaining 3ph theory to someone and how it relates to ultimate efficiency etc and they challenged me as to why therefore don’t we use six phases to be even MORE efficient. As a curiosity experiment I later fired up Excel and expanded the formulae that you’ve quoted above which we all know and turned it into the six phase combinations (considerably more) at which point excel fell over. Anyone know why? (I don’t - genuine question!)

 

[Calebp43]

Okay so if there is current at A and not B then where does the current from A go? I know it "balances out" but I don't understand how though. I'll watch that video now and see if it tells me

 

[James]

Through the lamps on yellow and blue phases
if the lamps are all the same rating, theoretically all of current A would travel down this route and the system would be balanced and N current would be zero.

imagine though the lamp on blue phase blew and became open circuit.
a portion of the current would then go down the N, some from the lamp on red phase, some from lamp on yellow phase.

 

[timhoward]

I wonder whether another mathematical way of thinking about it is helpful.
When generated 3 phase is 3 separate waveforms. They are not all positive and negative at the same time.
Divide 360 degrees by 3 = each phase is 120 degrees apart.
The current at any given phase angle is found by multiplying the current by Cosine (phase angle)

So at any given moment, a balanced 5 amp load on each phase is
5Cos(0) +
5 Cos (120) +
5 Cos (240) [or -120 normally]

= 0

The key point is that at the point one phase is positive, the other two are negative.

So while “where does the current go” is the right question, the answer is that it goes nowhere but is offset by negative currents form the other phases.

 

[snowhead]

and they challenged me as to why therefore don’t we use six phases to be even MORE efficient.

Anyone know why? (I don’t - genuine question!)

I don't know the official answer but would guess it's practicality of not having to have 6 lines instead of 3, alternators with more complex connections, transformers with more coils / connections, isolators and switchgear not having 6 poles etc.
Not forgetting physically bigger cables due to the additional, though smaller cores.

 

[Rockingit]

I don't know the official answer but would guess it's practicality of not having to have 6 lines instead of 3, alternators with more complex connections, transformers with more coils / connections, isolators and switchgear not having 6 poles etc.
Not forgetting physically bigger cables due to the additional, though smaller cores.

I think you misunderstood me there (or more likely I did a bad job of explaining myself....) What I was referring to was simply the maths - I understand all the practical engineering reasons.

If we had a theoretical six phases then if we stick to the principle of the given formula we'd end up with..

SQRT((A^2, B^2, C^2, D^2, E^2, F^2) - (AB, AC, AD, AE, AF, BC, BD, BE, BF......)) however it doesn't work in practice.