neutral currents

Discussion in 'Electrics UK' started by jamescook, 27 Sep 2016.

1. jamescook

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I can't quite understand the neutral.

Its the return path after the load (do correct if I'm wrong)

Its at 0 volts, but if a break in the neutral happens, the neutral will rise to mains voltage, the circuit has become a series circuit, and anything that comes into contact with both ends of the broken neutral will become the new load

Single phase

An RCD would see the current come in though the line conductor, it would then power its load and leave by the neutral conductor.
If 10 amps came in, 10 would leave, and the RCD is happy, if only 9 left, it would trip.
This is where I start to get confused.

So 230 volts is pushing the current to the 10A load,
The load takes the 10A and returns on the neutral.
How come its still 10 amps on the neutral, with 0 volts?

The neutral is at 0 volts, so no push, but I know the 'power' really wants to get back to the start, but what is pushing it.

3 phase

Clamp on the neutral

Brown phase switched on, 50A say.

so 50 Amps on neutral

Black phase switched on 50A

How many amps through the neutral?

Grey 50A

Amps through the neutral?

The three phases all being 50A, would they cancel each other out, so the neutral return would be around zero? but what if only one phase used?

Thanks for anybody who could get through that, and be inclined to reply.

2. JohnW2

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Voltage only has meaning as the voltage difference ('potential difference') between 2 points. When you say that neutral is 'at 0V' what you really mean is that is is roughly at 0V relative to earth (since neutral is connected to earth at the substation). Relative to the line/live conductor, neutral is 'at 230V'.

The electricity source (i.e. the 230V coming out of the transformer at the substation) is 'pushing current all the way around a loop involving any load. So, that loop starts at the transformer, goes along the L wires, through the load and then all the way back to the transformer through the N wires. If you cut any of those wires (L or N), the current everywhere in that loop would stop.

Think of it in terms of a circulating pump. The pump is analogous to the electricity source, and pushes water all the way around the 'loop' and back to the pump - e.g., in a CH system, along the flow pipework (analogous to L wires), through radiators (analogous to the electrical load) and then back through the return pipework (analagous to neutral wires) back to ('the other side of') the pump.

Does that help at all?

Kind Regards, John

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3. EFLImpudence

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You're probably best Googling lots of articles.

Meanwhile, this won't help.

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4. 333rocky333

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If as you say 10 amp enter the load via the rcd, 10amp will always leave the load, however if say in a fault only 9amp goes back via the N on the rcd and the other 1 amp goes through your body or some means to earth, then
That RCD with 10 amp leaving it and only 9 returning, is what trips it, however the return from the load will always somehow total the 10.
However mains current is AC to further complicate things.
each phase is 120 degrees out of phase with each other, so its not a case of adding the 3 phases.
1 phase as you say would be L and N the same
2 phase would need a formulae to work out
3 phases all at 50 amp would as you say have 0 current on the N, usually only relevant to moters.
3 phases at various currents would again need a formulae

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5. flameport

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Current isn't something that is consumed by a load, it just represents the quantity of electrons moving in the circuit.
Voltage is the force applied to move those electrons, higher voltage = more electrons and more current.

Basic concepts: http://www.bbemg.be/en/main-emf/electricity-fields/electrical-concepts.html

For the 3 phase neutral current, it depends on the current in each phase. It's not a matter of cancelling out, but rather that the direction of current changes all the time so if it flows 'out' of one phase, it will flow 'into' the other 2 phases.

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6. ColJack

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three phase is a little more difficult to explain in terms of water as above but I'll give it a go..

There are 3 "circulating pumps"

The neutral is a bucket
At any given time 1 pump is filling the bucket while the other two are emptying it at a slower rate..

In a balanced load, the amount of water going into the bucket is exactly equal to the amount of water being sucked out by the other two pumps combined.. ( zero neutral current )

If you turn one of the other 2 pumps off ( or down ) then the bucket would fill up slowly ( smaller neutral current )
If you turn both of the other 2 pumps off then the bucket would fill up faster ( higher neutral current )

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7. 333rocky333

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Hi coljack, you not been around for a while

8. endecotp

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Yes

50A. The total phase current is 100A; half of that returns through the other phase and half through the neutral in this case. This is best explained using a diagram.

Zero.

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9. JohnW2

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With (180°) 2-phase (although some here don't like the term) you don't need 'a formula'. In analogy with the 3-phase scenario, if such a 2-phase supply has 50A in both phases, then (just as with balanced 3-phase) it would also have "0 current on the N". If the currents in the two phases were different, then the neutral current would be the difference between the two phase currents.

Kind Regards,

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10. 333rocky333

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I take it he meant 3 phase loaded at
L1 50 amps
L2 50 amps
L3 0 amps
I dont agree the answer is either 0 Amps or 50 Amps as someones stated

11. endecotp

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It's undoubtedly 50A. Draw a diagram; it's equilateral triangles.
What do you think it is?

12. JohnW2

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OK. I think it was you who first mentioned 2-phase, and I thought you actually meant 2-phase
I think you'll find that the answer is 50A. You could use a formula, or trigonometry to work that out, but in that simple case I think all you have to do is note that (as you know) the current you would have to have add in L3 to make things balanced (hence neutral current = 0) would be 50A. In a less simple case (e.g. L1=50, L2=40, L3=0), you definitely would need to do some maths!

Kind Regards, John

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13. 333rocky333

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still dont get how you get 50, for some reason
Though I concede I was wrong, apoligies

14. JohnW2

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Well, if you want to use 'a formula' then if you use A, B and C to represent the currents in L1, L2 and L3, then:

neutral current = sqrt(A² + B² + C² - AB -AC - BC)

... which, with A=50, B= 50, C=0 becomes:

= sqrt(50² + 50² + 0² - 50*50 - 50*0 - 50*0)
= sqrt(2500 + 2500 + 0 - 2500 - 0 - 0)
= sqrt(2500)
= 50

Any help?

Kind Regards, John

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