1. Visiting from the US? Why not try DIYnot.US instead? Click here to continue to DIYnot.US.
    Dismiss Notice

neutral currents

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

  1. jamescook

    jamescook

    Joined:
    27 Sep 2016
    Messages:
    1
    Thanks Received:
    0
    Country:
    United Kingdom
    Could someone please help me or point me in the right direction.

    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

    JohnW2

    Joined:
    28 Jan 2011
    Messages:
    41,976
    Thanks Received:
    2,638
    Location:
    Buckinghamshire
    Country:
    United Kingdom
    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
     
    • Thanks Thanks x 1
    • Like Like x 1
  3. EFLImpudence

    EFLImpudence

    Joined:
    7 Jul 2010
    Messages:
    29,759
    Thanks Received:
    3,206
    Location:
    Retired to:
    Country:
    Portugal
    You're probably best Googling lots of articles.

    Meanwhile, this won't help.
     
    • Like Like x 3
  4. 333rocky333

    333rocky333

    Joined:
    12 Jan 2008
    Messages:
    7,316
    Thanks Received:
    699
    Location:
    Essex
    Country:
    United Kingdom
    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
     
    • Thanks Thanks x 1
    • Like Like x 1
  5. flameport

    flameport

    Joined:
    10 Mar 2007
    Messages:
    8,230
    Thanks Received:
    1,511
    Location:
    Poole, Dorset
    Country:
    United Kingdom
    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.
    https://upload.wikimedia.org/wikipedia/commons/4/48/3-phase_flow.gif
     
    • Thanks Thanks x 1
  6. ColJack

    ColJack

    Joined:
    16 Feb 2007
    Messages:
    11,793
    Thanks Received:
    409
    Location:
    West Midlands
    Country:
    United Kingdom
    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 )
     
    • Thanks Thanks x 1
  7. 333rocky333

    333rocky333

    Joined:
    12 Jan 2008
    Messages:
    7,316
    Thanks Received:
    699
    Location:
    Essex
    Country:
    United Kingdom
    Hi coljack, you not been around for a while :)
     
  8. endecotp

    endecotp

    Joined:
    2 Dec 2013
    Messages:
    3,134
    Thanks Received:
    386
    Country:
    United Kingdom
    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.
     
    • Thanks Thanks x 1
  9. JohnW2

    JohnW2

    Joined:
    28 Jan 2011
    Messages:
    41,976
    Thanks Received:
    2,638
    Location:
    Buckinghamshire
    Country:
    United Kingdom
    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,
     
    • Thanks Thanks x 1
  10. 333rocky333

    333rocky333

    Joined:
    12 Jan 2008
    Messages:
    7,316
    Thanks Received:
    699
    Location:
    Essex
    Country:
    United Kingdom
    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

    endecotp

    Joined:
    2 Dec 2013
    Messages:
    3,134
    Thanks Received:
    386
    Country:
    United Kingdom
    It's undoubtedly 50A. Draw a diagram; it's equilateral triangles.
    What do you think it is?
     
  12. JohnW2

    JohnW2

    Joined:
    28 Jan 2011
    Messages:
    41,976
    Thanks Received:
    2,638
    Location:
    Buckinghamshire
    Country:
    United Kingdom
    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
     
    • Thanks Thanks x 1
  13. 333rocky333

    333rocky333

    Joined:
    12 Jan 2008
    Messages:
    7,316
    Thanks Received:
    699
    Location:
    Essex
    Country:
    United Kingdom
    still dont get how you get 50, for some reason :)
    Though I concede I was wrong, apoligies
     
  14. JohnW2

    JohnW2

    Joined:
    28 Jan 2011
    Messages:
    41,976
    Thanks Received:
    2,638
    Location:
    Buckinghamshire
    Country:
    United Kingdom
    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
     
    • Thanks Thanks x 2
  15. DIYnot Local

    DIYnot Local

    Joined:
    3 Sep 2019
    Country:
    United Kingdom

    If you need to find a tradesperson to get your job done, please try our local search below, or if you are doing it yourself you can find suppliers local to you.

    Select the supplier or trade you require, enter your location to begin your search.


    Are you a trade or supplier? You can create your listing free at DIYnot Local

     
Loading...

Share This Page