The IET is carrying out some important updates between 17-30 April and all of our websites will be view only. For more information, read this Announcement

Rules and Guidelines
Want to participate in the discussions? Please read our Community Rules and Guidelines. Need some help? Visit Support and Answers.

Options
This discussion is locked.
You cannot post a reply to this discussion. If you have a question start a new discussion

Neutral Voltage Question

fiftyhertz
Hi all, 


Hoping someone can help with this, as it bothering me that I dont know this. 


This is quite tricky to explain without an image. Essentially, if we have a circuit supplying one luminaire. The line conductor has a voltage of 230v from earth potential supplying the light. To complete the circuit, on the return leg a neutral is required (<50v from earth potential). Where is the point where the neutral is no longer at mains voltage, is it at the neural terminals? 


Is it a case of the luminaire will "use up" the supplied mains voltage? 


Any assistance is appreciated. 


Thanks
Register to replyAlready registered? Log in and reply
  • 0
    Mapj1 - Forum closing soon, so if we get pushed off, please start again in the new place, same title. 


    OK, so the voltage between L and N is 240V or whatever at the substation, and so long as current is flowing, then as we walk along the wires towards the load, voltage is dropped, I * R but R in the wires is very small - we make the wires big enough this is alwys true. As voltage is dropped in the live, the voltage is falling as we move from the source and is nearest earth towards the load end, even if that is only a volt or two lost out of 230. 

    The current in the neutral is flowing the other way, so the voltage drop is upwards - rising a volt or two nearer towards the live at the load end. The load sees a difference relative to the origin, due to both Live and Neutral voltage drops. 


    (the exact voltage may be complicated as the 0v refernce, the N-E bond, may not be at the substation, but the slope is always that way, but some voltages may be below zero as it were - i.e. the phase of the sine wave is inverted.) 

    Hopefully most of the voltage is dropped in the load; and if you could cut the load in half and stick a meter half way along - and with an old fashioned bar type electric heater maybe you could, that point would be equally far from the original L and N, and so at 120 V to earth or so.
  • 0
    Welcome to the new place, glad you made it OK ?

    More importantly, did my answer actually make sense?

    3 phase gets more complex, as the currents from the three single phases to some degree will cancel, but the concept remains the same, just the neutral voltage slopes uphill towards the phase with the greatest load (unless phase balance is perfect - then you can cut the neutral wire at both ends and walk off with it to weigh it in for scrap without affecting the overall voltages.)
  • 0

    Mapj1:

    ... if you could cut the load in half and stick a meter half way along - and with an old fashioned bar type electric heater maybe you could, that point would be equally far from the original L and N, and so at 120 V to earth or so. 




    Sounds like a good old-fashioned potentiometer!

  • 0

    mapj1:

    Welcome to the new place, glad you made it OK :D

    More importantly, did my answer actually make sense?

    3 phase is more complex, as the currents from the three single phases to some degree will cancel, but the concept remains the same.




    Thanks Mike!


    Thanks for the detailed response, so essentially the load resistance is the seperating line between the line voltage and the neutral voltage. Which does make sense as if two common points were measured on either line or neutral it would be 0v (+-1) for volt drop. Is there an instance where the load would not drop all the voltage, and therefore there would be a case where there was a higher voltage on the neutral side - potentially in the three phase case, where there could be large currents in the neutral?


     

  • 0
    Actually if you have a 3 phase plus neutral supply and  three loads each connected L-N, then the voltage across any one load may be pulled around by the current on the other two either up or down by at most the drop in the neutral wire resistance times the uncancelled fraction of the 3 load currents. But except if there is a high resistance fault in the neutral wiring the voltage seen by any one load is never pulled around by more than a few %, because in a well designed system, the voltage drop in the wires is always supposed to be single figure percents, as dropped voltage is just really lost energy in the wiring, and we'd very much like it to run cool.

  • 0
    Given that V = IR (volts = amps times resistance), there are only two cases where the voltage on the neutral is ever likely to be significant:
    1. There is a high-resistance fault in the neutral (R is higher than it should be), or

    • The current is excessively high (I is higher than it should be), for instance during a short-circuit condition.

  • 0
    Back on the old forum I asked what is the highest acceptable voltage on a consumers domestic installation earth terminal installed by a DNO on a TNC-S supply, I cannot remember the exact answer, but from memory it was around 33 volts.


    Andy.
  • 0
    Watch out though if you break the neutral and leave the phases on  reg 471 from 17th from memory...
  • 0
    Maybe this image will help clarify the worst case Neutral voltage you should ever find without a fault.
  • 0

    Dave Newton:

    Maybe this image will help clarify the worst case Neutral voltage you should ever find without a fault.




    Not with PME - N-E voltage is zero at the origin.

1 2 3

Join us to get the best from IET EngX.

Joining EngX lets you personalise your experience so you stay up to date on the topics that interest you, plus you’ll be able to make connections who are looking to collaborate, exchange ideas and more.

Join us Not now

Community Rules & Guidelines