Discuss neutral-earth voltage in the UK Electrical Forum area at ElectriciansForums.net

I have been having a quite night but after trying to get my mind round this I think it's time to reach for a bottle.

The way I understood this phenomenon was simple physics:
A is a measure of the number of electrons passing a point per second.
V is the potential difference so the force pushing the electrons along.

In plumbing speak the A is the volume of water and V is the pressure or head.

In a circuit the electrons will take the easiest route to ground. The electrons are all being squeezed through a small section of copper and therefore encounter resistance - V=I R. The back up due to resistance is the V measured between E and N. It is 0 at the consumer unit as this is near where E and N are merged tnc (not tt) so there is no relief of V by shorting N and E.

This leaves vd. Is this not caused by the resistance in the cable v=ir. E.g. If 10a of current flows through cable with 1 ohm of resistance the vd is 10v?


I am now going to pour a stiff one. Would be grateful for some clarity on this.




Physics????

A lot of mumbo Jumbo if you ask me!!!!!!

Iv'e never seen any electrons or even been shown any at the college.

Its not electrons, its positrons that flow when a current flows.
 
Thats marvellous, but I still don't quite understand!!!!

Think of it as 3 resistances in series, 1 for the Line conductor, 2 for the load and 3 for the Neutral.
As you well know (!) voltage is dropped across a resistor.
In this case the load resistance is much greater than the conductor resistances, so the vast majority of the voltage is dropped across the load and small amount on the conductors...

Clearer??? :)
 
no that the waters are well and truly muddied, i realise why i neve rdrink water ( apart from what fish do in water)
 
Physics????

A lot of mumbo Jumbo if you ask me!!!!!!

Iv'e never seen any electrons or even been shown any at the college.

Its not electrons, its positrons that flow when a current flows.

Maybe when Moses was a lad but when I went to school an Amp was a coulomb of electrons passing a given point in a second. Tbh though I haven't seen an electron or proton either!
 
I thought lineProducts comparison with pipework and water was a good descriptive post

Voltage drop causes electricians to have nasty traits
It also tends to diminish with understanding
 
If you think of the N cable as having some resistance (as will the L conductor).

it will be possible to pull the N conductor at the end nearest the load towards the load (supply) slightly (raising it by 4V in this case) as it (the N conductor) has a slight VD along it's length, which is proportional to the current it is passing.

the E conductor meanwhile has no load on it, as it is not passing any current, hence the slight discrepancy.

Tony may be able to post the vector/phasor drawings better than I can.
 
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You are a magician!!!!

The OP does not tell us the 242V figure.

Where did you get it from????????

You sure that's r8????

If you have 4V drop in the neutral then it stands to reason there will be 4V drop in the line. So 234V across the load plus 4VD on each line equals 242V at the source.

Based on a good N/E connection.

I’ve realised I made a cock up in my first post so I’ve highlighted it in red. Sorry for any confusion, but all the figures and the drawing are correct.
 
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Physics????

A lot of mumbo Jumbo if you ask me!!!!!!

Iv'e never seen any electrons or even been shown any at the college.

Its not electrons, its positrons that flow when a current flows.

And there was I, believing my old lecturer when he taught us it was all about electron-hole-pair movement....
 
guys

thanks for all the responses and a lot of it makes sense. I can see the VD situation been the most probable cause now but if this is the case I dont recall seeing this on a lighting circuit at the same site when I replaced some fittings few month ago?
Of course I shall check when im there next week again

I love this weather, im off to pick up some beers :)
 
If you think of the N cable as having some resistance (as will the L conductor).

it will be possible to pull the N conductor at the end nearest the load towards the load (supply) slightly (raising it by 4V in this case) as it (the N conductor) has a slight VD along it's length, which is proportional to the current it is passing.

the E conductor meanwhile has no load on it, as it is not passing any current, hence the slight discrepancy.

Tony may be able to post the vector/phasor drawings better than I can.


Surely if you pull the N conductor it will stretch and get thinner and increase its resistance and all your maths will be wrong and to be done again.
 
And there was I, believing my old lecturer when he taught us it was all about electron-hole-pair movement....

What are you saying now, we have to fit the non-existent electrons in to Holes now?
This is just silly and a farce!!!!
How can you put something into a hole when you can't see it in the first place.:flowers:
 
Last edited by a moderator:
And the dish ran away with the spoon!
 

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