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Hi, I'm not an electrician but a keen amateur. I understand the testing on RFC and how it's done but couldn't get my head round the (r1+r2)/4 calculation. I've trawled this site (and others) and it seems to me that a lot of people struggle to explain this fully. The truth is that this equation only holds up when the CSA of the conductors are equal - which is not the case with the line and cpc. I attach a document that I think explains it reasonably but I'm sure more experienced professionals can comment (which would be appreciated)
 

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Ian1981

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Having been an electrician for 19 years I can categorically say that the formula is correct.
If I have say 0.30 end to end for my line conductor and 0.50 end to end for my cpc’s (2.5mm T+E) then when I measure at each socket outlet with the cross connection method, then I will measure 0.20 ohms for my R1+R2 measurement.
The calculation/formula is indeed correct.

For same size conductors say 2.5mm then my R1+R2 will be half of my measured r1 ,rn and r2 end to ends, in this example 0.15ohms.
It’s a simple resisters in parallel calculation clearly explained in GN3.

We are splitting hairs here maybe I’ll measure 0.21 at some sockets maybe 0.20 at some and maybe 0.19.
 
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telectrix

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basically, your end-end is the resistance of 1 conductor. add a 2nd conductor in parallel, the resistance is divided by 2. R1+R2 is 1/2 of this so in total you divide by 4. ( ending up with the ansewr Tuesday).
 
(R1+R2)/4 = Expected R2

Its divided by 4 because its half the length (divided by 2) of R1+R2; then divided by 2 again because your eliminating R1. The result is R2
 
D

Deleted member 26818

Hi, I'm not an electrician but a keen amateur. I understand the testing on RFC and how it's done but couldn't get my head round the (r1+r2)/4 calculation. I've trawled this site (and others) and it seems to me that a lot of people struggle to explain this fully. The truth is that this equation only holds up when the CSA of the conductors are equal - which is not the case with the line and cpc. I attach a document that I think explains it reasonably but I'm sure more experienced professionals can comment (which would be appreciated)
No it stands up whether the CSA is equal or not.
It even stands up if the CSA of each conductor varies along the length of that conductor.
 
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D

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(R1+R2)/4 = Expected R2

Its divided by 4 because its half the length (divided by 2) of R1+R2; then divided by 2 again because your eliminating R1. The result is R2
No.
We halve it once, because we effectively have two conductors going to the same point, then we halve it again because the two conductors have reduced the resistance by half.
 
Yeh that's what I said :confused:
 
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Des 56

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That has completely baffled my little brain.
On the second attempt reading the post my brain managed to get itself unfried o_O
 
D

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Er no it was a reply to me spin haha
That was my second post, which was a reply to you.
My first post was a reply to the OP which I started before you, Ian, tel and Ferg replied.

There I’ve altered my first post to make things simpler.
 
Post #7 you disagree with
 
D

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We would be if you need R2
Not if the CSA or resistance of R1 was different to R2.
If we have a resistance of 1 Ohm for R1 and a resistance of 0.5 Ohms for R2, adding them together and dividing them by 2 would not give us the resistance of R2.
 
I do enjoy my Saturdays lol.
r1+r2 is halved cause its half the length (agreement on that), then the formula needs another division of 2 to satisfy (division of 4). So in that case its half the remainder
 

telectrix

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remainder of what you bin drinking will do.
 
D

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I do enjoy my Saturdays lol.
r1+r2 is halved cause its half the length (agreement on that), then the formula needs another division of 2 to satisfy (division of 4). So in that case its half the remainder
Yes, which will give us the R1 + R2.
The second halving is required because we have two conductors in parallel which reduces the resistance by half.
 

happysteve

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I wrote a Matlab script a while ago to determine the deviation from equality between R1+R2, and r1+r2/4, when the cpc and line conductors are different sizes.

For the different sizes of T&E available, the maximum discrepancy at the very edges of the ring was 20% (with 4+1.5mm cable); for 2.5+1.5mm, the maximum discrepancy was about 6%. If you've got old 2.5+1mm, you might get as much as 18%.

r1_plus_r2_over_4.png

For all practical purposes, it's as near as Phuket is to swearing.
 

Midwest

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Some very strange things happen on this forum of late. I'm perturbed.
 
From my understanding it's r1 +r2/4 because by making the figure of 8 you are effectively halving the length of the conductors and doubling their CSA and when you measure at each socket you are effectively measuring a quarter of the circuit at any point.
 
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Thank you all for taking the time to respond. "happysteve" in post 23 is agreeing with my original post.....ie the formula does through up discrepancies/variations if the live conductors and cpc are of different csa (as his graph clearly shows)
 

telectrix

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was already strarrrtiiing to see doublle .happpy stevesss graph finnishhed time for bed seddd zebeddeee. boiiing. .
 
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