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Discuss (r1+r2)/4, the myth expelled? in the Electrical Forum area at ElectriciansForums.net

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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

Deleted member 26818

(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.
 

telectrix

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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.
 

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