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(r1+r2)/4, the myth expelled?

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KeenPensioner

KeenPensioner

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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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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
 
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haptism said:
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
Click to expand...
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.
 
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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%.

[ElectriciansForums.net] (r1+r2)/4, the myth expelled?


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