R1 + Rn

[R32]

Anybody able to explain how to calculate this? From my notes I have for 2.mm[SUP]2[/SUP] T&E 30m - ( R1+Rn) A1 is size of ring circuit 30m and A2 is line conductor 2.5mm.
R1= (Rt×A1)/(A1+A2) = (0.453492×30)/(30+2.5) =0.418608Ω
Rn =(Rt×A2)/(A1+A2) = (0.453492×2.5)/(30+2.5) = 0.034884Ω
R1+Rn = 0.418608+0.034884=0.453492.Ω

Only issue is I cannot find a webpage or post anywhere stating whether this is the correct method to use...

 

[telectrix]

use table 9A in OSG. 2.5/2.5 gives 14.82mOhms/m. multiply by L = 0.4446 Ohms

your R1 and Rn should be equal as they are both 2.5mm

 

[johnboy6083]

the easy way to work it out is using the resitance tables from the OSG. the resitance per m of the appropraite conductor size x 2 will equal R1+ Rn. Unless a reduced neutral system has been used, but you probably wont come across that

 

[johnboy6083]

quicker fingers tel!