Ive used both ohms law R = V / I and used the values of mohms per meter out of table 4D5 in BS 7671.
From the OSG 2.5 T&E has a resistance of 0.01951 ohms per meter, so R = 0.01951 times m (meters) that is
R = 0.01951 * m
Ive taken V as volt drop at 11.5V and I as 32A, the rating of the PD, giving:
0.01951 * m = 11.5 / 32 ohms law
or m = (11.5 / 32)/0.0195 = 18.43m ??
Using the mV/A/m from table 4D5 of BS 7671, we have a value of 18mV, therefore
(18*32*m)/1000 = 11.5V
or m = 19.96m ??
Ive got this all wrong i know i = p/u. Think my perceptions of how a RFC handles voltage and current is off the scale!! But now ive had a thought from a post earlier.
Is from jamesbrownlive's post above, pg2, and how you both mention about dividing by 4. It would seem logical that when considering a RFC's construction, we have to consider that as a ring the CSA is doubled and the length effectivly halved, which would reduce the resistance to 1/4 of its standard value of 0.0195 ohms per m, in this case that would become 0.004875 ohms per m. Recalculating we have:
0.004875 * m = 11.5 / 32
m = 73.7m
which now seems ample length of cable to construct an A1 ring
What you think?