Example 2 If the spur is wired with 1.0 mm2 cable, using the go and return column of
table 1, the Max cable length will be 33.3/36.2= 0.9 Km , for 1.5 mm2the Max
cable length is 33.3/24.2 = 1.38 Km and finally for 2.5 mm2the cable length is
33.3/14.8= 2.25 Km.
5.1.2 Loop Circuit;
Calculation as above but assume the loop is broken at one end.
5.1.3 Loop with spurs;
Calculate all individual spurs including maximum return to Panel.
Example 3
As the above example with 250 fittings and using 10 spurs, each having 25
fittings with a main loop length of 100 Metres as shown in Fig1.
Assuming worse case with all load lumped in one place, the maximum
distance of each spur using (Equation 1) will be;
Vdrop = [ RL x n x I ] x [ Rspur x ns x nfs I ] (4)
Where RL = Main Loop cable resistance as shown on Fig 1.
n= Total number of fittings on the loop.
I= Average current of an addressable interface = 0.0012 A
Rspur = Spur resistance
ns = Total number of spurs
nfs = Total number of fittings per spur
Re-arrange (Equation 4) so that the spur resistance is;
Rspur = [ Vdrop - [ RL x n x I ] ] (5)
ns x nfs x I
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