The Lollipop Circuit
In my design ill assume that at the far end of the circuit there is a 20 amp load and an additional 12 evenly distributed around the circuit, average current = 32 + 20/2 = 26 amps.Providing load current in any part does not exceed 26amps we may use an Iz of the below.
Iz Equal > 26 amp, 6mm, 2.5 so both are o.k
Volt drop = Distribution Circuit 10 meters 6mm, Ring/Parallel circuit 60 meters
Radial = 2.336 volts Ring equals = 7.02 volts = 9.356 volts so less tan the 11.5
Fault Protection Maz Zs(32 Amp type B) = 1.44 Ohms Ze = .35 Ohms(TNC-S) 100 amp 1361 type II
Distribution Circuit = .13 Ohms Ring = .35 Ohms = ZS = .83 So comply's
Fault current Protection Front end using adiabatic and energy let throughs Minimum size = . 38mm
far end minimum size .21mm so comply's
Short circuit Front end .38mm Far end = .30mm so comply's32 Type b 60898, fusing factor 1.45 = Must trip within 1 hour at 46.4 amps
If i place a 46.4 load between the two legs at the mid point of a ring the load will be split between the two legs.
So we know it will trip within 1 hour at that load and the split load will equal 23.2 amps per leg, now 2.5 is 27 amps.
Is the cable protected yes.
The key is even distribution of load.
Now common sense tells us that if i place some 6mm in front of this ring it not going to have a major impact on the current distribution of the circuit.
Omit overload Protection at JB Using 433.3.1
Utilizing Appendix ten and using kirchoff's law to evaluate the load in each section, it can be seen that the cable is protected.