End of Fault current for Parallel lines in Ligting

If I've understood your setup correctly then the you don't have anything in parallel - they're just three branches of the same circuit. A single fault can only occur in one place - if it's at the end of one of the branches then the fault current won't be flowing in the other branches, so they can be ignored. So yes, just pick the worst case - i.e. the one with the highest impedance - usually the longest.

Cables in parallel are connected together at both ends - so current is shared between the distinct cables (sometimes in strange ways if there's a fault part way along just one of the cables) - but that's a different kettle of fish.

If the cable is of the same type throughout, the calculation is usually somewhat simpler - just use the overall length (to the furthest point) and the per metre characteristics of the cable - usually no need to split it down into each individual section.

   - Andy.

Thank you! your understanding is correct. I just need to split it down with individual section. The question arise because I'm working on the spreadsheet created by some one. What is baffling me is, that some one averaged the resistance and reactance in the equation. I could not paste the excel equation here as it runs long. In simple terms, they summed up the resistances based on the individual section and divided it by the total length (summing up all the lengths from individual sections). Similarly done to reactance values. Why would someone do it?

Also a top up question is, do I calculate  Voltage drop for each branch separately by taking individual section's resistance and reactance. (or) again the reduced resistance and reactance due to parallel branches?

I am a bit confused by what you are trying to do here. Are you trying to find out what the maximum fault current should be? In which case you need to consider a fault close to the supply rather than at the far end.

If you are considering voltage drop you need to consider the voltage at the far end load so the furthest light from the supply.

These ought to be two completely separate calculations, though they will use the same basic information of cable resistance/reactance.

There is almost never a case  to average the resistances of the branches- this sounds like the work of someone better with Excel than with circuit theory. Only if the branches interlink, or if there are short branches on the end of long shared sub-main do the loads on one branch affect the other.

I suggest ditching the excel and drawing it as resistors on the whiteboard, at least the simplest example.

Only the first 10m has the volt drop due to the total current, there-after each brack carries its own, but as far as PSSC is concerned, this is small - with kA flowing to the faulted branch, an amp or 3 down another is not going to change much.

As sanity check that does not need the excel, you can assume that 1m of 1mm2 wire is 16 milliohms when cold, or more like 19-20 milliohms when hot, and this scales as you expect with cross-sections. (so 10m of 10mm2 is the same as 1m of 1mm2 etc. ) It then simplifies  to some resistors in series for the wire, and in /// for the loads. (cable normally has equal live and neutral current (lets ignore 3 phase for now) so the volt drop to the load is the live voltage falling on the way to the load, and the neutral voltage rising.)

Mike.

Alasdair Anderson said:

I am a bit confused by what you are trying to do here. Are you trying to find out what the maximum fault current should be? In which case you need to consider a fault close to the supply rather than at the far end.

I might also add that you need to calculate the PSSC at the far end to ensure that it is high enough to trip the OCPD.

Hi Anderson,

I'm trying to calculate at the fault current at the far end of the circuit to verify whether the breaker trips or not at that fault current and also trying to find out the voltage drop