To be frank, no, not really.

I'd like to know more about what you have in mind please.

To be frank, no, not really.

I'd like to know more about what you have in mind please.

It's hard to explain but pretty much when your calculating R1+R2 when designing circuits you obviously take into account the resistance of the line conductor and the resistance of the protective conductor and added together but when the cables are in parallel and obviously you'll have an extra line and earth conductor which would reduce the total resistance. So to calculate the resistance in parallel would you need to divide by 4 with there been 4 total conductors. 

I hope that clears it up it's the only way I can think to put it 

I see what you mean. The answer is no. Let's call the 4 conductors R1a, R2a, R1b, and R2b. R1a and R2a (and R1b and R2b) are in series: each is half of the (fault) circuit so their values are added. However, (R1a + R2a) and (R1b + R2b) are in parallel and if their values are equal, the total resistance is halved.

You may have  4 conductors, but there are only 2 circuits (strictly earth fault loops).

Ah okay that makes alot of sense I was comparing them to ring circuits which is where my value of 4 came from.

So what if the cables where ran in parrallel but only one of the cables where used as the earth would you just treat the R1+R2 value as normal and not divide it by anything? 

I see what you mean. RFCs are a strange beast.

First, think of the line conductors, call them R1a and R1b. Their resistance is 1/(1/R1a +1/R1b) and given that R1a = R1b, their resistance is R1a/2. However, the resistance of the circuit (or earth fault loop) is now ½R1a + R2. Unsurprisingly, taking out one of the earth conductors increases the overall EFLI.

Ah okay I get what you mean so when you have 2 R1 conductors and only 1 R2 conductors you treat it as half of the total R1 conductor + your total R 2 conductor that makes perfect sense thank you for the response. 

Would this remain as a constant if you was to wire in trefoil but obviously divided by three for everything. 

you are both making this sound far harder then it needs to be and also appear to be slightly at cross purposes.

If you have multiple paths for the live to get to the point of fault, the resistances of all these are in parallel and make up your R1.

Now if the lengths are the same, this simplifies to just adding all the cross sections. and the resistance of that total cross section. If not you may care to mis use the rule of 16 to aid a rapid estimation. (* see below)

If you have multiple paths back to earth from the point of fault, again the resistances are in parallel and make up your R2.

Now  as before if the lengths are the same, this simplifies to just adding all the cross sections.

hope this helps.

regards Mike

* Rule of 16, or more like 19 for cables running at 70C

One meter of copper of cross section 1mm2 is 16 milliohms,

So is 2,5m length of 2.5mm2,  10m length of 10mm2 etc.

The resistance rises with length - so say 5m of 1mm2 is 5* 16 milliohms, - 0.08 ohms,  while half a meter would be 8 milliohms etc.

Do not forget that current flows out and back, these figures are one way only.

Other rules of 16, 1.6mm is almost exactly 1/16 of an inch, and the thickness of 16SWG metal sheet. (not however American gauge, be warned.)

I am no where near as smart as Mike so even the nice round number of 16 can be a strain on the grey cells. I use 20 with the excuse that it is closer to the resistance of a 1mm2 conductor at 70C.

Way back when I was a student at QUB, a busy Friday night in the Students Union meant long waits at the bar. Back then a pint was 18p. There were usually five of us that bought a round each. I recall one fellow, (BA of course), suggesting that there would be some serious advantage in raising the price of a pint to 20p, that way the maths would be easier and you wouldn’t have to wait at the bar for your change from a £1. 

So if we had let's say two 50mm conductors running in parrallel would you treat this as a 100mm conductor and use the resistance values for this to calculate the R1 value. 

You can see it at work here in a voltage drop table

Take the 1mm2 which has a volt drop of 44mv/A/m so by extrapolation a resistance of 22mOhms (22 for the phase and 22 for the neutral). So effectively 44mOhms per 2m length. Divide 44 by 4 and you get 11 which you can see is the volt drop (or resistance) of 4mm2. Similarly, 44 divided by 10 is 4.4 or by 400 which is 0.11.