# (R1 + R2) / 4

Discussion in 'Electricians' Talk' started by DSOTM, Dec 7, 2007.

1. ### DSOTMMember

Can anyone explain why (R1 + R2) / 4 is obtained at any ring socket between phase and earth when the interconnections have been made?

I can prove it mathematically for the case when the socket is exactly half way round the ring but it doesn't work for any other position - so my thinking must be wrong, although I don't see how.

2. ### *welsh.wizard*New Member

Ye olde figure of Eight.

3. ### Ripped OffNew Member

yep. think ring

4. ### DSOTMMember

Would you like to elaborate?

Essentially it's a big loop... measuring the resistance between two points means that those two points are joined in two ways, effectively by 2 parallel resistances. Adding the resistances doesn't always give (R1+R2)/4

5. ### jay_tNew Member

r1 + r2 / 4 is only applicable on a dead test when the conductors are cross connected (incoming phase to outgoing neutral and vice versa) and only really applies when r1 and r2 are the same size conductor (i.e. not for t&e) as you are connecting 2 resistors in parallel. the value of 2 resistors in parallel if they are the same vaue is half of any one of them, or r1 + r2 / 4

6. ### DSOTMMember

I was assuming that it was obvious that I was talking about a dead test with interconnections made as in jay t's post above.

What you say is quite true, jay t, apart from it only applying if conductors are of equal csa.

(R1 + R2) / 4 holds if csa is different and that's the bit I'm not managing to prove.

The proof is simple where csas are the same.

7. ### jethroGuest

Did you sit last nights 2391 exam by any chance?

8. ### wklivesvtimeNew Member

The best way to demonstrate this is to draw out the loop with some rectancles like fuse symbols to represent the coppers resistance. Then place a link between any point and you wil see equal amount of resistance at any point. **** explanation i know

9. ### SWNew Member

i did (2391 that is) - thought it was reasonable exam

10. ### sparkin!New Member

I can prove it mathematically for the case when the socket is exactly half way round the ring but it doesn't work for any other position - so my thinking must be wrong, although I don't see how.
No you think correct!

11. ### sinewaveScrewfix Select

r1 + r2 / 4 is only applicable on a dead test when
the conductors are cross connected (incoming phase to
outgoing neutral
and vice versa)

Hmmmmmmmmmmmm!

12. ### DSOTMMember

wklivesvtime (if I've got your name right...)

If you draw the thing you will see that resistances are only the same for a socket half way round the ring.

If you take 2 conductors of different csa, a Phase and cpc, i.e. R1 and R2, perhaps 2 foot long each, hold them parallel to each other in a circle as if they were in a ring, make the cross connections, then pick a point somewhere that isn't halfway round the loop (taking the cross connections to be the start). Hold one conductor in each hand at this point and then let the 'ring' open out. You will be holding a loop made up of R1 and R2. You will also notice that what is joining your two hands is not the same. One 'leg' of the loop will have more Phase than CPC and the other will have more CPC than Phase. Two different resistances in parallel. Add them up and it won't come to (R1 + R2)/4. It should actually come to R1R2/(R1 + R2). The mathematically inclined amongst you will notice that when R1 = R2, R1R2/(R1 + R2) is the same as (R1 + R2) / 4. Takes a few intermediate steps to prove this but quite straightforward.
I don't have my tester to hand to see if R1R2/(R1 + R2) matches measured results for PE test on a ring but I'm confident... at the moment... and if it turns out to be the case I wonder why the IET don't give R1R2/(R1 + R2) as the formula which works for equal or mixed csa.

btw this isn't me trying to be smart. When I started the thread I was hoping that someone could explain it... but since then I've examined it further and cannot see that I am wrong. I don't like to think that the IET is wrong so I hope that someone will point out the error that I might be making... but I can't see it. I've derived the formula from first principles and it seems to cover equal and non equal csa nicely.

jethro - yes, I did have a go at 2391 last night. I wonder what they were looking for when they asked to explain, briefly, why (R1 + R2)/4 is expected result for PE test on a ring? Obvioulsy for 3 marks they weren't after a long, mathematical proof (which as far as I can see actually proves that (R1 + R2)/4 is inaccurate), so who knows.

Pity I didn't do 2391 a few years ago when the questions were about inspection and testing instead of now when they seem to ask a lot of obscure regulations questions... but that's another story.

13. ### DSOTMMember

i did (2391 that is) - thought it was reasonable exam

Do you work on caravan sites?

14. ### HomerNew Member

Dave, think you are confusing two different tests here the one that is done to calculate R1+R2 is done by measuring the resistance of the cpc end to end then divide by 4 to give R2 repeat for the phase conductor to give R1. What you are referring to is the test to determine that there are no interconnections in the ring circuit.

15. ### losingitNew Member

The way I see it,when you link ph and earth conductors,you create a large circle,with the ohm meter across any point on the diameter,so you will get the same reading where ever you place the meter.When you measure ring end to end(L and E)and add them together you are adding 2 resistors in series.When you cross connec t and measure.you are measuring the res in parallel,so they are 1/4 of the series reading,I think

16. ### JP.Screwfix Select

On the assumption that the CSA of the earth is in fact the same as phase and neutral CSA..?

18. ### HomerNew Member

The reason you divide by 4 is that the resistance end to end of the cpc is divided by 2 because the cpc is then joined in the cu. but to calculate R2 you have to remember that the earth fault will travel down each leg of the cpc and it is assumed that equal fault current will travel down each leg (not true in practice)

19. ### HomerNew Member

No jp you do not assume that the csa is the same you measure the resistance of the cpc (end to end) and the resistance of the phase (end to end) you do not assume that they are the same resistance

20. ### seneca2New Member

`The way I see it,when you link ph and earth conductors,you create a large circle,with the ohm meter across any point on the diameter,so you will get the same reading where ever you place the meter.When you measure ring end to end(L and E)and add them together you are adding 2 resistors in series.When you cross connec t and measure.you are measuring the res in parallel,so they are 1/4 of the series reading,I think `

Yes,That`s why we cross-connect at the c/u then measure at the sockets to find r1+r2 to enter on the test cert.The reading does vary slightly througout the ring but we`re only looking for the highest value to enter.Can`t see what all the fuss is about on this one, 0.0 something of an ohm!
regards.