Continuity of ring final circuit conductors

[ringer]

Why is it when you cross connect the conductors at each leg and then test continuity, the result is one quarter of the combined resistances of the legs? I am usually pretty good at seeing how/why calculations work out, but just cannot in this case get my head around why the value measured is a quarter of the combined resistance of the legs. Can someone please explain in plain and simple terms why we divide by four, and not by two?

 

[Paul.M]

If you were testing just one leg let's say the line conductor, at the middle of the ring would be the highest point of impedance so therefore its half of the measured result.

If your testing line and neutral your in affect going around twice and therefore divide it by 4 to get the true value of impedance.

 

[Deleted member 26818]

If your two conductors were of the same CSA, you would expect them both to have the same resistance, say for example 1Ω.
If you then connected the two conductors in parallel, you would expect the resistance to halve to 0.5Ω.
However we don't do that, first we add the two resistances together, which doubles the resistance, but the resistance hasn't really doubled, we've artificially doubled it for the purpose of this test.
The reason we add the two values together, is because they are very rarely the same, especially with T&E where the conductors are of different CSAs.
I hope that explains it?

 

[ringer]

Thanks for answering guys. Unfortunately I still don't get it. I understand that by cross-connecting the legs we have effectively created two parallel conductors. I'm now not understanding how when we connect our low resistance ohmmeter, we get a reading of one quarter of the combined resistances. Surely the length of each of the parallel conductors is the length of the ring? Now, because we have two conductors going from L to N, we are halving the resistance because we have effectively doubled the csa from L to N? So I'm halfway there.

Let's work through a scenario - 2.5mm conductors (L and N). The ring is 8 metres long. End to end over 8m measures 0.06ohms on each conductor. So we cross connect the legs at the DB and go to the socket at midpoint. We measure between L and N at that socket. The path is 4m of L1 (0.03 ohms) + 4m of N2 (0.03 ohms) and in parallel to that we have 4m of L2 (0.03 ohms) + 4m of N1 (0.03 ohms). So we have two conductors, each 8m in length, each of 0.06 ohms resistance, in parallel.

By calculation method product over sum
(0.06 x 0.06) / (0.06 + 0.06) = 0.03

or

By calculation 1/Rt = 1/R1 + 1/R2
1/Rt = 1/0.06 + 1/0.06 = 16.67 + 16.67 = 33.34
Rt = 1/33.34 = 0.03

Wait - I'm starting to see the sums working out.
If we take the resistance of the L plus the resistance of the N we get 0.06 ohms + 0.06 ohms = 0.12 ohms
divide by four 0.12 /4 = 0.03

So I now see how the maths works out. I'm just having trouble looking at the conductors and 'seeing' how this works (if you know what I mean).
I think I need to sleep on it.

 

[Paul.M]

Ok ringer I'll try my best to explain in writing, so much easier when its face to face.

Scenario:- You have a 2.5mm ring that is 12 meters in length and has 5 sockets on it, one every 2m starting and ending at the c/u. The mid point of the ring has the highest resistance which in this case would be at 6m the 3rd socket. Therefore the longest part of the circuit is the 3rd socket.

From the c/u
Skt 1 = 2m
Skt 2 = 4m
Skt 3 = 6m
Skt 4 = 4m
Skt 5 = 2m

When we cross the legs at the c/u we create one continuous conductor that goes around the ring twice. A total of 24m.

24m / 4 = 6m. That is the highest point of resistance on the ring.

Hope that helps mate.

 

[Octopus]

Ok ringer I'll try my best to explain in writing, so much easier when its face to face.

Scenario:- You have a 2.5mm ring that is 12 meters in length and has 5 sockets on it, one every 2m starting and ending at the c/u. The mid point of the ring has the highest resistance which in this case would be at 6m the 3rd socket. Therefore the longest part of the circuit is the 3rd socket.

From the c/u
Skt 1 = 2m
Skt 2 = 4m
Skt 3 = 6m
Skt 4 = 4m
Skt 5 = 2m

When we cross the legs at the c/u we create one continuous conductor that goes around the ring twice. A total of 24m.

24m / 4 = 6m. That is the highest point of resistance on the ring.

Hope that helps mate.

But, not all sockets on a ring are directly on the ring so when the values rise you could well be testing a spur or a socket with loose connections

 

[durhamspark]

When you test and add together the 2 ring resistance values you are effectively measuring 2 "laps" around the ring on a single conductor.

By cross connecting the conductors you have now measured only 1 "lap" which will halve the value.

At the same time you have doubled the cross sectional area which will halve the reading again.

As the circuits are connected into a figure of 8 the reading is the same no matter where on the ring you measure it (spurs excluded obviously) .This is also a good indicator of a spur on a system if you were testing an existing installation.

 

[Silly Sausage]

When you test and add together the 2 ring resistance values you are effectively measuring 2 "laps" around the ring on a single conductor.

By cross connecting the conductors you have now measured only 1 "lap" which will halve the value.

At the same time you have doubled the cross sectional area which will halve the reading again.

As the circuits are connected into a figure of 8 the reading is the same no matter where on the ring you measure it (spurs excluded obviously) .This is also a good indicator of a spur on a system if you were testing an existing installation.

That's only true when both the Line and cpc have the same csa.
(r1+r2)/4 will be highest at the midpoint point of ring (both legs equal length) and be lowest when measured at the CU. But with the low values of r1 & r2 on a RFC the difference is so small that it'd be dismissed as measurement error.

If Line and cpc have the same csa, then (r1+r2)/4 will be same where ever you measure it.

 

[durhamspark]

That's only true when both the Line and cpc have the same csa.
(r1+r2)/4 will be highest at the midpoint point of ring (both legs equal length) and be lowest when measured at the CU. But with the low values of r1 & r2 on a RFC the difference is so small that it'd be dismissed as measurement error.

If Line and cpc have the same csa, then (r1+r2)/4 will be same where ever you measure it.

GN3 2.7.6 step 3 : the values obtained at each of the sockets will be substantially the same and 1/4 of the line and cpc resistances combined.

This is because as you are connecting the circuit into one big figure of 8 it makes no odds where you measure it, the result wil be the same.

 

[Silly Sausage]

GN3 2.7.6 step 3 : the values obtained at each of the sockets will be substantially the same and 1/4 of the line and cpc resistances combined.

This is because as you are connecting the circuit into one big figure of 8 it makes no odds where you measure it, the result wil be the same.

Note the high lighted phrase.

I was bored all right and was messing about with the maths to prove the (r1+r2)/4 result.
With r1 = 1ohm and r2 = 1.67 the difference is only about 0.04ohm!

I need to get out more.

 

[ringer]

...
I was bored all right and was messing about with the maths to prove the (r1+r2)/4 result.
With r1 = 1ohm and r2 = 1.67 the difference is only about 0.04ohm!...

Without going into detail, if it is complicated, can you explain why there is this (miniscule) difference with a smaller cpc?

 

[Silly Sausage]

Without going into detail, if it is complicated, can you explain why there is this (miniscule) difference with a smaller cpc?

I'll try!

It's resistors in parallel.
@ mid point it's (r1/2 + r2/2) & (r1/2 + r2/2)
@ CU it's r1 & r2

do the maths and you'll come up with 2 different looking equations.
mid point (r1+r2)/4
CU r1r2/(r1 + r2)

Use r1 = 1 & r2 = 1.67 and you'll see the difference
r1 = r2, same result from both.

 

[PC Electrics]

That's only true when both the Line and cpc have the same csa.
(r1+r2)/4 will be highest at the midpoint point of ring (both legs equal length) and be lowest when measured at the CU. But with the low values of r1 & r2 on a RFC the difference is so small that it'd be dismissed as measurement error.

If Line and cpc have the same csa, then (r1+r2)/4 will be same where ever you measure it.

Not so. On an RFC where the line and cpc are the same length and take the same route around the ring (as they should for it to be a true RFC), then the R1+R2 (or (r1+r2)/4 if you prefer) will be exactly the same at all points. It must be, as you are measuring the same amount of cable at every point - it's why it's connected as a 'figure of 8'.

Ringer: try thinking about it like this:
lets say you have a 'ring' of 2.5 which has an end to end (r1) of 0.6 ohm.
if you now cut the 2.5 at its centre and measure one length (half the original length) you will get 0.3 ohm
now connect the two lengths together at each end. because you have two lengths of the same csa the resistance is halved. So you will get 0.15.
Imagine you hadn't actuallly cut the cable at its midpoint, but went straight to connecting the ends together and measuring the resistance from the ends to the midpoint you are measuring the two lengths as shown above.
You should now be able to see that R1 = (r1)/4
by applying the same working to the cpc (r2) you will end up with R1+R2=(r1+r2)/4

 

[ringer]

My head hurts, but I think I get it now. Have drawn it out and stretched out the points to make it two conductors in parallel. Each conductor is (0.5 r1 + 0.5 r2) = 0.5(r1+r2)
Because the two conductors are in parallel, the resistance is then halved.
0.5(0.5(r1+r2)) = 0.25(r1+r2)
The acid test will be to see if I can remember this in the morning!
I still don't quite get why, with a smaller cpc, the midpoint resistance will be (minimally) lower, but will leave that thought for another day.
Thanks for your help guys.