Crossover in ring circuit

[KeenPensioner]

HI, as my profile says I'm not a sparky and know my limitations. I'm trying to learn (for the sake of it only as I'll never do it) what tests for "Ring Final Circuits" are. I understand how the Tests 1, 2 and 3 are done but in one of the videos I watched the presenter (John Ward) mentions a "crossover" see image and says that this "could lead the current being shared in an undesirable fashion. I don't understand that.

 

[Upton Sparks]

Its an equation at the end of the Day,

The whole idea of the ring is at any socket outlet the resistance should be the same back to the Fused board.
Therefore the current will travel both way back to the fuse board at an equal 50% each way down each leg of the ring. by adding a connection between the sockets and creating a figure 8 the resistance will now be imbalanced. therefore the current will now be unevenly splite. ie 40-60% 30 -70%, it can even get to 10- 90%, depending on the resistance of the Legs, it would be a very difficult job to access this and require a level of mathematics beyond just looking at tables.

and the reason we don't see any damage is most domestic house don't use a lot of current of their ring mains.
But should they try to use the 32 amps on offer then they will most certainly start to get issue,

 

[Rpa07]

I can't see why though!

I’m sure most ‘dumb’ ratings are people scrolling through on iPhone - too easy to do. Anyway - don’t let it get you down!

 

[HandySparks]

The whole idea of the ring is at any socket outlet the resistance should be the same back to the Fused board.
Therefore the current will travel both way back to the fuse board at an equal 50% each way down each leg of the ring.

What???

The resistance of a section of cable between the origin of the ring and the load will be proportional to the length of that cable. So a socket near the origin will pull more current along the 'short' route than the 'long' route, in inverse proportion to the resistance.

 

[Deleted member 26818]

The R1 + R2 should be the same at each socket (except spurs) when testing, because the conductors are joined together into a big circle.
Current sharing in normal use, will be as Handyspark says proportional to the resistance of each leg.
The longer the leg, the greater the resistance, the lower the current.

 

[Upton Sparks]

Ok Boys back to the testing books for you I'm afraid.
After doing your ring continuity, I hope you then do your Figure 8, You then test each socket to obtain your R1+ R2, your expecting the resistance reading R1 + R2 of the Ring to be the same at every point!.(Give or take 0.05)
therefore the current will flow equally back to the board regardless of where you are on the Ring,( its the same resistance either way its a ring) thats the science behind and why you do the Figure 8 testing and check every socket..
If you have high reading its normal a spur, or you've connected it up wrong, (normally only happens on Single..... unless your a really idot). Or the circuit is not a ring.

 

[DPG]

Upton, the resistance of the conductors cannot possibly be the same from every socket back to the board. Basic Ohm's law will confirm that.

 

[Deleted member 26818]

Ok Boys back to the testing books for you I'm afraid.
After doing your ring continuity, I hope you then do your Figure 8, You then test each socket to obtain your R1+ R2, your expecting the resistance reading R1 + R2 of the Ring to be the same at every point!.(Give or take 0.05)
therefore the current will flow equally back to the board regardless of where you are on the Ring,( its the same resistance either way its a ring) thats the science behind and why you do the Figure 8 testing and check every socket..
If you have high reading its normal a spur, or you've connected it up wrong, (normally only happens on Single..... unless your a really idot). Or the circuit is not a ring.

Yes when testing the resistance will be about the same at each socket (except spurs), because the conductors are connected together at the CU into a big circle.
In normal use the conductors are separated into individual circles (hopefully).

 

[Upton Sparks]

Its a Ring, they are connected at the board in the MCB.
Explain. how does basic ohm law explain that? your going to need to put a bit more meat on the bones for me.

But
If the resistance is the Same at each socket( as we've tested), then the equation V =I X R, we should all be familiar.
Therefore I = V / R
V is constant at 230V, and as we've said above R is the same at Each location,
therefor I will be divided equally around the circuit.

 

[Deleted member 26818]

Its a Ring, they are connected at the board in the MCB.
Explain. how does basic ohm law explain that? your going to need to put a bit more meat on the bones for me.

But
If the resistance is the Same at each socket( as we've tested), then the equation V =I X R, we should all be familiar.
Therefore I = V / R
V is constant at 230V, and as we've said above R is the same at Each location,
therefor I will be divided equally around the circuit.

Yes when cross connected for testing not when uncross connected for normal use.

 

[Pete999]

Its a Ring, they are connected at the board in the MCB.
Explain. how does basic ohm law explain that? your going to need to put a bit more meat on the bones for me.

But
If the resistance is the Same at each socket( as we've tested), then the equation V =I X R, we should all be familiar.
Therefore I = V / R
V is constant at 230V, and as we've said above R is the same at Each location,
therefor I will be divided equally around the circuit.

Doesn't Mr KIRCHOFF have a say in this debate, far to far back for me to remember, but it does ring a bell.

 

[DPG]

Uptonsparks, in post #16 you state "The whole idea of the ring is at any socket outlet the resistance should be the same back to the Fused board"

 

[Upton Sparks]

It should be its a ring, looks like this ( O), if you cut in to the Ring and measure ring continuity( R1 for example) it would be the same at the Fuse board and if we measured it at every socket as well it should be the same, providing you close it back to a ring at the Fuse board. When the Current hits the Ring it see the same resistance in the conductors and Flows accordingly,

My understanding is that the current is divided equally around the ring, therefore allow us to use a 2.5mm cable to be used on a 32amp MCB, if it took the shortest leg back, then surely you would have over current issue on the shortest section of of the Ring?..

 

[Upton Sparks]

Uptonsparks, in post #16 you state "The whole idea of the ring is at any socket outlet the resistance should be the same back to the Fused board"

He did Peter well done for remembering his name.,,something about what flow in to a junction must flow out, I think. I cant remember that far too long ago, I'm afraid I've Kill a few brain cell since those day.....
LOL that will be the Beer.

 

[Pete999]

He did Peter well done for remembering his name.,,something about what flow in to a junction must flow out, I think. I cant remember that far too long ago, I'm afraid I've Kill a few brain cell since those day.....
LOL that will be the Beer.

Wrong post maybe? Ah I see now, yes it was many moons ago, now White man speak with forked tongue, or he don't understand diddly squat, I must refer to Medican Man @Hughes Electrical Technology book, he speaks words of wisdom, or he did when I was a mere Brave in this world of magic, kemosabi.

 

[Midwest]

Yayyy! I got a dumb!

I got a 'dumb' from a bloke that joined in 2010, never posted or started a thread. Mind you, he must be a good judge of character.

In fact looking at his/her profile, he's done 6 dumbs, closely followed by 3 bad spellings & 3 olds.