Ring r1r2 test

[OnOff]

Rob,

You need to interpret the result.

If r1=r2, then the last line equals 0 and RE does indeed equal RM, both conductors with same csa, so R1+R2 is the same wherever you measure it, this is undisputed.  Look at it as a special case.

However, if r1<>r2 (different csa cables), the last line of the proof is greater than zero, hence proving RE<RM

It might be clearer if you say ... for r1<>r2, I claim that RE<RM, then the last line would be (r1-r2)2>0, proving they are not equal. (Just remove the equals sign out of the proof)

It's all academic anyway, as we both know, the difference is almost negligible!

Conflicting statements there, comes across as "....there is a difference.....but it's the same".  I'll dispute that R1+R2 is the same wherever you measure it. I'll dispute it all day long in fact. It's simple parallel resistance calcs. I'd accept substantially the same.....or "as near as makes no difference".....or....."'cos my meter will only read to....."  :tongue in cheek

 

[Andy™]

here's a (very bad) drawing or a ring, with L & E connected as they should be, making one nice big loop



you will always test at exact opposite sides. ive even marked (badly) 3 test points as though there are 3 sockets. so, since there is always the exact same amount of 1.5 & 2.5 between them, please explain how they will have different readings

 

[Lusk]

OK Andy,

At 9 and 3 o'clock you have red in parallel with black, r1 || r2, which is (r1*r2)/(r1+r2)

At 12 and 6 o'clock you have (r1/2 + r2/2) || (r1/2 + r2/2) = (r1+r2)/4

Stick the numbers for a 100m of 2.5mm T&E and see what you get.

 

[Andy™]

you dont need any formula. you will always have the exact same amount of cable between all points, so you cannot have a different resistance

 

[Rob_the_rich]

you dont need any formula. you will always have the exact same amount of cable between all points, so you cannot have a different resistance

you appear to be adding the resistances, eg

test point 1, to the left 3 red and 1 black, to the right 3 black and 1 red = 4 red and 4 black

test point 2, to the left 2 red and 2 black, to the right 2 black and 2 red = 4 red and 4 black

That works out fine if they were in series, but you have to use a formula if you wish to work out parallel resistances.

Test point 1 to the left 3x 1.21 and 1x 0.741 =4.371 =R1

to the right 3x 0.741 and 1x 1.21 =3.433 =R2

(4.371x3.433)/(4.371+3.433)=15.005643/7.804=1.9228

Test point 2 to the left 2x 1.21 and 2x 0.741 =3.902 =R1

to the right 2x 0.741 and 2x 1.21 =3.902 =R2

(3.902x3.902)/(3.902+3.902)=15.225604/7.804=1.951

 

[Sharpend]

I'm lost wasn't this about R1+R2? 

The thing with a circle is that it has no beginning and no end so it matters not what or where you measure it will always be equal or as close as damn it??

You keep me spinning round and round and that's alright with me!!

 

[James1978.]

Is everyone saying that the origin of the circuit for a ring main makes no difference for the R1+R2 but it will for the impedance test. Do I need to make a clear distinction in the head.

 

[Andy™]

if R1R2 is the same, then Zs will also be the same. you may get some variation in reading on both, but theoretically they should be the same

 

[steptoe]

The thing with Zs tho is that you will have parallel paths of differing lengths unless you are dead centre of the ring,

That's why it will gradually increase then fall again as you move away from source then towards it again,

if R1R2 is the same, then Zs will also be the same. you may get some variation in reading on both, but theoretically they should be the same

No,

R1 R2 is constant, 

Zs will rise then fall again