Discuss Line to CPC ratio. in the UK Electrical Forum area at ElectriciansForums.net

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Just curious. I'm trying to find a poor connection on a ring circuit CPC by comparing resistance between the lines and cpc. I've chocced them all together at one end at the CU. I'm looking for a divergence between the CPC to line and line - line ratio. What is the ratio of CPC - line compared with line to line on 2.5 twin and earth? I know end to end it's 1.67. I tried halving 1.67 to 0.83 but I'm not sure this is right?

Cheers
 
use a wander lead with a break out box. just measure L then cpc. should be 1.67x that.
 
It's the ratio of the CSA of the cables.. so 2.5/(2.5+1.5) = 0.625 or 1.6 if you want in the inverse... if I've understood what you're trying to do.

Or you could use the resistance of conductors (in mΩ/m) from the back of GN3 ?
 
OK, so for every ohm on L, you have 1.67 ohms on E.
 
It's the ratio of the CSA of the cables.. so 2.5/(2.5+1.5) = 0.625 or 1.6 if you want in the inverse... if I've understood what you're trying to do.

Or you could use the resistance of conductors (in mΩ/m) from the back of GN3 ?
Thanks. So now I'm back home and had a chance to think about it. It must be an average of the 2? ie r1 and r2. So it's 1 + 1.67 / 2 = 1.335.

So the ration of r1 to r1 + r2 is 1.335 Does anyone concur? But yes I can see that I could use a table to measure resistances against each other.
 
Thanks. So now I'm back home and had a chance to think about it. It must be an average of the 2? ie r1 and r2. So it's 1 + 1.67 / 2 = 1.335.

So the ration of r1 to r1 + r2 is 1.335 Does anyone concur? But yes I can see that I could use a table to measure resistances against each other.
I suggest that you draw it out... each conductor will be a resistor... it might then become obvious.
 
If I've understood correctly, you want to:

Disconnect both legs of the ring at the CU
Connect together L,N and E of one leg only
Test continuity at each point, L to N, and L to E
Compare results, checking that they are in proportion, in order to find a loose connection on the CPC

If so, the L to E result should be 1.333 times the L to N result:

((2.5/1.5)+(2.5/2.5))/2 = 1.333
 
If I've understood correctly, you want to:

Disconnect both legs of the ring at the CU
Connect together L,N and E of one leg only
Test continuity at each point, L to N, and L to E
Compare results, checking that they are in proportion, in order to find a loose connection on the CPC

If so, the L to E result should be 1.333 times the L to N result:

((2.5/1.5)+(2.5/2.5))/2 = 1.333
Great. Yes that's what I was trying to do.
 

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