Global Insulation resistance <1MΩ but individual circuits are all > 1MΩ

[HappyHippyDad]

With regards the title is the reason for this because the circuits are all in parallel?

I am talking in relation to a L/N - E test.

If Rt = global resistance for whole CU and R1, R2, R3 = L/N - E resistance for each circuit then am I right with the following:

1/Rt = 1/R1 + 1/R2 + 1/R3

So basically even if the global L/N-E IR test is < 1MΩ, that doesn't automatically mean that an individual circuit is < 1M Ω?

 

[telectrix]

correct. bear in mind also that you may have parallel paths due to earthing and bonding etc. simplified, if you have 3 circuits, each 3 Meg, then your global will be 1 Meg,

 

[sinewove]

With regards the title is the reason for this because the circuits are all in parallel?

I am talking in relation to a L/N - E test.

If Rt = global resistance for whole CU and R1, R2, R3 = L/N - E resistance for each circuit then am I right with the following:

1/Rt = 1/R1 + 1/R2 + 1/R3

So basically even if the global L/N-E IR test is < 1MΩ, that doesn't automatically mean that an individual circuit is < 1M Ω?

If the global reading is 1M ohm,no cct could be less than that if they are all in parallel,if that's the question. Regards,S

 

[Andy78]

I seem to recall as a rule of thumb that the total resistance will always be lower than the lowest individual value. you are right that the reason is because they are all in parallel with each other.

Compliance wise, the >1MOhm is for the installation, not individual circuits.

 

[DPG]

For simple resistances in parallel, the total will always be lower than the lowest individual resistance.

 

[HappyHippyDad]

I seem to recall as a rule of thumb that the total resistance will always be lower than the lowest individual value. you are right that the reason is because they are all in parallel with each other.

Compliance wise, the >1MOhm is for the installation, not individual circuits.

So if I had 10 circuits and each of the circuits had 9MΩ of resistance L/N-E, the global IR would then be 0.9MΩ and therefore a fail?

 

[sinewove]

So if I had 10 circuits and each of the circuits had 9MΩ of resistance L/N-E, the global IR would then be 0.9MΩ and therefore a fail?

No,because you can,t divide by 10,use 1/R =1/r+1/r2+.......etc.S

 

[DPG]

No,because you can,t divide by 10,use 1/R =1/r+1/r2+.......etc.S

The reciprocal formula does work out at 0.9M though.

 

[telectrix]

So if I had 10 circuits and each of the circuits had 9MΩ of resistance L/N-E, the global IR would then be 0.9MΩ and therefore a fail?

it would not be a fail. as long as each cct. is > 1Meg, it's a pass, but it is advised to investigate any circuit < 2 Meg.

 

[HappyHippyDad]

No,because you can,t divide by 10,use 1/R =1/r+1/r2+.......etc.S

That is the formula I used. It works out to 0.9M Ω . I'm not sure where you got divide by 10 from, that has nothing to do with this formula.

 

[DPG]

Your answer is correct HHD.

 

[HappyHippyDad]

it would not be a fail. as long as each cct. is > 1Meg, it's a pass, but it is advised to investigate any circuit < 2 Meg.

So if each of the 10 circuits tested as 1MΩ, which is a pass. The global IR would be 0.1M Ω .
Would you still class this as a pass?
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Reading reg 643.3.2 it does seem to state that the consumer unit AND each circuit have to be >1M Ω . Therefore 9MΩ on each of 10 circuits would be a fail! Blimey, doesn't sound right.

 

[DPG]

All the circuits at 9M would be OK, but whatever was causing the <1M reading would be an issue.

 

[HappyHippyDad]

All the circuits at 9M would be OK, but whatever was causing the <1M reading would be an issue.

But that's just it DPG...

1/9 x 10 = 1.1111 recurring

therefore Rt = 1 / 1.111 recurring = 0.9 MΩ

There is nothing specifically causing the 0.9MΩ, there is nothing wrong, this is just what the resistance is with 10 circuits in parallel each with 9M Ω IR.

 

[Andy78]

This theoretical situation would be extremely unlikely in my experience, to have all circuits over 1MOhm but the total under. More likely that one or more circuits would be pulling the total readings down due to faults.