Shower Cable in wall

EFLImpudence · 11 Feb 2013

Scoby_Beasley said:
TBH. something pulled out of GN3 which I now can't find.

0.05Ω in GN3 is the accepted value for negligible impedance so would be more the result of bonding rather than the need for it.

Electrifying · 11 Feb 2013

Scoby_Beasley said:
Have you a test/multimeter?

If, from your main earth terminal (usually in the CU) to structural metalwork (pipework) has a resistance of greater than 22ohms then 'earthing' is not required.

John, this statement is 10 x worse than the one about the '0.05' ........you're slipping.

And 'Scoby_Beasley', you really need to make sure of your facts before posting advice.

JohnW2 · 11 Feb 2013

EFLImpudence said:
0.05Ω in GN3 is the accepted value for negligible impedance so would be more the result of bonding rather than the need for it.

Indeed ....

JohnW2 said:
Hmmm. You seem to be talking about two things that are already very effectively 'bonded'.

Kind Regards, John

JohnW2 · 11 Feb 2013

Electrifying said:
Scoby_Beasley said:

If, from your main earth terminal (usually in the CU) to structural metalwork (pipework) has a resistance of greater than 22ohms then 'earthing' is not required.

Click to expand...

John, this statement is 10 x worse than the one about the '0.05' ........you're slipping.

Whoops

Indeed! I confess that, in haste, I read that at 22kΩ - which didn't sound too far off the mark.

Electrifying said:
And 'Scoby_Beasley', you really need to make sure of your facts before posting advice.

Very much so - that was crazy 'advice'.

Kind Regards, John

Scoby_Beasley · 12 Feb 2013

Electrifying said:
Scoby_Beasley said:

Have you a test/multimeter?

If, from your main earth terminal (usually in the CU) to structural metalwork (pipework) has a resistance of greater than 22ohms then 'earthing' is not required.

Click to expand...

John, this statement is 10 x worse than the one about the '0.05' ........you're slipping.

And 'Scoby_Beasley', you really need to make sure of your facts before posting advice.

yup that busy finding the ohm symbol, which I didn't I forgot the (capital) K as in 22KΩ

Using an MFT:
I = V / R;
I = 500V / 22000Ω = 0.023Amp

If the fault was on a single phase 230V supply the current that would flow between the conductive parts would be only 0.0115Amp or 11.5 mA. Just above the 'let go' of 10mA.

This is unlikely to give a fatal shock.

EFLImpudence · 12 Feb 2013

Scoby_Beasley said:
I = 500V / 22000Ω = 0.023Amp

It's the other way round.
10mA is used to determine the resistance.

230V / 10mA = 23,000Ω

Never see the point of deducting 1,000Ω for the human body.
It's not going to be 21,999 or 23,001 is it?
More likely to be 240V anyway, so 24,000Ω

Scoby_Beasley · 12 Feb 2013

EFLImpudence said:
Scoby_Beasley said:

I = 500V / 22000Ω = 0.023Amp

Click to expand...

It's the other way round.
10mA is used to determine the resistance.

230V / 10mA = 23,000Ω

Never see the point of deducting 1,000Ω for the human body.
It's not going to be 21,999 or 23,001 is it?
More likely to be 240V anyway, so 24,000Ω

True. Sometimes my head hurts. Where's my medication???

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Shower Cable in wall

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