Touch voltage confusion

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I get a bit confused about touch voltage calculations. As this ramble will show.

Especially when it involves supplementary bonding.

So the resistance between exposed conductive parts and Extraneous conductive parts has to fulfil this formula

RA x Ia < 50 v

So in a bathroom.

Ia is the operating current of the protective device.

But which protective device?

I presume the highest rated device (A) that provides power to the bathroom

This is for an installation with no RCD
So in a bathroom you could have the following directly in the room such as a 6A lighting circuit
An electric shower protected by a 50 A MCB

Hot water which comes from a boiler,not located in the bathroom, but has a conductive path to the bathroom via the copper pipe work

So what would be the protective device for the boiler? Which usually has a 3A fuse.
It could be taken from a ring final circuit protected via a 32A mcb

So would the protective device be a 3 amp fuse or the 32A MCB?

(As a side matter)

I understand the pipework is an Extraneous conductive part.
A boiler would be and exposed conductive part, so in my mind that would make any metallic pipe work coming from it also exposed?
If a fault happened on the boiler, that fault current would appear on the pipe work.
But pipe work , radiators etc are classed as Extraneous?

But back to the formula.

So if my maths is correct if it was just a 6A lighting circuit - 50/30 =1.6ohms

shower 50/250 = 0.21ohms

So if the resistance between a lighting fixture cpc and a towel rail is < 1.6ohms supplementary protection is not needed?

But if you had a class 2 fitting the resistance would be infinite

And if 0.21 was the resistance between the shower and a towel rail, supplementary protection is not needed?

But then I read about maximum fault current being 0.01A
Then take into account 1000 ohms fingertip resistance??
So we are looking at a resistance of 22000Ohms

Which needs an IR test

So if a known earth and and Extraneous part is < 22000ohms it needs bonding, but if its more bonding not required

The voltage and current touch voltage requirements seems to be in opposite directions

I hope that makes some sense


Thanks
 
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That would be decided by which Live had come into contact with the Earthed metal work/pipe.
Well, yes, that's obviously the case - but that doesn't help an electrician (unless he/she has a crystal ball) undertaking 'touch voltage calculations'.

Kind Regards, John
 
So the resistance between exposed conductive parts and Extraneous conductive parts has to fulfil this formula
RA x Ia < 50 v
It's usually written as R≤50/Ia as you have done in your calculating.

So in a bathroom.
Ia is the operating current of the protective device.
But which protective device?
I presume the highest rated device (A) that provides power to the bathroom
Yes.

This is for an installation with no RCD
So in a bathroom you could have the following directly in the room such as a 6A lighting circuit
An electric shower protected by a 50 A MCB
Yes, so the 50A MCB will be the highest.

Hot water which comes from a boiler,not located in the bathroom, but has a conductive path to the bathroom via the copper pipe work
So what would be the protective device for the boiler? Which usually has a 3A fuse.
It could be taken from a ring final circuit protected via a 32A mcb
So would the protective device be a 3 amp fuse or the 32A MCB?
Does it matter? Both are lower than the 50A.

If there were no other circuits in the room then I would think the 3A.

(As a side matter)
I understand the pipework is an Extraneous conductive part.
It can also be an Exposed-c-p.

A boiler would be and exposed conductive part, so in my mind that would make any metallic pipe work coming from it also exposed?
Yes.

If a fault happened on the boiler, that fault current would appear on the pipe work.
But pipe work , radiators etc are classed as Extraneous?
It would be an Exposed-c-p at the appliance making it an Extraneous-c-p to the bathroom.

However, if it were no other circuits in the bathroom then it would have to be treated as an Exposed-c-p in relation to any other Extraneous-c-ps.

But back to the formula.
So if my maths is correct if it was just a 6A lighting circuit - 50/30 =1.6ohms
shower 50/250 = 0.21ohms
Well, 1.66 and 0.20 but otherwise correct.

So if the resistance between a lighting fixture cpc and a towel rail is < 1.6ohms supplementary protection is not needed?
Correct but between the light Exposed-c-p and towel rail.
701 just says the bonding shall be connected to the cpc.

But if you had a class 2 fitting the resistance would be infinite.
It would have no Exposed-c-ps so nothing to measure.
701 only says a Class 2 fittings shall have its cpc bonded for spurious future proofing; it's not really necessary.

And if 0.21 was the resistance between the shower and a towel rail, supplementary protection is not needed?
Correct ≤0.20

But then I read about maximum fault current being 0.01A
Yes 10mA - see below.

Then take into account 1000 ohms fingertip resistance??
I think that is not worth considering as people vary and the figures are arbitrary anyway.



So we are looking at a resistance of 22000Ohms
Which needs an IR test
So if a known earth and and Extraneous part is < 22000ohms it needs bonding, but if its more bonding not required
You are confusing two things.
22000 or 23000 is the measurement between Extraneous-c-ps and the MET to determine in the first place if the part is an Extraneous-c-p to the bathroom.

That is what I mean by the figures being arbitrary.
230/10mA = 23000
10mA being the recognised current above which muscles will contract and you will not be able to let go.
However, the voltage might actually be 240V so that makes the figure 24000

Some people think this current should be limited to less than 10mA.
Say 5mA so that makes the figure 46000 or 48000 and a touch voltage of 25V maximum.
 
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