Supplementary Bonding Revisited

[JohnW2]

What I am trying to do is help with where I think Studentspark's confusion lies.

Yes, I realise that.

I am trying to distinguish between just getting a shock when a person is only earthed by the ground/floor and the situation when a person touches a live expo-c-p and another item which is also connected to the MET where bonding might help.

I continue to be a bit unsure about what distinction you are trying to make.

If the 'another item' is 'connected to the MET' it surely IS already 'bonded', isn't it? I would have thought the situation we should be considering is when the 'item' is not 'connected to the MET' in which case, as you say, making that connection (i.e. 'bonding' it) will improve the situation (I think probably inevitably, to at least some extent).

Kind Regards, John

 

[EFLImpudence]

To put it another way:

To determine the 'touch Voltage' at the expo-c-p when the person is earthed, it is not necessary to calculate the current flowing in the circuit - even though it is dependent on the same numbers.

 

[EFLImpudence]

If the 'another item' is 'connected to the MET' it surely IS already 'bonded', isn't it?

Not necessarily; it might just be connected by a CPC.

 

[JohnW2]

To put it another way: To determine the 'touch Voltage' at the expo-c-p when the person is earthed, it is not necessary to calculate the current flowing in the circuit - even though it is dependent on the same numbers.

Indeed not, but you take us back to the very earliest discussions you and I had about SB.

As I explained back then, there are two (mathematically equivalent) ways of determining the potential difference between two points in a circuit:

1. use the total loop impedance (and voltage) to determine the current, and then multiply that current by the resistance between the two points of interest
2. determine the ratio of the resistance between the two points of interest and the total loop impedance, and then multiply that by the voltage

For example, using the 'with bonding' figures from what you illustrated:

1. total loop impedance = 1.035Ω, hence If = 230V/1.035Ω = 222.22A, then multiply this by R2 (0.145Ω) to get touch voltage of 222.22 x 0.145 = 32.22 V
2. ratio of R2 to total loop impedance = 0.145Ω/1.035Ω = 0.140. Multiply this by 230 V to get touch voltage of 0.140 x 230 = 32.22 V

If I recall correctly you preferred (and/or found easier to understand) approach (1) (with calculation of the current), which is also what studentspark has been using, whereas I found approach (2) (using ratio of resistances/impedances, without calculating current) more simple. Is that correct and, if so, have you moved in the direction of my 'preference' since then?

Kind Regards, John

 

[JohnW2]

Not necessarily; it might just be connected by a CPC.

Well, if I understand you correctly, that's just a matter of CSA.

A conductor connecting the part to the MET is a conductor connecting the part to the MET - but, sure, one can reduce the voltage across it (in the presence of a very high fault current flowing through it) by increasing the CSA - for example by putting another conductor (whether called a CPC or 'bonding conductor') in parallel with the existing CPC.

Kind Regards, John

 

[EFLImpudence]

I am obviously not explaining myself very well.

Indeed not, but you take us back to the very earliest discussions you and I had about SB.

As I explained back then, there are two (mathematically equivalent) ways of determining the potential difference between two points in a circuit:

1. use the total loop impedance (and voltage) to determine the current, and then multiply that current by the resistance between the two points of interest
2. determine the ratio of the resistance between the two points of interest and the total loop impedance, and then multiply that by the voltage

For example, using the 'with bonding' figures from what you illustrated:

1. total loop impedance = 1.035Ω, hence If = 230V/1.035Ω = 222.22A, then multiply this by R2 (0.145Ω) to get touch voltage of 222.22 x 0.145 = 32.22 V
2. ratio of R2 to total loop impedance = 0.145Ω/1.035Ω = 0.140. Multiply this by 230 V to get touch voltage of 0.140 x 230 = 32.22 V

If I recall correctly you preferred (and/or found easier to understand) approach (1) (with calculation of the current), which is also what studentspark has been using, whereas I found approach (2) (using ratio of resistances/impedances, without calculating current) more simple. Is that correct and, if so, have you moved in the direction of my 'preference' since then?

Yes, that's it. I feel Studentspark might be in the same boat as I was; not realising the numbers are the same either way but the actual current does not play a part where there is no bonding.

Apologies if Studentspark is not in that boat.



Also, when I discussed with you that Volt-drop was a week point in my understanding.

I now feel that this was (mistakenly) because I could never make sense of "R≤50/Ia" between exposed- and extraneous-c-ps in relation to supplementary bonding no matter how hard I tried.
I had assumed the regulation must be correct and so I must be wrong.

Since then you have discovered/realised that the regulation itself is incorrect in the situations to which they state it should be applied.

 

[JohnW2]

I am obviously not explaining myself very well. .... Yes, that's it. I feel Studentspark might be in the same boat as I was; not realising the numbers are the same either way but the actual current does not play a part where there is no bonding.

Fair enough, but I think you are perhaps at risk of introducing confusion (maybe for Studentspark) by saying "... the actual current does not play a part where there is no bonding", since the current actually plays a crucial part in the actual situation -i.e. if there were no fault current, there would be no touch voltage.

What I think you are trying to say/explain (with which I would agree) is that one does not have to explicitly calculate the fault current in order to determine the touch voltage [i.e.one can use my approach (2)] - although, even then, one has 'implicitly' determined the fault current, since that would just be the touch voltage (determined by other means) divided by R2.

Also, when I discussed with you that Volt-drop was a week point in my understanding.
I now feel that this was (mistakenly) because I could never make sense of "R≤50/Ia" between exposed- and extraneous-c-ps in relation to supplementary bonding no matter how hard I tried. I had assumed the regulation must be correct and so I must be wrong. Since then you have discovered/realised that the regulation itself is incorrect in the situations to which they state it should be applied.

Quite so. As you say, unless I've been missing something important, the reason you could make no sense of the "R<50/Ia" thing is because it actually makes no sense!

Kind Regards, John

 

[EFLImpudence]

Yes, that's it - in so much that if the resistances of the circuit were all twice as much, the touch voltage (w/o bonding) would be the same but the fault current would be halved (although all the values are related).

Much like Mr.Ohm's discovery - isn't it lucky that V, I, and R were all exactly 1? .

 

[JohnW2]

Yes, that's it - in so much that if the resistances of the circuit were all twice as much, the touch voltage (w/o bonding) would be the same but the fault current would be halved (although all the values are related).

Indeed - that's why I've always thought that my 'ratio of resistances' way of looking at it was simpler and (I thought!) easier to understand - albeit, if I understood correctly, you seemed to initially find the 'current-based' approach to be easier to understand.

Kind Regards, John

 

[EFLImpudence]

Indeed - that's why I've always thought that my 'ratio of resistances' way of looking at it was simpler and (I thought!) easier to understand - albeit, if I understood correctly, you seemed to initially find the 'current-based' approach to be easier to understand.

It was as if I thought the two ways were separate and one did one thing and one did another.

It's hard to remember exactly now since you have explained it all. Thank you very much.


That's why I thought Studentspark might be in a similar position and missing something obvious.

 

[JohnW2]

It was as if I thought the two ways were separate and one did one thing and one did another.

Fair enough. I think I probably convinced you otherwise a fair while ago, but perhaps some other readers (if there are any!) will have seen and understood from my explanations that the two different approaches (calculating current OR working with ratios of resistances/impedances) are mathematically identical.

It's hard to remember exactly now since you have explained it all. Thank you very much.

You're very welcome.

That's why I thought Studentspark might be in a similar position and missing something obvious.

Very few (if any) things are "obvious" (to me or anyone else) until one understands them! ... but, yes, maybe Studentspark had/has a similar 'problem'. Since, from the start, he's been working by first calculating fault current, I've stuck with that approach in most/all of my exchanges with him, but maybe I should (or should have) tried introducing him to the alternative approach!

Kind Regards, John

 

[studentspark]

OK.
I fear I'm going backwards, but I need to get something clear in my head please John, before I unleash my next confusing drawing.

MET potential, and how that affects the connected conductors...

Im getting a bit confused by the volt drop in a conductor when no current is flowing.

So in CPC 1 we have 222.222 Amps ( as in previous examples)

The voltage at the fault was 121v, by the time it gets to the MET the voltage has dropped to 88.88v , due to the resistance of the CPC.

So has the METs voltage to earth has risen by 88.88 v ???

But there is not a fault on CPC2

You said ---Since there will be no current flowing in the CPC of the 'other circuit' (without a fault), there will be no VD across the length of that other CPC, so that the potential at the end of it will be exactly the same as the potential of the MET.

But is the MET at 88v or 0v

So what would be the voltage at the pipe (MPB) or the light switch (CPC)?

So basically what is the voltage to earth at the MET and at the connected ECPs

I think if I understand that, it will all drop into place.

 

[JohnW2]

OK....I fear I'm going backwards, but I need to get something clear in my head please John, before I unleash my next confusing drawing. ... MET potential, and how that affects the connected conductors...

OK.

...Im getting a bit confused by the volt drop in a conductor when no current is flowing.

If there is no current through a conductor, then there is no 'volt drop' across it - so the potentials(relative to earth or anything else) at the twop ends will be identical.

Looked at the other way around, if there IS a voltage across a conductor, then there WILL be a current flowing through the conductor (the magnitude determined by Ohm's Law).

Does some of the confusion perhaps result from this term "voltage drop", I wonder? What/all we are talking about is the voltage across a conductor (of non-zero resistance) when a current flows though it - i.e. again just Ohm's Law. The difference between potentials(relative to earth or anything else) at the two ends of the conductor will then be equal to that 'voltage drop' (voltage across the conductor).

... So in CPC 1 we have 222.222 Amps ( as in previous examples) ... The voltage at the fault was 121v, by the time it gets to the MET the voltage has dropped to 88.88v, due to the resistance of the CPC.

Yes, that's one way of looking at it, but, also ...

So has the METs voltage to earth has risen by 88.88 v ???

Again, yes - as above, you could look at it that way (i.e. that the potential of the MET is the potential at the fault MINUS the voltage across ('voltage drop') the CPC. However, you could also look at it as being due to the fault current flowing through the 'conductor' from MET to earth (i.e. your R2e) ...
.... - i.e. 222.222 A x 0.4Ω = 88.88v

... But there is not a fault on CPC2 ... You said ---Since there will be no current flowing in the CPC of the 'other circuit' (without a fault), there will be no VD across the length of that other CPC, so that the potential at the end of it will be exactly the same as the potential of the MET.

Indeed (in the absence of SB)

But is the MET at 88v or 0v

88v (relative to earth). Connecting other CPCs (just 'bits of wire which go nowhere', unless there is a fault in their circuit) to the MET obviously does not alter the situation in which the MET is at 88V relative to earth due to a fault on another circuit.

So what would be the voltage at the pipe (MPB) or the light switch (CPC)? So basically what is the voltage to earth at the MET and at the connected ECPs

You may be confusing yourself by referring to both the light switch and pipe as "ECPs" - the light switch is an Exposed-c-p, but the pipe is an Extraneous-c-p - and that difference is 'important' (albeit maybe fairly trivial)....

The exposed-c-p (light switch) is connected (by CPC) to MET, but no current is flowing through the CPC, hence no 'voltage drop' across it, hence is at exactly the same potential (88.88 V wrt earth) as the MET.

The extraneous-c-p (pipe) is connected (by bonding conductor) to MET. However, in this case, since the pipe provides a path to earth, there will be some current flowing through it, hence there will be some 'voltage drop' across it, hence the potential of the pipe (relative to earth) will be a bit less than the potential of the MET (relative to earth) - i.e. a bit less than 88.8 V. In practice, since the resistance of the bonding conductor will be very low, and (usually) only a small proportion of the fault current will flow through the extraneous-c-p to earth, the voltage drop in bonding conductor will usually be very low, hence potential of extraneous-c-p very close to MET potential.

Does any of that make sense to you?

Kind Regards, John

 

[studentspark]

Yes this is making sense, but I am still pondering the CPC to the light switch

The exposed-c-p (light switch) is connected (by CPC) to MET, but no current is flowing through the CPC, hence no 'voltage drop' across it, hence is at exactly the same potential (88.88 V wrt earth) as the MET.

Oh dear ..Sorry John

So, how does the voltage appear on the light switch plate - It has to be by the CPC. So there is a voltage on it.
So we have a voltage, we have a resistance, so we must have some current. ?

The other bits on the side were to do with EFLImpudence's question. a, being the formula I have been using, B, EFLI and C your ratio method.

I will come back to that but, I am confused how the voltage is on the light switch. If in this example the CPC to the light had a resistance of 0.45Ω, and the only way a voltage could appear at the switch, was via the CPC's connected to the MET. which is subject to a rise in potential , due to a fault.

 

[JohnW2]

Yes this is making sense, but I am still pondering the CPC to the light switch.... So, how does the voltage appear on the light switch plate - It has to be by the CPC. So there is a voltage on it. So we have a voltage, we have a resistance, so we must have some current. ?

I think you are getting confused about these 'voltages', which is why I suggested that maybe this term 'voltage drop' is part of what is causing your confusion.

There is no such thing as ('absolute') "voltage" - only potential differences between two points., When you talk about "voltages" (e.g.~122V at fault, ~88V at MET etc.) you are usually talking about potential differences to earth.

You will only get current through a resistance (per Ohm's Law) if there is a potential difference between it's two ends. If you had a resistor with, say, a PD of 1,000V relative to earth at both ends, the potential difference across the resistor would be zero, hence no current would flow through the resistor.

... and so it is with the exposed-c-p of your light switch. The CPC connecting it to the MET has a resistance, but the voltage at both ends of that resistance is ~88 V, so there is no PD across the resistance, hence no current flowing through it.

Does that help?

Kind Regards, John