Is that perhaps a bit confusing?
The voltage is the same at both ends because there is no current flowing.
Supplementary Bonding Revisited
- Thread starter JohnW2
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The voltage is the same at both ends because there is no current flowing.
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Any of it can be confusing ungtil one actually understands.
Ohm's law is symmetrical.
Yes, one can say (and I did say) that the voltage (potential difference relative to something) is the same at both ends because there is no current flowing ... but it is equally true to say that there is no current flowing because the potential (relative to something) is the same at both ends.
I suppose it's down to 'the reader' to decide which way of thinking of it they find easier to understand. If one takes a hydraulic analogy, with two water tanks filled to the same height connected by a pipe, I am personally more likely to say that "no water flows through the pipe because the pressure is the same at both ends" than that "the pressure is the same at both ends because there is no water flowing through the pipe", wouldn't you?
Kind Regards, John
Yes, I agree but thought it might be easier for SS as he said that was what he was confused about.
Yes, maybe. One of the greatest problems with any teaching is that those who understand something very well can find it almost impossible to understand what it is that others struggle to understand.
One of my maths teachers was a dramatic example of this. He was, in an academic sense, a brilliant mathematician, but totally useless as a teacher. When asked something like "why is that the answer to that bit of the calculation" he would answer "because it is" or "because it is obvious" - and he really didn't understand why this was not equally apparent to others!
Over the years I've learned that when trying to help people like SS, one has to try to get into their minds and work out why things are not as 'apparent', 'clear' or 'obvious' to them as they are to oneself, but that can be very difficult.
Kind Regards, John
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Yes, I have heard that is why some geniuses appear aloof if not actually miserable.
Well, you'll be pleased to know that I predicted all your answers correctly this afternoon.
Well, you'll be pleased to know that I predicted all your answers correctly this afternoon.
I fear I will drive you chaps mad....
I do understand the concept of potential difference. Voltage needs to be different at two points for current to flow.
What I am struggling with is.
With no fault, all the Extraneous and Exposed CP are at the same potential. MET voltage, as they are all connected together.
Now I understand this all happens at the speed of light. But if we could slow it down.
That current from the fault travels down the CPC in question, through the MET and back to the
transformer, as this is the least path of resistance.
The voltage also pushes along the CPC to the MET, and will gradually lessen, as it gets to the MET because of the resistance of the CPC. Voltage drop. or voltage use ? or voltage left?
In this case the EMF has about 88 volts left in it, by the time it gets to the MET.
But a exposed CP is also connected to the MET. But that is some distance away.
A millionth of a second ago (sort of) That EXP -CP was 0v. Now its at 88V. But there is a resistive path to the EXP -CP
That 88v had to travel to the EXP C-P. Why has that not dropped, as it did, it the original CPC.
I get that they are both at 88V and as so no current will flow, but they was a time when they were not both at 88V.
I do understand the concept of potential difference. Voltage needs to be different at two points for current to flow.
What I am struggling with is.
With no fault, all the Extraneous and Exposed CP are at the same potential. MET voltage, as they are all connected together.
Now I understand this all happens at the speed of light. But if we could slow it down.
That current from the fault travels down the CPC in question, through the MET and back to the
transformer, as this is the least path of resistance.
The voltage also pushes along the CPC to the MET, and will gradually lessen, as it gets to the MET because of the resistance of the CPC. Voltage drop. or voltage use ? or voltage left?
In this case the EMF has about 88 volts left in it, by the time it gets to the MET.
But a exposed CP is also connected to the MET. But that is some distance away.
A millionth of a second ago (sort of) That EXP -CP was 0v. Now its at 88V. But there is a resistive path to the EXP -CP
That 88v had to travel to the EXP C-P. Why has that not dropped, as it did, it the original CPC.
I get that they are both at 88V and as so no current will flow, but they was a time when they were not both at 88V.
Don't worry about that. It's 'frustrating', not 'maddening', that I'm not yet doing well enough in helping you to understand.
Right. That's a good, and crucial, start.
Not only that. Rather ".... as they are connected together by conductors, none of which have any current flowing through them".
If there were more than one path (i.e. 'parallel paths') back to the transformer, the current would not only pass through "the path of least least resistance" - it would be shared between all the parallel paths, each carrying a proportion of the current which as inversely proportional to its resistance - so, for example, if there were two paths in parallel, one with twice the resistance of the other, twice as much current would pass through the latter (i.e. two-thirds would go through that path, and one-third through the other).
Your language, together with the concepts of voltage 'moving', 'pushing'and 'being left' are all rather 'quaint' - but, yes, I think what you are saying is roughly correct!
I think you are probably at risk of confusing yourself even further by bringing speed and time into the discussion. All of the changes we're talking about effectively happen 'instantaneously' and, hence, 'simultaneously'
As above, you seem to be over-thinking this by bring time/speed into it in other words, there will not, once the fault has arisen, really be "a time when they were not both at 88V" - once the fault arises, both would 'immediately' rise to 88V.
Perhaps it would be easier for you think about this in terms of circuits around which current flows. If (as in the simplest cases we are considering) it is a 'simple' circuit (i.e. without branches) then, at any point in time, exactly the same current must flow through every part of the circuit (that's essentially 'obvious', but you could also call it Kirchoff's Law if you wanted). The changes in potentials (relative to earth or whatever) we're talking about are all due to changes in the current flowing through resistances of various parts of the circuit- and, since exactly the same current must be flowing through every part of the circuit at any point in time, then the changes in potential (relative to anything) at any point in the circuit must happen simultaneously.
Does that get us any nearer?
Kind Regards, John
Yes, I have heard that is why some geniuses appear aloof if not actually miserable. Well, you'll be pleased to know that I predicted all your answers correctly this afternoon.
Maybe my recent comment about the current everywhere within a simply circuit having to be exactly the same at any point in time will help SS a bit?
However, all this recent interaction with SS, has made me realise that I may have been somewhat misleading you (and myself) with some of what I wrote about this topic 'way back then' - for which I apologise.
To the best of my knowledge, there was nothing actually incorrect about what I was saying/writing, my calculations or the figures (for 'touch voltages' etc.) those calculations produced BUT they were (I probably thought 'for convenience/simplicity') prefaced by "assuming that the supply voltage was 230V during the fault" - "supply" voltage being the L-MET PD at the origin for a TN-C-S installation.
With that assumption (and assuming TN-C-S), using the 'ratio of resistances' approach, with SS's most recent (perfectly credible) figures for R1 (0.09Ω) and R2 (0.145Ω), that would mean that the potential of the exposed-c-p would, during the fault, be 230 x 0.145 / (0.09 + 0.145) = 141.9 V above MET potential, so the PD between that exposed-c-p and a bonded extraneous-c-p (AT MET potential, by virtue of bonding) would also be 141.9 V.
However, as SS has illustrated, with his values for Ze (R1e and R2e) the (L-MET) 'supply voltage' falls to dramatically less than 230V during the fault, leading to his guesstimated 'touch voltage' of only about 34V (~122V- ~88V).
I say 'guesstimated' since, in practice, that's the best one can do because, even with TN-C-S (when one knows that "R1e"="R2e"), one doesn't know what the voltage at the transformer is. SS assumed 230V (or sometimes slightly above that), but that is presumably far lower than would normally be the case. Also, the simplistic calculations ignore the path to earth via the extraneous-C-P. As I recently illustrated with my graphs, if that path to earth is of very low impedance (less than "R2e") it becomes impossible to get a very low 'touch voltage', since that path 'pulls down' the potential of the MET, thereby increasing 'touch voltage'.
Kind Regards, John
I'm getting it John
Been reading up on kirchhoff's law.
As you say John. This 50V is not really achievable.
Its interesting that the very low EFLI which we are encouraged to obtain, increases the touch voltage.
And I can see more clearly, the importance of supplementary bonding, which again as you say has all been all but done away with, because RCDs are there to save us. When they work. But thats disconnection times. Touch voltage will be over 50V until a protective device operates
Back in a bit with some parallel path questions
Been reading up on kirchhoff's law.
As you say John. This 50V is not really achievable.
Its interesting that the very low EFLI which we are encouraged to obtain, increases the touch voltage.
And I can see more clearly, the importance of supplementary bonding, which again as you say has all been all but done away with, because RCDs are there to save us. When they work. But thats disconnection times. Touch voltage will be over 50V until a protective device operates
Back in a bit with some parallel path questions
I put this together.
I was listening to the Footie and drinking beer, whilst doing but I think my sums are correct
I was listening to the Footie and drinking beer, whilst doing
but I think my sums are correctGreat!I'm getting it John
Good
In fact, although it gets a bit more complicated when there are branches, if you have a single, simple (unbranching) circuit (i.e. everything simply 'in series') then it is unbelievably simple, since K's Law then merely means that, at any particular point in time, the current everywhere in the circuit must be identical (I suggest you avoid confusing yourself by trying to think about 'how long it takes the current to get around the circuit!)That's not quite true. As I illustrated with my graphs, and as you discovered yourself when I suggested you tried re-doing your calculations with a Ze of 0.8Ω (0.4Ω + 0.4Ω), it may be possible to get a touch voltage <50V (or any other arbitrary figure). As I illustrated, the point is that for any final circuit (i.e. any R1 and R2), the lowest touch voltage achievable will depend upon the Zs (particularly the "R2e" component thereof) and the resistance to earth of the extraneous-c-p (which I will call Zecp). If Zs and/or Zecp are too low, then it becomes impossible to get a 'touch voltage' as low as 50V.
That makes sense. If R2e and/or Zecp reduce, the EFLI decreases, so the fault current increases. Since (with bonding) 'touch voltage' is simply fault current multiplexed by R2, the touch voltage will therefore increase. If you consider the extreme situation of Ze=0,then that takes us back to the situation in which touch voltage (the voltage across R2) will (if voltage at transformer is 230V) simply be 230 x R2 / (R1+R2), which is about 140V for 2.5mm²/1.5mm² T+E.
Indeed -as above.
All true, at least 'in my opinion' (although most of the electrical industry appears to ignore this -or, at least, decide to just 'rely on RCDs').
I think you are, indeed, 'getting there'- well done!
Kind Regards, John
Indeed yes. But thats at the maximum (Tns) Ze. I generally never find it that high. Usually less than 0.2Ω.
Agreed. Of course, TT is a whole different story - and I wonder if that might be where some of the traditional ideas about 'touch voltages' (and bonding) came from?
Kind Regards, John
With Supplementary bonding all but done away with (for now) and the additional protection of the RCD, now being found to be wanting, because the AC types we have been installing, can't deal with the DC modern equipment uses. You wonder if installations have been made less safe.
Maybe. However, as you will be aware if you read some of my mutterings here, I have yet to see the evidence to convince me that the protection provided by Type AC RCDs is necessarily significantly 'wanting'. All I know about it is 'the theory'.
As you probably also know, I do think that domestic installations are being "made less safe" (electrically) by the installation of metal CUs etc. If it hasn't already happened, it is (in my opinion) only a matter of time before that results in deaths!
Kind Regards, John
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