John, check what the caching settings in your browser are set to. Some browsers cache things to varying degrees and this maybe effecting what you're seeing. It can make setup of some equipment with a web interface complete madness if its turned on.
Question About Volt Drop
- Thread starter KayBull
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John, check what the caching settings in your browser are set to. Some browsers cache things to varying degrees and this maybe effecting what you're seeing. It can make setup of some equipment with a web interface complete madness if its turned on.
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That's a good point, and I certainly will check, but if the settings have change, they've changed without my assistance - and this is the first time I've experienced this behaviour in this forum in the past couple of years or whatever.
Kind Regards. John.
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I would agree John that with a 20A radial to consider it at 20A at furthest point is going to ensure the volt drop at all points is within that limit.
The 32A radial however raises another question. Now I have read the old red book cover to cover and I find nothing but when I took my 2382 the on site guide had not been published so never got a copy maybe answer in in there?
At a IET instruction evening on the 17th Edition it was stated we could have 106 meters of 2.5mm cable in a ring and comply with volt drop.
However getting home and trying to emulate the results I failed so next time I questioned why. I was told we don't calculate at 32A but at 26A and fair enough once I used 26A the answer was correct.
As to why 26A it seems that some agreement has been made that likely one would have only 20A draw at centre with the rest of the 12A being evenly distributed over the whole lenght. So the average would be 6A hence the 26A figure.
I would guess there is something similar for a radial circuit? So I would say up to 20A we use the full 20A for calculations. Above 20A we halve the value over 20A so a 25A MCB would be calculated at 22.5A.
However I have not seen where this is written. Only been told about it in IET meetings.
Your right likely we are looking too deep for the answer. I do remember at college being given the question why does a battery voltage go down. Until the question I had not thought about it so did some research and wrote a two page essay one the chemical reactions and why the voltage dropped. Only to find all he wanted was because the battery was discharged.
As to question of i or j since the calculator always shows it as i using i seems a sensible option. However even A level maths does not teach imaginary numbers one needs to do further maths or mechanical maths to be taught it. It was not until I did my degree was I taught it. Yet my son did it in school but he did further maths.
The 32A radial however raises another question. Now I have read the old red book cover to cover and I find nothing but when I took my 2382 the on site guide had not been published so never got a copy maybe answer in in there?
At a IET instruction evening on the 17th Edition it was stated we could have 106 meters of 2.5mm cable in a ring and comply with volt drop.
However getting home and trying to emulate the results I failed so next time I questioned why. I was told we don't calculate at 32A but at 26A and fair enough once I used 26A the answer was correct.
As to why 26A it seems that some agreement has been made that likely one would have only 20A draw at centre with the rest of the 12A being evenly distributed over the whole lenght. So the average would be 6A hence the 26A figure.
I would guess there is something similar for a radial circuit? So I would say up to 20A we use the full 20A for calculations. Above 20A we halve the value over 20A so a 25A MCB would be calculated at 22.5A.
However I have not seen where this is written. Only been told about it in IET meetings.
Your right likely we are looking too deep for the answer. I do remember at college being given the question why does a battery voltage go down. Until the question I had not thought about it so did some research and wrote a two page essay one the chemical reactions and why the voltage dropped. Only to find all he wanted was because the battery was discharged.
As to question of i or j since the calculator always shows it as i using i seems a sensible option. However even A level maths does not teach imaginary numbers one needs to do further maths or mechanical maths to be taught it. It was not until I did my degree was I taught it. Yet my son did it in school but he did further maths.
Even so one has to know the difference between a dis-charged battery and battery not is not dis-charged.
The voltage from the electro-chemical re-action is almost constant, it is the increase of internal resistance in the cells that is the main cause of voltage drop when current is taken. A dis-charged battery has a higher internal resistance than when charged ( or new ).
Some types of battery have a near constant voltage until the last molecules of "power" have reacted as their construction ensures a constant internal resistance.
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The Daniell cell from memory was used to test voltage as it was 1 volt. I remember is as doing the CSE (Before GCSE) the question was on the exam paper and had not been covered during the year.
Yes I did the research and know the answers. Oddly I could not find any books at the time detailing lead acid batteries but assumed the action would be similar to all the batteries I did find details about in the library.
However the lecturer never gave 10 out of 10 so I got 9. However as a result of my paper the rest of the class got 1 out of 10. I was not flavour of the month with other students.
The exam end of year I got distinction Rest of the class all failed. Clearly the teaching was at fault and we were not being taught to correct standard. However 1972 there were no league tables. So lecturer pointed I that I passed and claimed the class was full of low achievers and got away with it.
Yes I did the research and know the answers. Oddly I could not find any books at the time detailing lead acid batteries but assumed the action would be similar to all the batteries I did find details about in the library.
However the lecturer never gave 10 out of 10 so I got 9. However as a result of my paper the rest of the class got 1 out of 10. I was not flavour of the month with other students.
The exam end of year I got distinction Rest of the class all failed. Clearly the teaching was at fault and we were not being taught to correct standard. However 1972 there were no league tables. So lecturer pointed I that I passed and claimed the class was full of low achievers and got away with it.
ericmark said:
Oh how standards have slipped!
When I did A-level maths (1969 - 1971) complex numbers were definitely taught - and the most oft heard comment from us pupils (sixth form students were still very new) was "It's a fiddle!"At about the same time, I taught myself how to analyse reactive circuits using complex numbers and caused myself plenty of confusion as a result, as in "How can you have an imaginary current?"
Many years later, an electronics engineering lecturer put us all on the right track from day one:"Some people will tell you that "j" is the square root of minus one. It is not! "i" is the square root of minus one. "j" is an operator."
But to get back to the original question, I think bernardgreen provided the most concise answer:
The many comments about unknown currents are surely not relevant. You're looking for a worst case drop here so work out the maximum current that could flow in each section. This is the total current that could be drawn from all the sockets it feeds or the breaker rating, whichever is less.
Yep, and also in 1964-1966.Oh how standards have slipped!
Indeed - and, as I said early in this discussion, I've always thought that students would find it much easier to get their heads around this if they were first taught about the vector operator and then told that applying that operator twice (i.e. 'squaring it') had the same effect as multiplying by -1 ... rather than rushing straight in and talking about "the square root of -1" - which, as you say, most students regard as either a 'fiddle' or incomprehensible!
Yes, I think that we (at least I) have always accepted that. What I was not initially convinced about (but think I now am) is the fact that the intuitive 'worst case' (of distributing the total available current amongst the available sockets, when the OPD In is the limiting factor) is, indeed, the case which results in the greatest VD. However, I think that I have now convinced myself that the 'obvious' worst case really is!But to get back to the original question ... The many comments about unknown currents are surely not relevant. You're looking for a worst case drop here so work out the maximum current that could flow in each section. This is the total current that could be drawn from all the sockets it feeds or the breaker rating, whichever is less.
Kind Regards, John
And I have convinced myself that this topic should not have needed to go on so long for something which is one of the more fundemental aspects of electrical engineering !!!
But apparently not a fundamental aspect of the training or education of some who do electrical work.
It would appear so !
68-70, and ditto, I'm sure.
TBH I don't remember doing it, but I feel I must have because:
a) I did do A level maths (& physics)
b) Although I did more maths in my first year at Uni, it was all new, not a recap of A level
c) I have never studied maths as a hobby
d) I have encountered complex numbers
I don't regard i as a fiddle - seems perfectly valid as a symbol with a value, or an operator, to use during calculations. Not really different in concept from using symbols which represent transcendental numbers - although they may be more 'real' they can't be replaced by actual numbers unless you start approximating.
All true. However, as I've said, I think the problem arises when (as I'm pretty sure happened when we first encountered it in Maths, but not in Physics) it is first introduced as a symbol which represents "the square root of -1" - which, in itself, is obviously a difficult/impossible concept to grasp.
Kind Regards, John
Sorry for the late response. I think I may have confused u all. No, I bought the book to teach myself as a head start before I embark on a lifetime of studying. I thought it would be easier to learn by mysef first to save stress later on.
Anyway, besides this, thanks for the replies.
The currents, allbeit unknown are irrelevant to the question. I just wished to know the correct principle of working out the potential max volt drop liable to be present in the circuit, not know what the actual overall volt drop is as that would be impossible to know without knowing the current drawn at each 'leg'.
The regs state the the max permitted volt drop if I remember correctly is 11.5V @ 230V (5%) for a power circuit. So I assume whatever is plugged into any given socket on the diagram, I work out the volt drop for each load and add together to achieve the max volt drop for the circuit. The book only seems to cover obtaining the volt drop for the circuit from a single load, but not from multiple loads on a circuit. Hope that makes sense.
Thank you all.
Right, you just need to remember that the first leg of the circuit carries the sum of all the currents, some current flows in the branch to the first socket so is subtracted from the total for the next leg, and so on. Just draw it out and the current in each conductor will be obvious.
You don't actually need anybodies theorem, or any understanding of vectors quantities and imaginary numbers etc.
You don't actually need anybodies theorem, or any understanding of vectors quantities and imaginary numbers etc.
KayBull said:
No they're not. Voltage drop is proportional to current. That's ohm's law and you really need to know it.
and also said:
The principle is straightforward. For any length of cable:
Voltage drop = current (amps) x length (metres) x ohms per metre.
The ohms per metre you can get from a book. The length you can measure. And the current?
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