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Bright Sparks
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(S)he has now more-or-less acknowledged that (although (s)he didn't work out the answer before posting, hence didn't realise (s)he'd created an impossible set of numbers!), in mathematical terms, the answer given by several of us (first by Spark123) is the coirrect one - and (s)he has certainly indicated that (s)he knows what mathematical process has to be undertaken to get an answer. However, as you will have noticed, I'm interested to know how far his/her mathematical skills actually go.
Kind Regards, John.
I'm not brilliant at maths. I think I got a grade C at GCSE, but Carol Vorderman I aint.
Does that mean I'm a bad electrician?
Does that mean I'm a bad electrician?
Of course it doesn't. To be a competant electrician, you need to be able to do simple arithmetic, nothing more; even working with the infamous adiabatic equation (an adiabatic equation, as I always have to remind people), only requires the most basic of understanding of the concept of algebra and the ability to plug values into an equation.
I certainly didn't start this maths thing - and, as you imply, even if it were true that BAS were mathematically incompetent (which he certainly isn't), that would in no way mean that, should he wish to become an electrician, he would be a bad one! It's no secret that I'm not BAS's greatest fan, but I'll defend him as much as I have to on this one!
Kindest Rehards, John.
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ooh, that would be toasty - wouldn't want to be standing too near when "testing" it
No, because you realise the limits and so are able to deal with it - whether that involves using the books a bit more, checking with a colleague, or using some ready made tables, or whatever. Mind you, I can't think of anything where the actual calculations are that difficult (doubt if a GCSE is needed) - as long as you know what they are. The example that kicked this off is a bit contrived and I can't see any real reason for having to do it for real - work out the total length of cable perhaps, but in general all you are interested in is R1+R2 and R1+Rn not exceeding certain limits.
It's only dangerous when someone doesn't know the maths and either doesn't realise or doesn't care that they don't.
Basic mathematical skills are needed for some domestic work - most circuits are generic anyway and it's all been worked out for you by the heads at the IET. Most domestic electricians only have to read a meter and copy the results.
When/if you progress into more complex engineering and dealing with imaginary numbers and 'j' notation is when you can scratch your head a bit.
When/if you progress into more complex engineering and dealing with imaginary numbers and 'j' notation is when you can scratch your head a bit.
I can do cable calcs and ohms law dead easy, and the adiabatic if I sit down and think about it, and that's about as much as I've ever needed to know.
Exactly.
That's really electrical engineering. 'An electrician' does not really need to deal with any of that, although I'm sure that many are quite capable of it.
Kind Regards, John.
Exactly - and no electrician should normally ever need to do anything much more mathematically complicated than that. As others here are hinting, dealing with 3-phase currents is perhaps the most 'mathematical' thing which will be required, and even that virtually never for those who work only on 'ordinary' domestic installations - even my house, which, in a literal sense, has a 3-phase supply is, in reality, just three single-phase installations.
Kind Regards, John.
No, but I bet you look c*** in a tight dress.
Can the load have a negative resistance?
Hey, you're giving him/her clues about the magnitude of the answer
Indeed - but that is equal true of every relevant aspect of electrical work, not just (the small amount of) maths.
This discussion reminds me of the fairly common belief that accountants, or even those who work in banks etc. "need to be good at maths"!
Kind Regards, John.
Insufficient information - it's not possible to calculate from individual current magnitudes only. However, if you make assumptions about power factor for each load then that changes.
Without calculating it ('walking' phasor diagrams with my fingers), I'd guesstimate about 25A if all currents are at the same phase angle to their voltages.
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