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Discuss Any tips ?? Remembering formula in the UK Electrical Forum area at ElectriciansForums.net
transposing a formula is easy if you follow this rule.
whatever you do to one side of the equation, do same to the other. e.g:
P = IR. easy. now suppose you want to find if I as you don't know it's value, but you know the value of P and R.
so.. divide each side by R, thus maintaining the equality.
so you have
P/R =I
ssimplesss.
same principle for more complex eaquations,
Well don't leave us all in suspense Jay, what's the answer?Yes the example you gave is very easy to understand
I mean like for example.....
A 5kw unit is connected to a 230 supply and draws a current of 24 Amps
Work out the effienciency of the unit ?????????
A 5kw unit is connected to a 230 supply and draws a current of 24 Amps, Work out the effienciency of the unit ?????????
Even though you may think they are a bit advanced just have a bash. a wise man once told me you learn best by making an a**e of it. get it wrong and we can show you where you went off the tracks get it right and that’s all good. How will you know if you never even try!Hey paddy I believe we are in the same position !
I started my level 2 in September and have been going over and over the theory as telectrix stated and believe it or not it does start to sink in
Maybe we could have some theory questions based on formula suitable for “level 2 students”
From some of the more experienced guys!
Just a thought.....
So far we have covered powers, ohms law triangle, power triangle, working out energy, work done, but the hardest for me is deffo transposition!
I have looked at some of the electrical design questions on the trainee section! But they are a little advanced for me at the moment!
But I still read it to get a flavour for real life scenarios
Even though you may think they are a bit advanced just have a bash. a wise man once told me you learn best by making an a**e of it. get it wrong and we can show you where you went off the tracks get it right and that’s all good. How will you know if you never even try!
Is it bad that I read that with the accentConfucius say, "If it too easy now, it bite you on arse later"
You need to take more care over details in engineering. 5kW what? Input? Output? a '5kW' motor, that would usually mean the mechanical output rating. A 5kW heater, that would be input power. Presumably 230 means 230V single-phase AC but you don't say that. 24A with what power factor or is this DC?
If I rephrase the question as:
A device uses 24A at unity power factor from a 230V single-phase AC supply and outputs 5kW useful power. What it its efficiency?
Now it can be calculated because I've filled in the missing info.
Input power = V.I.cos(phi)
Efficiency = useful output power / input power
=5000 / (230 x 24 x 1)
=91%
Reply to Any tips ?? Remembering formula in the UK Electrical Forum area at ElectriciansForums.net