View the thread, titled "Any tips ?? Remembering formula" which is posted in UK Electrical Forum on Electricians Forums.

Paddy4444

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Trainee
Hey I’m studying 2365 at college and I’m at the science part now .. any tips on remembering formula or is it just a case of going over and over it till it’s in your head .. any help much appreciated
 
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,
 
I think time would be better spent learning where to find them rather than learning them off by heart.

There is always a possibility that you will remember them incorrectly but if you can find then in the book then it is there in front of you.
 
Rather than trying to memorise formulae by rote, it's better to try to understand where they come from. Look carefully at the underlying principles and try to get a real handle on what each term in the formula represents. An engineer can derive many of the formulae he/she uses, out of nowhere, by remembering how things behave.

A simplified example. Suppose I had forgotten that P=V²/R. I would say to myself, more volts across a known resistance pushes proportionately more amps through. More volts makes more power and more amps also makes more power, so V must appear on top, squared. But more resistance means less amps get through per volt, so that appears on the bottom. Presto, P=V²/R.

Once you start using something every 10 minutes of the working day, it becomes automatic and very much faster, so it's good to practice. When you see electrical stuff, apply what you know to calculate things you don't know. What's the total current used by all the lights in the supermarket? What temperature would 2.5mm² T+E reach if you powered your whole house from it? How many coulombs would have to pass through your uncle's mobility scooter motor to carry him up a 1/2 mile hill at 1 in 10 gradient?

After a while, literally, it becomes second nature. I did the last one in my head in about 30 seconds, it's somewhere between 5 & 10kC for a 24V scooter depending on weights, efficiencies, road surface etc.
 
Last edited:
We were told to make formula sheets and refer back to them whenever we needed to, mine was riddled with reminders of what the various symbols meant.

In our college moodle was a great place for pdf books full of useful information to be downloaded, it's amazing just how much help is there if you just look. I say that because I was always trying to point the younger students towards it, but if it wasn't an app or some crap they could view on their phone they wouldn't seem to take it on board. Mobile phones should be banned in college time in my book, you seriously need to concentrate to do the work and phones are the antithesis to concentration!
 
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,



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 ?????????
 
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 ?????????
Well don't leave us all in suspense Jay, what's the answer?
 
A 5kw unit is connected to a 230 supply and draws a current of 24 Amps, Work out the effienciency of the unit ?????????

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%
 
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 :D

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 :eek:
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!
 
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!

Confucius 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%



Thanks it was a question I remembered from my exam, as everybody got it wrong! Hence the reason I relayed it incorrectly.

The formula we learnt to work out efficiency
Was

Output power
————————. X 100
Input power


But we never knew how to work out input power from volts and amps

Using power triangle I understand

P
I V

But I don’t understand the p = v2 part


And where did you get 1 from

5000/ (230 x 24 x 1?)
 

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Paddy4444

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