Electrician's Notes

Devon Sparky · Apr 5, 2013

True...the equation you put up isn't a difficult sum...but it doesn't give the right answer. I was looking for the 80% figure for a 6amp breaker on a C curve (the breaker that the manufacturers of the LED downlighters I use recommend for circuits with more than 14 fittings). My Electrician's Notes (and the 7671) tell me this is 3.06 - I make your sum to be 4.79. This sort of backs up my liking for a quick reference guide in the toolbox or handy in the van.

brman · Apr 15, 2013

Devon Sparky said:
True...the equation you put up isn't a difficult sum...but it doesn't give the right answer. I was looking for the 80% figure for a 6amp breaker on a C curve (the breaker that the manufacturers of the LED downlighters I use recommend for circuits with more than 14 fittings). My Electrician's Notes (and the 7671) tell me this is 3.06 - I make your sum to be 4.79. This sort of backs up my liking for a quick reference guide in the toolbox or handy in the van.

Sorry, I've been on my hols and only just seen your reply.

I think you might have had finger trouble with your calculator

230/(6*10)*0.8=3.0666 according to mine :lol:

I think you might have done 230/(6*10*0.8) which gives 4.79.....

Mind you, that might have just proved your point. I guess a table of numbers is less likely to go wrong that working it out on a calculator all the time

Devon Sparky · Apr 15, 2013

brman said:
Sorry, I've been on my hols and only just seen your reply.

I think you might have had finger trouble with your calculator

230/(6*10)*0.8=3.0666 according to mine :lol:

I think you might have done 230/(6*10*0.8) which gives 4.79.....

Mind you, that might have just proved your point. I guess a table of numbers is less likely to go wrong that working it out on a calculator all the time

What can I say....I've had several sleeps since my 'A' Level maths! Also in 30 years in the trade I've never used that equation for a breaker so this old dog has learnt something new today!

Devon Sparky · Apr 19, 2013

I think I still prefer my Electrician's Notes but I like your equation. I've had a look through my old course work and can't find any reference to the equation you're using - I'd be interested to know the algebraic equation if you have and if you have the figures for a C rated breaker? Cheers :43:

brman · Apr 19, 2013

The equation is based on the curves for BS breakers. If you look at the curves in the BGB you will see that a type B breaker needs 5 times its rated current for a 0.1 to 5 second break time. A type C needs 10 times the rated current and a type D is 20 times I think (but I don't use them much! )

So, if your breaker rating is I Amps. Nominal voltage (Uo) is 230V and derating for temperature (d) is 0.8 then the equations is:
Max Zs = d * Uo / (I * 5) for type B or
Max Zs = d * Uo / (I * 10) for type C etc

Hence, a type C 6A is 0.8 x 230 / (6 * 10) or 3.066 ohms.

I hope that makes sense!