Work out breaker size?

[YaHandys]

Hi, I'm a first year apprentice... just trying to get an edge on theory.

Calculating breaker size:

W ÷ V = A

So say appliance needs 4000 watts

4000w ÷ 230v = 17.39amps

Then, from research I seen a rule that the breaker shouldn't be running above 80% of it specific ampacity, so you multiply your amps by 1.25.

17.39amps x 1.25 = 21.73amps

So that would mean a 25amp breaker would be needed... like I said I'm only learning, if this is wrong, advice?

Many thanks

 

[TheConduitKing]

What you’re referring to is known as “Design Current” or “Ib”

This is the amount of current that an item of electrical equipment will use under normal use. It’s the maximum current that will flow to operate the equipment.

Using the Power Triangle. P I and V (Look it up for a picture) You can work out the design current.

Here is an example for a purely resistive load such as an electric heater. Let’s say its 3000 watts or 3 kilo watts.

Ib(Design Current) = 3000 / 230 = 13amps

So, you would need a MCB or (BS EN 60898) with a rating of a minimum of 13 amps. A 16amp breaker could be used.

Once you understand this you can learn about derating cables and diversity.

I haven’t heard that 80% rule if I’m honest.

I would suggest buying a copy of the “Student's guide to the IET wiring regulations” it’s a great wee book and I’ve really enjoyed reading though it.

 

[Lucien Nunes]

Your location shows as 'London', please could you confirm that's London, England? (It could be Ontario, CA!)

You mention ampacity, which is an American term not used in British english, and the 80% rule for continuous loads is also part of the USA's NEC which is their equivalent of our BS7671 wiring regs.

It sounds like you have been reading American sources, which although correct in themselves, will lead you astray when studying how we do things in the UK. There are quite different ways of achieving the same goals in different countries' regulations. What is correct in one place might be forbidden in another, like driving on the left hand side of the road.

FWIW as @TheConduitKing hints, there is sometimes more to working out Ib than dividing power by voltage. I=P/V works for resistive loads like heaters, but will be too low for inductive loads like induction motors, due to power factor. I won't bog you down in that for now, just be aware of it.