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daiplayer

When calculating max demand for eic's......the NIC dvd shows them switching on as much as possible and meusuring the current ....

However the house in question has nothing in it other than a few light bulbs.

How do i calculate this ?

Or am i just being dull
 
However the house in question has nothing in it other than a few light bulbs.

Dont use that method then
[ElectriciansForums.net] Max demand


If you search the forum this has been on here quite a lot
There are other methods available
 
come accross this , deos it look correct to you ? also....the osg keeps saying rated current of sockets...is this 32A ? ( ring) many many thanks.

'The formula is actually 100% current demand of largest circuit + 40% current demand of every other circuit.

In your case it would be..

32A
(Largest circuit) + (3 x 32 x 0.4 = 38.4A) + (4 x 6 x 0.4 = 9.6A)

32 + 38.4 + 9.6 = 80A '

 
The on site guide tends to have much higher results with their method than you will almost certainly find in the real world,because extra circuits does not always = extra load

Without looking in that book,I believe the ring circuit is 100 percent of the first ring and 40% of any others
 
i seem to remember ,...when doing my level 2 practical , we did conduit spacing factors, max demand , diversity , etc etc.....

It turned out the OSG ...guides....determind that we needed a 10mm twin and earth for a 3kw motor !!!!!
 
for maximum demand on a PIR, i was taught to put a clamp meter on the live at a busy time of the day, or in the case of a factory, take the reading of the ammeters built into the switch panels

Agreed mate, but as the originall post states, the house is not lived in at the mo, and only contains two light bulbs ,.....cant out about 1A as maz demand can i. lol
 
come accross this , deos it look correct to you ? also....the osg keeps saying rated current of sockets...is this 32A ? ( ring) many many thanks.

'The formula is actually 100% current demand of largest circuit + 40% current demand of every other circuit.

In your case it would be..

32A
(Largest circuit) + (3 x 32 x 0.4 = 38.4A) + (4 x 6 x 0.4 = 9.6A)

32 + 38.4 + 9.6 = 80A '


Thats the way i have calculated it for years.
 
I have to say that I believe the diversity calculation is out of date as we are being encouraged to install more circuits than before. Sure if the househas a couple of electric showers the load will be high but if not ...........
 
the NIC dvd shows them switching on as much as possible and meusuring the current

That is complete NIC nonsense, for the average uk house md is 14kw. Just put down 14kw, despite what the regs say the dno decide how much power their system can supply, and its got nothing to do with BS7671.

Or as others have said add up cpd's and multiply by 0.4.
 

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