How do you figure out how many breakers you can fit on a 100,150,200 amp panel?
Thanks for any info.......

Isn't it printed on the labels of your panels? It should be.

let me reword it,if you have a 100 amp panel, how do you know how many 15 or 20 amp breakers you can use?

12/24 + 12 full size, or 24 1/2 size. Same for 20/30, 8/16, 30/36, 30/42

If not marked in that way it would be marked with how many spaces or (30/30 all full size), as would be any panel that half size aren't made for like bolt ons. i.e. a 24, 30 or a 42.

I heard that in Canada you can get 84?

Oh how many can you fit of 15 or 20's. You can fill it up! Even throw some 50's or larger and feed another panel from it.

So long as the load calc is OK, maybe that is what you're really looking for.....

NO MORE THAN 42 OTHER THAN THE MAINS
408.15

Local3Jr never more than 42 overcurrent devices in one enclosure (excluding the main)

But you can have as many breakers as you want connected to any service.

100 amp service go ahead and install ten 42 circuit sub panels and fill them up with breakers.

The only thing that matters is that the calculated load is less than the service size.

Say you are planing to wire a house in the 'normal' way with the code minimum circuits and the calculations show a 100 amp service will do the job.

Now the future homeowner says they want every receptacle, switch and appliance on it's own circuit.

To do this you will need a few sub panels and lots of breakers but the calculated load remains the same. The 100 amp service will still be fine.

Bob

e57...I recently read the label on a 200 amp Cutler Hammer loadcenter that did specify 84 overcurrent devices in Canada.

But why would the same panel used in Canada be able to handle 84 breakers with out failure?

shortcircuit

note...408.15 has disappeared in the 2005 NEC...

wait a minute...its at 408.35 now

shortcircuit

Sorry for the confusion guys,lets say you have a 100 amp 3 wire panel(a phase,b phase and a neutral. Each leg is 100 amp right? You wouldn't be able to load it up with 42 20 amp breakers right? Maybe I am looking for some kind of calculation. Thanks again

The answer is above. I think iwire's post is the answer you were actually seeking.

Think about what is connected to it, and forget about the number of breakers for a minute. If you've got 21 amps on each leg, you will have 21 amps on each leg, period.

Now, split it up however you'd like.
4-15 amp 1 polebreakers with 10.5 amps on each of them.
42-20 amp breakers with 1 amp on each of them.

The panel still sees a load of 21 amps per leg.

Get it? It's the LOAD, not the # of breakers.

[This message has been edited by electure (edited 02-19-2005).]

the amount of breakers has nothing to do with connected load.

Think about it this way, you have a 20 amp single pole circuit feeding receptacles in any given room, you put a recording amp meter on it for an extended period of time and see the peak is 7 amps with an average of 2 amps over this period. This would probably be exaggerated for many circuits, so the face value of an OCPD is not really relative when computing a service or panel feeder.

Roger

Hello Electure, I appologize, we must have been typing at the same time.

Roger

That is the concept behind that pesky term "load diversity". Not all loads will be operating fully and simultaneously. This means you can put (40) 20 amp breakers in a 200 amp, 40 ckt panel.

what you seem to be trying to do is a load calculation.
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