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Doug Miller

In article , wrote:
On Sun, 25 Oct 2009 12:13:18 GMT, (Doug Miller)
wrote:

In article ,
wrote:
On Sat, 24 Oct 2009 22:37:59 -0700 (PDT), gwandsh
wrote:

This weekend I had planned to add a 1000 watt electric heater to an
existing circuit with three similar heaters on it. The other heaters
are 750 watt, wired 240, and draw 2.7 amps each. The circuit is 30
amps, so I expected to pony the new heater (4.5 amps @ 240V) off the
wire to one of the existing heaters, and still not challenge the
breaker.
[...]

You are talking about 10.4a or so. (2500w @ 240v = 10.42a)

As I pointed out in another post, his numbers are wrong, but so are yours.
3 * 750W + 1000W = 3250W = 13.5A @ 240V.

You do need to size at the smallest conductor on the circuit.
I would just put a 2 pole 15a breaker on that circuit and get on with
your life being perfectly legal.

This would be good advice if your numbers were right. They're not, and he will

create a hazard if he adds that 1000W heater to the existing circuit.

Hazard?, What hazard? I agree I read two 750w not 3 but the 15a
breaker will still protect the 14 ga wire.
You size the largest load at 125% and the rest at 100%

Code cite on that, please?

I'm specifically questioning the use of the 125% factor for _only a portion_
of a continuous load, as opposed to the _entire_ continuous load.

so he just squeaks in.
1000 x 1.25 = 1250 / 240 = 5.3a
750 x 3 = 2250 / 240 = 9.43a
Calculated load 14.7a