wrote in message
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On Oct 27, 2:50 pm, "Steve N." wrote:
wrote in message

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On Oct 26, 7:18 pm, (Doug Miller) wrote:

In article
,
wrote:

On Oct 25, 2:45=A0pm, (Doug Miller) wrote:
Also, you only get those 400 amps if the load is balanced so that it
appears as a series load.

You're right. If the the two 120V loads are perfectly balanced, then it's
the equivalent to two 120V 0.6 ohm loads in series across 240V each
pulling
200 amps, and you can disregard the neutral or even disconnect it. The
neutral is there to
hold the voltage on each leg or side to 120V when the loads aren't
perfectly
balanced.

The 200 amp current flows in one hot and

out the other.

True

If you had a single 120V 400 amp load, it would sit
between one hot leg and neutral, where the capacity is limited to 200
amps and the cables would melt. Gee, I wonder why? Could it be
because the actual current in a 200 amp service circuit is only 200
amps?

No, the main breaker would open long before any melting

Yes, I agree, assuming there is one. In our hypothetical case I was
ignoring any breakers.




You can divide and get any answer you want. I could divide 48KVA by
10volts and get 4800 amps. So a 200 amp service could support a
total 4800 amp, 10 volt load too. But how much max current is
actually flowing in the service cable entering the house?

A whole bunch for a few microseconds. A breaker isn't instantaneous.

We have been discussing continous loads at the service max, not
transients. That 200 amp service can support a total load of 4800
amps at 10 volts, or 2400 amps at 20 volts. As I said before, you
can slice it and dice it anyway you want, but you still have 200 amps
max of current flowing in the service.

200A thru each main-breakered leg - yes. But I'm not sure where the 4800A @
10V is coming from.
Do you think the transformer secondary windings will sustain at 4800A? Are
you now taliing about continuous or transient??