In article , Sam E wrote:
[snip]


You have a SERIES circuit (considering that the neutral is effectively
disconnected).

[snip]

Wrong. The neutral is "effectively disconnected" *only* if the loads on the
two legs are exactly the same.

Which they are (either in the 200A+200A example or the 1A+1A one).

And that almost never happens in real life, either....

The two legs function as two parallel circuits
with respect to 120V loads.

In a parallel circuit BOTH ends of the loads are connected together
(or at least to identical voltages). Neither is true here.

Wrong -- both are true.

Obviously they are indeed in series WRT 240V
loads.

Strangely, I get the idea that you actually know this stuff.

In this 200A service there are THREE current-carrying conductors.
Each of these conductors is of the proper size to carry 200A. OK so
far?

OK

You say (when this service is fully loaded) that two of these
conductors is carrying 200A (for a total of 400A, as you say).

400A @ 120V, or 200A @ 240V, yes.

Then where is that 400A going? The only remaining conductor is the
neutral, a big enough conductor for 200A (yes, this 400A was at 120V
but current is still current and voltage doesn't change the
conductor's current capacity).

Somehow I'm imagining a bridge that can handle 200 cars per minute,
but that can be 400 if half the cars are blue :-)

Cute. Just answer these questions; assume a 240V 200A service.

What is the maximum power that service can provide?

If all the loads supplied by that service are 120V loads (e.g. blender,
toaster, light bulbs, range hood, stereo, TV, computer, etc.) what do you get
when you divide that maximum power by 120V?