Tony wrote:
"mm" wrote in message
...
On Sat, 04 Aug 2007 20:38:19 -0700, terry
wrote:

The reference 'is only capable of carrying 55 amps "at 120V", is
superfluous. We are considering current here, not voltage.
Absolutely. The heat, if any, generated by the current flowing
through the wire is totally dependant on how much current is flowing,
and unrelated to the voltage applied to the whole circuit.

That is not quiet right!
"voltage drop" in the line times current equals watts
which generate heat in or on the line
so voltage does have some effect.
....

Well, yes and no...depends on whether it is a constant current or
constant voltage source (or another way to consider it is in regards to
the chicken and egg )...

For a power circuit fed from the grid, one can consider it a constant
voltage source.

There is current flow through the circuit only because there is a
voltage drop from that source to the neutral.

How much voltage drop is in the wiring depends on the resistance of the
wiring, that is correct. But how much current flows through the wire
depends on the _total_ resistance from the source back to the neutral
(assuming a series circuit, the current through each component is the
same), not just the wire. So, given that there is a fixed voltage
supply and for a particular piece of equipment on the circuit in the
shop, the current will also be (nearly) constant and the voltage drop
across the wiring will be determined by that current. Consequently, of
the wiring loss, the situation looks more like a current source rather
than a voltage source.

But then again, one can analyze it as if there were a fixed voltage
impressed over the cable and arrive at the same numerical result...

V = IR

P = VI = (IR)I = I^2 R

Which is more fundamental; the VI or I-squared R form? All depends on
point of view...

....