"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.


I think this will help a few people and won't help others at all.

E=IR (Remember that. ) In other words:
EMF=Current times resistance.

EMF= Electro Motive Force

In other words, in this case: IxR=E
Voltage = 55 amps x R (resistance)

Power = Voltage x Amps. (Remember this.)In other words, in this case:
Watts = voltage x 55 amps.

But the voltage in the formula above is not the 110 or 220 that comes
from the wall. It's the voltage drop from one end of the wire to the
other, which is small, because the resistance of the wire is small.
The total voltage drop of an appliance plugged in to a wall is 110 or
220 or whatever, but almost all of that voltage drop is in the
appliance, and not in the cord going to it. It's the saw or washer or
water heater or mixmaster that is designed to do the work, and the
wire to it is designed to have low resistance and not impeded the
electrficity getting to the appliance. But, in this case we're
talking about running 6 gauge wire with a 60 amp breaker, even though
the wire is rated at only 55 amps. So how much heat (power or watts,
iirc) will be generated in that wire if the current goes up to just
shy of 60 amps, even though it is rated at 55 amps.

Look at the formulas at the top, and then continue he

Since voltage = 55 amps x R, then:
Watts = 55 amps x R x 55 amps.

In other words:can you make simple
(E=Volts) x (I=Curent) =(W=Watts)

Watts = (55 amps) squared x R.

Where R is the resistance of the wire feedinn the appliance.

This is not the number of watts given off by a lightbulb, if there is
a lightbulb in the appliance, or used by the motor if there is a motor
in it. It's the number of watts given off by the wire feeding the
appliance. And it is totally unrelated to the voltage applied to the
appliance**

**Except insofar as if higher voltage were applied, more current woudl
go through the same resistance wire and appliance.

And for any General use #6 wire three in conduit is rated for 65 Amps.