On Monday, May 30, 2016 at 1:57:47 AM UTC-4, Diesel wrote:
Sam E
Mon, 30 May 2016 03:14:39 GMT
in alt.home.repair, wrote:

On 05/29/2016 09:23 PM, Diesel wrote:

[snip]

1500/232.1=6.5 amps *rounded up*
1500/235=6.39 amps *rounded up*

And that's due entirely to distance with no loads present on the
line yet.

I'm often surprised by how little some people know about
electricity.

I sometimes feel the same way.

Voltage drop depends on current flow. There's no
power (wattage) without current flow either.

If you put 120volts AC on a 12/2 wire, and run that wire, 100 ft out,
and take a measurement at the end of the run, the available voltage
at the end of the run isn't going to match the volts going into the
wire. Some is lost during transit due to the wires resistance and the
length.


BS. There is no "transit", the current flow is zero, the voltage across
the wires is the same at both ends. The voltage drop for each wire is
V = I * RW, where I is the current, RW is the resistance of the wire.
Set I to zero, and what do you get? This is a very simple application
of Ohms's Law. And if you think this is wrong, explain what the correct
formula for the voltage drop is.



If you try this experiment using DC power, it's even worse. DC really
doesn't travel distance well.


Sigh, more FUD.


https://en.wikipedia.org/wiki/Voltage_drop

For example, an electric space heater may have a resistance of ten
ohms, and the wires which supply it may have a resistance of 0.2
ohms, about 2% of the total circuit resistance. This means that
approximately 2% of the supplied voltage is lost in the wire itself.
Excessive voltage drop may result in unsatisfactory operation of, and
damage to, electrical and electronic equipment.


The simplest way to reduce voltage drop is to increase the diameter
of the conductor between the source and the load, which lowers the
overall resistance. In power distribution systems, a given amount of
power can be transmitted with less voltage drop if a higher voltage
is used. More sophisticated techniques use active elements to
compensate for the undesired voltage drop.


More wandering in the wilderness. Just apply Ohm's Law to the actual
circuit, like I did above.




So you have a load which is not present. How is it drawing power?

You aren't losing power due to the load, You've already lost it in
route to the load via wire resistance and the distance said
electricity has to travel on the wire. Thicker wire, less resistance,
less voltage drop. Simple concept, really.


Yes, it's simple, unfortunately you have it all fouled up. With no load,
there is no loss, because no current is flowing, period.




Like you wrote though, it does sometimes amaze me how little people
know/understand about electricity.



Even more amazing is how people that don't understand Ohm's Law make
comments like that.