On Monday, May 30, 2016 at 1:57:48 AM UTC-4, Diesel wrote:

Mon, 30 May 2016 02:46:57 GMT in alt.home.repair, wrote:

On Mon, 30 May 2016 02:23:31 -0000 (UTC), Diesel
wrote:


7.9 volts lost 12/2 at 100ft
4.97 volts lost on 10/2 at 100ft

240-7.9=232.1 on the 12ga
240-4.97=235 on the 10ga


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

With no load, there is no voltage drop at all.

The voltage drop is due to the wire's own resistance, and the
distance the voltage must travel. The wire we're using isn't a super
conductor. It has a certain amount of resistance to it. As a result,
some volts are no longer available to us, we spent them getting the
rest down the line. Nothing for free, you know.


Good grief, you need to take electricity 101. Gfre is correct, with no
load there is no voltage drop, the full 240V is available and can be
measured at the end of the 100 ft of wire. The only way you get voltage
drop is with a load and current flowing. Let RL be the load connected to
a voltage source V using wires that have a resistance RW. With RL not
present the voltage at the open wires where RL would be is equal to V.
With the load RL inserted, the voltage across RL is V- RW*V/(RL+RW).
You should know this just from common experience. Measure the voltage at
a receptacle on a long circuit with no loads on it, you get the same
voltage as you do at the panel. Turn all the appliances, lights, etc on
and you get some number of volts less, due to the voltage drop.




Watts is watts, man. Available voltage determines how many amps it's
going to take to get them, though. Lower voltage=more amps to do the
same job.


Resistor is a resistor man. And a heating element looks more like a
resistor that an intelligent widget that draws more amps. Here's a
though experiment. I take a 3500W heating element that draws 14.6 A at
240V. Now I hook it up instead to a 24V source. Does it draw 146 A?