Re: welding machine as a power supply

>
> Right. Now figuring that the current is ten times on the

secondary side what

> it'd be on the primary side for the same effect. The formula is
> P=I<squared>R, so for putting the resistor on the secondary side

it'll have

> to handle ten times the current, or 100 times the power

dissipation...

>
> This sort of thing when considering transmission line losses, is

exactly why

> transformers are used to step the voltage way up in the first

place.

> Currents go down proportionally to the turns ratio, and power

losses go down

> proportionally to the square of the turns ratio.

Roy: Draw it out on paper. Use any turns ratio, primary voltage
and max current number you want. Calculate the resisitor values you
would need on either side of the transformer given the turns ratio.
Apply any of the watts formulas (E*I, E^2/R, I^2*R) you will come
out with the same answer.

If what you are saying is true you can generate more heat from a
load on one side of a transformer than the other given a fixed
current ratio....more power out than in.

This has nothing to do with transmission line theory. In that case
you are starting with a fixed resistance per foot. At a given load
(in Watts) the voltage drop across the series resistance will remain
the same regardless of the voltage. So at 120V the voltage drop
across twenty miles of line might be 50V. At 10,000 volts the same
load (in watts) still has 50V which is a much smaller percentage of
the total.

This probably needs to be moved to DIY-CNC as it is quickly
progressing to being OT on this group.