If your asumption that the winding was shorted was correct, than a heavier gauge wire would indeed cause the breaker to trip even faster. But if you actually looked into what could be an actual problem with the use of this machine or just some basic electrical math and trouble shooting instead of just jumping right to the welder being FUBAR, then you would see that there is a simple solution here.
With out having all the unknown factors like actual wire length and even conductor size your cant calculate actual circuit voltage drop. Not to mention if the circuit goes around the house twice and has 10 less than stellar splices in it. You could have big problems getting the minimum 115 volts need by this machine.
But then again if you look at the user guide, even at best if you had a dedicated 20 amp circuit to only the plug that your plugged into and NO EXTENSION CORD, and had that thing cranked to the max you would still be popping breakers.
basic formulas for everyone
V = I x R (Voltage = Current multiplied by Resistance)
R = V / I (Resistance = Voltage divided by Current)
I = V / R (Current = Voltage Divided by Resistance)
Voltage drop calculator if your interested
http://www.electrician2.com/vd_calculator.htm
This here is directly from the users manual.
http://www.lincolnelectric.com/assets/servicenavigator-public/lincoln3/im725.pdf
Requirements For Rated Output
A
power cord with a 15 amp, 125 volt, three prong
plug (NEMA Type 5-15P) is factory installed on the
SP-135 PLUS. Connect this plug to a mating grounded
receptacle which is connected to a 20 amp branch
circuit with a nominal voltage rating of 115 to 125
volts, 60 Hertz, AC only.
The rated output with this installation is 90 amps, 18
Volts, 20% duty cycle (2 minutes of every 10 minutes
used for welding).
Requirements For Maximum Output
In order to utilize the maximum output capability of the
machine, a branch circuit capable of 25 amps at 115
to 125 volts, 60 Hertz is required. This generally
applies when welding steel that is equal to or greater
than 12 gauge, 0.105
” (2.5 mm) in thickness.
So at the end of the day get a dedicated circuit capable of handling what your doing and you will be fine. And (old man) Im not trying to be a prick or bust your balls, but with so many unknow factors and trying to trouble shoot stuff over the interweb you cant allways be right so dont ASS-U-ME that your the only one on here that knows anything about electricity/electronics.
/end rant
OK, a civilized discussion works for me. How do you explain how you are going to pull more current with a higher resistance in series with the machine? Current in a welder is based upon a few things. (1) voltage delivered to the welder (2) the inductance (simplified) of the transformer (3) strength of the arc ...ie the wire size.
The current through the transformer is really a function of the saturability of the core, the reluctance of the core, the Q of the transformer, the size of the wire, and the load, etc.
The load on the output of the transformer is really dependant on the arc. The arc depends on mainly on the voltage output, which is a function of the windings being selected by the switch and to a lesser amount the size of the wire/arc. Less voltage means less arc current.
Given the fact that it once worked and now doesn't, what changed? My first thought was the breaker, but it was said that the breaker was changed out. That pretty much leaves the welder pulling more current. What would cause that? There are several things; shorted windings on the transformer, a bad diode in the bridge. All of these things cause more current to be drawn. It would be interesting to see the welder put on a much bigger breaker and then seeing if it overheated and blew the thermal breaker.
I had a Craftsman (Century) 120v welder for years. I welded a lot of super thin sheetmetal material. To lessen the output power, I ran it on a large variac and dropped the voltage. Before that I used a wet pile resistor to limit the current. Using a small power cord acts like a voltage divider and also drops the voltage. Doing these things caused a reduction in current, not an increase.
Using the formula I=V/R, what happens to the current when you increase R? Since the R is in series, what happens to the voltage across the welder as the R is increased?
The standard house type breaker is referred to as a thermal-magnetic circuit breaker. They have what are called trip curves. A breaker does not instantly trip when its rated current is exceeded. Typically at a load of about 1.5 times the rated current,it can easlily take up to 2 minutes to trip. The OP said he could only weld an inch or two before it tripped. Lets call that 30 seconds of weld time. To get it to trip in that amount of time, you need to more than double the current. What can cause a doubling of current. Like I said, there are basically only a couple of things chat could double the current; a shorted winding on the transformer or a blown diode.
Having said all of this, it all depends on an accurate statement of the conditions by the OP. Without that, all bets are off.