How Old microwave turned into a machine welding metals?

"You can welding metal stuff"

I am not sure whether this person knows how much danger they are in. They could have mains voltage transformed up to thousands of volts if the are using a low primary, high secondary transformer. Just holding it with nothing but wire for protection is one of the most idiotic acts I can think of, other than building this in the first place.

If you wish to weld metal, I suggest you buy a professional welder.

IF YOU DID THIS WRONG IT COULD KILL YOU.

 

OK, if it is theory, would you like to know more?

If so, I would-be happy to explainthe operation.

 

OK, the method used to heat the metal is resistive heating. The metal is not a perfect conductor and so dissipates electrical energy as heat, this is done so according to
P=I^2r
power dissipated = current squared * resistance

With AC, the current is RMS in the equation.

This can also be expressed as
P=IV
power dissipated = voltage across the component * current

This is for ohmic loads, assuming resistance is the only factor. This energy is dissipated as heat. There are other factors due to the AC input, including inductive impedance, but these are negotiable next to resistive heating.

The equation is not solvable with the information given as we do not know the voltage across the transformer, nor do we know the resistance of the metal.

The explanation for resistive heating lies in the electron-ion collisions, the motion of the electrons become random and lose electrical energy, releasing heat.

The reason that high voltage is required in the circuit is due to the fact that the current flowing through an ohmic load is proportional to the voltage across it. This is useful, until the power created by the electrical input becomes capable of outputing less current than is flowing through the ohmic load due to the raised voltage.

A 2.5 volt source with a 2A max draw will put 0.5A through a 5 ohm resistor. (2.5 watts)
We step this up with a transformer with 1:2 primary:secondary proportionality, we produce a:
5 volt source with a 1 A max draw, this will put 1A through a 5 ohm resistor (5 watts)
We step this up again, with a transformer with 1:2 primary:secondary proportionality, we produce a:
10 volt source with a 0.5A max draw, this will only put 0.5A through the 5 ohm resistor (1.25 watts)

The problem is, this does not factor in internal resistance, so can not be taken as exact in a real life situation. A source with internal resistance will decrease voltage output as the load current draw increases, is as the resistance of the load decreases.

A voltage source would ideally have an unlimited current draw and zero internal resistance. The above factors in limited current draw, but not non-zero internal resistance.

I hope this answers any questions (and that it is all correct! If you spot any words that are mispelt, tell me, my autocorrect is really annoying!)