Wall supply gives as much power as a component uses. I'm trying to connect a 800W transformer to produce an arc of electricity. But I can't exceed 10A because home fuse will triger (so every device will be instantly cut from electricity etc). So I want to know, what will be the amperage, should be 800W/230V = 3.48A right?
However what causes the transformer to be 800W - it doesn't drain energy directly as a bulb does. It produces electromagnetic field and the "amount" of this field is indicated by number of scrolls I guess, but is it also increased by it's resistance?
I want to be sure that adding 1M*2 resistance would make it drain P=I^2*R, I=(800/2M)^(1/2) = 1/50 = 0.02 A and that wall supply won't somehow try to compensate 2M resistance and increase amperage to like 800A.
To do so, I connected an analog amperage meter 0-1A and a 47k ohm resistor, it should be I = (800/47000)^(1/2) = 0.13 A. And U = 6132V. When I connected 2M ohm, nothing has happened, it should have been U = (800*2M)^(1/2) = 40k V right? Maybe then increased 10 times as it's a MOT, 1000:100, so 400k V... But the resistors didn't burn.
Why did 47k burn and 2M not, in spite of 6k vs 40k voltage?
Used MOT: MD903-EMR-1 Class 220, 230V 50Hz
I didn't count the turns on each coil, but because it's supposed to give over 2k V output, I guess it has 100:1000 ratio, considering one of the coils is really thicc and the other is rather thin, eye-estimation that is.
Also it might be a 900W not 800W