Transformer - I connected 50k resistance which burnt - why? AC

Question

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?

^{simulate this circuit – Schematic created using CircuitLab}

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

Answer 1

It's hard to understand exactly what you are asking, but I will go for the question in the title of the question.

The power dissipated in a resistor is $$P_R = V_R^2/R$$ where \$V_R\$ is the voltage across the resistor and \$R\$ is the resistor value. So, if you put \$47\mathrm{k}\Omega\$ directly across 230 V the resistor will dissipate 1.13 W. That's a lot for a common resistor, which is probably rated for 0.125 W or 0.25 W. It's no surprise that the resistor would burn up. On the other hand, if you do the same thing with a \$2\mathrm{M}\Omega\$ resistor the power dissipated is only 26 mW...it probably won't even get warm.

Answer 2

As @EugeneSh mentions in comments, the rating is a maximum rating.

Power Supplies can be theoretically grouped in 3 categories

1. Constant Voltage

   A supply that is operating in constant voltage mode will vary the current that is needed to maintain a fixed voltage for load. The total (V_SP * I_LOAD < P_MAX)

2. Constant Current

   A supply that is operating in constant current mode will vary the voltage needed to maintain the fixed current for the load , (V_LOAD * I_SP < P_MAX)

3. Constant Power

   A supply that is operating in constant power mode will vary the voltage until the current into the load is the desired power (V_LOAD * I_LOAD = P_SP < P_MAX)

The vast majority of off-the-shelf AC power supplies operate as Constant Voltage Supplies, and the current is determined by the load. Often the max current of the device is specified as well as the power.

Many lab power supplies will let you switch from CC/CV mode

Constant power supplies are often used in applications like resistive heating, where the load is high power and can tolerate a large range of input voltages (not unusual to run resistive heating wire at 100VAC)

A passive transformer can be thought of as a constant (AC) voltage device in this context. The output voltage is fixed as a ratio of the input voltage. When the load is large the current will be high, if the load is small the current will be small The amount of current depends only on the connected load. If the current exceeds the transformer rating it will heat up and melt the wires. If the current (multiplied by transformer ratio) exceeds the input supply rating it will shut down, droop, or trip a breaker/fuse.