I'll put a set of ''right's''numbered based on my current assumptions, so if one of them is wrong, you can just point a couple out.
Let's say I have a device that is labeled ''660W/220V'', so It delivers 660W on a jack of 220V, right(1)? Then, let's move on: Based on Ohm's law: P = U²/R, applying the values, we get an INTERNAL RESISTANCE of the device of about 73,3 ohms, right(2)?
With U = R * i, the device draws a current of about 3A, right(3)?
So, if the internal resistance doesn't change, plugging the same device on a 110V jack now will only produce 165W and will only draw 1,5A right(4)?
Thing is, I've seen a video of a dude plugging 2x 60W lightbulbs on 2x outlets each, one in 110V and one in 220V. The one in the 110V drawed twice the amount of current compared with the one in 220V. That doesn't make sense.
Maybe the bulbs were different and designed to operate on the voltage he plugged? So in that case, a 60W/110V bulb have a smaller internal resistance, so it need to draw way more current to produce 60W, is that it?
Extreme stupid question: I don't understand the logic behind P = V * i, The amount of voltage in an outlet should only determine the capacity of current it can provide. More voltage should equal more current. More voltage equals less current makes sense in the formula, but not in my head lol.