3 ms·
220 V is already in RMS [1] so its just 220V * 16 A = 3.5kW (current is also rms) Of course, thats truly only correct for a resistive load (but, without knowin
by tagoregrtst 5y ago
220 V is already in RMS [1] so its just 220V * 16 A = 3.5kW (current is also rms)
Of course, thats truly only correct for a resistive load (but, without knowing much about induction stoves, its probably within 10% - modern power electronics are amazing)
[1] Root Mean Square. Basically the measure needed to make a simply multiplication give you the power. Other ways to measure line voltage are
Amplitude. This would be the A in V(t) = Asin(wt). A = sqrt(Vrms)
Peak to peak. This is 2xA.
They’re all useful ways to thinking of the same thing. If you’re designing insulation you’re designing for A, or peak to peak. If you’re designing a water kettle, you’re interested in Vrms.
- pandaman 5y agoGreat, let's say it can draw 4Kwatt from the wall. It's still nowhere close to a 100K BTU stove (~28Kwatt). To have this much of electric power in the US you'd need a 120A 220V (actually 240 nowadays) outlet. I don't even know if codes allow this much current inside the house.
- bayindirh 5y agoThe heat of the fire is not completely transferred to the cooking utensil due to air and losses, on induction and resistive stove tops, the heat losses are much lower since you don't heat anything in inductive (magnetic field heats the metal directly), and electric stovetops have no air-gap to transfer heat. So, in the end, 90% of the time you're looking at the same cooking duration at the end of the day. So, you don't need that 120A load. In other words, if the box says 5 minutes on medium, it's 5 minutes on medium. Regardless of gas, electric or induction.
- pandaman 5y agoYeah, I know that, I've even mentioned it above as I mentioned that gas does not lose nearly as much heat as necessary to make a 3.5Kwatt stove as efficient as a 28Kwatt one.