Induction stovetop design: am I using these equations correctly?

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SUMMARY

The discussion centers on the design of an induction stovetop, specifically the calculations required to achieve an effective power output of 1000W. The user employs the formula for energy stored in an inductor, E = 0.5*I²*L, and derives that an inductance (L) of 1.2mH is necessary when using a maximum current of 0.2A. The user also calculates the required ratio of turns and area for the coil, concluding that a coil with 10 turns would need an area approximately 10 times its length. However, the user realizes that achieving 5000V from a standard wall outlet is impractical.

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  • Understanding of inductance and energy storage in inductors
  • Familiarity with electrical power calculations (Watts = Volts * Amps)
  • Knowledge of coil design and magnetic field principles
  • Basic electrical engineering concepts related to induction heating
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  • Research the principles of induction heating and its efficiency
  • Learn about coil design for induction applications, focusing on turns and area ratios
  • Investigate safe voltage levels for residential electrical systems
  • Explore alternative power sources for induction stovetops
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Electrical engineers, hobbyists designing induction heating systems, and anyone interested in the principles of electromagnetism and power calculations in practical applications.

nagyn
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TL;DR
I am trying to make an induction stovetop using some magnet wire. I understand how induction stovetops work but am less confident on the math I've done to design one. Can you double-check that I'm approaching this correctly?
I've been told an effective induction stovetop needs to deliver about 1000W of power. I have magnet wire that can tolerate at most about 0.2A of current, and was planning on using a 60Hz wall outlet as my source (obviously I'll need to bring down the outlet current quite a bit).

So the energy stored in an inductor is given by:

E = 0.5*I2*L

Which I assume would be roughly the same energy transferred to the pot when on the stove. So the power delivered would be

P = E / t = E*(60 Hz)

Then substituting P = 1000W and I = 0.2A I get that L = 1.2mH.

Then given that L = μ0*N2*A / ι , I find that I need a ratio close to

N2*A / ι = 952

So if I had a coil with 10 turns, the area would need to be about 10 times the length to deliver the power I need.

Am I doing this right? Did I mess up anywhere in my understanding?
 
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Power in Watts = Volts * Amps
Watts / Amps = Volts
1000 / 0.2 = 5000V.

:cry: Sorry, I don't know of anyplace that has 5000 Volts at their wall outlets.
 

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