Did i calculate the V/A/R of a plasma arc properly?

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SUMMARY

The discussion centers on calculating the voltage, amperage, and resistance of a plasma arc produced by a Tesla lighter. The user estimated a voltage of 150,000 volts based on a 5 cm arc length and air's breakdown voltage of 3×10^6 V/m. They calculated the resistance to be 2x10^12 ohms using air's resistance of 4x10^13 ohms per meter, leading to an amperage of 7.5x10^-8 amps via Ohm's Law. Feedback highlighted that the resistance of the arc decreases significantly once ionization occurs, and the arc does not follow a straight line due to Paschen's law.

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  • Understanding of Ohm's Law and its application in electrical calculations.
  • Familiarity with plasma physics and the behavior of electrical arcs.
  • Knowledge of Paschen's law and its implications on electrical discharge.
  • Basic concepts of voltage, current, and resistance in electrical circuits.
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Thomas Conway
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Hello there! So i am currently sitting in my EME class and have nothing to do, so i decided to try spit-balling the V/A/R of the plasma arc my Tesla Lighter produces. note that i did not physically take it apart, and my goal was just to get a rough estimate for the arc itself or at least get a minimum rating, not the circuit inside. Please tell me if i did something wrong or improperly assumed something, I am only afew months into this class and am still learning but here's how my logic follows:
Voltage- To get the minimum voltage i had to determine the length between the positive and negative ends that the arc stretches across which is 5 centimeters. I then did some research and found that it takes 3×10^6 V/m of air for an arc to jump across. I then divided that by 100 and multiplied by 5 to get 150,000 volts to stretch an arc 5 cm.

Resistance- To get the overall resistance i did some more research and found that air has a resistance of 4x10^13 ohms per meter (it varies a lot but this is probably a good range) and simply divided by 100 and multiplied by 5 again to get 2x10^12 ohms.

Amperage- Finally to find the amperage i used ohms law for I=E/R. I= 150,000 / (2x10^12) to get I=7.5x10^-8 amps.
So that sums up the process i used, please let me know if i did anything incorrect and if my reasoning was flawed. Just to re-iterate this is just a thought experiment and i am not going to use this line of educated assumption for anything besides food for thought. Thanks again!
 
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