Energy utilized by Electricity Powered Nuclear Fusion?

[CNBA3]

This is based on the item called Fusor which is a nuclear fusion device that uses electrical fields to move ions for collision.

So I was doing a bit of calculation with how Fusors utilize Voltage for heating up ions to achieve nuclear fusion and this is what I found. I will be utilizing the Ohm's Law as well.

First we convert Electron-Volts (30000) or 30 KeV to that of Voltage with using an elementary charge as Coulombs (1.60217663e-19) during the process which 30000 Voltage.

We would need to find the Ohms from the Ohms meters of the material, in this case I will be using Tungsten 4.9e-8 Ohms meters which is suggested to be one of the preferred materials for Fusors.

For the Ohm's calculator, I will be using a length of .07 meters and the Cross Section Area = 0.0012566 m^2:

(4.9e-8 / .07 length) * 0.0012566^2 cross section area = 1.1053305e-12 Ohms

Then I will be finding the Amps from the KeV and the Ohms:

Amps = 30000 / 1.1053305e-12 = 2.7141203e+16 Amps

Then we shall find the number of Watts from multiplying Amps and Voltage which is:

814236090000000024576.00 Watts

Thoughts?

@DMUA @Psychomaster35 @Damage3245

 

[ImmortalDread]

This needs to be moved to calculation discussion board.

 

[CNBA3]

This needs to be moved to calculation discussion board.

This can be closed.