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Coil firing
- Thread starter sprojam34
- Start date
gopher_byrd
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There are numerous tutorials on Ohm's Law on YouTube and other places. You will learn more than what I can tell you in a short post. That being said, here is the basic theory:
E=IR where E= volts I= current(Amps) and R= resistance
Your coil is the resistance so 2.0 ohms. The battery puts out 3.7 volts (actually 4.2V when fresh off the charger) with a max current of 20 Amps but that is not the current you will be drawing with your coil. We have to solve for current so we will divide both sides of the equation by R to find I.
I=E/R
Using your numbers: 3.7/2 = 1.85. Your 2 ohm coil will pull 1.85 amps from your battery when running at 3.7 volts.
Power expressed in Watts is found with this P=IE Using the result from above: P=1.85*3.7=6.845 Watts.
Again go find a full tutorial and fully understand it before you begin making your own coils!
@Huckleberried maybe you should move this to a more appropriate forum, Mahalo...
E=IR where E= volts I= current(Amps) and R= resistance
Your coil is the resistance so 2.0 ohms. The battery puts out 3.7 volts (actually 4.2V when fresh off the charger) with a max current of 20 Amps but that is not the current you will be drawing with your coil. We have to solve for current so we will divide both sides of the equation by R to find I.
I=E/R
Using your numbers: 3.7/2 = 1.85. Your 2 ohm coil will pull 1.85 amps from your battery when running at 3.7 volts.
Power expressed in Watts is found with this P=IE Using the result from above: P=1.85*3.7=6.845 Watts.
Again go find a full tutorial and fully understand it before you begin making your own coils!
@Huckleberried maybe you should move this to a more appropriate forum, Mahalo...
