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Voltage Drop and Resulting Watts

InMyImage

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Ok, so I know that using the Ohm's Law calculator at steam-enging.org I can input the Ohm value of my atomizer and the current test voltage of my battery/mod to find out the effective watts that I will be vaping at.

The question is... an inline voltage to test the voltage of my battery in the mod and then with my atomizer attached to get the voltage drop using the same battery I get the following.

FUHattan by Unknown from 101vape.com
  • Voltage: 4.01
  • With Atomizer: 3.83 average
  • Effective drop = .18
Apollo Clone by Tobeco
  • Voltage: 4.03
  • With Atomizer: 3.88 average
  • Effective Drop = .15
So, when using the Ohm's Law calculator should I be using the base voltages of 4.01 or 4.03 or the voltage drop values of 3.83 or 3.88 when trying to determine the resulting watts vaping at?

Thanks
 

RoccoV

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Always use the under load number. Enjoy the day!:cool:
 

InMyImage

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That is what I figured, but thought that it was good to ask and verify.

Thanks!
 

Eric DeCastro

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is that with the same atomizer and build? yes use the voltage under the load. it will give you the actual power out.
 

InMyImage

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Yes, it was with the same atomizer and battery with a couple of minutes and no use of the battery between readings, other than what was required to get the readings of course ;)
 

RoccoV

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That is under load. Enjoy the day!:cool:
 

BoomStick

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Voltage is additive in series, so measuring it between the mod and atty don't tell you exactly what's hitting the coil.
 

Zamazam

Evil Vulcan's do it with Logic
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Most Li-ION batteries will drop to their rated voltage under load, 3.8-3.9V is pretty decent under load.
 

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