Battery theory, calculating power and current

Hello guys,

I haven’t been around too much here lately, been hanging over on some ecig forums. Still using plently of lights and have collected an absurd number of hi-drain 18650’s. The level of knowldege of my new community isn’t not even close to what I’m used to over here, especially in the battery subject. So I would like to pose a question if I may.

In my new hobby, many are using mechanical devices with hi-drain 18650 batteries of all various chemistries and manufactures. IT’s just tube containing a spring and silver/chrome plated contacts.

To check power and amperage coming from the battery, obviously using ohm’s law. We know resting voltage, voltage under load and resistence and that’s what they use to calculate amps from battery

Volt^2 / ohms = Watts = Volts * Amps

or Amps = Volts/Ohms

Most people use resting voltage 4.2V in this equation. To make sure we’re staying under current limits of a certain cell, To be most accurate to the actual amperage coming from the battery, would you use resting voltage or voltage under load? It makes a pretty big difference in calculations

Just as an example, resting voltage 4.2V and .5ohms gives 8.4A. Using a working voltage for example, 3.3V and .5ohm gives 6.6A. Thanks guys!

You are correct in your reasoning; voltage under load is what you should use. As you have noticed, most of the cells we work with sag far below 4.2 volts under any sort of a heavy load.

HKJ’s website is awesome for his graphs:
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