Awsome info there. I have a question though. If at 1ohm/12.25w it's drawing 3.5a and at 2.2ohms/14.5w its drawing 2.5a then the latter is actually making more power using less battery life? I'm assuming how much current is being drawn is relative to battery life regardless of power/volts
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Power is specifically Voltage times Current. (Voltage, current, resistance & power are all related, you can mathematically solve any of those four values by using any pair of the remaining, but that doesn't mean it's an accurate description of what they are/how they work - just for clarification)
Since P = E * I, increasing either E (voltage) or I (current) is going to result in a higher amount of power. (P = E * I) = (P/E = I) = (P/I = E), meaning it's all the same equation just written differently. The latter 2 are important since P is a constant in your question. If you raise current, P/I = E, meaning you're drawing at a lower voltage. If you raise the voltage, P/E = I, meaning you're drawing less amps.
So technically, yes, you can raise the voltage, get the same power, all while drawing less amperage.
HOWEVER... this is in a pure state - chalkboard mathematics. We're not accounting for the losses in components or the copper connecting them. Sending 3.7v into a circuit with a 3.7v output is going to have very minimal losses. Sending 3.7v into a circuit with a 6v output is going to have substantially more losses... which is going to eat up some of your current. I don't know how the circuit board is specifically designed or what the losses are. You very easily could be breaking even. The numbers in the chart are simple mathematics, not real world data. No circuit is perfect.
My advice... it's good to understand how this all works, but ultimately it doesn't matter. Try some different coils, try them at different settings, VV or VW, find what you like & vape it.
ETA:
P.S. It was an excellent observation & question, regardless.
You get 2 points for the day.
