internal resistance

[j122068]

i know that curent isnt the only factor when vaping. its the power measured in watts wich is curent times the voltage but now im learnig that a larger bat has less internal resistance and vapes better my question is

is there another ohms law fomula besides power = curent x voltage that factors in the internal resistance of a bat

 

[DonDaBoomVape]

Power = Voltage(squared) / Resistance​

Or: watts = volts X volts / ohms

But I don't know from internal battery resistance. The only resistance that I am used to figuring on is that of the atomizer. For example, we've heard that 5V is the "sweet spot" of vaping. In fact, the sweet spot really is a matter of power, approximately 10 watts.

• A 2.2 ohm Joye510 atty on a 3.2V Joye510 battery = 3.2 X 3.2 / 2.2 = 4.65 watts. Not very impressive. [No wonder I don't get TH from a 510. Yeah, I know a lot of people do.]

• A 2.2 ohm Joye510 atty on a 5V battery mod = 25 / 2.2 = 11.36 watts. Very impressive.

• A 1.5 ohm Low-Resistance 510 atty on a 3.8V Kr8 battery = 9.63 watts. Also very impressive, without stacking batteries.

I acknowledge that more technically astute vapers with voltmeters might quarrel with my resistance and voltage examples (and point out that the atty resistance and battery voltage are not static), but the principle remains intact: you don't have to go mega voltage (e.g., stack unprotected CR2 batteries) to get a powerful vaping experience.