Volts don't drain a battery, amp load does:
P = power (watts)
V = volts
R = resistance
I = current (amps) I forgot why it's "I"
Voltage x voltage ÷ resistance (ohms) = power (watts)
Voltage ÷ resistance (ohms) = I (current/amps)
Hi- res:
4.9v²/3.0Ω = 8.0 watts
4.9v/3.0Ω = 1.6 amps
Lo-res:
3.7v²/1.7Ω = 8.0 watts
3.7v/1.7Ω = 2.2 amps
Even though a 3.0Ω atty at 4.9v and a 1.7Ω atty at 3.7v are producing the same amount of watts, when vaping lo-res the amps increase, which draws more from the battery.
Since 8 watts is 8 watts, an average vaper probably isn't going to notice a difference in the temperature of their vape, but they may notice the battery charge doesn't last as long.
EDIT-reread again and sry, I think I did flip it accidentally. Now gotta read again!!! LMAO!
well we were talking about two different things, my fault, I wasn't following the thread very closely.
on a VV, assuming the electronics are perfect with no losses (which is impossible), more watts will drain the battery faster. and the same watts will drain the battery in the same time.
1 amp at 10 volts == 10 amps at 1 volt == 10W.
But with the VV, it might be less efficient producing higher voltage, so even though in your examples, both being 8W output, in theory they will drain the battery in the same time, in practice, they probably won't. But that's not a function of ohms law, it's because the electronics aren't perfect and have losses.
practical experience trumps theoretical equations, if you say a high ohms coil on a VV drains the battery faster, then it does. you can't prove that with a simple equation.