That's not really correct... Because you won't get a 10 watt differential in similar setups.... And that goes to prove what I was saying.
A large gauge wire will have lower resistance and not get as hot at the same wattage as a thinner wire...
See your 20 watts will pop the 1/4" of 40 gauge wire you're pumping 6 volts through...
My 10 watts will heat my 2" of 30 gauge nicely.
But since the question was same watts..
17 watts
0.8 ohms
2" of 28 gauge kanthal
3.7 volts
Cool vape, big flavor, huge plumes of vapor
Or
2.1 ohms
3/4" of 36 gauge kanthal
6 volts
HOT vape, burnt flavor, low vapor
But one thing is missing from this equation and that is current, measured in amps.
A few simple formulas where V=voltage, R=resistance, W=watts, A=amps (sqrt = square root):
W = V x A
W = (V x V)/R
V = sqrt (W x R)
A = V/R
So if we look at the two examples, which are both producing (roughly) the same 17 watts:
3.7 volts / 0.8 ohms (R) = 4.625 amps of current
and
6 volts / 2.1 ohms (R) = 2.857 amps of current
Since the resistance itself consumes power and the rate of consumption increases as current (amps) is increased there is a benefit in using higher voltage to achieve the same wattage because the overall system uses less amperage and is therefore more efficient, which should translate into a longer battery life....
Just to make things EXTRA confusing!