A 50 W light bulb will dissipate 50 Watts of power at the voltage it is rated
at.
The voltage is the force needed to provide the circuit with enough current to
do what it was designed to do. Using this force requires power. The power lost
using the force of voltage to supply the circuit with enough current to accomplish
its work is not available to be radiated as heat.
The higher the resistance the more force(voltage) necessary to dissipate the
same amount of power. The extra force required comes with a price tag.
The use of more power. Watt = force + work done + loss.
Regards
Mike
Let me try another angle. Consider a basic electrical resistor. And there are wire wound resistors that electrically and mechanically are essentially identical to our atomizer coils, except they have some potting material around them to keep them mechanically stable.
Resistors have two basic specs: resistance and max power dissipation. A class of resistors with varying resistances will have the same power dissipation. In my example, let's use a 1 watt resistor.
If you put 1 volt across a 1 ohm resistor then 1 amp of current flows, and the resistor will dissipate one watt.
If you put 10 volts across the same *CLASS*resistor, then only 0.1 amp flows through it. But the resistor will dissipate the same one watt. And if you put your finger on both resistors they will be identically hot.
In terms of power dissipation the resistor does not care how many volts are across it or how many amps. It only cares that the
sum product of the voltage and current is less than one. Otherwise it burns up.
This applies similarly to vape coils. I have built two coils for two subtank minis. I built one to 0.5 ohms and the other to about 1.5 ohms. As far as I was concerned they performed nearly identically, as they should. But I did try to configure them with similar turns and diameter (same wire length), trying to make them as similar as possible, it being impossible to make them "identical" of course. One might have ramped up a bit quicker than the other. So I am convinced that the basic Ohm's Law/Watt's Law equivalence actually can carry through even when considering the fluid dynamics and thermodynamic complexities are are added into the mix when trying trying to compare the actual vape performance of dissimilar coils dissipating the same power.
(eta: in the above comparison I vaped both coils at the same power level. Different volts and amps but same power level)
Please also read what I said to
@WattWick above. Look at any definition of a watt. It is one joule of energy dissipated for one second. Or one joule per second if you like. A watt does not care how much voltage or how much current, it only cares about the product of the two. A watt is a watt, regardless of the resistance of the coil.
"Watt = force + work done + loss."
In the above you are over complicating things. There is no "work done" here, in the context of mechanical work (could be if this were a motor but it isn't). All the power is "lost" here, to the extent that effectively all the power is dissipated as heat through the coil. Assuming no power is dissipated via light, which it is effectively not.
