Just to add to the already good replies, here is a cut-and-paste from an older post of mine (from
http://vapingunderground.com/threads/voltage-question.281497/page-2#post-1421268) that explains the basics of what “voltage sag” actually is and what causes it. At the end of the day, it’s not really adding a whole lot to what has already been said in this thread, but maybe puts it in some sort of context that might be helpful to some.
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"Voltage sag" is simply the result of the division of voltage in a circuit amongst its various resistors.
When any battery is put under a load, its voltage will drop (or "sag"). This is because no battery is a perfect, ideal voltage source since they all have some degree of "internal resistance." And in any circuit, the total voltage potential of the voltage source (the battery in this case) is divided amongst the various loads (resistors) that form the circuit, and that division is proportional to their resistance. In the case of a typical vaping device, the total voltage potential is divided primarily amongst (i) the atty/coil/mod and (ii) the internal resistance of the battery itself. And the higher the resistance of the external load (primarily, in our case, the coil), the higher proportion of the battery's total voltage potential that will be applied to it. The rest of the voltage potential is applied to the other resistor in the circuit (the internal resistance of the battery itself). As an aside, when you use a multimeter to measure the open circuit voltage of a battery, you're really just applying the battery across a big resistor built into the multimeter, and the resistance is so high, that practically all of the battery's voltage potential is applied to the external "load" (the multimeter), yet practically zero charge will flow through the circuit (i.e., no measurable current will result) because the resistance is so high, thus yielding a measure of the battery's open circuit voltage (or, electromotive force, or "emf").
If you had a perfect, ideal voltage source (one that, among other things, had no internal resistance) that could supply a constant voltage potential across a given load, then if you measured the voltage across the external load, you would find it to be the exact voltage of the voltage source itself. But if you were using a battery as your voltage source, let's say two fully charged Li-ion cells in series at 8.40v resting voltage, and you applied that across a load with a resistance of, say, 0.5 ohms, you might expect, from Ohm's law, that we would have a current of exactly 16.8 amps (8.4 divided by 0.5), and that we could measure the voltage across the load at exactly 8.40v. Unfortunately, reality is not that simple, nor convenient. Because the battery itself introduces a small amount of resistance to the total circuit, some of the battery's voltage potential will be allocated to the resistance of the battery itself, leaving something less than 8.40v allocated to the external load. If each cell's internal resistance is, say, 0.025 ohms, then the total resistance of the circuit is actually 0.55 ohms (0.50 plus 2*0.025), in which case Ohm's law tells us that we will have only 15.3 amps (8.4 divided by 0.55) (as opposed to 16.8 amps) of current in the circuit. And 15.3 amps through a 0.5-ohm external load means there is only 7.6v (15.3 multiplied by 0.5) across the external load. This is what we mean by "voltage sag." Our example battery, with an open-circuit voltage of 8.4v and internal resistance of 0.05 ohms, will put only 7.6v across our 0.5-ohm external load. From this, we can readily see why batteries with relatively high internal resistance can supply less current than those with relatively low internal resistance
