If for example the Mech mod has an internal resistance of .1Ω (which would be really high) then if you had a .1Ω build, 50% of the voltage would be developed inside the mod and 50% would be delivered to the coil. So using a 4.2V battery the coil would see a 2.1V drop across it. The battery would be providing 4.2V/.2Ω or 21A. The power developed would be equally split between the mod and the coil. Each would dissipate 44 Watts. As higher resistances are used, greater percentages of the voltage applied would be developed by the coil and the mod would rob a lesser percentage of the total wattage developed. Conversely, batteries loose voltage as load increases. A battery's voltage under a .1Ω load will be less than a battery's voltage under a 1Ω load. Vapers call this voltage sag. In the past it was simply called voltage under load. The internal resistance in a mechanical mod is dependent upon materials used, wire used, surface area of connection points e.g. switch contacts, wire lengths, the connectivity of any threading in the electrical path, etc. This can best be measured by measuring the total voltage across the batteries and the total voltage across the coils simultaneously. The difference in the voltages is the voltage developed inside the mod. From there the resistance of the mod and atty deck can be calculated by calculating the amperage V (coil) / R (coil) and dividing that into the differential of the two measured voltages or in other words the voltage developed within the mod and atty.
