I came across this, written by Manu, it really helps clarify things for anyone trying to gauge a mech mod and how it performs and if after modding it you have a better device.

MANU, 06/02/2013, 1:27 PM:

A lower atomizer resistance translates into higher current = more power = more vapour. However, the higher the current, the higher the voltage drop on the internal battery resistance and mod resistance.

Remember that voltage drops at high currents cannot be completely eliminated. The presence of voltage drop only means that the efficiency is decreased. With lower resistance coils, the power on the atomizer increases, but a higher percentage of it is lost on the way.

Let’s get to an example:

[Joe]

Joe has an 18500 non-IMR battery. He vapes with an RBA that has a 1.2 Ohm coil. His battery is fully charged, so he decides to measure the voltage drop of his Roller.

Joe measures the battery voltage at open-circuit, and finds it to be 4.2 Volts.

Then, Joe measures the voltage across the atomizer terminals under load, and finds it to be 3.79 Volts. He assumes that his mod is giving him a voltage drop of 0.41 Volts and frowns unhappily. He is convinced that his mod is not giving him a good hit, although it vapes like a train.

[Jay]

Jay has a fresh 18500 IMR battery. He vapes with an RBA that has a 2.2 Ohm coil. His battery is fully charged, so he decides to measure the voltage drop of his Roller.

Jay measures the battery voltage at open-circuit, and finds it to be 4.2 Volts.

Then, Jay measures the voltage across the atomizer terminals under load, and finds it to be 4.02 Volts. He assumes that his Roller is giving him a voltage drop of 0.18 Volts and goes on to say how amazing it is and how great it vapes.

One minor detail I forgot to mention is that Joe sold his Roller to Jay because of the “voltage drop issue”, so it’s the *same* device.

On to the explanation:

The voltage across a resistance is given by the rather simplified formula V = I * R, where I is the current supplied by the battery and R the resistance we are examining.

The current I flowing through a mod is roughly equal to: I = Voc / ( Ra + Rm + Ri ), where Voc is the open circuit voltage of the battery, Rm is the equivalent mod resistance, Ri is the internal battery resistance and Ra is the atomizer resistance.

A typical value for Ri would be around 0.08 Ohms for a non-IMR battery that is in *OK* condition, while a newer, larger capacity, non stressed, high-drain battery might be better. Take this value with a grain of salt, since each battery is different.

Now, the mod’s equivalent resistance is again not a static quantity, since it depends on how tight the different components are screwed, how clean they are, and many other variable factors. A typical equivalent value would be around 0.05 Ohms, perhaps even less.

So, with the same mod (Rm = 0.05) in the same condition and configuration, Joe was vaping at a current of:

I_joe = 4.2 / (0.08 + 0.05 + 1.2) = 3.16 Amps,

which gives an atomizer voltage of

Va_joe = 3.16 * 1.2 = 3.79 Volts.

In the case of Jay, these values were:

I_jay = 4.2 / (0.05 + 0.05 + 2.2) = 1.83 Amps

Va_jay = 1.83 * 2.2 = 4.02 Volts

So – the voltage drop says nothing. In fact, Joe’s kit was putting out many more watts than Jay’s, because of the low atomizer resistance. Joe was vaping at:

P_joe = Va_joe * I_joe = 3.16 * 3.79 = 12 Watts

while Jay is vaping at:

P_jay = Va_jay * I_jay = 1.83 * 4.02 = 7.36 Watts

Bottom line:

If you want to vape with a low resistance coil, the best thing to do in order to maximize efficiency is to use a high-drain, high-energy battery (18500/18650 IMR).

If Joe had a good IMR battery, his current and voltage would be:

I_joe’ = 4.2 / (0.04 + 0.05 + 1.2) = 3.26 Amps,

which would give an atomizer voltage of

Va_joe’ = 3.16 * 1.2 = 3.91 Volts

and a power of

P_joe’ = 12.73 Watts,

which is much better than the 3.79 Volts and 12 Watts he got with the non-IMR battery.