Correct way to determine amp draw from battery

[bfrie]

So this question has been bothering me for some time now. what is the correct way to measure amp draw from a battery through a mechanical mod?
a. fully charged battery, 4.2v, fires a 1 ohm coil, drawing 4.2 amps from the battery
b. fully charged battery, 4.2v, fires a 1 ohm coil, except .2v are lost in the circuit (voltage drop), sending 4v to the coil, drawing 4 amps from the battery

which one is right and why? i have been assuming A. because better safe then sorry, and would not plan on pushing batteries further to the limit if B. was true, but i just like to know answers to questions like this

 

[Hoosier]

Current from the power source is equal to the power source voltage divided by the total resistance of the circuit.

Voltage comes from current flowing through a resistance and all voltages in a circuit add up to the total voltage.

So, the voltage drop only occurs because current is flowing through a resistance. Technically, "b" is correct, but it is being looked at sideways. It would make it a straight view if the voltage drop was expressed as a resistance in series with the coil resistance.

Taking your scenario, your mod has a 0.05 Ohm resistance in series with the coil. The battery at 4.2 Volts sees the 1 Ohm coil and the 0.05 connector + switch resistance in series. I=V/R=4.2/4.05=4 amps.

The big problem with this approach is that the mod's resistance would have to be measured to make sure the voltage drop is linear. (It should be, but there is a possibility that it is not.) "b" is technically correct, but I would stick with "a", even though I know it is wrong.

 

[steved5600]

here is a spread sheet that does the work for you.
View attachment ecig calculators.xlsx

 

[Ryedan]

Taking your scenario, your mod has a 0.05 Ohm resistance in series with the coil. The battery at 4.2 Volts sees the 1 Ohm coil and the 0.05 connector + switch resistance in series. I=V/R=4.2/4.05=4 amps.

The big problem with this approach is that the mod's resistance would have to be measured to make sure the voltage drop is linear. (It should be, but there is a possibility that it is not.) "b" is technically correct, but I would stick with "a", even though I know it is wrong.

And to complicate things a bit more, the battery has internal resistance which is significant also (and different batteries have different resistance), so it's not just the mod at play here. In the end measuring the voltage at the coil under load is the best number to use for calculating amp draw because that's what the battery sees, so this is a second vote for 'B' if accuracy is what you are looking for.

Now, about how hard to push a battery, I don't get too near the continuous amp rating. I actually stay quite far away. I'm drawing about 8.5 amps (taking voltage drop with a freshly charged battery into account) on MNKE 20A and Sony 30A IMR batteries. I would be comfortable using as much as 70% of the battery rating with good IMR or high drain hybrid batteries that have not seen too many cycles and have not been abused.

On another note, for a good explanation of voltage drop in mech mods I like Manu's post here. Explains the importance of the battery in the equation very well and also the significance of different coil resistances. Not a lot of people seem to understand this.