IF you have a higher battery ir (20 m ohms vs. 10 m ohms) what will happen? Will my amps go up and voltage to the motor drop? Will the pack heat up more?

Thanx,
Marc

Skyhawk,

Think of the IR of a battery as a little resistor you place in series between the battery and what ever you connect the battery to. The current that flows from your battery pack to your load will flow through this resistor causing what is known as a voltage drop. This drop is equal to the current times the resistance.

For example, let's say you have a 7 cell pack with an open circuit voltage of 8.4 volts and the cells have an IR of 10mohms. This means the pack has a total IR of 7*10 = 70mohms. If you connect this pack to a motor/prop/ESC combo that pulls 10 amps, the IR will drop a voltage of 10*0.07 = 0.7 volts. This means that you will see the pack voltage drop from 8.4 to 7.7 volts.

Also, this IR will dissipate power in the form of heat. Increasing the resistance at the same current levels means the pack will get hotter.

What this all boils down to in actual practice is that using cells with a higher IR means that you will not be able to pull as much current from that pack as a pack made with cells that have a lower IR. Trying to do so will cause the voltage of the pack to drop. This causes the current to go back down and you have a self defeating setup.

I hope this helps,
Joe