Measuring the IR of your battery requires a special toolset. You either need a charger that will measure it for you or a tool that specifically measures internal resistance. Given that the only tool I have found for this (at least in the hobby world) is almost as expensive as a charger that does this for you, I'd go with a charger for this process. Some chargers measure each cell's IR separately, and some measure the entire battery pack as a whole. Since internal resistance is a cumulative effect, and the cells are wired in series, if you have a charger that does each cell independently, you need to add up the IR values of each cell, like this:

Suppose we have a 3S (3-cell) LiPo battery, and the measuring the cells independently yields these results.

Cell 1: 3 mΩ Cell 2: 5 mΩ Cell 3: 4 mΩ

To find the total internal resistance for the battery pack, we would add up the values for the three cells.

3Ω + 5Ω + 4Ω = 12 mΩ

For a charger that measures the pack as a whole, all you would see is the 12 mΩ - the rest would be done for you - behind the scenes, as it were. Either way, the goal is to have the IR for the entire pack.

The first reason internal resistance is important has to do with your battery's health. As a LiPo battery is used, a build up of Li2O forms on the inside terminals of the battery (we'll go more in depth on this later in the Discharging section). As that build up occurs, the IR goes up, making the battery less efficient. After many, many uses, the battery will simply wear out and be unable to hold on to any energy you put in during charging - most of it will be lost as heat. If you've ever seen a supposed fully charged battery discharge almost instantly, a high IR is probably to blame.

To understand how Internal Resistance works in R/C applications, first we have to understand Ohm's Law. It says that the current (Amps) through a conductor between two points is directly proportional to the difference in voltage across those two points. The modern formula is as follows: Amps = Volts / Resistance. In the formula, the resistance is measured in Ohms, not milliohms, so we'd have to convert our measurements. If we use our previous 3S LiPo, and plug it into the equation along with a 1A draw, we can find out how much our battery pack's voltage will drop as a result of the load. First, we have to change the equation to solve for volts, which would look like this: