I'm experimenting with how fast you can discharge a lipo. I have an 2200 mAh battery. Its peak discharge is 100C. I hooked up some copper tubing to act as a resistor. Its resistance is 0.02 ohms. I connected the battery to a DC contactor rated to 200 Amps with 10 gauge wire. I'm measuring the current with a clamp meter and the cell voltage with a 6.5 digit meter. I charged the cell to 4.15 V and then turned on the contractor. The cell's voltage immediately dropped to 2.3 V and continued to drop to 1.5. The weird thing is when I turned the contactor off the voltage went back up to 3.5V. I am confused about how the voltage dropped so quickly and how it went back up after the current stopped flowing. Am I just seeing the voltage drop of the leads coming from the cell or is this a result of the batteries internal resistance? Here is a video of the experiment https://photos.app.goo.gl/tpBxalM5EUMwJZed2 Thanks!
Most likely both te internal resistance and leads resistance take part, but I'd guess the internal resistance dominates.
Note how the voltage dips to below 2V, but then improves to >2V with the load still connected. The likely explanation for this is that your battery heated in the process. Like any battery out there, your LiPo runs a chemical reaction inside to produce the voltage across its terminals. The higher the current, the faster the reactions needs to happen. There are limits to how fast the reaction can run and this is one of the reasons for the internal resistance. However, as a battery heats up, its reaction rate improves (just like any chemical reaction), and the effective internal resistance decreases.