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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!

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    \$\begingroup\$ Internal resistance. \$\endgroup\$ – Marcus Müller Feb 13 '18 at 0:18
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    \$\begingroup\$ Ooh this sounds like a real fun way to start a fire :) \$\endgroup\$ – Selvek Feb 13 '18 at 0:22
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    \$\begingroup\$ Also be aware that peak is not the same as sustained, and you probably dont want to be holding your battery when it explodes \$\endgroup\$ – BeB00 Feb 13 '18 at 0:30
  • \$\begingroup\$ Is this a Darwin Award submission? As in "this is not a good idea"? \$\endgroup\$ – Solar Mike Feb 13 '18 at 5:02
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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.

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  • \$\begingroup\$ The OP said the voltage increased when the load was disconnected, which invalidates your conclusion. \$\endgroup\$ – Neil_UK Feb 13 '18 at 5:50
  • \$\begingroup\$ My conclusion is from what I saw in OP's video. I can't see the readings too well, but when the voltage rises above 2V, the load is still present, around 110A if I read correctly. \$\endgroup\$ – anrieff Feb 13 '18 at 7:40
  • \$\begingroup\$ That's a very good reason why most people, myself included, don't bother to click on links to videos. You could well be right about the timings of the readings. I am right about what the OP said. It seems the OP was majoring on why the voltage rose rapidly off load, and the answer is internal resistance. You are correctly describing an effect which is probably present in the video, that the OP didn't appear to be asking about. \$\endgroup\$ – Neil_UK Feb 13 '18 at 8:26

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