# Vehicle battery (NMC) max discharge rate

I wanted to understand a few things related to secondary batteries:-

1. Do the secondary batteries not have any limit on how much current we can draw from them? a. I understand that we have a limit on that due to battery's internal resistance. b. and the battery datasheets (for example) mentions the max discharge rate (maybe 5C or something depending on battery).

But what will happen if i try to load a NMC 12V/15Ah battery pack (4 cells each with 3.6V/15Ah rating connected in series) with a 1000A load for 1s or 700A for 3s?

Will it only heat up badly for that duration? But will it be able to support that huge current without any voltage dip?

• What does the datasheet say? Can you add a link in your question? "But will it be able to support that huge current without any voltage dip?" Of course not. Commented Nov 4, 2020 at 7:27
• Thanks. I have added a link to one of the secondary batteries. This is just for an example. They have specified the max discharge current. Commented Nov 4, 2020 at 7:34
• Please put light on this statement "of course not". I want to understand what will happen if i load my battery with such huge current (1000A for 1s and 700 for 3s). Commented Nov 4, 2020 at 7:34
• Why did you link to that datasheet, is it the battery you want to know about? Commented Nov 4, 2020 at 7:50
• 4 cells, 4 * 70 mOhm (internal impedance from datasheet) = 280 mOhm. Even into a short circuit (i.e. voltage dip to the tune of the whole pack V) you may not get much more than 16V/0.28 about 60 Amps.
– user16324
Commented Nov 4, 2020 at 15:06

Parameter 5.3 on the datasheet you linked specifies that the internal impedance is <= 73 mΩ. At 1000 A that would introduce a voltage drop of 73 V. Since the battery is 3.7 V this is impossible. If the impedance remained at 73 mΩ at high current then you would have a 50% voltage reduction (1.85 V) at $$\ I = \frac V R = \frac {1.85}{0.073} = 25 \ \text A \$$.