Calculating Battery Current

[Canton]

140621-1938

Canton:

If I assume a linear drop in voltage over a 15 minute period from 2.00 to 1.67 V, then average voltage is 1.835 V over this time period. Now assume the given power at this point is the 506 W, and assume the current at this point is the desired average current, then current is 506/1.835 = 275.7 A. Close to your 271. If you use a voltage of 1.867 the the result is 271.02 A.

384 V with an internal resistance of 2.8 ohms calculates to a short circuit current of 384/2.8 = 137 A. This does not correlate with your stated short circuit current.

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Yes...that is correct but it is for one battery 12 volts/.0028 ohms = 4285 amps.....no sure how they got to 4510..?

I guess they used 12.6 volts/ .0028 = 4500 amps....spec sheet says 4510 amps....I guess that will work

 

[Canton]

OK.
I really didn't think that it was 2.8 Mohms as in your previous post.
Nor the 2.8 ohms in your earlier posts.

Lol....ya I guess Meg-ohms would make the calculation even worse...