I'm confused about how to understand battery capacity by its ratings. I though I understood Amp-Hours, as a nominal rating where they usually specified a period of time where that rating was approximately valid. So below is the rating a particular batter, where nominal capacity for 20 hours rate-Ah is stated as 74 and the 10 rate-Ah . I assumed means that for 20 hours it can sustain the 74 Amp-hour rating, so I would expect that the amp-hour rating for 10 hours would be as high or higher but it's 65. So i'm not sure it means what I thought it means. So does it mean that I can operate the battery for 20 hours at 74 Ah /20 h = 3.7 amps, but if I use it for 10 hours, then it can run at 65 Ah /10 h = 6.5 amps for each hour? Why this reasoning doesn't seem quite right to me is that I thought the Amp-hour rating was high if you specify shorter periods of time.

Below is an example of a battery rating. And the link to the published chart.

| Volts| PHCA | CCW | HCA | MCA |       Nominal Capacity          | Reserve Cap. |
|      | 5 sec|     |     |     | 20 HR Rate-Ah  | 10 Hr Rate-Ah  | in Minutes   |
| 12   | 1750 | 950 | 1350| 1070|       74       |       65       |        145   |

Extreme Series Battery Specifications

  • \$\begingroup\$ electronics.stackexchange.com/questions/122732/… I think this might help \$\endgroup\$ – muyustan Dec 6 '19 at 20:29
  • \$\begingroup\$ you have it sort of backwards ... if you draw 3.7 A, then the battery will last 20 hours before the output voltage starts to drop below 11 V (not sure about the actual voltage ... check graphs in datasheet) .... if you draw 6.5 A, then the output voltage will start to drop after 10 hours \$\endgroup\$ – jsotola Dec 6 '19 at 21:04

Believe the specification.

All chemical batteries will have reduced capacity at higher discharge rates.

I thought the amp-hour rating was high if you specify shorter periods of time.

No, you can draw a higher current for a shorter time but the total energy the battery provides will be lower at higher discharge rates. There will be several reasons for this including the higher operating temperature at the electrodes due to higher current and the I2R internal heating losses due to the internal equivalent series resistance.

Worth reading:

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  • \$\begingroup\$ also, things like less uniform metal deposition on the electrodes leading to reduced "available" chemical energy, but very much depending on chemistry. \$\endgroup\$ – Marcus Müller Dec 6 '19 at 20:36
  • \$\begingroup\$ OK. So if I'm powering a load of, say 65 amps I can expect much less total energy, I guess due to heating energy, etc. \$\endgroup\$ – Frank Dec 6 '19 at 21:49
  • \$\begingroup\$ By the way, is anyone aware of a good battery engineering textbook, or comprehensive online course specifically for battery technology . In my EE courses everything was an ideal voltage or current source. Not a very useful for designing battery systems it seems. \$\endgroup\$ – Frank Dec 6 '19 at 21:53
  • \$\begingroup\$ At 65 A much, much less total energy. You might find batteryuniversity.com useful. \$\endgroup\$ – Transistor Dec 6 '19 at 21:56

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