I have some confusion regarding c-rate of batteries and its capacity. For example, a battery of 1Ah is discharging at 1C rate. This means it will discharge 1A in an hour. If the same battery is discharging at 2C, does this mean that it that it will discharge 2A in 30 mins? If thats the case why would the capacity of the battery decrease?

  • 3
    \$\begingroup\$ Because batteries aren't perfect. Like my 73mpg car only gets 65mpg on the motorway, it's less efficient at higher power. So your 1Ah battery will give 1A for an hour, or 2A for 27 minutes ( 0.9Ah), or 6A powering a tiny drone for 5 minutes ( 0.5 Ah). YMMV (literally). \$\endgroup\$
    – user16324
    Commented Sep 1, 2020 at 16:35
  • \$\begingroup\$ @BrianDrummond So the problem with efficiency lies with the internal resistor of the battery? \$\endgroup\$
    – ffriends_
    Commented Sep 1, 2020 at 16:37
  • \$\begingroup\$ With chemistry. Which appears to us as the internal resistance (and, heat). \$\endgroup\$
    – user16324
    Commented Sep 1, 2020 at 16:41
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    \$\begingroup\$ Not an accurate formula, no. But your battery's datasheet will have a graph showing that for a range of conditions. If it doesn't, get a battery that does. (aka No datasheet, no sale.) \$\endgroup\$
    – user16324
    Commented Sep 1, 2020 at 18:45
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    \$\begingroup\$ @ffriends_, Also when continuously sourcing large amounts of current, the internal temperature of the battery increases. With the temperature increases the battery chemistry starts breaking up faster, causing the internal resistance to increase. As a result the life of the battery decerases (Mostly for primary cell batteries) \$\endgroup\$ Commented Sep 1, 2020 at 20:46

1 Answer 1


Yes, twice the current discharge means half the time to battery depletion in the ideal case.

The capacity (at least to a first order) is the same in both cases.

A battery's capacity is the energy stored, measured in amp hours, ergs, joules, or whatever unit you like.

Watts are volts*amps or in your cases battery voltage times 1A, or battery voltage * 2A. So twice the power for half the time is the same amount of energy drained from your battery.

EDIT: If the question is why would the battery capacity decrease over the expected ideal, then Brian's comment is the answer. The internal battery impedance means more power dissipation at higher currents. The chemical reactions in the cell are non-ideal as well and become less efficient at higher discharge rates.


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