# Why is battery C rating not a function of battery topology/voltage?

In a battery, there is a 'C' rating, which appears to be a function of its capacity. a 5 Ah battery charged or discharged at 1C charges or discharges at 5A.

However C capability is generally spoken of with respect to battery chemistry only.

If I have four cells, each 3 V, 0.8 Ah, then I can make a battery with three different topologies:

1. 1S 4P: 3 V
2. 2S 2P: 6 V
3. 4S 1P: 12 V

Each variant has a 3.2 Ah capacity*, and 1C in each case suggests charging or discharging at 3.2 A. However in 1), each cell will experience 0.8 A of charge or discharge. In 2) each cell will see 1.6 A, and 3) each cell will see the full 3.2 A. Battery 3 at 3.2 A will charge 4x faster than battery 1, corresponding to four times the power input, totally changing how the chemistry might be expected to behave. C rating sure surely then include the topology, or else it would be a nonsensical metric. What is the reality?

*Incorrect. Amp hour rating varies as inverse of voltage from topology. Amp hour rating alone is insufficient to characterise energy capacity, voltage is required also leading to a common Watt hour capacity. Series topology is cancelled and the Amp hour rating of the battery is equal to the Amp hour rating of the smallest parallel arrangement of cells in the battery.

• If the battery charges at 1C that's 0.8A, not 3.2A. May 11 at 17:48
• But your assumption that four 0.8 Ah batteries in series is 3.2 Ah is wrong. Putting batteries in series increases voltage, not current capability and thus does not increase Ah, and putting batteries in parallel increases current capability and thus Ah but not voltage. May 11 at 17:51
• I am working on the assumption that an Amp hour is a unit of energy capacity in a cell, that the hour is as arbitrary a choice of time as speaking of speed in miles per hour. 1 Ah could be consumed in 30 min at 2A. If this is the case then all of my configurations have a 3.2 Ah capacity. May 12 at 9:25

You are miscalculating, or neglecting to calculate, the effect on the capacity of the entire pack of connecting cells in parallel. When you connect two cells in parallel, the current going into or out of them splits. Assuming that the split is equal, then you get:

1. 1S 4P: 3 V
• 3 V 3.2 Ah (9.6 Wh) 1C charge = 3.2 A. Split evenly over four cells, 3.2 A = 0.8 A/cell
2. 2S 2P: 6 V
• 6 V 1.6 Ah (9.6 Wh) 1C charge = 1.6 A. Split evenly over two strings of two cells 1.6 A = 0.8 A/cell
3. 4S 1P: 12 V
• 12 V 0.8 Ah (9.6 Wh) 1C charge = 0.8 A. There's just one string, so 0.8 A = 0.8 A/cell

(As a practical note, you can't trust that a parallel arrangement will split the current evenly -- just be aware).

• I believe I may be confusing energy capacity for current capacity. In that case, My configuration 1 is 3.2 Ah and could produce 3.2 Amps for 1 hour. In configuration 2, if you're telling me the battery current capacity is 1.6 Ah, then it can produce 1.6 Amps for two hours. Same with config 3, will make 0.8 Amps for four hours. Is this more accurate? May 12 at 9:29
• Aaah I think I see the key issue. The current drops as voltage rises, meaning the above comment is incorrect. The configuration 3 will discharge at 0.8 A for 1 hour at 12 V, amounting to the same power as configs 1 and 2. Same power for same time is same total energy. May 12 at 9:33