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If I have two identical batteries (capacity: 300 mAh, voltage: 7.4 V, C rating: 25 C).

If I connect those two batteries in parallel or in series, does this affect the C rating of the combination of the two batteries?

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    \$\begingroup\$ Does this answer your question? Batteries connected both in series and in parallel \$\endgroup\$ Commented Sep 22, 2020 at 13:49
  • \$\begingroup\$ when connecting the 2 batteries in parallel it's equivalence to offering a higher capacity battery for the same voltage the C rating is the maximum current the battery can source without a series damage to it's performance with respect to it's capacity so 300mah battery can source 300 milliamps of current for an hour but it can source a current of up to 300mah * 25 = 7.5 amps continuously for around 2 minutes Now when connecting 2 of them in parallel you are doubling the c rating for the combination or the equivalent battery that is formed. \$\endgroup\$ Commented Sep 22, 2020 at 13:58
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    \$\begingroup\$ @MahmoudSalah See my answer - notice the need to specify what the C rating is relative to. \$\endgroup\$
    – Russell McMahon
    Commented Sep 23, 2020 at 2:09
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    \$\begingroup\$ @MarcusMüller This is a trickier question than it looks and the cited answer addresses mAh ratings and largely NOT C ratings. For identical cells. If both have eg 25C capacity then in parallel they are 50C rated compared to ONE CELL. BUT if each is 300 mAh then in parallel capacity is 600 mAh and C rating is 25 x 600 mAh Ie you need to specify if the new C is relative to the per cell capacity or the new combined capacity. Roughly: Paralleling adds C ratings relative to one cell. Series gives C rating of lowest Ah cell. \$\endgroup\$
    – Russell McMahon
    Commented Sep 23, 2020 at 2:15

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There is a "trap" in understanding this question.
It is NOT the same as asking about mAh ratings.

This is a trickier question than it looks and the cited answer addresses mAh ratings and largely NOT C ratings. For identical cells:

For parallel cells:

If both have eg 25C capacity then in parallel they are 50C rated compared to ONE CELL.

BUT if each is 300 mAh then in parallel capacity is 600 mAh and C rating is 25 x 600 mAh
ie the C rating relative to the new combined mAh capacity is the same as the C rating of each cell individually.
But the mAh capacity of the battery is the sum of the mAh rating of the two cells.

ie When specifying battery C rating you need to specify if the new C is relative to the per cell capacity or the new combined capacity.

For series cells the mAh capacity of the battery is the same as each cell and the C rating does not change.

Roughly: Paralleling adds C ratings. Series gives C rating of lowest Ah cell.

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It is presumed that the batteries are identical (7.4 V, 300 mAh, C 25), new, good and fully charged.

Case No.1 - Batteries in series

The rating of the combination would be 14.8 V, 300 mAh, C 25

Case No.2 - Batteries in parallel

The rating of the combination would be 7.4 V, 600 mAh, C 25.

The yardstick is the same irrespective of the batteries being single or combined.

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  • \$\begingroup\$ Your case No.1 new mAh is incorrect. If one cell is 300 mAh then two cells gives 600 mAh. <- Do you agree? ||| " This battery then has C = 25 relative to 600 mAh or C=50 relative to 300 mAh. See my answer. \$\endgroup\$
    – Russell McMahon
    Commented Sep 24, 2020 at 10:49
  • \$\begingroup\$ No Russel, it's correct. Energy / battery / 25 h = 7.4 * 0.3 = 2.22 Wh . Energy / 2 batteries in series or parallel / 25 h = (14.8 * 0.3) = (7.4 * 0.6) = 4.44 Wh . It figures. \$\endgroup\$
    – vu2nan
    Commented Sep 24, 2020 at 14:36
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    \$\begingroup\$ Trying again: Please follow this carefully - it's important. Really. || No BECAUSE. C rate is NOT proportional to energy - it is proportional to mAh capacity. When you double the cells - I agree that in parallel and series you double the energy. BUT: In parallel you double the mAh but do not change the voltage. In series you double the voltage but do not change the mAh. In BOTH cases V x mAh = energy = the same. But in parallel the C rate doubles. \$\endgroup\$
    – Russell McMahon
    Commented Sep 25, 2020 at 2:24
  • \$\begingroup\$ Hi Russel, Case 1: A 300 mAh C25 battery will continuously deliver 300/25 = 12 mA for 25 hours. Case 2: A 600 mAh C25 battery will continuously deliver 600/25 = 24 mA for 25 hours. Case 3: Two 300 mAh C25 batteries, connected in parallel, will continuously deliver (12 + 12) = 24 mA for 25 hours. Since Case 2 & Case 3 behave identically, would it not be right to conclude that the rating of the parallel combination is 600 mAh C25? \$\endgroup\$
    – vu2nan
    Commented Sep 25, 2020 at 10:31
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    \$\begingroup\$ I agree with you. My 1st comment has a typo (or brain fade) and should have referred to your case 2. || What you have written is correct in both cases, but it may not be obvious to some (even though in plain sight) that your C25 in parallel is relative to the new 2 cell 600 mAh battery. It's C25 wrt battery and C50 wrt one cell. ie even though the C25 figure is unchanged the C25 current has doubled. So I agree that what you have written is correct. What I wrote is also correct (apart from the typo :-). || I suggest that adding a comment that C25 is now double may avoid misleading a newcomer/ \$\endgroup\$
    – Russell McMahon
    Commented Sep 25, 2020 at 10:56
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I believe none of the answers above are correct. C rate's are important because of the peukert constant. C Rates tell you the AH capacity of batteries at different loads. C Rates are incomplete if you do not specify an End Of Discharge Voltage as well so in reality you need a few tables or a pivot table with CRate and EODV. The reason will become clear in a moment.

The EODV and the capacity of the battery will vary depending on the load placed on the battery. A higher load causes a greater current and wattage per cell. The greater current will cause more heat build up than a lower current. This will cause the batteries to lose some of their energy capacity to heat loss (higher power draw has less efficiency converting the capacity energy to load energy being used).

The higher power draw also causes more voltage sag. So when the batteries are placed under a load they lower the voltage that the batteries immediately drop to and the voltage they will drop to when the batteries fail to provide adequate power. A higher power draw will need a lower end of disconnect voltage to save the batteries from over heating and presenting a threat to device health, over-heated battery danger and other reasons. Losing some efficiency to heat and having to use a different EODV will cause the batteries to have lower capacity ratings at different loads.

When you put batteries together in series their capacity rating at a specific C rate will be close to the capacity rating of a single battery but the wattage per cell, current across each cell will change even though the voltage per cell remains the same and as a result you will need a different C rate rating to find the best estimated capacity of your series-wired batteries.

For ex. 12V battery with 100AH capacity at C rate 10. battery will be drained over the course of 10 hours. Take 100AH * 60 * 60 to get Amp Seconds, * 12 to get the capacity in wattage. This wattage is used over the course of ten hours so we divide that wattage by 10. This is the wattage used per hour. We divide that by 60 and again by 60. We now have the wattage used per second.

We divide that wattage per second by 6 (because there are 6 cells in a 12V battery wired in series and we have the wattage per cell needed to drain the batteries in 10 hours). In this case this would be 120W per second divided by 6 which is 20W/cell.

If we put two of these batteries in series we are now operating at 24 nominal Voltage.

Let's assume the load is the same because as we know, the load and c rate determine the capacity because of the Peukert Constant (higher loads are less efficient because of energy lost in the form of heat).

In the c rate given by the manufacturer in this example (100AH capacity at C rate 10 for a 12v battery) our load is at 120W per second in order to drain the battery in 10 hours and thus has a capacity rating of 100AH at that C rate. This load is causing 20W per cell load and each cell is experiencing 2V and 10A.

If we put two of these batteries in series and put this same load on we are still at 120W per second, but we have 12 cells now. The wattage per cell is now 10W per cell. Each cell is at 2V still but each cell is now seeing 5A per second.

The efficiency of the series-wired batteries will be greater because they share the load between more cells and experience less energy lost in the form of heat. So in order to have an accurate AH capacity you would need to perform a test or use a more accurate C Rating provided by the manufacturer for your load. (It will be very close to the AH capacity listed at C rate 10 but depending how much of a variance there is between the wattage per cell at C Rate 10 for ONE battery and your wattage per cell depending on your custom load you're estimated capacity will vary accordingly. On this 12V example, the difference between C rate 10 and C rate 1 is about a 30 percent difference in capacity. Putting batteries in series changes how well they handle the same load so if the load stays the same the capacity for one battery at that c rate will not be perfectly accurate. You would have to find the capacity rating at a similar WPC to get the most accurate estimated capacity and also consider the effect of your EODV on the capacity after doing that if you will be utilizing one.


Clarifying my position further.

The original poster's question doesn't give enough information to be able to give a correct response.

A C Rate on it's own only means how long the battery can run before dying. On it's own it isn't very useful. When we combine multiple different factors with a C Rate then we are able to clearly define the capacity of a battery.

In order to correctly describe the capacity of the battery we need the C Rate, AH at that C Rate, EODV used, and nominal Voltage per cell.

The purpose of defining battery capacities in this way is to show that the measurement of resistance of our battery powered device is going to determine the amount of power we draw from the batteries. Batteries have different efficiency rates at different power draw measurements. This means that there isn't one single AH (capacity) rating that can accurately describe a battery. Depending on your purpose for the battery the capacity will be different. In the same vein the End of Disconnect Voltage used will also impact the capacity rating of a battery.

Thus, we need C Rate, AH Capacity, Voltage per Cell and EODV to accurately describe the capacity of a battery. (There are other factors that can impact the capacity as well such as temperature, age of the battery, etc) but at a minimum we need those 3 factors.

Using the C Rate and the AH Capacity of a battery and it's nominal voltage per cell we can determine the Wattage Per Cell which will be instrumental in answering the original poster's question.

If the resistance of the device that is being powered by the batteries will remain a constant then we can not accurately answer the question (meaning the device can operate within a wide enough voltage range that covers the EODV of a single battery and up to the float voltage of two batteries in series). The WPC will be different when using two batteries as compared to just one. Because the WPC will be halved when using the two batteries the run-time capability will be doubled plus some change because the batteries will be able to handle that same load more efficiently. We would have to calculate the Wattage Per Cell that our load will place on two batteries and find a C Rate that matches that WPC for one battery, and then double the AH rating within the C Rate (for two batteries in parallel) or keep the same AH rating and double the nominal V (for two batteries in series).

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