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I was intrigued by the answer and comments on this question about connecting a second battery in parallel to increase capacity. The original battery has obviously lost capacity and is nearing its end of life. The OP wanted to add a second in parallel to increase capacity, which might make sense if 1) the UPS was compatible with doing so, and 2) the original battery was not faded/damaged.

Marcus indicated in the answer that adding a second battery in parallel would add "but 25% more capacity than just the new one."

I'm unsure how that figure was arrived at. If two identical batteries in parallel are twice the capacity of one alone, how does one calculate that capacity when the batteries differ in capacity or age? It seems to me that a deficient cell would obviously contribute less than 100%, but could also parasitically drag down the newer battery if they attempt to equalize. (That's my assumption.)

Is the 25% figure given based on a rule of thumb, a best-estimate based on the limited information in the original question, or something else?

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  • \$\begingroup\$ I assumed that because charging is shared, the resulting full capacity is degraded for both batteries etc. etc.. \$\endgroup\$
    – Andy aka
    Commented Jun 1, 2020 at 16:12
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    \$\begingroup\$ I don't think you read that right... he said "25% more than just a new one", meaning that the original had a 30 minute runtime, down to 10 minutes, so adding a new one would give 40 minutes of runtime, or only 25% more than the original 30 minutes (percentages are a little off, but you get the idea). \$\endgroup\$
    – Ron Beyer
    Commented Jun 1, 2020 at 16:13

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I think the point of the original answer was that the two batteries are not identical. They had the same specifications when new, but the OP said that one of the batteries was dying and had lost most of its capacity. So Andy was trying to say that there was little point in using both batteries; just use the new battery and recycle the dying one.

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    \$\begingroup\$ Exactly. If we assume the overall runtime to be 40 minutes when using both batteries, then only the last quarter (=25%) of that is actually from the old battery. That's what I meant. \$\endgroup\$
    – Marcus Müller
    Commented Jun 1, 2020 at 16:26
  • \$\begingroup\$ Thanks for the clarification, @MarcusMüller I think that's what I was missing. \$\endgroup\$
    – JYelton
    Commented Jun 1, 2020 at 16:51
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I think that is just a simple estimate - not even a "best" one.

There is no reason for spend time with an exact estimate (and measurements would be needed).

OP said that the old cell was supporting 10 min, it's about 1/3 of the 30 min of the new one.

But it's an old battery in parallell with a new one, with a charger/inverter circuit that has losses and possibly will not handle the two batteries well.

Because of this, he estimated that only 25% (1/4) of the capacity of the old battery will be available in the whole system (instead of 1/3 of the old battery), this number was found just rounding to the next whole number (1/3 to 1/4).

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  • \$\begingroup\$ as the "he" from your answer: nope, that would've been clever, but it's not what I did. I did. "even if your switchover idea worked, you then would have 10+30 = 40min of battery life. So, for the last 25% of that overall runtime, you're even thinking about a technical solution." \$\endgroup\$
    – Marcus Müller
    Commented Jun 1, 2020 at 17:46

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