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I am confused. Some sources say that connecting batteries in series also doubles the amperage (not only voltage) while maintaining the same mAh rating, while connecting them in parallel only increases the capacity mAh, while other sources say the amperage remains the same.

To put it simple, say I want to power an electrical motor that is 48V/1000W. I will first need to connect 12 (3.7 hi-drain 20A 3000mAh li-ions) in series output 44.4V / 3000 mAh. But since the motor drains 20.8Amps and 3000mAh means the battery can provide 3 Amps per hour, this essentially means the battery will get drained for what. 7 minutes? On max power.

This means I will also need to connect some batteries in parallel, to provide more capacity. So if I connect those 12 batteries in parallel with another battery.. does this make 14 minutes?


Edit, according to this, two batteries connected like that output 6A. If measured separately they are 3A.

enter image description here

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    \$\begingroup\$ Wikipedia bans expressions like "Some sources" for a very good reason. Which sources? \$\endgroup\$
    – pipe
    May 11, 2018 at 19:45
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    \$\begingroup\$ Some sources must be wrong. \$\endgroup\$
    – Andy aka
    May 11, 2018 at 19:47
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    \$\begingroup\$ for instance this: electronics.stackexchange.com/questions/189310/… \$\endgroup\$
    – Malina
    May 11, 2018 at 19:54
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    \$\begingroup\$ According to the comments of the top answer, the poster says that the current will be increased. \$\endgroup\$
    – Malina
    May 11, 2018 at 19:54
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    \$\begingroup\$ Don't confuse maximum discharge current with discharge current. Notice at maximum, the total energy drawn from a typical battery is less than (maybe 1/3 of) what one could draw at, say, half the maximum \$\endgroup\$
    – mike65535
    May 11, 2018 at 20:10

4 Answers 4

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"Some sources" are correct - but only in certain circumstances. If you are driving a fixed resistance, connecting two batteries in series will, in fact, double the current. Well, approximately. It won't be an exact doubling, since batteries have a volt/amp curve which produces less voltage for more current. For very low currents and some high-current battery chemistries, two batteries in series may come very close to twice the current.

For high currents, such as a level which will discharge the battery in 10 hours or less, you can count on a noticeably smaller capacity when the current is increased.

Conversely, to your specific situation, you've done the calculations correctly, and 7 minutes/14 minutes is about right. However, since each battery string is providing half the current when in parallel, you might reasonably expect greater run time. Like maybe 15 minutes instead of 14.

Your figure, on the other hand, makes no sense at all, and I have no idea where you got the idea.

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  • \$\begingroup\$ I like your answer. So why the other guy says that batteries don't have current? What did he mean? Give an insight of what happens if I connect another 12 batteries (each connected in parallel) ? How significantly this would increase the duration? Add up this and your answer will be golden \$\endgroup\$
    – Malina
    May 11, 2018 at 23:00
  • \$\begingroup\$ @Malina - "So why the other guy says that batteries don't have current?" Sorry? Where does it say that? You have badly misunderstood whichever answer you are referring to. \$\endgroup\$ May 12, 2018 at 2:04
  • \$\begingroup\$ "You cannot short the batteries to get their "current". Batteries don't have "current" per se." – mike65535 \$\endgroup\$
    – Malina
    May 12, 2018 at 9:55
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Batteries connected in series will raise the effective voltage of the battery pack. Batteries connected in parallel will raise the effective current capacity of the battery pack.

A few examples. My base battery is 3volts and 1 Ah of capacity. If I put two in series, I will have a 6 volts (3 + 3)/1Ah equivalent battery. Two in parallel will yield a 3V/2Ah equivalent battery. If I combine 4 of them into two series group put together in parallel, you will get a 6V/2Ah equivalent battery.

Pushing further, you can combine many cells to get the desired voltage and Ah caracteristics.

Regarding discharge, a nominal 2Ah battery will be delpeted in one hour if you continously drain 2 amps. This is a 1C discharge. 2C discharge means that you discharge it at 4 amps...therefore it will last half an hour.

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Yes you are confusing the information.

Adding cells increases the total mWh capacity if they are reasonably equal in parallel. If in series then the weakest cell limits current.

Thus putting in parallel will add up the milliamp hour capacity, while stringing in the series Adds up to cell voltages. Thus an array is often the optimum configuration to balance the voltage and increase the milliamp hours and again when balanced the total of mWh will be the sum of each component cell or pack.

When taking the total milliwatt hours and dividing by the mW of the load results is less than 10 hours then the total capacity the may be reduced. This will be significant if it is much less than one hour, But depends greatly on the cell’s quality factor for max discharge rate. Eg C/10 or C/40 etc

Always referred to the OEM data sheets for best calculations. If none are available then test and verify

Also keep in mind that A DC motor will draw 10 times the rate at maximum current if starting at full acceleration or full voltage and then reduce as the speed increases.

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  • \$\begingroup\$ What do you mean "equal" in parallel? Number, year, voltage value, discharge rate, mAh? I read text like that everywhere in the internet and it continues to be unclear to me. Perhaps I am not familiar with the electronics terminology in English (I am not native speaker). I understood everything else, just not the batteries. \$\endgroup\$
    – Malina
    May 11, 2018 at 20:03
  • \$\begingroup\$ Also No one talks about the amps of the battery and what difference does it do. I got that the mWh increases in parallel. \$\endgroup\$
    – Malina
    May 11, 2018 at 20:04
  • \$\begingroup\$ In parallel the cell voltages must be matched before putting them in parallel then the milliamp rating adds. The maximum Amps of a battery is always given on the data sheet and is affected by the effective series resistance of the cell \$\endgroup\$ May 11, 2018 at 20:08
  • \$\begingroup\$ The best way to learn this is to use tinkercad and connect batteries differently and use the virtual multimeter. \$\endgroup\$
    – Malina
    May 11, 2018 at 20:10
  • \$\begingroup\$ the best way is the one which you know how to understand. Like Thevenin equivalent circuit voltage and resistance using ESR of each cell and ESR or DCR or motor load for surge current,. Each Lion cell is equivalent to about 10kFarads + 50 mOhms while your Tinkcad assumed 500 mOhms for each AA cell which is about right. ESR is inverse to Wh capacity and brand quality \$\endgroup\$ May 11, 2018 at 23:53
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Amps are drawn from a power source, whether the source is a battery or a wall outlet. A 3.7 V 30 mAh battery supports 30 milliamperes an hour to a 3.7 V load (or device drawing the amps). Two 3.7 V 30 mAh batteries connected in series can safely push 7.4 V to a 3.7 V load for probably a little less than an hour because 7.4 V will cause the load the draw more amps (Ohms law states voltage is directly proportional to amperes over a constant resistance). If two 30 mAh batteries exhausted their power in 1 hour or less on the same load, then batteries connected in series support a greater current simply because of the combined voltage.

The Mah, and Ah limits on batteries co-exist with the C-rating for proper and safe use of the battery. Those limits do not conflict with Ohms law even in the least. Amperes are drawn from the conductivity of the load.

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    \$\begingroup\$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. \$\endgroup\$
    – Community Bot
    Sep 8, 2023 at 12:39

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