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I have basic question about connecting two batteries of same type (ie of same voltage and same capacity, say 12V and 100Ah) parallel.

A usual rule of thumb says that if with connect two batteries of same type (Y V, X Ah) in

-series, the "composed" battery twice as big voltage \$2Y\$, but same capacity \$X\$
-parallel, then the "composed" battery has same voltage \$Y \$, but twice as big capacity \$2X \$

I prepared some calculations in order do derive the predicted rule of thumb for the series case based on elementary Kirchhoff's circuit laws. while the addition of voltages results as an immediate consequence, I not see how I can conclude from consideration below the prediction that the capacity of the two batteries in series is the same as for single battery.

Setting: to apply Kirchhoff laws we should assume that our battery V has small internal resistence \$ Ri \$. let moreover R the a test / load resistance. For a single battery we have:

enter image description here

By kcl's we obtain

$$ U_R= V- U_{Ri} \approx V $$ and

$$I =U_R/R= (V-U_{Ri})/R \approx V/R $$ since \$Ri\$ and \$ U_{Ri} \$ are considered to be "small" (\$ R_i << R, U_{Ri} << V, U_R \$)

Let now think about serial case with two batteries 1 & 2 of same type and same load resistance \$R\$:

enter image description here

We conclude with kcl's

$$ U_{R,S} = U1+U2 -U_{Ri,1}- U_{Ri,2} \approx U1 +U2 =2V$$

and

$$ I_S = U_{R,S}/R \approx (U1+U2)/R = U1/R + U1/R = 2 V/R \approx 2I $$

since the voltage of the parallel bank battery \$U_{BB,S} \$ is nearly \$ U_{R,S} \approx U1+U2 =2V\$ that's consistent with serial thumb rule.

But how can we derive from \$I_S=2I\$ that the capacity of serial bank battery stays the same as for a single battery?

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    \$\begingroup\$ In your diagram, batteries 1 and 2 are in series, not in parallel. I did not read through your question in great detail because this glaring problem likely renders the whole question invalid. \$\endgroup\$
    – mkeith
    Dec 25, 2021 at 20:46
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    \$\begingroup\$ Mmm can't see anything parallel here \$\endgroup\$
    – Mitu Raj
    Dec 25, 2021 at 20:50
  • \$\begingroup\$ Yes, of course I had the series case in mind. everywhere where a wrote "parallel" should be replaces by "series"! Thanks! \$\endgroup\$ Dec 25, 2021 at 21:36
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    \$\begingroup\$ Geez, you're making things complex. With 2 identical batteries in series: The voltage doubles and so does the series resistance (realize that the series resistance of the batteries are in series with eachother). The battery capacity in Ah stays the same as the currents stay the same as the batteries are in series. The battery capacity in Whr doubles as at the same current stays the same (capacity in Ah) but the voltage doubles. So one battery: 12 V, 1 A = 12 W but two batteries 2 x 12 V = 24 V, 1 A = 24 W. My suggestion: don't immediately start with KCL, instead think what happens. \$\endgroup\$ Dec 25, 2021 at 21:46
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    \$\begingroup\$ When two devices are in series, the current must be the same. If you don't understand this / need proof then you should go back to studying parallel and series circuits. \$\endgroup\$ Dec 25, 2021 at 22:08

3 Answers 3

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In this example I'm going to assume an ideal world just to make the math easy, real batteries will behave differently.

Suppose you have two 10V batteries, each with a rating with 10Ah. If you connect one battery to a 10 ohm load the current will be 1A (10V / 10 ohms), so theoretically it would last for 10 hours.

If you connect two of them in parallel the voltage will remain the same, so connected to the 10 ohm load the current will still be 1A, but each battery will be providing half of that, or 500mA. 10Ah / 500mA = 20 hours.

Now put those same batteries in series. You will have 20V. If you connect them to the same 10 ohm load the current will be 2A. 10Ah / 2A = 5 hours. If you connect them to a 20 ohm load so that the current is 1A the time will be 10 hours.

For a single battery into 10 ohms the power delivered would be 10V * 1A * 10h = 100Wh. For the parallel case, the power delivered to a 10 ohm load would be 10V * 1A * 20h = 200Wh. In series with 10 ohm load, it would be 20V * 2A * 5h = 200Wh. In series with 20 ohm load it would be 20V * 1A * 10h = 200Wh

In each case each battery will supply it's voltage times it's amp hour rating, 10V * 10Ah = 100Wh. The difference in discharge time will depend on the configuration and load.

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Suppose we have two batteries with a capacity of 100 Ah. Then suppose that those batteries are in series, connected to a load. Then, because of Kirchhoff's circuit law, we know that all of the following quantities are equal:

  • the current through the first battery,
  • the current through the second battery, and
  • the current through the load.

Now suppose that the load has consumed 100 Ah of charge. Because of all of the above, we know that the first battery has produced 100 Ah of charge. Since the amount of charge the first battery has produced is equal to its capacity, the first battery has been depleted.

By the same logic, the second battery has been depleted, too.

The voltage and resistance calculations that you've done seem to be correct, but they're not really relevant to the question you're asking.

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But how can we derive from \$I_S=2I\$ that the capacity of serial bank battery stays the same as for a single battery?

What you can derive from \$I_S=2I\$ is that two batteries in series will discharge into a given load resistor at a rate twice as fast as one battery alone (using the same load resistor).

Remember current is the rate at which charge flows through a circuit element. By itself, the rate of discharge does not tell you anything about the capacity.

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