If you connect an uncharged battery to a charged battery in series (+ to - and - to +) there will be a large current flow between the batteries and it will heat up as if it's being short circuited, but if you do this with two charged batteries it will just double the voltage that the system has. Why? It seems that the electrons would flow from the negative of one to the positive of the other for each battery if they were connected in series bringing both of them to 0 volts, but why does this not happen?

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    $\begingroup$ please draw for us the circuits you describe. Include the load resistances in each case. $\endgroup$ – niels nielsen Apr 15 '18 at 18:51
  • $\begingroup$ @nielsnielsen the circuit I am thinking of is just two batteries connected to each other, positive of one connected to negative of the other and vice versa. $\endgroup$ – user180969 Apr 15 '18 at 19:46

What you have described is what I would call shorting two batteries out. In both cases you will get a very high current that is only restricted by the internal resistance of the two batteries (and the wires you used). A dead battery will have a much higher resistance than a changed battery. Thus, contrary to what your question stipulates, the circuit with the two good batteries will have a much higher current than a circuit with one good and one dead battery. A circuit with two dead batteries will have an even lower current.

The correct way to describe the circuit is $I = V_{total}/R_{total}$, where $R_{total}$ is the sum of the internal resistances of the two batteries + the total resistance of the wires. $V_{total}$ is the sum of the voltages of the two batteries with no load—i.e. as measured by a high-impedance voltmeter while not connected to any load (this is just what we normally call voltage).


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