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Why do batteries in series add voltage and not current ? How does the first battery increase the voltage of the second,i.e. do we model it as the first pushing an electron that actually goes through the second battery appearing at the second's anode ?

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Voltage is a potential difference between two points. So, the first battery just creates a higher starting point for the 2nd one. I guess an analogy for this would be two identical step stools placed on top of one another. It is not that the 2nd one is suddenly getting taller, it is just placed on top of the 1st one producing a taller total height.

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Why do batteries in series add voltage and not current?

It's the law. Specifically, Kirchhoff's Voltage Law (KVL).

How does the first battery increase the voltage of the second

But, I don't think you're correctly interpreting what it means to "add voltage". One battery does not increase the voltage of the other.

If you place a voltmeter across either series connected cell, you'll still measure the nominal cell voltage.

Only by placing the voltmeter across the series combination will you measure the sum and, again, this must be so by KVL.

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    \$\begingroup\$ Whilst KVL is true, I find it a bit circular when offered as an explanation: K noticed that voltages in series add. K wrote this down. Therefore voltages add because K noticed this fact. I would hope it is possible to explain the phenomenon in terms of more fundamental concepts. \$\endgroup\$ Oct 16, 2013 at 13:02
  • \$\begingroup\$ @RedGrittyBrick, KVL follows from Faraday's Law of Induction in the electroquasistatic approximation such that the electric field is conservative. Otherwise, KVL is an approximation. \$\endgroup\$ Oct 16, 2013 at 14:12
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An ideal battery is a dc voltage source and an internal resistance in series. This resistance determines the maximum current it can provide. If you add them in series they can deliver the same current as you add the resistors and the sources at the same time. If you put them in parallel you get the same voltage but the resistance gets reduced (Ohm's law), so you get more current.

you can read more here

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  • \$\begingroup\$ This is a great explanation of the current changes. I am still stuck on how to model the voltage increase. Using the electron model, an electron leaves the anode of battery1 with emf x. It arrives at cathode of 2. Now we cross an electrochemical barrier. Are we somehow driving the redox rx harder ? So that the emf at the anode of 2 is higher than if the first battery was absent ? The redox Rx should not be involved since the same occurs with inductive voltage sources. \$\endgroup\$
    – user30520
    Oct 16, 2013 at 22:49

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