I've been poking around looking for an answer for this, mostly because my intuitions have been wrong.

What happens to the voltage and the resistance when 2 identical real batteries are connected in parallel or in series?

For a parallel pair, my intuition says that the voltages should stay, because they're technically capacitors in parallel, and parallel capacitors have have the same voltage. The inverse of the resistances are added together, and the inverse found because they are also in parallel, technically.

So, the voltage of the two batteries with an EMF $E$ and resistance $R$ would be $E$ and $1/(1/2R)$.

For a pair in series, I would do the usual operations for capacitors in series and resistors in series. So the equations would be $1/(1/2E)$ and $2R$.

Where am I going wrong?

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    $\begingroup$ In series the resistances add and the voltages add, in series resistance doubles and voltage doubles. In parallel the resistance is 1/2 and the voltages do not change. $\endgroup$ – S. McGrew Apr 30 '18 at 13:26

In the case of the parallel batteries, you can image it as one battery with plates twice as big, so your intuition is partially correct. However you said

parallel capacitors have have the same voltage

but you should note that anything in parallel has the same voltage drop across it. Also, you got the parallel resistance formula wrong. $$R_{parallel}=\frac{1}{\frac{1}{R}+\frac{1}{R}}=\frac{1}{\frac{2}{R}}=\frac{R}{2}$$

For the series case, you got the right answer, for the wrong reason. Note that your $1/(1/2E)$ simplifies to just $2E$, but that you got that from an erroneous calculation. If you had used the right formula for capacitors in series, you would still get the wrong answer, because you would get $C_{series}=\frac{1}{\frac{1}{C}+\frac{1}{C}}=\frac{1}{\frac{2}{C}}=\frac{C}{2}$. The problem with your reasoning here is that voltages does not add like capacitance. For a capacitor, $V=\frac{Q}{C}$ so they have an inverse relationship, they are not analogues. So while for series capacitors you have $\frac{1}{C_{total}}=\frac{1}{C_1}+\frac{1}{C_2}$ you have for voltage, which is the inverse relationship, just $V_{total}=V_1+V_2$. This is kind of the long way of arriving at this relationship, as you should just know that voltages in series always add.

One further point to consider, is that if you already know the voltage at the external terminals of the battery, then it is unlikely that you have any use for the internal resistance of the battery. The current flowing in the circuit will be defined by the resistance in all the other elements in your loop.

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