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I have a doubt on the difference between series and parallel connected capacitors. Suppose that two capacitors $C_1$ and $C_2$ are charged at two potentials $V_1$ and $V_2$, then they are connected in two different ways

$(a)$ respective positive and negative sides are connected (they are connected in parallel)

$(b)$ positive side are connected with negative sides (they are connected in series)

The picture shows the situations $(a)$ and $(b)$

enter image description here

Now situation $(a)$ is quite clear to me, since they are connected in parallel, they will reach the same $\Delta V$ and the initial charge will be restributed in such what that $\Delta V$ is the same for both capacitors.

But what about situation $(b)$? On textbook it is said that such situation is the same as $(a)$ (same $\Delta V$ for both capacitors) except that the numerical values are different. How is that possible?

I was convinced that if two capacitors are connected in series they will not have the same $\Delta V$ but the same $Q$. But here this seems to be wrong.

So I would like to know what does happen precisely in situation $(b)$. Do capacitors have the same $\Delta V$, do they have the same $Q$, and why?

I really appreciate any help, since this is confusing me quite a lot.

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    $\begingroup$ Just a quick note and a hint: when there are just two circuit elements connected to form a circuit, the parallel / serial distinction is 'degenerate'. That is, there is no distinction. $\endgroup$ – Alfred Centauri Oct 10 '16 at 23:53
  • $\begingroup$ You should show some effort to solve this homework problem yourself. Only a hint: In the end, both capacitors are in parallel. In case a with same polarity charges, in case b with opposite charges. What will happen? $\endgroup$ – freecharly Oct 11 '16 at 2:26
  • $\begingroup$ Try using the Kirchhoff's voltage law. Do you know how the series and parallel connection formulae for capacitors are derived? If you do, use that method on both problems. Applying the general concept of series and parallel here will unnecessarily increase confusion (atleast, it confuses me). $\endgroup$ – FreezingFire Oct 11 '16 at 7:29
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With the circuits which have been drawn there is no point in referring to series or parallel capacitors as it gets you nowhere.
The use of formulae for capacitors in series and parallel is to simplify the problem.
There is no such advantage here.
Just deal with each configuration from first principles with the important constraints that charge is conserved and that after the capacitors are connected together the voltage across each of the capacitors is the same.

In both cases a and b the initial charges on the capacitors are $q_1=C_1V_1$ and $q_2=C_2V_2$

Now when joining the capacitor the total change shored is $Q= q_1 \pm q_2 = C_1V_1 \pm C_2V_2$ where the positive sign is case a and the negative sign is case b.
In case a after joining the capacitors together the polarities of the plates stay the same whereas in case b if $q_1 > q_2$ then the polarity of the plates of capacitor $C_2$ will be reversed and if $q_2 > q_1$ then the polarity of the plates of capacitor $C_1$ will be reversed.

Assume $q_1>q_2$ and that the new charges on the two capacitors are $Q_1$ and $Q_2$.

Since the potentials across both capacitors must be the same $\dfrac {Q_1}{C_1} = \dfrac{Q_2}{C_2 }= \dfrac{Q-Q_1}{C_2}$

You can then find the new charge on each of the capacitor and the new voltage across both capacitors

Note that if at the start for case b the charge on each capacitor was the same then the final voltage across the capacitors would be zero as there is no charge on the capacitors.

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See, if a capacitor is charged at a potential V then the charge on the positive plate will be CV (-CV for the negative plate). Accordingly, the charges in series will be opposite and will cancel each other. Any excess charge left will be distributed such that the potential across both capacitors is the same. Eventually, you will reach a condition same as part (a) with both capacitors in parallel and with same Delta V across them.

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