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Let's consider a capacitor precharged to 5V.

Now immagine to put it in parallel with a capacitor which has no charge. What does it happen? They reach a voltage in the middle? Does it depend on their capacitances?

In theory they are both in series and in parallel, so we will get Q1 = Q2 and V1 = V2, but this will mean C1 = C2, that is an absurd.

Kinka-Byo
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    What do you think might happen? – User323693 Aug 17 '19 at 16:25
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    I am hesitant to answer as this feels very much like home work. – Oldfart Aug 17 '19 at 16:25
  • I do not know if they have the same charge (Q1=Q2) or not. In theory they are both in series and in parallel, so we will get Q1 = Q2 and V1 = V2, but this will mean C1 = C2, that is an absurd – Kinka-Byo Aug 17 '19 at 16:29
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    If you connect a voltmeter across C1, aren't you also connecting the voltmeter across C2? – Chu Aug 17 '19 at 16:38
  • You start with an initial charge which in the ideal case does not get lost. Afterward the charge will still be there but it will be spread over your two capacitors. – Oldfart Aug 17 '19 at 16:39
  • Consider it as parts in parallel. Both units are connected across them. And derive the outcome – User323693 Aug 17 '19 at 16:40
  • The electrons cannot escape. But there is another parameter that also describes what is stored. – analogsystemsrf Aug 17 '19 at 17:05
  • And what does it happen in the transient? – Kinka-Byo Aug 17 '19 at 17:29
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    In the transient, a mighty big current pulse flows, theoretically infinite. It’s unhelpful to regard them as both simultaneously series and parallel. Regarding them as parallel is sufficient. – Andy aka Aug 17 '19 at 17:31
  • So you have 100 amps, for 100 nanoSeconds. What happens? – analogsystemsrf Aug 17 '19 at 17:40
  • They will reach a voltage in the middle. How quickly they reach that voltage depends on many things, for example the combined ESR of the two capacitors. They will not necessarily have the same charge. The initial conditions for the two capacitors are very different, so the charge after voltage attains steady state need not be the same. – user57037 Aug 18 '19 at 01:20

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At the current state of our universe, charge is conserved. (This wasn't necessarily always the case. See this article on dark matter, for example, discussing the possibility that charged particles created shortly after the Big Bang lost their electric charge during the inflationary period.) This means that the sum of two relative charges held by the two capacitors before being connected to each other must be the same as the relative charge of the combined capacitor after being connected. When you place two capacitors in parallel, the total charge of the final system is the sum of the two original charges on the two earlier systems. In short, \$q_{_{tot}}=q_{_1}+q_{_2}\$.

Since \$q_{_{1}}=C_{_{1}}\cdot V_{_{1}}\$ and \$q_{_{2}}=C_{_{2}}\cdot V_{_{2}}\$ and also that \$q_{_{final}}=C_{_{final}}\cdot V_{_{final}}\$; and since we also know that when two capacitors are placed in parallel that the total system capacitance is the sum of the two original system capacitances, or \$C_{_{final}}=C_{_{1}}+ C_{_{2}}\$; it then follows directly that:

$$q_{_{final}}=q_{_1}+q_{_2}=C_{_{1}}\cdot V_{_{1}}+C_{_{2}}\cdot V_{_{2}}=\left(C_{_{1}}+ C_{_{2}}\right)\cdot V_{_{final}}$$

It's easy then to re-arrange this so that:

$$V_{_{final}}=\frac{C_{_{1}}\cdot V_{_{1}}+C_{_{2}}\cdot V_{_{2}}}{C_{_{1}}+ C_{_{2}}}$$

Note that the energy of the final system is not the sum of the energy of the original two systems:

$$\frac12 C_{_{final}}\,V_{_{final}}^{^2}\ne \frac12 C_{_{1}}\,V_{_{1}}^{^2}+\frac12 C_{_{2}}\,V_{_{2}}^{^2}$$

In fact, the total energy will be smaller by this factor:

$$\Delta W = W_{_{final}}-\left(W_{_1}+W_{_2}\right)=-\frac12 \left(V_{_1}-V_{_2}\right)^2\frac{C_{_{1}}\cdot C_{_{2}}}{C_{_{1}}+C_{_{2}}}$$

This is true regardless of how that difference is expended. You may want to read this hyperphysics page on where the energy loss goes when charging up a capacitor. This lost energy can be through realistic resistance dissipation as heat. But as you assume a smaller and smaller connection resistance, so that the current is very much larger and the transfer time much shorter, then an increasing amount of the lost energy goes into electromagnetic radiation.

jonk
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The scenario you describe is nonsensical and cannot be analyzed using normal circuit analysis techniques.

Suppose you have two ideal capacitors with two different voltages across them. The voltage across a capacitor cannot change instantaneously because an infinite current would be required. So if you connect the two capacitors together with ideal wires then at that instant the two capacitors will still have their original, different voltages. But they are connected in parallel, so by definition they must have the same voltage across them. Therefore, the circuit presents a contradiction and is not consistent with normal circuit design rules.

The same situation occurs if you took two ideal voltage sources and connected them in parallel.

Of course, in the real world the wires are not ideal and they have some finite resistance. This resistance limits the maximum current when the capacitors are connected. Furthermore, since there is a "resistor" element between them the capacitors are no longer connected in parallel.

Elliot Alderson
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  • The little details of nature that are often negligible come into heavy play in the edge cases to eliminate paradoxes in the real world. That's why ideal things don't exist and ideal models have to be taken with a grain of salt. – DKNguyen Aug 17 '19 at 20:13
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When you connect \$C_1\$ and \$C_2\$ in parallel they will share the same voltage \$V'\$.

A portion of the initial charge from \$C_1\$ (\$Q_{1_0}\$) will flow to \$C_2\$, so that in the end we'll have:

$$ Q_1'+Q_2' = Q_{1_0} \text{ (conservation of charge)} $$

$$ \frac{Q_1'}{C_1}=\frac{Q_2'}{C_2} = V' \text{ (same voltage for two components in parallel)} $$

So, yes, the final voltage will be somewhere in the middle between the initial voltage on the pre-charged capacitor and the voltage on the discharge capacitor (zero Volts in this case). The exact value will depend on the ratio between the two capacitances.

Yes, you're right when you say they are both in series and parallel. However, the fact they are in series makes the currents (\$\frac{dQ}{dt}\$) the same absolute value, not the charges themselves. Actually, if you assume the current is positive as it leaves the capacitor, the currents will be the same but with opposite polarity (\$I_2=-I_1\$) as the charge leaves one capacitor and enters the other. Also, when the charge transfer is finished, both currents will be zero.

joribama
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