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If you have some capacitors in series then their EC (= equivalent capacitance) means the capacitance of the capacitor which will store the same charge as any individual capacitor in the series when the same voltage which is applied b/w first and last plate in the series is applied to it. If the capacitors were in parallel then EC is defined as the capacitance of the capacitor to which for the same voltage as applied to the parallel capacitors then it stores the same total charge.

  1. But what does EC mean for a general circuit ie circuit consisting of capacitors in series and in parallel?

You may say that it is the capacitance got by simply the circuit by replacing all the capacitors in series and parallel with their equivalent and doing the same until we reduce the circuit either to capacitors in series or parallel, then just use the above definition. Well well well.

  1. What if we couldn't reduce the circuit to capacitors in series or parallel? For example, what does EC physically mean for the circuit in this question? Note that I am not asking how to find EC. I am asking for its physical meaning as I gave for capacitors only in series or parallel.

  2. A circuit can be reduced to a simple circuit in many ways. How do you prove that for all these ways the EC calculated will be the same?

I have the same question for equivalent resistance.

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    $\begingroup$ It means that a capacitance meter connected to a 'black box' with your capacitors inside will measure a single value of capacitance, and many arrangements of various capacitors will yield that same value. You can always reduce an arrangement of just capacitors to one equivalent capacitor - measuring across your two-terminal black box will yield one value of capacitance. $\endgroup$
    – Jon Custer
    Commented Apr 14, 2022 at 13:19
  • $\begingroup$ @JonCuster Is there any mathematical definition like, as I said in the question, I gave for series and parallel connections? $\endgroup$
    – Osmium
    Commented Apr 14, 2022 at 13:32
  • $\begingroup$ Are you asking if any network of capacitors can be reduced to an equivalent capacitor, then the answer is yes. In the question you reference, given two connection points that circuit can be uniquely reduced using capacitors in series and parallel. As for the uniqueness of the solution, again, for linear elements there must be a unique lumped response (but given a lumped response there are infinitely many ways to produce it). $\endgroup$
    – Jon Custer
    Commented Apr 14, 2022 at 14:05

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Stick the capacitor network you want to measure inside a black box. Connect the box in parallel with a DC voltage source with voltage $\Delta V$. Disconnect the box from the source and connect it across a resistor in series with an ammeter. Record the current $I$ through the ammeter as a function of time. Integrate the current to get the total charge $Q = \int dt\, I(t)$ discharged from the capacitor. The equivalent capacitance is $C_{\text{eq}} = |Q / \Delta V|$.

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  • $\begingroup$ Thanks for your answer. Do you have a reference? I have searched through many books but didn't find anything related to my question. $\endgroup$
    – Osmium
    Commented Apr 15, 2022 at 1:57
  • $\begingroup$ @Osmium I don't have a specific reference for this, but all I've really done is restate the definition of capacitance you can find in any undergraduate textbook (e.g., Halliday and Resnick) in a slightly more operational way. $\endgroup$
    – d_b
    Commented Apr 15, 2022 at 2:13
  • $\begingroup$ So how do you use this definition to solve the question linked in the post? $\endgroup$
    – Osmium
    Commented Apr 15, 2022 at 2:37
  • $\begingroup$ @Osmium You can use conservation of charge and the loop rule to find the total charge on the bridge, the same way you'd find the current in the resistive version of the circuit using the junction and loop rules. (You can also use fancy things like Y-$\Delta$ transforms that I am less familiar with.) I may try to work it out if I have time, but if you know how to do the resistive version with Kirchhoff's rules, it's just a matter of translating the same work over to capacitor language. $\endgroup$
    – d_b
    Commented Apr 15, 2022 at 2:48
  • $\begingroup$ Can you do the same for equivalent resistance? Thanks. 🙏 $\endgroup$
    – Osmium
    Commented Apr 15, 2022 at 9:17
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It means if you replace all capacitors with the equivalent capacitor the circuit's operation won't change.

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  • $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Commented Apr 14, 2022 at 13:17
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    $\begingroup$ You should define "operation" precisely. $\endgroup$
    – Osmium
    Commented Apr 14, 2022 at 13:23

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