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Current-voltage relation for series association of variable capacitors

  1. Nov 1, 2013 #1
    [itex][/itex]Hello,

    I have an issue with the problem below.

    I have a series connection of two variable capacitances [itex]C_{1}(t)[/itex] and [itex]C_{2}(t)[/itex]. I want to establish the differential equation between the current i and voltage V on the ports of the series connection.
    Cs1.jpg

    The capacitance of the series connection of the two capacitors is:

    [itex]C_{s}=\frac{C_{1}C_{2}}{C_{1}+C_{2}}[/itex]​

    The charges on the electrodes of [itex]C_{s}[/itex] are ±[itex]C_{s}V[/itex] and the current i is then

    [itex]i=\dot{(C_{s}V)}=\dot{C_{s}}V+C_{s}\dot{V}[/itex]​

    I use the dot superscript for the time derivative. Using the expression of [itex]C_{s}[/itex] given above, I get the differential equation relating the voltage and current, which is what I am looking for:

    [itex]i=\frac{C_{1}^{2} \dot{C_{2}}+C_{2}^{2} \dot{C_{1}}}{(C_{1}+C_{2})^{2}}V+\frac{C_{1}C_{2}}{C_{1}+C_{2}}\dot{V} \; \; \; \; \; \; \; \;Eq.1[/itex]​

    However, because the problem I describe is just a step in a more complex system that I am studying, I need to understand the method to obtain Eq.1 without knowing in advance that the equivalent series capacitance of [itex]C_{1}[/itex] and [itex]C_{2}[/itex] is equal to [itex]C_{1}C_{2}/(C_{1}+C_{2})[/itex]. To do that, I first express the charge on the top electrode of [itex]C_{2}[/itex] as:

    [itex]Q=C_{2}V_{2}[/itex]​

    and the charge on the bottom electrode of [itex]C_{1}[/itex] as:

    [itex]-Q=-C_{1}V_{1}[/itex]​

    For the current, I get:
    [itex]i=C_{2}\dot{V_{2}}+V_{2}\dot{C_{2}}=-C_{1}\dot{V_{1}}-V_{1}\dot{C_{1}} \; \; \; \; \; \; \; \;Eq.2[/itex]​

    And I get stuck here. I am unable to get back to Eq.1 from Eq.2, even by introducing [itex]V=V_{1}+V_{2}[/itex]. Any ideas of how I can do that?
     
  2. jcsd
  3. Nov 2, 2013 #2

    UltrafastPED

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    Science Advisor
    Gold Member

    The voltage doesn't change when passing through a capacitor; only the phase changes. Perhaps you should use phasors: http://en.wikipedia.org/wiki/Phasor

    So the question is about your voltage/current: is it DC, AC, or something else? If it is DC, then you only have voltage=V.
     
  4. Nov 2, 2013 #3
    The voltage and current are AC, as are the variations of the capacitances.
    Unfortunately, a method based on complex analysis does not work well here.
     
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