Current-voltage relation for series association of variable capacitors

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Barloud
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[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.
Circuit schematic of two variable capacitors C1(t) and C2(t) connected in series, with voltages V1(t) and V2(t) across each, total voltage V(t), and current i(t) entering and leaving the series pair.


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?
 
Barloud said:
even by introducing [itex]V=V_{1}+V_{2}[/itex]. Any ideas of how I can do that?

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.
 
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.