Current-voltage relation for series association of variable capacitors

AI Thread Summary
The discussion focuses on deriving the differential equation relating current and voltage for a series connection of two variable capacitors, C_{1}(t) and C_{2}(t). The user successfully establishes an equation for the current in terms of the capacitances and voltage but struggles to derive this without prior knowledge of the equivalent series capacitance formula. Suggestions include using phasors to analyze AC voltage and current, as well as clarifying whether the system operates under DC or AC conditions. The complexity of the system and the variations in capacitance are acknowledged as challenges in the analysis. The conversation emphasizes the need for a clear method to connect the derived equations effectively.
Barloud
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Hello,

I have an issue with the problem below.

I have a series connection of two variable capacitances C_{1}(t) and C_{2}(t). 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:

C_{s}=\frac{C_{1}C_{2}}{C_{1}+C_{2}}​

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

i=\dot{(C_{s}V)}=\dot{C_{s}}V+C_{s}\dot{V}​

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

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​

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 C_{1} and C_{2} is equal to C_{1}C_{2}/(C_{1}+C_{2}). To do that, I first express the charge on the top electrode of C_{2} as:

Q=C_{2}V_{2}​

and the charge on the bottom electrode of C_{1} as:

-Q=-C_{1}V_{1}​

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

And I get stuck here. I am unable to get back to Eq.1 from Eq.2, even by introducing V=V_{1}+V_{2}. Any ideas of how I can do that?
 
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Barloud said:
even by introducing V=V_{1}+V_{2}. 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.
 
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