Current-voltage relation for series association of variable capacitors

In summary, the conversation is about establishing a differential equation between the current and voltage in a series connection of two variable capacitances. The equation is obtained by expressing the charges on the electrodes and the current in terms of the time derivatives of the capacitances and voltages. The question is about finding a method to obtain the equation without prior knowledge of the equivalent series capacitance. The suggestion is to use phasors, but a method based on complex analysis does not work well in this case due to the AC nature of the voltage and current.
  • #1
Barloud
17
0
[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?
 
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  • #2
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.
 
  • #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.
 

FAQ: Current-voltage relation for series association of variable capacitors

What is the concept of series association of variable capacitors?

The series association of variable capacitors refers to the connection of multiple capacitors in a circuit in a series arrangement, where the positive plate of one capacitor is connected to the negative plate of the next capacitor and so on. This allows for the overall capacitance of the circuit to be varied by adjusting the individual capacitance values of each capacitor.

How does the current-voltage relation change in a series association of variable capacitors?

In a series association of variable capacitors, the current-voltage relation follows the same principles as any other series circuit. The total current in the circuit is the same at all points, and the voltage drop across each capacitor is dependent on its individual capacitance value. However, the total capacitance of the circuit is determined by the inverse sum of the individual capacitance values.

What factors affect the current-voltage relation in a series association of variable capacitors?

The main factors that affect the current-voltage relation in a series association of variable capacitors are the individual capacitance values and the external voltage being applied to the circuit. The current and voltage can also be affected by the properties of the dielectric material used in the capacitors, as well as the frequency of the applied voltage.

How can the current-voltage relation be manipulated in a series association of variable capacitors?

The current-voltage relation in a series association of variable capacitors can be manipulated by adjusting the individual capacitance values of each capacitor. By increasing or decreasing the capacitance of one or more capacitors, the overall capacitance of the circuit can be changed, which in turn affects the current and voltage in the circuit.

What are some practical applications of series association of variable capacitors?

The series association of variable capacitors has various practical applications in electronic circuits. It is commonly used in tuning circuits, such as in radio receivers and filters, to adjust the frequency response. It is also used in electronic circuits that require precise control of voltage or current, such as in power supplies and amplifiers.

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