- #1
Barloud
- 16
- 0
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.
The capacitance of the series connection of the two capacitors is:
The charges on the electrodes of C_{s} are ±C_{s}V and the current i is then
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:
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:
and the charge on the bottom electrode of C_{1} as:
For the current, I get:
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?
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.
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?
