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Capacitance Calculation under the Electric Currents interface of AC/DC Module
Posted Dec 14, 2011, 11:06 p.m. EST 3 Replies
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Hi Everyone,
I am a beginner of Comsol. I need to figure out how to calculate the capacitance and resistance of a Polymer Block at first.
I can easily get the capacitance from the Electrostatic interface under AC/DC Module same as the theoretical calculation C[F] = epsilon[F/m]*epsilonr[1]*A[m^2]/d[m]. My process is as follow:
a. Build a block.
b. Set the materials properties values, the epsilon and epsilonr.
c. Under physics, apply terminal V on the top face and set the bottom as ground.
d. Mesh.
e. Study. And do the global calculation to get the capacitance value.
But, I also need to know the resistance of the block, which I cannot get any useful information from this situation. For the resistance, I want to use the equation, R=V[v] /I [A]. The voltage is already set, but I need know the current, which I cannot get here. My purpose is that I want to compare it with the theoretical calculation value R=L[m]/(A[m^2]*Sigma[s/m].
Therefore, I change my physics to Electric currents interface, where most of parameters setup in Electrostatic are same. After study, I can get the current from the global evaluation. After divided by voltage, I get the resistance same as theoretical calculation value. But there is not capacitance evaluation. I really want to calculate both of them together at the same time, under the same physics interface.
I know that maybe I can use energy method to calculate the capacitance indirectly, W[j]=0.5*C*V^2. But I don't know how to get the energy. I just see that there is Total Power Dissipation Density in the global evaluation, I don't understand it either. Now, I am stuck. Please Help!
Is there anyone that can help me? I am really appreciate for it!!!
From:
Lost Beginner.
I am a beginner of Comsol. I need to figure out how to calculate the capacitance and resistance of a Polymer Block at first.
I can easily get the capacitance from the Electrostatic interface under AC/DC Module same as the theoretical calculation C[F] = epsilon[F/m]*epsilonr[1]*A[m^2]/d[m]. My process is as follow:
a. Build a block.
b. Set the materials properties values, the epsilon and epsilonr.
c. Under physics, apply terminal V on the top face and set the bottom as ground.
d. Mesh.
e. Study. And do the global calculation to get the capacitance value.
But, I also need to know the resistance of the block, which I cannot get any useful information from this situation. For the resistance, I want to use the equation, R=V[v] /I [A]. The voltage is already set, but I need know the current, which I cannot get here. My purpose is that I want to compare it with the theoretical calculation value R=L[m]/(A[m^2]*Sigma[s/m].
Therefore, I change my physics to Electric currents interface, where most of parameters setup in Electrostatic are same. After study, I can get the current from the global evaluation. After divided by voltage, I get the resistance same as theoretical calculation value. But there is not capacitance evaluation. I really want to calculate both of them together at the same time, under the same physics interface.
I know that maybe I can use energy method to calculate the capacitance indirectly, W[j]=0.5*C*V^2. But I don't know how to get the energy. I just see that there is Total Power Dissipation Density in the global evaluation, I don't understand it either. Now, I am stuck. Please Help!
Is there anyone that can help me? I am really appreciate for it!!!
From:
Lost Beginner.
3 Replies Last Post Dec 15, 2011, 11:02 a.m. EST
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Posted:
1 decade ago
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Posted:
1 decade ago
Hi, John,
Thanks for your advice. I really appreciate it.
I didn't use both of the interfaces at the same time before, because I thought they might affect each other. I am a newer, and not clear whether there would be a coupling effect between them.
As you suggested, I just put them together and got the values I want. Does this mean that the separated calculation method and the combination calculation way are same ? Is the calculation of combination method solving the two physics interface situations one by one? Both are based on the same initial conditions?
Actually, this is the first step. Later, I need to calculate a changing capacitance and resistance under various frequency. So, I'd better use the Electric Currents interface. I think I still have to find a direct or indirect way to calculate capacitance under the Electric Currents interface.
Yes, the dimension of the block is 20*5*0.5 [mm]. It is a very large and thin plate.
Thanks.
Thanks for your advice. I really appreciate it.
I didn't use both of the interfaces at the same time before, because I thought they might affect each other. I am a newer, and not clear whether there would be a coupling effect between them.
As you suggested, I just put them together and got the values I want. Does this mean that the separated calculation method and the combination calculation way are same ? Is the calculation of combination method solving the two physics interface situations one by one? Both are based on the same initial conditions?
Actually, this is the first step. Later, I need to calculate a changing capacitance and resistance under various frequency. So, I'd better use the Electric Currents interface. I think I still have to find a direct or indirect way to calculate capacitance under the Electric Currents interface.
Yes, the dimension of the block is 20*5*0.5 [mm]. It is a very large and thin plate.
Thanks.
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Posted:
1 decade ago
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