yeah, you're right. I know that it has no usage. But I'm not the one who write the rules, I'm just expected to find an answer for this question! BTW, do you have any hint about it?You are describing a lossy capacitor, which is no use to man nor beast. Why would you try to find out its value?
Must admit I never came across a question like this at college, and it's been decades since I had to work out capacitive value from plate area etc.... But I'm not the one who write the rules, I'm just expected to find an answer for this question! BTW, do you have any hint about it?
It's still a series circuit of complex impedances, that can be calculated. So what's your particular problem with the setup?but now that each material has conductivity, too, we cannot treat them as series capacitors.
...the material in the right side (orange one) has a really good conductivity (but not as good as the cap plate), so what are we supposed to do?
you say that it's made of two series capacitors that each one has a parallel resistor?It's still a series circuit of complex impedances, that can be calculated. So what's your particular problem with the setup?
Tnx for the links. I read them but I think those formulas don't work here. Since materials used between cap plates have conductivity too, and it's not been consideed in those formulas, we cannot find the model by using those equations.Take a look at equations close to Fig.10 on followiong page, witch covers the issue you´re asking for :
Calculation of electrostatic forces in presence of dielectrics
dear KerimF, this is exactly my question! If I solve this part of question, solving rest of it will be so easy!So how could we model a component made by two plates covering a layer of orange material?
Yes, i think so. And capacitance and resistance of both "blocks" can be calculated independently by applying simple relations.you say that it's made of two series capacitors that each one has a parallel resistor?
I'll be so grateful if you could give me any ref about it (if you have).Yes, i think so. And capacitance and resistance of both "blocks" can be calculated independently by applying simple relations.
C = ε0*εr*F/d
R = d/(F*σ)
I think we cannot use those formulas for finding capacitance and resistance. I read the link you posted and I also took a look at these two pages:The relations for R and C are directly referring to the defintion of σ and ε, and can be found in many text books.
I don't get it! why there is not frequency dependency? As I wrote in the former post, the conductivity is a frequency dependent value (σ(ω)).But you explicitely stated resistivity and permittivity parameters, which implicitely involves that there's no further frequency dependancy. In other words, there's exactly one resistor and one parallel capacitor in the equivalent circuit.
what does f stand for? frequency?f -> 0 and f -> ∞
It's the first time, that you mention frequency dependancy in this thread! In this case, it would be good, if you can be more specific about what you want to achieve. If other threads of yours are necessary to understand your intention, please post a link.I don't get it! why there is not frequency dependency? As I wrote in the former post, the conductivity is a frequency dependent value (σ(ω)).
In a parallel RC circuit, capacitance won't depend on conductivity, I think, Where did you get it would?It doesn't make any sense for me that capacitance value is a function of conductivity.
I did what he said, and I found the value of the capacitance (please take a look at the post#14). Do you think that I made a mistake in this way?Write the Capacitance (C) using the complex dielectric function ϵ=ϵ1−iϵ2, and then calculate the capacitive impedance Zc=1/(iωC). You'll see that the impedance turns out to be not purely reactive - there's a resisitive term as well. It is this resistance that dissipates power. The ratio of the resistive component to the reactive component (or real to imaginary terms) is called the loss tangent.
Which capacitance? Generally, it's a complex impedance. You can represent it by different equivalent circuits. If you choose a series (ESR + ideal cap) circuit, than the originally parallel circuit parameters have to be transformed. But you didn't yet tell a reason, why one of both representations should be preferred. I suggested the parallel one, because it corresponds to your original problem formulation (σ and ε given).I did what he said, and I found the value of the capacitance
I think if you want to use parallel circuit, you should only use Xc (as opposed to the C in series equivalent circuit) in parallel with resistor.
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