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Disclaimer: I'm an engineer, not a mathematician

Below is the derivation of the voltage across one capacitor with two capacitors in series, so C1 and C2 are greater than zero. $\omega$ is the frequency.

$ V_{C2} = \dfrac{Z_{C2}}{Z_{C1} + Z_{C2}}V1 = \dfrac{\dfrac{1}{j\omega C2}}{\dfrac{1}{j\omega C1} + \dfrac{1}{j\omega C2}} V1 = \dfrac{\dfrac{C1}{j\omega C1 C2}}{\dfrac{C2}{j\omega C1 C2} + \dfrac{C1}{j\omega C1 C2}} V1 = \dfrac{C1}{C1 + C2} V1 $

I understand that the last step is not allowed if $\omega$ = 0, and that in that case I should take the limit:

$ V_{C2} = \displaystyle \lim_{\omega \to 0} \dfrac{\dfrac{C1}{j\omega C1 C2}}{\dfrac{C2}{j\omega C1 C2} + \dfrac{C1}{j\omega C1 C2}} V1 = \dfrac{C1}{C1 + C2} V1 $

That makes sense to me, but I wonder if I didn't skip a step for the limit.

I remember when I was a student I found most steps in a proof "trivial", much to the professor's vexation :-)

So, are there additional steps I shouldn't have skipped?

I'm not a mathematician, so please type slowly :-)

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Makes sense to me -- don't see how it can be any clearer. –  Peter Grill Jun 11 '12 at 6:45

2 Answers 2

Edit: This answer has been revised to answer why taking the limit is justified.

The short answer is that essentially all the parameters of an electric system are continuous with respect to frequency (even smooth, i.e. infinitely differentiable). Even circuits made up of op amps with high slew rates have output frequencies which are continuous functions of the input voltages, the derivative is just very high. In particular, the voltage across a single capacitor in a circuit with two capacitors in series is a continuous function of the frequency of the input. It is a property of continuous functions that any continuous function $f$ satisfies $$\lim\limits_{x\to a} f(x) = f(a)$$ for all $a$. In this particular case, you are making use of the fact that $$\lim\limits_{\omega\to 0} V2(\omega)=V2(0)$$ in order to calculate $V2(\omega)$.

As for why the parameters of an electrical circuit are continuous with respect to frequency, you'll have to consult a physicist.

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Thanks for the quick reply. But my problem is that in the application frequency is zero, I apply the limit for that case, as I mention in my question. –  stevenvh Jun 11 '12 at 6:42
    
@stevenvh Ah, you wan't to know why taking the limit is justified, rather than why the limit is that value. I misunderstood. Let me think on that one. –  Alex Becker Jun 11 '12 at 6:44
    
@stevenvh Does that answer your question? –  Alex Becker Jun 11 '12 at 6:52
    
Yes, thanks. +1. It's too early for an accept, there might come other answers. –  stevenvh Jun 11 '12 at 6:59
    
@stevenvh No problem. For the record, why these functions are continuous bugged me all through electronics class. –  Alex Becker Jun 11 '12 at 7:00

Whether taking the limit is justified or not is not really a mathematical question. Strictly speaking the equation you use is invalid for $\omega=0$, so no value is defined there, and it makes no sense to ask whether the limit as $\omega\to0$, which exists in this case, corresponds to the value at $\omega=0$. One could define the value at $\omega=0$ by the limit, but whether that definition corresponds to the physics of the situation has to be judged separately.

Normally one would expect that the behaviour of the system depends continuously on the frequency, and therefore that the solution for direct current is given by the limit of the solution for alternating current as the frequency goes to $0$. However, it is conceivable that the solution for alternating current only describes the stable sitaution reached after a transient period of duration proportional to the inverse of the frequency, and in that case the stable sutation would never be reached for the case of direct current; this might lead to a discontinuity of the solution at $\omega=0$. Whether this is the case here I cannot say.

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