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I know that capacitors block DC and low-frequency AC, and their reactance decreases as the frequency of the AC circuit increases. As such they are useful as high-pass filters and so forth.

I had learned this a while back, but I didn't have the mathematical understanding to explore deeper. Since then I have refreshed my memory on algebra and studied a but of calculus. I read that the current through a capacitor is equal to the rate of change of the voltage across it, or...

$$i=C\frac{dv}{dt}$$

I figured I'd try to make sense of the capacitors behavior by considering what would happen if I applied a simple sine wave:

$$v=Asin(\omega t)$$

The derivative of which is...

$$\frac{dv}{dt}=A\omega cos(\omega t)$$

This gives the current

$$i=C\frac{dv}{dt}=CA\omega cos(\omega t)$$

This shows that the current will be 90° out of phase with the voltage, and that higher frequencies will give larger magnitude currents, which makes sense. Firstly, have I reasoned the above steps properly?

Secondly, I was hoping to be able to derive the capacitive reactance formula $$Xc=\frac{1}{\omega C}$$

I got this far:

$$Xc=\frac{v}{i}=\frac{sin(\omega t)}{C\omega cos(\omega t)}$$

but I don't know how to get rid of the sin and cos. Is the only way to use imaginary numbers (which I have studied once upon a time) or LaPlace (which is beyond my current ability level) as mentioned below, or is there a simpler solution? I reviewed my trig ids but other than replacing sin/cos with tan, I got nowhere.

How to Derive Capacitive- and Inductive Reactance Formula

nuggethead
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    good start, you're nearly there. Both sin and cos have max magnitude 1. As reactance deals with magnitude, not phase, you have already written down all you need in your question. You have an expression for magnitude voltage, and magnitude current, involving the terms C and w. Remember w is per radian, so for Hz you need a 2pi in there as well. – Neil_UK Nov 27 '22 at 13:43
  • I took it one step further by substituting in my expressions for v and I. I'm stuck now with a sin and cos that i don't know hiw to eliminate though. Help? – nuggethead Nov 27 '22 at 14:32
  • You can write sine and cos in terms of complex exponentials, or you can recognize that sin/cos = tangent. – Designalog Nov 27 '22 at 14:34
  • I recognized that sin/cos is tangent, but now I have that tangent hanging out there. @ErnestoG how do I get rid of that tan. – nuggethead Nov 27 '22 at 15:03
  • @nuggethead, you want the magnitude of the reactance (sorry, I missed that earlier), so, you need get the RMS of the sine and cosine, which is 1/sqrt(2) or ~0.707. Obviously they cancel out and you're left with 1/(omega*C). – Designalog Nov 27 '22 at 15:50
  • @ErnestoG are you able to post that as an answer and show the math for how to get from my last step to the 1/omega*c – nuggethead Nov 27 '22 at 15:58
  • @nuggethead, will do. – Designalog Nov 27 '22 at 16:02
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    If you know the rule that if \$v(t)\leftrightarrow V(f)\$, then \$\frac{dv(t)}{dt}\leftrightarrow j\omega V(f)\$, you're done after your first equation. – The Photon Nov 27 '22 at 16:33
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    @nuggethead for the magnitude of the impedance, you need the magnitude of each term. So replace both sine, and cosine, by 1! – Neil_UK Nov 27 '22 at 17:25

3 Answers3

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You have done everything correctly. Now apply the definition of impedance which is the ratio of voltage to current in a device. Use the equations you have derived.

Barry
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I got this far:

$$Xc=\frac{v}{i}=\frac{sin(\omega t)}{C\omega cos(\omega t)}$$

Almost right. The reactance is the ratio of magnitudes, not the complete functions, as shown in the link provided in the question.

RussellH
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From your last step:

$$ X_c = \frac{v}{i} = \frac{\sin(\omega t)}{C\omega \cos(\omega t)} $$

We would like to obtain the magnitude of the reactance, i.e.:

$$ |X_c| = \frac{1}{\omega C} $$

To get there, since we do not need information about the phase, we can obtain the RMS (root mean square) of this fraction, which allows to represent a periodic function with a single number. The RMS value of a constant is the same constant, obviously. The root mean square is given by:

$$ \lim_{T\to \infty} \sqrt{\frac{1}{2T}\int_{-T}^{T} [f(t)]^2\;dt}$$

I'll excuse myself from not proving this for sines and cosines since they're pretty standard, you can find the proof elsewhere on the internet. However, I'll mention the result here for completeness:

$$ RMS(A\sin(\omega t)) = \frac{A}{\sqrt{2}} $$.

The same holds for a cosine.

Using this it is trivial to prove that $$ |X_c| = \frac{1}{\omega C} $$

EDIT:

An alternative, which is the usual path of derivation taken in textbooks, is to represent your periodic function (sine or cosine) as the real part of a complex exponential:

$$ A\sin(\omega t) = \Im\{A\exp{j\omega t}\} $$

which simplifies the derivation a great deal. Bode himself recognized this in his book.

Designalog
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