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I'm trying to figure out the impedance of a capacitor. My textbook tells me the answer is $\frac{-i}{\omega C}$ and plugging that into the equation does work but I wanted to come up with that answer myself. So I wrote out the equation with what I know:

$$-V_0\omega C\sin\omega t = Re\left( \frac{V_0(\cos\omega t + i\sin\omega t)}{x} \right)$$

This is where I get stuck. I don't know how to isolate $x$ given that it is inside the $Re()$ function. Trying to get somewhere, I tried this:

$$x = \frac{V_0(\cos\omega t + i\sin\omega t)}{-V_0\omega C\sin\omega t} = \frac{\cos\omega t}{-\omega C\sin\omega t} - \frac{i}{\omega C}$$

Seeing $-\frac{i}{\omega C}$ makes me feel like I'm on the right track. Now I just need to figure out how to get rid of the first part of that answer. And I'm guessing that if I knew how to isolate $x$ from the first equation, that would do the trick. So how can I isolate $x$ when it is included in the $Re()$ function?

Aust
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    Is $x$ a real number? – José Carlos Santos Jan 11 '18 at 18:48
  • Does $\operatorname{Re}(y) = (y+\bar{y})/2$ not give you what you want? – Dfrtbx Jan 11 '18 at 18:50
  • Mathematically, if $a$ is a real number and $z$ is a complex number, then $a={\rm Re}(z/x)$ does not have a unique complex solution $x$. Perhaps your knowledge of what $x$ represents should tell you it's imaginary and then you can use the substitution $x=iy$ for some real number $y$? (Or is one of $\omega$ or $C$ also imaginary? I know zilch electrical engineering.) – anon Jan 11 '18 at 18:53
  • $V_0$, $\omega$, $C$, and $t$ are real numbers. I'm not too sure about $x$. I assume it's complex/imaginary because I know the answer is $-i/\omega C$. – Aust Jan 11 '18 at 19:10

2 Answers2

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Re() is a projective map; Re(a+bi) = a. Thus Re(z) = z-iIm(z). So given RHS = Re(z), we have that RHS+bi = z for some real b. Note that Re(z) = a does not yield a single value of z as a solution, but instead gives a vertical line in the complex plane. Each point on that line will give a different value for x.

$$-V_0\omega C\sin\omega t + bi = \frac{V_0(\cos\omega t + i\sin\omega t)}{x} $$

In terms of b, x will be:

$$\frac{V_0(\cos\omega t + i\sin\omega t)}{-V_0\omega C\sin\omega t + bi} $$

Assuming that $\omega$, $V_0$, and C are real numbers, they can be "absorbed" into b; b is an arbitrary real number, so dividing by a real number just gives another arbitrary real number. So the above can be rewritten as

$$\frac{(\cos\omega t + i\sin\omega t)}{-\omega C(\sin\omega t + bi)} $$

Factoring an i out of the numerator, we get

$$\frac{i(\sin\omega t-i\cos\omega t )}{-\omega C(\sin\omega t + ib)} $$

Again, this describes a solution set, not a particular x. But if you take $b = -\cos\omega t$, then you recover the given expression. Any motivation for that choice will have to come from further facts about the capacitance rather than mathematical properties.

Acccumulation
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For a capacitor, there is the relation:

$$\text{I}_\text{C}\left(t\right)=\text{C}\cdot\frac{\text{d}\text{V}_\text{C}\left(t\right)}{\text{d}t}\tag1$$

Considering the voltage signal to be:

$$\text{V}_\text{C}\left(t\right)=\text{V}_\text{p}\sin\left(\omega t\right)\tag2$$

It follows that:

$$\frac{\text{d}\text{V}_\text{C}\left(t\right)}{\text{d}t}=\omega\text{V}_\text{p}\cos\left(\omega t\right)\tag3$$

And thus:

$$\frac{\text{V}_\text{C}\left(t\right)}{\text{I}_\text{C}\left(t\right)}=\frac{\text{V}_\text{p}\sin\left(\omega t\right)}{\omega\text{V}_\text{p}\cos\left(\omega t\right)}=\frac{\sin\left(\omega t\right)}{\omega\text{C}\sin\left(\omega t+\frac{\pi}{2}\right)}\tag4$$

This says that the ratio of AC voltage amplitude to AC current amplitude across a capacitor is $\frac{1}{\omega\text{C}}$, and that the AC voltage lags the AC current across a capacitor by $90$ degrees (or the AC current leads the AC voltage across a capacitor by $90$ degrees).

This result is commonly expressed in polar form as:

$$\text{Z}_\text{c}=\frac{1}{\omega\text{C}}\cdot e^{-\frac{\pi}{2}\cdot\text{j}}\tag5$$

Or, by applying Euler's formula, as:

$$\text{Z}_\text{C}=-\text{j}\cdot\frac{1}{\omega\text{C}}=\frac{1}{\text{j}\omega\text{C}}\tag6$$

Now for $\text{X}_\text{C}$:

$$\text{X}_\text{C}=\left|-\text{j}\cdot\frac{1}{\omega\text{C}}\right|=\frac{1}{\omega\text{C}}\tag7$$

Where $\omega=2\pi\text{f}$

Jan Eerland
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