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Can someone explain the math behind the following relationships between the equalities in a capacitor's reactance?

$$X_C = \frac{1}{2\pi f C} = \frac{-j}{\omega C} = \frac{1}{j \omega C}$$

For instance the second member, $$\frac{1}{2\pi f C}$$ It has no complex component, how can be equal to the others when $$2\pi f = \omega$$ The complex component is missing right?

And the two last ones, they are alike, but the complex component is tossed around a bit. I don't understand what is happening there either.

The individual expressions/members is fine, but according to my textbook they are supposed to be equal but, I don't see how. $$$$ EDIT: The two last equalities are from my textbook, the second one is from the Electronic Tutorials webpage. All under the X_C symbol. Link: http://www.electronics-tutorials.ws/filter/filter_1.html

Alright. So what I've derived from the comments, is that different definitions of $$X_{LC}\ \ and\ \ Z_{LC}$$ are used in different places. I was under the impression that X, in a mandatory way, always had the complex unit inside it and was always a pure imaginary quantity. Not true though. The examples in my textbook now makes more sense now because it would seem it uses Z as the imaginary part and X as the mixed im. and real part.

Dave Tweed
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E. l4d3
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    The question about equality of the formulas with j in denominator and -j in numerator can easily be explained by j²=-1. Just expand the fractions by j and you get the other formula. – Curd May 10 '17 at 07:36
  • Convcerning the other equality, you have to tell something about the context. Where did you find *exactly* this equality? In two different contexts (books)? Then you have to look closely how \$X_C\$ is defined in each of them. – Curd May 10 '17 at 07:39
  • And the missing j in the first formula can also easily be explained - it should be there. All the j shows is that the scalar value you obtain from the formula is on the imaginary axis. That's because it's a reactance - not a resistance. – Tim M May 10 '17 at 07:41
  • Adding the info to OP – E. l4d3 May 10 '17 at 07:46
  • @Tim Mottram: Not neccesarrily; it depends on the definition. E.g. [here](https://en.wikipedia.org/wiki/Electrical_reactance) \$X_C\$ is defined as a real quantity (that needs to be multiplied by \$j\$ to form the imaginary part of a complex quantity called impedance). That's why I was asking for the context of **both** formulas. – Curd May 10 '17 at 07:49
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    I thought it was mandatory to write R as real resistance, Z as impedance, and X as reactance? – E. l4d3 May 10 '17 at 07:51
  • @user80556: ok, but how do you define X? Z = R + X (where X is an imaginary quantity) or Z = R + jX (where X is a real quantity)? It seems to me that you are mixing both variants of definitons of X. – Curd May 10 '17 at 07:57
  • @user80556: you provided a link where you found the first formula. But where did you find \$X_C= \frac{-j}{\omega C}\$? – Curd May 10 '17 at 08:04
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    \$X_L = \omega L\$; \$X_C = \frac{1}{\omega C}\$. In complex form: \$jX_L\$ and \$-jX_C\$, or \$0+jX_L\$ and \$0-jX_C\$ if you want to be pedantic. Also \$R=R+j0\$. – Chu May 10 '17 at 11:53
  • Of course; but that's not what OP has given. That's why I'm (still) asking: where did he get \$X_C=\frac{-j}{\omega C}\$ from? Nobody can explain "relationship between equalities" if those "equalities" are wrong (or at least are based on contradicting definitions). – Curd May 10 '17 at 14:01
  • @Curd, is the above intended for me? I was just responding to the OP, which clearly contains errors. – Chu May 10 '17 at 14:29
  • @Chu: it was intended for you (and indirectly also for OP). If your comment is intended for OP then I understand and fully agree :-) – Curd May 10 '17 at 15:19

1 Answers1

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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:

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

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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