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.

  • 1
    \$\begingroup\$ 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. \$\endgroup\$
    – Curd
    May 10, 2017 at 7:36
  • \$\begingroup\$ 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. \$\endgroup\$
    – Curd
    May 10, 2017 at 7:39
  • \$\begingroup\$ 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. \$\endgroup\$ May 10, 2017 at 7:41
  • 1
    \$\begingroup\$ I thought it was mandatory to write R as real resistance, Z as impedance, and X as reactance? \$\endgroup\$
    – E. l4d3
    May 10, 2017 at 7:51
  • 1
    \$\begingroup\$ \$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\$. \$\endgroup\$
    – Chu
    May 10, 2017 at 11:53

1 Answer 1


For a capacitor, there is the relation:


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:


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


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

  • \$\begingroup\$ Still doesn't explain or define \$X_C\$. \$\endgroup\$
    – Curd
    May 10, 2017 at 8:05
  • \$\begingroup\$ @Curd Now, it does! \$\endgroup\$ May 10, 2017 at 8:15
  • \$\begingroup\$ Missing C in the denominator of the middle expression in (4). \$\endgroup\$
    – jonk
    May 10, 2017 at 19:27
  • \$\begingroup\$ @jonk edited, so it does now ;) . \$\endgroup\$ Mar 1, 2018 at 21:24
  • 1
    \$\begingroup\$ @Jan Well, I already +1'd your answer. So all I can say is "thanks." \$\endgroup\$
    – jonk
    Mar 1, 2018 at 21:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.