Please help me out a little bit, I got really stuck here with somethings and I just can't figure it out.

So, I need to calculate the complex impedance, the impedance and admittance of a capacitor with : C = 33 nF f = 100 Hz and knowing that X = -1/2*pifC.

So far I figured out that the pulsation is ω = 6,28 * 10^2 and that X = -48,2532 * 10^-3 [Ohm].

Right now I need to calculate the impedance Z. However I found more than one formulas for it and I don't know which one to use. So far I was thinking of going with Z = -jX or Zc = 1/jωC. However I don't really understand what to do with that "j". I know it shows that this is a complex number and that it represents a change of phase with pi/2. However I don't know what to do with it in calculus.

  • 1
    \$\begingroup\$ j is electrical-engineer-speak for i. \$\endgroup\$
    – Hearth
    Commented May 6, 2018 at 11:59
  • \$\begingroup\$ Try asking a specific question rather than sy you don't know what to do. This is a Q and A site. \$\endgroup\$
    – Andy aka
    Commented May 6, 2018 at 12:00
  • \$\begingroup\$ Sure. My bad. Basically I'd like to know which one of those formulas I should use and if it's correct at all to use any of them. \$\endgroup\$
    – Edward B
    Commented May 6, 2018 at 12:01
  • \$\begingroup\$ Both the formulae you mentioned for z are correct and you should keep j as it is for calculus. z = impedance, admittance = Y = 1 / z. And if you want the magnitude of either of them, just do square root(real part squared + imaginary part squared). Here there is only imaginary part, so it should be the magnitude as well. \$\endgroup\$
    – akshayk07
    Commented May 6, 2018 at 12:21
  • \$\begingroup\$ Since you're saying that both formulae are correct, that would mean -j = 1/j? That's the part I didn't get about that coefficient. \$\endgroup\$
    – Edward B
    Commented May 6, 2018 at 13:15

3 Answers 3


The most direct answer to your question is this, I think:

$$\begin{align*}\frac{1}{j}&=\frac{1}{j}\cdot\frac{j}{j}=\frac{j}{-1}=-j\\\therefore\\\frac{1}{j}&=-j \end{align*}$$

From that fact, it follows that:

$$\frac{1}{j\:\omega\: C}=-\frac{j}{\omega\: C}$$

They say the exact same thing. So either one can be used. Some people will prefer one over the other, when writing out their ideas. So you just have to get used to the fact that you may find either one of these in your reading experiences.

There is another area of confusion -- no, perhaps its better to say disconcerting, because it's the introduction of something powerful and also kind of beautiful when compared to something much more mundane.

Most anyone can get the idea about plotting complex numbers on the complex plane. You just turn the old \$y\$-axis into a new \$i\$-axis (or \$j\$-axis for electronics.) And then you plot complex numbers with ease. A young student can learn to do this, almost mechanically. It's just another way of plotting \$\left(a,b\right)\$ because the complex number isolates the two values in a brain-dead, newly substituted \$a+j\:b\$ way.

In one sense, a young student just thinks to themselves, "So what? This is just another way of writing the same silly thing. What makes complex numbers special? Why can't I just keep the two values as \$\left(a,b\right)\$ rather than being told about \$a+j\:b\$? Does it matter?"

Well, it does, actually. Partly, because of group theory (the concepts of the additive group, the multiplicative group, and by extension the idea of what raising to a power actually means.) Partly, because of something called Euler's: \$e^{\:j\theta}=\operatorname{cos}\left(\theta\right)+j\:\operatorname{sin}\left(\theta\right)\$.

In real number systems, you can see addition as a simple shift motion on the number line. In the complex number system (which uses a number plane instead of a number line), addition is also just a simple shift. Except that you can shift up and down, or right and left, or both of these where there is a combination of the two "shift motions." Simple enough, really. But I think you can also recognize that addition of complex numbers actually embodies two actions in a single operation.

Similarly, in real number systems, you can see multiplication as a simple stretching or shrinking (compression) of the number line. (Unlike addition which shifts.) In complex number systems, you might also guess that it will somehow embody two actions in a single operation. And you'd be right. But here, it embodies the stretching/shrinking as well as something new -- rotation. So multiplication of complex numbers can stretch/shrink, or rotate, or both stretch/shrink and at the same time also rotate. It takes a little getting used to, but this is the group theory perspective on the idea of multiplication taken into complex numbers.

Now, Euler's. You may recall that with the old \$\left(x,y\right)\$ plane there are a couple of ways you can represent a point. Either as the Cartesian \$\left(x,y\right)\$ method or in polar form with \$\left(r,\theta\right)\$. Either method has its advantages and disadvantages. It's kind of hard to describe a spiral, for example, using some \$y=f\left(x\right)\$. But it is trivial to do with \$r=f\left(\theta\right)\$. So there are reasons to want one, or the other, depending on the problem at hand.

It's also like that with complex numbers. There are times when a Cartesian \$c=a+j\:b\$ form is better and there are times when you'd rather somehow find a polar version of it. Here is where Euler's comes into play in a huge way. Suppose, rather than \$c=a+j\:b\$ we decided to apply Euler's with a special constant that everyone agrees to use, \$e\$. So rather than representing complex numbers as \$ c=a+j\:b \$ we decide to represent them as \$e^{a+j\:b}\$.

Since we've agreed upon \$e\$, there is a one-to-one mapping between these and therefore stating it one way or the other way can still be considered equivalent to each other. But we now gain something very important: \$e^{a+j\:b}=\left[e^a\right]\cdot\left[\operatorname{cos}\left(b\right)+j\:\operatorname{sin}\left(b\right)\right]\$. The second factor here always has the magnitude of 1, because \$1=\operatorname{sin}^2+\operatorname{cos}^2\$. So we can think of this second factor itself as a simple vector that always has a length of exactly 1, but where \$b\$ describes a rotation angle around the unit circle. The first factor is just a value that multiplies the length of this vector to stretch or shrink it to have a magnitude other than just 1.

So think for a second here. The \$a\$ now specifies the equivalent of \$r\$ and the \$b\$ now specifies the equivalent of \$\theta\$ in the real number polar coordinate system. The \$e^{a+j\:b}\$ form is the polar form of specifying something on the complex plane!!

So for complex numbers, we can either use the Cartesian form, \$a+j\:b\$, or else we can use the polar form, \$e^{a+j\:b}\$. These make up two different, but equivalent, ways of specifying any point on the complex plane! A Cartesian way and a polar way. Cool.

The only weirdness here is that there is a funny thing going on. In the old \$\left(r,\theta\right)\$ polar method, \$r\$ was the vector length. In this new complex polar form, \$e^a\$ specifies the vector length. So it's a little weird to get use to, at first, because the behavior of the vector length is an exponential behavior rather than a linear one we are more used to with real number polar coordinates. But... it's just something to get used to. It's not otherwise a fundamental problem to the idea.

In electronics, things have magnitude and things oscillate in time. Oscillation in time is the same as rotation on that unit circle (\$b\$) and voltage or current magnitudes is the same a vector length (\$e^a\$), so this means it's possible to encode both magnitude and angle in a single variable, if that variable is allowed to be complex and if we choose to use the complex polar notation discussed above.

So this is the big realization that made complex numbers far more important to electronics than they might have been. The Cartesian representation of a complex number is "mostly useless" in electronics. But the polar representation of a complex number? Now that is very useful, indeed!

This is where you are headed in thinking about complex impedances, voltages, and currents. It takes some getting used to, of course. But it makes everything easier to do. Just like "drawing a spiral" using \$y=f\left(x\right)\$ would be almost impossible to write out and use, but so easy when \$r=f\left(\theta\right)\$, choosing to use the polar form of representation of complex numbers makes things in electronics so much easier to express and manipulate. Ohm's law can still be the same, now, except that you are using this new polar complex notation instead. Sure, the details of the manipulations have to be "learned." But cripes, you had to "learn" a short-hand method used to multiply two, large integers by using this hard-learned "algorithm" of single digit multiplication and shifts before addition -- all that just to be able to multiply by hand. Of course, you can use a calculator now, right? Same thing here. There are short-hand rules that help you work with complex numbers more quickly. And they are algorithmic, not unlike that multiplication algorithm. But once you learn them, you just apply them. And you can handle "big stuff" by using the short-hand algorithms to cut through them more quickly.

It's actually not all that complex -- no pun intended!

External Tutorials

  • \$\begingroup\$ Thank you so much for the detailed answer. I will try to look into everything and hopefully comprehend everything. \$\endgroup\$
    – Edward B
    Commented May 7, 2018 at 9:11

First of all, you will need to learn the mathematical background of complex numbers. I will try to give you some basics:

We all are familiar with the real numbers that can be thought of lining up on a straight line from left to right (x-axis).

When dealing with complex numbers, you can imagine an additional straight line perpendicular to the real axis, crossing zero (y-axis).

The entity/unit of this complex axis is j. So j is a measure of how much up/down a complex number is on the perpendicular axis.

Complex numbers (like Z) on this "plane" thus have some sort of coordinate like e.g. Z = 4+5j. This means Z has a real part of 4 (4 units to the right) and an imaginary part of 5 (five units to the upper).

The shortest distance from zero to Z is called the absolute of Z which is the square root of its components to the power of 2, that is = 6.

I can not give you a complete lecture of complex numbers here, so you need to dig a bit further yourself.

Impedance, admittance, ... are just names for some geometric distances in the complex plane. An ideal capacitor has just imaginary but no real part.

  • \$\begingroup\$ Thank you very much. This info is much appreciated. I'll make sure to get some more background on complex numbers \$\endgroup\$
    – Edward B
    Commented May 6, 2018 at 13:10
  • \$\begingroup\$ @Neil_UK: Thanks for editing/fixing my typos. \$\endgroup\$ Commented May 6, 2018 at 13:15

When you get down to 50 mOhm Z(f) of the dielectric, realize that you are approaching the electrode resistance and inductance impedance too at higher frequencies. These are defined in datasheets when needed by ESR or DF and SRF (kHz or MHz) . Thus you have a Phasor angle of tan delta ( given for e-caps). The “apparent” impedance will be The hypotenuse of ESR and Zc.

For illustrations , see my other answers https://electronics.stackexchange.com/search?tab=newest&q=User%3ame%20SRF%20cap


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