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I'm having trouble understanding the maths for impedance.

Suppose we have a voltage of \$V(t)=V_{0}sin({\omega}t)\$ and we plug this into the capacitor equation \$I(t)=C\frac{dV(t)}{dt}\$. This would yield, $$I(t)=C\frac{d}{dt}V_{0}sin({\omega}t)$$ Pulling \$V_0\$ out (as it's a constant) and taking a derivative yields, $$I(t)=V_{0}C{\omega}cos({\omega}t)$$ Now we know that impedance is \$\frac{V(t)}{I(t)}=Z\$. Therefore If I try to calculate impedance by dividing the voltage by current, I would get, $$\frac{V(t)}{I(t)}=\frac{V_0sin({\omega}t)}{V_{0}C{\omega}cos({\omega}t)}$$ Simplifying this yields, $$\frac{V(t)}{I(t)}=\frac{1}{{\omega}C}*\frac{sin({\omega}t)}{cos({\omega}t)}$$ Therefore our \$Z\$ would be $$Z=\frac{1}{{\omega}C}*\frac{sin({\omega}t)}{cos({\omega}t)}$$ However capacitors impedance Z is equal to $$Z = \frac{1}{{\omega}C}*-j$$ So my question is How does \$\frac{sin({\omega}t)}{cos({\omega}t)} = -j\$? or since \$\frac{sin({\omega}t)}{cos({\omega}t)}=tan({\omega}t)\$, I can rewrite my question as How does \$tan({\omega}t)=-j\$?

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    \$\begingroup\$ For a similar discussion see: electronics.stackexchange.com/questions/719418/… \$\endgroup\$
    – Carl
    Commented Jul 17 at 13:15
  • \$\begingroup\$ I saw that @Carl! The problem is it doesn't answer why \$tan(ωt)=−j\$. I understand that they represented the voltage as a complex exponential and therefore derived \$-j\$ in the denominator of capacitors impedance. However that implies than \$tan(\theta)=-j\$. This is only true if \${\theta}=90{\text{degree's}}\$ which equals not defined. I just don't understand why is this tangent relation with iota true. \$\endgroup\$
    – Ayush
    Commented Jul 17 at 13:34

2 Answers 2

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The error is a matter of definition:

When you divide instantaneous voltage by current, you get an impedance, but you don't get the impedance as used in conventional terms.

Note that such a definition is non-constant, and divergent: the \$\tan\$ function has poles every \$(2n+1)\pi\$, so it's only sometimes meaningful, but when would you pick; and you can't average it, that's undefined.

The conventional definition is AC steady state, where we work in the Fourier domain, considering only the magnitude and phase of a periodic (repeating for all time) signal, not concerned about the moment-to-moment (time domain) details. We express this with complex numbers, as it's a number system well suited to the plane geometry of this problem.

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  • \$\begingroup\$ Thank you! I see why we switched to frequency domain and only consider time independent parts since time independent part (the phasor) for our \$tan\$ is defined and constant while the time dependent part for our \$tan\$ is sometimes undefined (at \$(2n+1){\pi}\$). However wouldn't ignoring a piece of signal (time dependent part) cause problems when determining the conventional (general) impedance since we're ignoring information? \$\endgroup\$
    – Ayush
    Commented Jul 17 at 19:37
  • \$\begingroup\$ That's the beautiful part: the Fourier operation is a transform, an equivalent way of writing a function. Because RLC circuits are linear and time-invariant, FT (or Laplace, a similar transform) are conservative, i.e. nothing is lost in the operation. And to get back to time domain, simply transform again (in fact, FT is self-dual down to a constant factor, but even if not, as long as an inverse transform exists, that's all you need). \$\endgroup\$
    – Tim Williams
    Commented Jul 17 at 20:33
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    \$\begingroup\$ @Ayush: The impedance “is” the capacitor. The ratio of voltage to current is used to reveal its value. But if you take the time varying signals away the capacitor still remains. So time dependent quantities must disappear. The capacitor and its impedance are still there whether the signals are or are not. \$\endgroup\$
    – user319836
    Commented Jul 17 at 21:31
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    \$\begingroup\$ Came here to say the same thing. +1. \$\endgroup\$
    – Rohat Kılıç
    Commented Jul 17 at 23:06
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    \$\begingroup\$ You can use an aggregating method in the time domain, like peak-to-peak or RMS (note that pp is only sensitive to the, well, peaks, a few points out of the whole waveform), and measure phase separately (typically, oscopes do it by measuring zero crossing -- again, only sensitive to narrow bands of pixels). This is equivalent for sine wave, but other waveforms (including noise, harmonics, modulation, etc.) need analysis to show a different impedance law (i.e. the ratio, for peak amplitude, for square vs. triangle (as in an inductor's V/I), has a different proportionality. \$\endgroup\$
    – Tim Williams
    Commented Jul 18 at 19:08
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The equality $$ \text{tan}(\omega t)=-j $$ doesn’t hold. The tangent is a real quantity while \$-j\$ is an imaginary quantity.

Recognize that the sine is associated with one signal and the cosine with another so the tangent doesn’t really make sense.

Impedance is a frequency domain quantity so is correctly expressed as $$ Z(j\omega)=\frac{V(j\omega)}{I(j\omega)} $$ independent of time. So the ratio of two time domain quantities won’t work.

A time domain to frequency domain transformation is required. This was done in Carl’s very good answer. Euler’s formula introduces j in a convenient way. Fourier or Laplace transformations can also be used.

Phasors can also be used. This will show that the ratio is actually -1.

In the graphic the green and the gray lines represent the sine and cosine that form the ratio. They are in different triangles so cannot form a tangent or cotangent. However it can be seen that their ratio is -1.

enter image description here

Here is the spread sheet (this made the graph) showing the cosine and sine quantities that give the concern. They are highlighted in yellow. The angle of rotation rotation is \$\omega t\$

enter image description here

So because the time variable disappears, the calculation in the OP finds the reactance $$X_C=\frac{-1}{\omega C}$$.

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  • \$\begingroup\$ Ah! The complex exponential (as was used in the derivation of reactance in conventional terms) gives us 4 quantities, two of which are time independent and two of which are time dependent \$Ae^{j{\phi}}e^{j{\omega}t}e^{{\sigma}t}\$ .Time independent part is the phasor \$Ae^{j{\phi}}\$. The time dependent part is \$e^{j{\omega}t}e^{{\sigma}t}\$. If I use phasors, my reactance comes out to be \$\frac{1}{j{\phi}C}\$ instead of the conventional \$\frac{1}{j{\omega}{C}}\$. From below @Tim Williams answer I can sort of understand why we switched to frequency domain and are only...(1/2) \$\endgroup\$
    – Ayush
    Commented Jul 17 at 19:32
  • \$\begingroup\$ ...considering time independent parts since tan is undefined at every \$(2n+1){\pi}\$ turns and it wouldn't make any sense to work on something that is undefined with respect to time specially when calculating a generalised impedance that works for all capacitors (what range do we pick, what to do on undefined parts etc.). This isn't a problem with phasor since it stays constant and defined regardless of time. Although, Why is my reactance different and wouldn't ignoring the time dependent part create problems? (2/2) \$\endgroup\$
    – Ayush
    Commented Jul 17 at 19:36
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    \$\begingroup\$ "my reactance comes out to be \$\frac{1}{j\phi C}\$" @Ayush, reactance cannot have a j involved. X is a real number. jX is an imaginary number. X is the reactance of a capacitor. jX is the impedance of the capacitor. You must keep these concepts straight. You will be frustrated if you don't. \$\endgroup\$
    – user319836
    Commented Jul 17 at 20:11
  • \$\begingroup\$ I apologise I mean't impedance and not reactance. This was a typo. Although, I understand this now so thank you so much Russell! : D \$\endgroup\$
    – Ayush
    Commented Jul 18 at 8:49

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