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Today I came across with Complex Capacitances and Inductances for the first time: $$L(j\omega)=L_{real}+jL_{imaginary}$$ $$C(j\omega)=C_{real}+jC_{imaginary}$$

So I started looking at their meaning. Let's start with the inductors.

Inductors:

I found this article, that states (pages 7 and 8):


Inductors - Article Part 1 Inductors - Article Part 2


After that, I found that this "Power Loss" is associated with "Core Losses", like hysteresis losses, Eddy/Foucault current losses. So, my first question is:

  • If $L_{imaginary}<0$, we have $Power\space Loss >0$, so we have a dissipation of energy by the magnetic core of the inductor due to (according to what the article says) eddy currents;
  • If $L_{imaginary}=0$, we have $Power\space Loss =0$, so no eddy current and no energy dissipation;

But, what if $L_{imaginary}>0$ ? Then we will have $Power\space Loss <0$, i. e., the magnetic core will supply power to the circuit! How is this possible? Or is the value of $L_{imaginary}$ somehow restricted by some equation, so that it cannot assume positive values?

(Another way to state this is considering the inductor impedance:$$Z_L=j\omega L=j\omega\cdot(L_r+jL_i\ )=-\omega L_i+j\omega L_r\ \ $$ The imaginary part of $L$ contributes to the real part of inductor impedance. This contribution is given in the form of a frequency dependent resistance $-\omega L_i$. So, if we have $L_i>0$ then $-\omega L_i<0$, a negative resistance!)


Capacitors:

I found this article. It states that (page 2) $C_{imaginary}$ is associated with dielectric loss.


Capacitors - Article


Another source states that this dielectric loss can be by Joule's heating effect, hysteresis losses and dielectric absorption. I guess we can model this dielectric loss in a analogous way to the losses in the inductor core: $$Power\space Loss=-\frac\omega2 C_i v^2$$

Thus, a similar question arises:

  • If $C_{imaginary}<0$, we have $Power\space Loss >0$, so we have a dissipation of energy by the dielectric;
  • If $C_{imaginary}=0$, we have $Power\space Loss =0$, so no energy dissipation;

But, what if $C_{imaginary}>0$ ? Then we will have $Power\space Loss <0$, i. e., the dielectric of the capacitor will supply power to the circuit! Again, how is this possible? Or is the value of $C_{imaginary}$ somehow restricted by some equation, so that it cannot assume positive values?

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closed as too broad by Dale, Jon Custer, GiorgioP, Kyle Kanos, ZeroTheHero Apr 7 at 12:20

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ Related: physics.stackexchange.com/q/109736/2451 and links therein. $\endgroup$ – Qmechanic Apr 2 at 9:00
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    $\begingroup$ Hi Vinicius ACP, When you edit a closed question, it automatically enters the review queue. $\endgroup$ – Qmechanic Apr 15 at 0:11
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    $\begingroup$ The post seems to be stuck in the review queue for more than 10 days since April 7...We need more reviewers. $\endgroup$ – Qmechanic Apr 18 at 6:07
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    $\begingroup$ The timeline log seems to indicate so. Mentioned in the hbar chat room here. $\endgroup$ – Qmechanic Apr 24 at 12:49
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    $\begingroup$ ${}$Never repost. $\endgroup$ – Qmechanic Apr 24 at 16:57