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You may be familiar with the surprising result one gets when calculating the equivalent of an infinite array of resistors. What if we change this circuit and replace the resistors with capacitors and inductors?

Following the notation given in the link I've provided above, let's replace $R_1$ with $C$ and $R_2$ with $L$ so that

$$R_1 \rightarrow Z_1=\frac{-j}{\omega C}$$ $$R_2 \rightarrow Z_2=j\omega L$$

where $j$ is the imaginary unit. Again using the result written in that link, we get the following equation:

$$Z_{eq}^2-\frac{j}{\omega C}Z_{eq}-\frac{L}{C}=0$$

Solving this quadratic equation we get that:

$$Z_{eq}=\frac{\frac{j}{\omega C}\pm\sqrt{\frac{-1}{\omega^2C^2}+4\frac{L}{C}}}{2}=\pm\sqrt{\frac{L}{C}-\frac{1}{(2\omega C)^2}}+j\frac{1}{2\omega C}$$

So an array of ideal capacitors and inductors lead to a complex (not imaginary) equivalent impedance if $L>\frac{1}{4C\omega^2}$. This means that, if the circuit was fed with a source, actual power would be dissipated, even though each of the individual impedances are purely reactive. How does this make sense?

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  • $\begingroup$ an infinite series of $LC$ is equivalent to a transmission line of constant wave impedance, hence its impedance is real and it is perfectly normal if you consider a pulse launched at one end that never gets reflected. $\endgroup$ – hyportnex Feb 14 '17 at 1:49
  • $\begingroup$ Think carefully about the definition of impedance in order to resolve this paradox. If you can't, I'll write an answer. $\endgroup$ – DanielSank Feb 14 '17 at 2:44
  • $\begingroup$ @DanielSank I have absolutely no idea of how to solve this. $\endgroup$ – Tendero Feb 14 '17 at 2:55
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Initially I thought you had just rediscovered the Telegrapher's Equations - but then I realized you had your capacitors and inductors "the other way around" from that more usual scenario (described here)

Even though your situation is unusual, there is a way to understand what is happening. The capacitors in your network are charging up - and while some of that charging is transient, some of it "goes on forever" because of the infinite extent of the network. This charging of the capacitors means there is a mechanism for storing energy - and I think that's what your equations are telling you.

This is much easier to understand when you switch your capacitors (to be $R_2$ in your diagram) and inductors (to be $R_1$). When you do that, you end up with an expression for the impedance that will tend to $Z=\sqrt{\frac{L}{C}}$ when you make $L$ and $C$ infinitesimal while preserving their ratio (which is what happens when you consider a transmission line as being made up of many small inductors and capacitors).

When you have an ordinary transmission line, a pulse will propagate, and energy will be stored per unit length. The storing of energy is indistinguishable from dissipation of energy (until you get a reflection or some other mechanism to extract the energy again).

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  • $\begingroup$ In the case of an ideal transmission line, the impedance is real. Here, the impedance is imaginary or complex, but never purely real. How can that be if the model used is the same in both cases? $\endgroup$ – Tendero Feb 14 '17 at 11:43
  • $\begingroup$ If you do your analysis with the components reversed you end up with an imaginary term in L. But when to realize that in the limit is infinitely many, infinitely small components, L tends to zero (while the ratio L/C does not) then you see how the imaginary component will become as small as you like in the continuous transmission line. $\endgroup$ – Floris Feb 14 '17 at 13:28
  • $\begingroup$ What about this paper? It seems to contradict the procedure I've done in my question, saying that the first move (taking one pair of impedances away) is illegal. I believe that it contradicts your answer, too. $\endgroup$ – Tendero Feb 14 '17 at 19:30
  • $\begingroup$ When you have a finite number of devices, of course the impedance changes when you add more devices? Also - if you operate the ideal network at the precise point of resonance, you have lots of infinities floating around; again, "normal" math doesn't work. I don't think the reasoning is solid for the general (non resonant) case. $\endgroup$ – Floris Feb 14 '17 at 21:59

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