# Infinite array of capacitors and inductors

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?

• 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. – hyportnex Feb 14 '17 at 1:49
• Think carefully about the definition of impedance in order to resolve this paradox. If you can't, I'll write an answer. – DanielSank Feb 14 '17 at 2:44
• @DanielSank I have absolutely no idea of how to solve this. – Tendero Feb 14 '17 at 2:55

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).