The capacitor and inductor are in series, which means that the equivalent impedance of these two elements is

$$Z_{LC}=\frac{1-\omega^2LC}{j\omega L}$$

So for the three cases I got the following results

$$\omega=0\ \ \ \ Z_{LC}\ =\infty $$

$$\omega=\infty\ \ \ \ Z_{LC}\ =\infty $$

$$\omega=\frac{1}{\sqrt{LC}}\ \ \ \ Z_{LC}\ =0 $$

If I were to plot these values I am not sure how the graph would look like.

In case the inductor and capcitor are in parallel,this is how the graph would look like. Link to the question and graph

  • 2
    \$\begingroup\$ Your equation is not impedance but admittance. \$\endgroup\$
    – Bart
    Feb 11 '20 at 0:56
  • \$\begingroup\$ be1995, @Bart is right. That's not impedance. You can easily see this by re-stating your equation as \$\frac1{s\,L}+s\,C\$ and noting that this is the sum of the inverse of each separate impedance -- namely, the sum of the admittances to get the total admittance in the parallel case. This is why "X J" below tells you that the denominator is wrong if you are talking about series impedance. It's hard to get the right plot if you start with the wrong expression. \$\endgroup\$
    – jonk
    Feb 11 '20 at 2:15

If you want to see what the graph would look like, you need to plot some more points. You picked only 3 obvious values. We know that as the frequency increases from 0 to the resonant frequency, the magnitude of the impedance goes from infinity to 0. As the frequency increases through resonance the magnitude of the impedance goes back to infinity. To get an idea of the actual shape, plot some in between points. For convenience, do it for integer multiples of the resonant frequency. Plot enough points so you can fill in the curve by interpolation. Note that the sign of the impedance is negative for frequencies below resonance (due to the capacitor) and positive for frequencies above resonance (due to the inductor).


First, the denominator should be jwC.

To give a meaningful graph, it is better to rearrange the |Z| with terms like,

enter image description here


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