#  Series RLC Resonance
When we have an Inductive and a Capacitive part in an ac circuit there is a certain frequency when the Impedance of those two parts are equal.

It can be calculated as follows

$\displaystyle f_{r} = \frac{1}{2\pi \sqrt{LC}}$

This formula is derived from 

$$
\begin{flalign}
\frac{1}{2\pi f C} = 2\pi f L \\
f^2 = \frac{1}{4\pi^4LC} \\
f = \frac{1}{4\pi^2LC} \\
\end{flalign}
$$

At this resonance frequency the impedance from capacitor and inductor cancel each other out, which means that $Z=R$ 

When we graph out the impedance vs frequency in a circuit, it could look something like this:

![](rlc-resonance.png)

This could cause problems because the voltage is really high at that point, which maybe bad for some other parts.

**Example:**

```circuitjs
$ 1 0.000005 30.13683688681966 45 5 43 5e-11
v 256 256 256 128 0 1 10000 5 0 0 0.5
l 432 128 432 256 0 1 0 0
c 256 256 432 256 0 0.000014999999999999999 0 0.001
r 256 128 432 128 0 10
o 1 64 0 4099 5 0.025 0 2 1 3
```

Lets calculate $f_{r}$ for this circuit.

$$
\begin{flalign}
f_{r} = \frac{1}{2\pi\sqrt{LC}} \\
L = 1H \\
C = 15\micro F = 0.000015F \\
f_{r} = \frac{1}{2\pi \sqrt{1*0.000015}} \\
\\
f_{r} \approx 41.093Hz
\end{flalign}
$$
As you can see in the following graph that result is correct:

<iframe src="https://www.desmos.com/calculator/esdrlexq4y?embed" width="500" height="500" style="border: 1px solid #ccc" frameborder=0></iframe>