51 lines
1.4 KiB
Markdown
51 lines
1.4 KiB
Markdown
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# Series RLC Resonance
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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.
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It can be calculated as follows
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$\displaystyle f_{r} = \frac{1}{2\pi \sqrt{LC}}$
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This formula is derived from
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$$
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\begin{flalign}
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\frac{1}{2\pi f C} = 2\pi f L \\
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f^2 = \frac{1}{4\pi^4LC} \\
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f = \frac{1}{4\pi^2LC} \\
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\end{flalign}
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$$
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At this resonance frequency the impedance from capacitor and inductor cancel each other out, which means that $Z=R$
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When we graph out the impedance vs frequency in a circuit, it could look something like this:
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This could cause problems because the voltage is really high at that point, which maybe bad for some other parts.
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**Example:**
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```circuitjs
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$ 1 0.000005 30.13683688681966 45 5 43 5e-11
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v 256 256 256 128 0 1 10000 5 0 0 0.5
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l 432 128 432 256 0 1 0 0
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c 256 256 432 256 0 0.000014999999999999999 0 0.001
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r 256 128 432 128 0 10
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o 1 64 0 4099 5 0.025 0 2 1 3
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```
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Lets calculate $f_{r}$ for this circuit.
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$$
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\begin{flalign}
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f_{r} = \frac{1}{2\pi\sqrt{LC}} \\
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L = 1H \\
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C = 15\micro F = 0.000015F \\
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f_{r} = \frac{1}{2\pi \sqrt{1*0.000015}} \\
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\\
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f_{r} \approx 41.093Hz
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\end{flalign}
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$$
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As you can see in the following graph that result is correct:
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<iframe src="https://www.desmos.com/calculator/esdrlexq4y?embed" width="500" height="500" style="border: 1px solid #ccc" frameborder=0></iframe>
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