# Resonant Frequency

simulate this circuit – Schematic created using CircuitLab

So, for series circuit to find resonance we have to have values of impedances (L and C) to be equal. And then resonant frequency can be calculated wl and 1/wc but this is combination of elements how to find resonant frequency here. Is there any formula for this? Zeq={(1*-jXc)/1-jXc} + jXl

• Do you mean maximum impedance or, impedance where the value is purely resistive? Commented May 5, 2023 at 10:49

Well, notice that the input impedance of your circuit is given by:

$$$$\begin{split} \underline{\text{Z}}_{\space\text{i}}\left(\omega\right)&=\underline{\text{Z}}_{\space\text{L}}+\underline{\text{Z}}_{\space\text{C}\space\text{||}\space\text{R}}\\ \\ &=\text{j}\omega\text{L}+\left(\frac{1}{\text{j}\omega\text{C}}\space\text{||}\space\text{R}\right)\\ \\ &=\text{j}\omega\text{L}+\frac{\displaystyle\frac{1}{\text{j}\omega\text{C}}\cdot\text{R}}{\displaystyle\frac{1}{\text{j}\omega\text{C}}+\text{R}}\\ \\ &=\text{j}\omega\text{L}+\frac{\displaystyle\frac{\text{j}\omega\text{C}}{\text{j}\omega\text{C}}\cdot\text{R}}{\displaystyle\frac{\text{j}\omega\text{C}}{\text{j}\omega\text{C}}+\text{j}\omega\text{C}\text{R}}\\ \\ &=\text{L}\omega\text{j}+\frac{\displaystyle\text{R}}{\displaystyle1+\text{CR}\omega\text{j}}\\ \\ &=\text{L}\omega\text{j}+\frac{\displaystyle\text{R}}{\displaystyle1+\text{CR}\omega\text{j}}\cdot\frac{\displaystyle1-\text{CR}\omega\text{j}}{\displaystyle1-\text{CR}\omega\text{j}}\\ \\ &=\text{L}\omega\text{j}+\frac{\displaystyle\text{R}\left(1-\text{CR}\omega\text{j}\right)}{\displaystyle1^2+\left(\text{CR}\omega\right)^2}\\ \\ &=\text{L}\omega\text{j}+\frac{\displaystyle\text{R}-\text{CR}^2\omega\text{j}}{\displaystyle1+\left(\text{CR}\omega\right)^2}\\ \\ &=\text{L}\omega\text{j}+\frac{\displaystyle\text{R}}{\displaystyle1+\left(\text{CR}\omega\right)^2}-\frac{\displaystyle\text{CR}^2\omega}{\displaystyle1+\left(\text{CR}\omega\right)^2}\cdot\text{j}\\ \\ &=\frac{\displaystyle\text{R}}{\displaystyle1+\left(\text{CR}\omega\right)^2}+\left(\text{L}\omega-\frac{\displaystyle\text{CR}^2\omega}{\displaystyle1+\left(\text{CR}\omega\right)^2}\right)\text{j}\\ \\ &=\frac{\displaystyle\text{R}}{\displaystyle1+\left(\text{CR}\omega\right)^2}+\omega\left(\text{L}-\frac{\displaystyle\text{CR}^2}{\displaystyle1+\left(\text{CR}\omega\right)^2}\right)\text{j} \end{split}\tag1$$$$

Where $$\\alpha\space\text{||}\space\beta:=\frac{\displaystyle\alpha\beta}{\displaystyle\alpha+\beta}\$$.

So, we can see that the amplitude of the input impedance is given by:

$$$$\begin{split} \left|\underline{\text{Z}}_{\space\text{i}}\left(\omega\right)\right|&=\left|\frac{\displaystyle\text{R}}{\displaystyle1+\left(\text{CR}\omega\right)^2}+\omega\left(\text{L}-\frac{\displaystyle\text{CR}^2}{\displaystyle1+\left(\text{CR}\omega\right)^2}\right)\text{j}\right|\\ \\ &=\sqrt{\left(\frac{\displaystyle\text{R}}{\displaystyle1+\left(\text{CR}\omega\right)^2}\right)^2+\left(\omega\left(\text{L}-\frac{\displaystyle\text{CR}^2}{\displaystyle1+\left(\text{CR}\omega\right)^2}\right)\right)^2} \end{split}\tag2$$$$

Solving:

$$\frac{\displaystyle\partial\left|\underline{\text{Z}}_{\space\text{i}}\left(\omega\right)\right|}{\displaystyle\partial\omega}=0\space\Longrightarrow\space\omega_0=\dots\tag3$$

Gives:

$$\omega_0=\frac{\displaystyle1}{\displaystyle\text{C}^2\text{R}}\cdot\sqrt{\frac{\text{R}}{\text{L}}\cdot\sqrt{\text{C}^5\left(2\text{L}+\text{CR}^2\right)}-\text{C}^2}\tag4$$

$$\omega_0=\sqrt{2\sqrt{30}-1}\approx3.15507\space\text{rad/sec}\tag5$$
$$\left|\underline{\text{Z}}_{\space\text{i}}\left(\omega_0\right)\right|=\frac{\sqrt{4\sqrt{30}-21}}{10}\approx0.0953364\space\Omega\tag6$$