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schematic

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

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  • \$\begingroup\$ Do you mean maximum impedance or, impedance where the value is purely resistive? \$\endgroup\$
    – Andy aka
    May 5, 2023 at 10:49

1 Answer 1

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

\begin{equation} \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 \end{equation}

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{equation} \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 \end{equation}

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:

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

Using your values, we find:

$$\omega_0=\sqrt{2\sqrt{30}-1}\approx3.15507\space\text{rad/sec}\tag5$$

And:

$$\left|\underline{\text{Z}}_{\space\text{i}}\left(\omega_0\right)\right|=\frac{\sqrt{4\sqrt{30}-21}}{10}\approx0.0953364\space\Omega\tag6$$

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  • \$\begingroup\$ Excellent derivation! A small adjustment: in (4) may be clearer to keep omega as 1/(RC) times radical, and it seems the C can be taken inside (as C^2, and then C^4 inside the second radical) to useful effect. Some discussion might also be illuminating: why this change (what units are omega and 1/(RC)?); for which R/Z_o do we get omega^2 ~= 1/(LC), or omega ~= 1/(RC); and removing the "~" from these, expressing it as omega_0 +/- omega_d (damped or difference). \$\endgroup\$ May 5, 2023 at 10:19
  • \$\begingroup\$ Understood its complex though. After finding equivalent impedance, Z=R1+(CRω)2 + ω(L−CR21+(CRω)2)j(1) i did separate real and imaginary parts, my Prof just said that equate that imaginary part to 0. So, ω(L−CR21+(CRω)=0 which gives ω=3 rad/sec which is given answer. But why? Why do we equate imaginary part to 0. \$\endgroup\$
    – Adyy
    May 5, 2023 at 19:23

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