Series Resonance in RLC Circuit

Demonstrative Video


Resonance


Series Resonance

\[\begin{aligned} \mathbf{Z}=\mathbf{H}(\omega) &=\frac{\mathbf{V}_{s}}{\mathbf{I}}=R+j \omega L+\frac{1}{j \omega C} \\ \mathbf{Z} &=R+j\left(\omega L-\frac{1}{\omega C}\right) \\ \operatorname{Im}(\mathbf{Z}) &=\omega L-\frac{1}{\omega C}=0 \end{aligned}\] image

\[\begin{aligned} X_L & = X_C \\ \Rightarrow~ \omega_{0} L &=\frac{1}{\omega_{0} C} \\ \end{aligned}\] \[ \begin{aligned} \omega_{0} &=\frac{1}{\sqrt{L C}} \mathrm{rad} / \mathrm{s} \\ f_{0} &=\frac{1}{2 \pi \sqrt{L C}} \mathrm{~Hz} \end{aligned} \]

\[\boxed{\omega_{0} =\frac{1}{\sqrt{L C}} \mathrm{rad} / \mathrm{s} } \quad \boxed{f_{0} =\frac{1}{2 \pi \sqrt{L C}} \mathrm{~Hz}}\]

At resonance:

series resonant circuit also called voltage resonant circuit image

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\[X_L = \omega L \qquad X_C = \dfrac{1}{\omega C}\]

\[\begin{aligned} & |\mathbf{I}|=\frac{V_{m}}{\sqrt{R^{2}+(\omega L-1 / \omega C)^{2}}} \\ & \text{Avg. power dissipated}~P(\omega)=\frac{1}{2} I^{2} R \\ & \text{Max. power dissipated}~P\left(\omega_{0}\right)=\frac{1}{2} \frac{V_{m}^{2}}{R} \end{aligned}\] image

\[\begin{aligned} &P\left(\omega_{1}\right)=P\left(\omega_{2}\right)=\frac{\left(V_{m} / \sqrt{2}\right)^{2}}{2 R}=\frac{V_{m}^{2}}{4 R} \\ &\text{Setting}~\sqrt{R^{2}+\left(\omega L-\frac{1}{\omega C}\right)^{2}}=\sqrt{2} R \end{aligned}\] \[ \begin{aligned} \omega_{1}&=-\frac{R}{2 L}+\sqrt{\left(\frac{R}{2 L}\right)^{2}+\frac{1}{L C}} \\ \omega_{2}&=\frac{R}{2 L}+\sqrt{\left(\frac{R}{2 L}\right)^{2}+\frac{1}{L C}} \end{aligned} \]

  • Resonant frequency is the geometric mean of the half-power frequencies \[\omega_0 = \sqrt{\omega_1 \cdot \omega_2}\]

\[\begin{aligned} Q&=\frac{\text { energy stored }}{\text { energy dissipated }} \\ &=\frac{\text { reactive power }}{\text { True or active power }} = \frac{I^{2} X_{\mathrm{L}}}{I^{2} R} =\frac{X_{\mathrm{L}}}{R}\\ &\boxed{Q=\frac{\omega_{0} L}{R}=\frac{1}{\omega_{0} C R}}\\ &\boxed{B=\frac{R}{L}=\frac{\omega_{0}}{Q}}\quad =\omega_{0}^{2} C R \end{aligned}\]

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\(\boxed{Q~\uparrow~\Rightarrow~B~\downarrow}\)


Problem

Given: \(R=2~\Omega,~L=1~\mathrm{mH},~C=0.4~\mathrm{\mu F}\). Determine

  • \(\omega_{0}, ~\omega_{1,2},~Q, ~B\)

  • Amplitude of the current at \(\omega_0, \omega_1, \omega_2\)

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Solution

\[\omega_0 =\frac{1}{\sqrt{L C}}=50 \mathrm{krad} / \mathrm{s}\]

Method-1 \[\begin{aligned} \omega_{1} &=-\frac{R}{2 L}+\sqrt{\left(\frac{R}{2 L}\right)^{2}+\frac{1}{L C}} \\ &=-1+\sqrt{1+2500} =49 \mathrm{krad} / \mathrm{s} \\ \omega_{2} &=1+\sqrt{1+2500}=51 \mathrm{krad} / \mathrm{s} \\ B &=\omega_{2}-\omega_{1}=2 \mathrm{krad} / \mathrm{s} \\ \text{or}~B &=\frac{R}{L}=\frac{2}{10^{-3}}=2 \mathrm{krad} / \mathrm{s} \\ Q &=\frac{\omega_{0}}{B}=\frac{50}{2}=25 \end{aligned}\] Method-2 \[\begin{aligned} Q &=\frac{\omega_{0} L}{R}=\frac{50 \times 10^{3} \times 10^{-3}}{2}=25 \\ B &=\frac{\omega_{0}}{Q}=\frac{50 \times 10^{3}}{25}=2 \mathrm{krad} / \mathrm{s} \\ \omega_{1} &=\omega_{0}-\frac{B}{2}=50-1=49 \mathrm{krad} / \mathrm{s} \\ \omega_{2} &=\omega_{0}+\frac{B}{2}=50+1=51 \mathrm{krad} / \mathrm{s} \end{aligned}\]

\[\begin{aligned} \omega=\omega_{0},~\Rightarrow \quad I&=\frac{V_{m}}{R}=\frac{20}{2}=10 \mathrm{~A} \\ \omega=\omega_{1}, \omega_{2}, ~\Rightarrow \quad I&=\frac{V_{m}}{\sqrt{2} R}=\frac{10}{\sqrt{2}}=7.071 \mathrm{~A} \end{aligned}\]