Parallel Resonance of RLC Circuits
Demonstrative Video
PARALLEL RESONANCE
also called Tank circuit
\[\begin{aligned} \mathbf{Y}&=H(\omega) =\frac{\mathbf{I}}{\mathbf{V}}\\ \mathbf{Y} & =\frac{1}{R}+j\left(\omega C-\frac{1}{\omega L}\right) \\ & \omega C-\frac{1}{\omega L}=0 \\ \omega_{0} & =\frac{1}{\sqrt{L C}} \operatorname{rad} / \mathrm{s} \\ \end{aligned}\] \[\begin{aligned} \omega_{1} & =-\frac{1}{2 R C}+\sqrt{\left(\frac{1}{2 R C}\right)^{2}+\frac{1}{L C}} \\ \omega_{2} & =\frac{1}{2 R C}+\sqrt{\left(\frac{1}{2 R C}\right)^{2}+\frac{1}{L C}} \\ B & =\omega_{2}-\omega_{1}=\frac{1}{R C} \\ Q & = \dfrac{\omega_0}{B} = \omega_0 RC = \dfrac{R}{\omega_0 L} \end{aligned}\]
Half-power frequencies \[\begin{aligned} \omega_{1} &=\omega_{0} \sqrt{1+\left(\frac{1}{2 Q}\right)^{2}}-\frac{\omega_{0}}{2 Q}, \quad \omega_{2}=\omega_{0} \sqrt{1+\left(\frac{1}{2 Q}\right)^{2}}+\frac{\omega_{0}}{2 Q} \end{aligned}\]
For high \(Q\) circuits \(Q \geq 10\) \[\begin{aligned} \omega_{1} & \simeq \omega_{0}-\frac{B}{2}, \quad \omega_{2} \simeq \omega_{0}+\frac{B}{2} \end{aligned}\]
Resonant RLC circuits
| Characteristic | Series circuit | Parallel circuit |
|---|---|---|
| Resonant freq, \(\omega_{0}\) | \(\dfrac{1}{\sqrt{L C}}\) | \(\dfrac{1}{\sqrt{L C}}\) |
| Quality factor, Q | \(\dfrac{\omega_{0} L}{R} \text { or } \dfrac{1}{\omega_{0} R C}\) | \(\dfrac{R}{\omega_{0} L} \text { or } \omega_{0} R C\) |
| Bandwidth, B | \(\dfrac{\omega_{0}}{Q}\) | \(\dfrac{\omega_{0}}{Q}\) |
| Half-power freq., \(\omega_{1,2}\) | \(\omega_{0} \sqrt{1+\left(\dfrac{1}{2 Q}\right)^{2}} \pm \dfrac{\omega_{0}}{2 Q}\) | \(\omega_{0} \sqrt{1+\left(\dfrac{1}{2 Q}\right)^{2}} \pm \dfrac{\omega_{0}}{2 Q}\) |
| For \(Q \geq 10, \omega_{1}, \omega_{2}\) | \(\omega_{0} \pm \dfrac{B}{2}\) | \(\omega_{0} \pm \dfrac{B}{2}\) |
Problem
Let \(R=8~k \Omega\), \(L=0.2~\mathrm{mH}\), and \(C=8~\mathrm{\mu F}\). calculate
\(\omega_{0,1,2}, ~Q, ~B\)
power dissipated at \(\omega_{0,1,2}\)
\[\begin{aligned} \omega_0 & = \dfrac{1}{\sqrt{LC}} = 25~\mathrm{krad/s} \\ Q & = \dfrac{R}{\omega_0 L} = 1600 \\ B & = \dfrac{\omega_0}{Q} = 15.625 ~\mathrm{rad/s} \\ \omega_1 & = \omega_0 - \dfrac{B}{2} = 24,992~\mathrm{rad/s} \\ \omega_2 & = \omega_0 + \dfrac{B}{2} = 25,008~\mathrm{rad/s} \\ \end{aligned}\] \[\begin{aligned} \text{at}~\omega_0 ~ \Rightarrow ~ Z & = R \\ I_0 & = \dfrac{V}{R} = \dfrac{10 \angle -90^{\circ}}{8000} \\ &= 1.25\angle -90^{\circ}~\mathrm{mA} \\ \text{at}~\omega_0~P & = \dfrac{1}{2}|I_0|^2R = 6.25~\mathrm{mW} \\ \text{at}~\omega_{1,2}~P & = 3.125~\mathrm{mW} \end{aligned}\]