Question 12.45: Refer to Fig. 12.63. Draw a qualitative phasor diagram for t...

Because \tilde{I} is the phase reference, we may write \tilde{I}=I \angle 0. At resonance, the admittance of the L C section equals zero (the impedance is infinite), so the current \tilde{I}_{L C}=\tilde{I}_L+\tilde{I}_C must be zero. It follows from Kirchhoff’s current law that \tilde{I}_R=\tilde{I} and then from Ohm’s law that \tilde{V}=R \tilde{I}_R=R I \angle 0. Because R is real and positive and \tilde{I}_R=\tilde{I} has zero initial phase, we have

 \begin{aligned} \tilde{I}_L &=\frac{R I \angle 0}{j \omega_r L}=\frac{R I \angle 0}{\omega_r L (\pi / 2)}=\frac{R I}{\omega_r L} \angle\left(-\frac{\pi}{2}\right), \\ \tilde{I}_C &=j \omega_r C R I \angle 0=\left[\omega_r C \angle(\pi / 2)\right][R I \angle 0] \\ &=\omega_r C R I \angle\left(\frac{\pi}{2}\right) \end{aligned} .

The resonant frequency is \omega_r=1 / \sqrt{L C} , so the expressions above reduce to

\begin{aligned} \tilde{I}_L &=\frac{R I}{L / \sqrt{L C}} \angle\left(-\frac{\pi}{2}\right)=R I \sqrt{\frac{C}{L}} \angle\left(-\frac{\pi}{2}\right), \\ \tilde{I}_C &=\frac{C R I}{\sqrt{L C}} \angle\left(\frac{\pi}{2}\right)=R I \sqrt{\frac{C}{L}} \angle\left(\frac{\pi}{2}\right) . \end{aligned}

At resonance, the currents \tilde{I}_L, \tilde{I}_C have equal magnitudes and are in phase opposition. Figure  12.64  shows a phasor diagram for the currents in the circuit at resonance.