 Solution

For the series-fed parallel LC voltage response, use phasors with time dependence $e^{i\omega t}$. The parallel branch has impedance $iX$, where $X=\omega L/(1-{\omega }^{2}LC)$. ThereforeAt the ideal parallel resonance ${\omega }_0=1/\sqrt{LC}$ its input impedance diverges. The input-current phase changes sign on crossing this pole; expressions there are interpreted by limits.

The capacitor voltage equals the parallel-branch voltage. Define $q=R(1-{\omega }^{2}LC)/(\omega L)$. ThenAt low frequency $H\sim i\omega L/R$; at high frequency $H\sim -i/(\omega RC)$; at resonance $H=1$. Thus it is a band-pass voltage filter or resonant frequency selector. Its phase decreases continuously from $+\pi /2$ through zero to $-\pi /2$. Half-power points satisfy $|q|=1$ and

For the supplied component values and $\omega =2\pi (50){\mathrm{s}}^{-1}$, $X=421.745\Omega$ and $V_{C,\mathrm{pk}}=99.3046\mathrm{V}$. The element peak currents areInductor and capacitor currents are opposite phasors; their magnitudes subtract here rather than add. The resistor carries the input current, withThese results assume ideal components and sinusoidal steady state.