= Solution
A parallel connection has the same <electric potential difference> across each element, so the <electrical admittances> add. With $G=1/R$ and $B=\omega C-1/(\omega L)$,
$$
Y_{\rm p}=G+iB,\qquad\boxed{Z_{\rm p}=\left[\frac1R+i\left(\omega C-\frac1{\omega L}\right)\right]^{-1}=\frac{G-iB}{G^2+B^2}}.
$$
For a prescribed sinusoidal current, $\widetilde V=Z_{\rm p}\widetilde I$. Thus the voltage-minus-current <phase> is
$$
\phi=-\tan^{-1}\left[R\left(\omega C-\frac1{\omega L}\right)\right].
$$
Below $\omega_0=1/\sqrt{LC}$ the circuit is inductive, so <electric potential difference> leads current; above it the circuit is capacitive, so <electric potential difference> lags current. At $\omega_0$, the reactive branch currents cancel, $Y_{\rm p}=1/R$ and $\boxed{Z_{\rm p}=R}$. This is the impedance maximum for positive finite $R$, whereas the series circuit has its impedance minimum at the same <frequency>. It does not mean that either individual reactive branch current vanishes.
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