= Solution
Define the phase difference as current phase minus voltage phase. Since $\widetilde I=\widetilde V/Z$,
$$
\boxed{\phi(\omega)=-\tan^{-1}\left(\frac{\omega L-1/(\omega C)}{R}\right).}
$$
It decreases continuously from $+\pi/2$ at low frequency, through zero at $\omega_0=1/\sqrt{LC}$, to $-\pi/2$ at high frequency. Thus the current leads in the capacitive limit and lags in the inductive limit. Defining voltage phase minus current phase would reverse the graph's sign.
\Image[/past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2014-rlc-phase.png]
{title=Current phase relative to the source voltage in a series RLC circuit}
{description=The example uses omega_0 L/R equal to one; the limiting phases and zero crossing hold for every positive R, L and C.}
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