Explanation

(c) In a series L – C – R circuit, potential difference leads the current by an angle $\phi$(let).
$\ \ \phi = \tan^{- 1}\left( \frac{X_{L} - X_{C}}{R} \right)$
or $\ \ \phi = \tan^{- 1}\left( \frac{\omega_{L} - \frac{1}{\omega C}}{R} \right)$
At resonance, $X_{L} = X_{C},\ \ ie,\omega L = \frac{1}{\omega C}$
Hence, $\phi = \tan^{- 1}{(0) = 0}$
Therefore, phase difference between current and voltage at resonance is zero.