Answer
The circuit contains an capacitor.
Work Step by Step
$\mathscr{E} = \mathscr{E}_m~sin~(\omega_d t-\frac{\pi}{4})$
$i(t) = I~sin~(\omega_d t+\frac{\pi}{4})$
The current leads the emf by $\frac{\pi}{2}$
Therefore, the circuit contains an capacitor.
