Answer
$i_C = -33.9~mA$
Work Step by Step
We can find $\omega_d~t$:
$\mathscr{E} = \mathscr{E}_m~sin~\omega_d t$
$sin~\omega_d t = \frac{\mathscr{E}}{\mathscr{E}_m}$
$sin~\omega_d t = \frac{-12.5~V}{25.0~V}$
$sin~\omega_d t = -0.5$
$\omega_d t = \frac{7\pi}{6}, \frac{11\pi}{6}$
Since the emf is increasing in magnitude, $\omega_d t = \frac{7\pi}{6}$
In part (a), we found that the maximum value of the current is $~~39.1~mA$
We can find the current:
$i_C = I_C~sin~(\omega_d t+\frac{\pi}{2})$
$i_C = (39.1~mA)~sin~(\frac{7\pi}{6}+\frac{\pi}{2})$
$i_C = (39.1~mA)~sin~(\frac{5\pi}{3})$
$i_C = -33.9~mA$
