Answer
$I = 0.170~A$
Work Step by Step
In part (b), we found that $f_d = 159~Hz$
We can find the impedance:
$Z = \sqrt{R^2+(-X_C)^2}$
$Z = \sqrt{R^2+(-1/2\pi~f_d~C)^2}$
$Z = \sqrt{(50.0~\Omega)^2+[-1/(2\pi)~(159~Hz)~(20.0\times 10^{-6}~F)]^2}$
$Z = 70.7~\Omega$
We can find the current amplitude:
$I = \frac{\mathscr{E}_m}{Z}$
$I = \frac{12.0~V}{70.7~\Omega}$
$I = 0.170~A$
