Answer
$\phi = 45.0^{\circ}$
Work Step by Step
In part (b), we found that $f_d = 318~Hz$
We can find the phase angle:
$tan~\phi = \frac{X_L}{R}$
$tan~\phi = \frac{2\pi~f_d~L}{R}$
$tan~\phi = \frac{(2\pi)~(318~Hz)~(40.0\times 10^{-3}~H)}{80.0~\Omega}$
$tan~\phi = 1.00$
$\phi = tan^{-1}~(1.00)$
$\phi = 45.0^{\circ}$
