Answer
a. $\phi=90^{\circ}$.
b. $f=193Hz$.
Work Step by Step
a. As seen in the caption to Figure 31.8b on page 1024, the phase angle is $\phi=90^{\circ}$. The source voltage leads the current by one-quarter cycle.
b. $X_L=\frac{V}{I}=\frac{45.0V}{3.90A}=11.54\Omega$
Solve $X_L=2 \pi fL$ for the frequency.
$$f=\frac{X_L}{2 \pi L}=\frac{11.54\Omega }{2 \pi (9.50\times10^{-3}H)}=193Hz$$
