Answer
a. $\phi=-90^{\circ}$.
b. $C=1.76\times10^{-4}F$.
Work Step by Step
a. As seen in the caption to Figure 31.9b on page 1027, the phase angle is $\phi=-90^{\circ}$. The source voltage lags the current by one-quarter cycle.
b. $X_C=\frac{V}{I}=\frac{60.0V}{5.30A}=11.3\Omega$
Solve $X_C=\frac{1}{2 \pi fC}$ for the capacitance.
$C=\frac{1}{2 \pi fX_C}=\frac{1}{2 \pi (80Hz)( 11.3\Omega)}=1.76\times10^{-4}F$
