Answer
a. $7400\Omega$
b. $0.38A$
Work Step by Step
a. Find the reactance of a capacitor with equation 21–12b.
$X_C=\frac{1}{2 \pi f C}=\frac{1}{2 \pi (720Hz) (0.030\times10^{-6}F)}=7400\Omega$
b. Find the peak value of the current using Ohm’s law.
$$I_{peak}=\sqrt2I_{rms}=\sqrt2\frac{V_{rms}}{X_C}$$
$$=\sqrt2\frac{2.0\times10^3V}{7368\Omega}=0.38A$$