Answer
(a) $5.9\mu A$
(b) $8.3\mu A$
Work Step by Step
(a) We know that
$X_c=\frac{1}{\omega C}$
$\implies X_c=\frac{1}{2\pi (52.0)(0.010)}=306\Omega$
$I_{rms}=\frac{V_{rms}}{X_c}$
We plug in the known values to obtain:
$I_{rms}=\frac{1.8}{3.06\times 10^5}$
$I_{rms}=5.9\times 10^{-6}=5.9\mu A$
(b) We can find the maximum current as
$I_{max}=\sqrt{2}I_{rms}$
We plug in the known values to obtain:
$I_{max}=\sqrt{2}(5.9\times 10^{-6})=8.3\mu A$
