Answer
a) $I_{rms}=36.2mA$
b) $\phi=26.0^{\circ}$
Work Step by Step
(a) We know that
$Z=\sqrt{R^2+(\frac{1}{\omega C})^2}$
We plug in the known values to obtain:
$Z=\sqrt{(3.35)^2}+(\frac{1}{2\pi(65)(1.50)})^2$
$Z=3.727k\Omega$
Now we can find rms current as
$I_{rms}=\frac{V_{rms}}{Z}$
$I_{rms}=\frac{135}{3.727}=36.2mA$
(b) We can find the phase angle as
$\phi=cos^{-1}\frac{R}{Z}$
We plug in the known values to obtain:
$\phi=cos^{-1}\frac{3.35}{3.737}=26.0^{\circ}$
