Answer
a) $I_{rms}=4.3\times10^{-2}A$
b) $\phi=2.7^o$
c) $P=5.1W$
d) Resistor: $V_{rms}=120V$
Inductor: $V_{rms}=5.6V$
Work Step by Step
$I_{rms}=\frac{V_{rms}}{Z}$
$Z=\sqrt{R^2+X_L^2}=\sqrt{R^2+(2\pi fL)^2}=2803.1\Omega$
$I_{rms}=\frac{120V}{2803.1\Omega}=4.3\times10^{-2}A$
b) $\phi=\arctan(\frac{X_L-X_C}{R})=2.7^o$
c) $P=I_{rms}^2R=5.1W$
d) V_{rms}=I_{rms}Z
Resistor: $Z=R$; $V_{rms}=(0.04281A)(2.80\times10^3\Omega)=120V$
Inductor: $Z=X_L=2\pi fL$
$V_{rms}=(0.04281A)(2\pi(60.0Hz)(350\times10^{-3}))=5.6V$