Question

What is the phase angle of the source voltage with respect to the current? Does the source voltage lag or lead the current?

Answer 1

$X_L=\omega L=100\Omega$
$X_C=\frac{1}{\omega C}=312.5\Omega$
$Z=\sqrt{R^2+(X_C-X_L{)}^2}$
$=\sqrt{(300{)}^2+(312.5-100{)}^2}=368\Omega$
Since $X_C>X_L$, voltage lags the current by an angle given by
$\varphi ={\cos }^{-1}\left(\frac{R}{Z}\right)$
$=co{s}^{-1}\left(\frac{300}{368}\right)={35.3}^o$