Answer
(a) increase
(b) $87.2Hz$
(c) $0.634A$
Work Step by Step
(a) We know that when the frequency increases, $X_L$ increases and $X_C$ decreases. Therefore, $tan\phi=\frac{X_L-X_C}{R}$ increases and hence the phase angle will also increase.
(b) We know that
$tan(-22.5^{\circ})=\frac{\omega L-\frac{1}{\omega C}}{R}$
$\implies \omega^2LC-1=-1R\omega C tan22.5^{\circ}$
We plug in the known values to obtain:
$(\omega ^2\times 90\times 10^{-3}\times 15\times 10^{-6})0+8(175\times 5\times 0.4142)-1=0$
This simplifies to:
$\omega=\frac{-1087.275\times 10^{-6}\pm \sqrt{(1087.275\times 10^{-6})^2-4(1350\times 10^{-9})(-1)}}{2\times 1350\times 10^{-9}}$
$\implies \omega=546s^{-1}$ (the negative value of $\omega$ is not possible).
$\implies f=\frac{\omega}{2\pi}$
$\implies f=\frac{546}{2\pi}=87.2Hz$
(c) We know that
$I_{rms}=\frac{V_{rms}}{\sqrt{R^2+(X_L-X_C)^2}}$
We plug in the known values to obtain:
$I_{rms}=\frac{120}{\sqrt{(175)^2}+(49-122)^2}$
This simplifies to:
$I_{rms}=0.634A$
