Answer
See explanation.
Work Step by Step
a.
At 60 Hz:
$X_L=2 \pi fL=2 \pi (60Hz)(0.450H)=170\Omega $
At 600 Hz:
$X_L=2 \pi fL=2 \pi (600Hz)(0.450H)=1700\Omega $
b.
At 60 Hz:
$X_C=\frac{1}{2 \pi fC}=\frac{1}{2 \pi (60Hz)(2.50\times10^{-6}F)}=1.06\times10^{3}\Omega$
At 600 Hz:
$X_C=\frac{1}{2 \pi fC}=\frac{1}{2 \pi (600Hz)(2.50\times10^{-6}F)}=1.06\times10^{2}\Omega$
c. Equate the reactance of the capacitor to that of the inductor and solve for the angular frequency.
$$\frac{1}{2 \pi fC}=2 \pi fL$$
$$f = \frac{1}{2 \pi \sqrt{ LC}} $$
$$f = \frac{1}{2 \pi \sqrt{ (0.450H)( 2.50\times10^{-6}F)}}=150Hz$$
