(a) Compute the reactance of a 0.450-H inductor at frequencies of 60.0 Hz and 600...

Step 1
Given
The inductance of the coil is ${\displaystyle L=0.45H}$.
Solution
(a) For ac source connected with a coil, the inductance reactance ${\displaystyle X_L}$ in the circuit is given by equation 31.12 in the form
${\displaystyle X_L=2\pi fL}$
This inductance reactance is related to the frequency f and the inductance L of the coil. Now, we want to get ${\displaystyle X_L}$ at ${\displaystyle f=60Hz}$ and 600 Hz.
For ${\displaystyle f=60Hz}$:
${\displaystyle X_L=2\pi fL=2\pi \left(60Hz\right)\left(0.45H\right)=170\mathrm{\Omega }}$
For ${\displaystyle f=600Hz}$ :
${\displaystyle X_L=2\pi fL=2\pi \left(600Hz\right)\left(0.45H\right)=1700\mathrm{\Omega }}$

Step 2
(b) The capacitance of the capacitor is ${\displaystyle C=2.5\mu F}$. The capacitive reactance ${\displaystyle X_C}$ in the circuit is given by equation 31.18 in the form
${\displaystyle X_C=\frac{1}{2\pi fC}}$
Now, we calcaute ${\displaystyle X_C}$ or 60 Hz and 600 Hz as next
${\displaystyle X_{C\left(60\right)}=\frac{1}{2\pi fC}=\frac{1}{2\pi \left(60Hz\right)(2.5\times {10}^{-6}F)}=1.06\times {10}^3\mathrm{\Omega }}$
${\displaystyle X_{C\left(600\right)}=\frac{1}{2\pi fC}=\frac{1}{2\pi \left(600Hz\right)(2.5\times {10}^{-6}F)}=106\mathrm{\Omega }}$

Step 3
(c) In this part, the two reactance $X_{C}X\text{and}{X}_{L}X$ are the same but at a specific frequency f. We are asked to determine the value of this frequency.
$X_C=X_L$
$\frac{1}{2\pi fC}=2\pi fL$
$f=\frac{1}{2\pi \sqrt{LC}}$
Now plug the values for LL and CC into equation (3) to get f
$=\frac{1}{2\pi \sqrt{LC}}=\frac{1}{2\pi \sqrt{(0.45H)(2.5\times {10}^{-6}F)}}=150Hz$