A) [tex]\omega = \frac{1}{\sqrt{LC}}[/tex]
The magnitude of the capacitive reactance is given by
[tex]X_C = \frac{1}{\omega C}[/tex]
where
[tex]\omega[/tex] is the angular frequency
C is the capacitance
While the magnitude of the inductive capacitance is given by
[tex]X_L = \omega L[/tex]
where L is the inductance.
Since we want the two reactances to be equal, we have
[tex]X_C = X_L[/tex]
So we find
[tex]\frac{1}{\omega C}= \omega L\\\omega^2 = \frac{1}{LC}\\\omega = \frac{1}{\sqrt{LC}}[/tex]
B) 7449 rad/s
In this case, we have
[tex]L=5.30 mH = 5.3\cdot 10^{-3}H[/tex] is the inductance
[tex]C= 3.40 \mu F= 3.40 \cdot 10^{-6}F[/tex] is the capacitance
Therefore, substituting in the formula for the angular frequency, we find
[tex]\omega=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{(5.30\cdot 10^{-3}H)(3.40\cdot 10^{-6} F)}}=7449 rad/s[/tex]
C) [tex]39.5 \Omega[/tex]
Now we can us the formulas of the reactances written in part A). We have:
- Capacitive reactance:
[tex]X_C = \frac{1}{\omega C}=\frac{1}{(7449 rad/s)(3.40\cdot 10^{-6}F)}=39.5 \Omega[/tex]
- Inductive reactance:
[tex]X_L = \omega L=(7449 rad/s)(5.30\cdot 10^{-3}H)=39.5 \Omega[/tex]
