Answer:
a)
[tex]z=561.7[/tex]
b)
[tex]z=214.1[/tex]
Explanation:
L = inductance of the Inductor = 3.14 mH = 0.00314 H
C = capacitance of the capacitor = 5.08 x 10⁻⁶ F
a)
f = frequency = 55.7 Hz
Impedance is given as
[tex]z=\frac{1}{2\pi fC} - 2\pi fL[/tex]
[tex]z=\frac{1}{2(3.14) (55.7)(5.08\times 10^{-6})} - 2(3.14) (55.7)(0.00314)[/tex]
[tex]z=561.7[/tex]
b)
f = frequency = 11000 Hz
Impedance is given as
[tex]z= - \frac{1}{2\pi fC} + 2\pi fL[/tex]
[tex]z= - \frac{1}{2(3.14) (11000)(5.08\times 10^{-6})} + 2(3.14) (11000)(0.00314)[/tex]
[tex]z=214.1[/tex]
