Answer:
Explanation:
Given;
current at resonance, I = 0.2 mA
applied voltage, V = 250 mV
resonance frequency, fâ‚€ = 100 kHz
capacitance of the circuit, C = 0.04 μF
At resonance, capacitive reactance ([tex]X_c[/tex]) is equal to inductive reactance ([tex]X_l[/tex]),
[tex]Z = \sqrt{R^2 + (X_ l - X_c)^2} \\\\But \ X_l= X_c\\\\Z = R[/tex]
Where;
R is the resistance of the circuit, calculated as;
[tex]R = \frac{V}{I} \\\\R = \frac{250 \ \times \ 10^{-3}}{0.2 \ \times \ 10^{-3}} \\\\R = 1250 \ ohms[/tex]
The inductive reactance is calculated as;
[tex]X_l = X_c = \frac{1}{\omega C} = \frac{1}{2\pi f_o C} = \frac{1}{2\pi (100\times 10^3)(0.04\times 10^{-6} ) } = 39.789 \ ohms\\[/tex]
The inductance is calculated as;
[tex]X_l = \omega L = 2\pi f_o L\\\\L = \frac{X_l}{2\pi f_o}\\\\L = \frac{39.789}{2\pi (100 \times 10^3)} \\\\L= 6.3 \ \times \ 10^{-5} \ H\\\\L = 0.063 \times \ 10^{-3} \ H\\\\L = 0.063 \ mH[/tex]
