Question

The current at resonance in a series L-C-R circuit is 0.2mA. If the applied voltage is 250mV at a frequency of 100 kHz and the circuit capacitance is 0.04 microfrad . Find the circuit resistance and inductance

Answer 1

Answer:

• The resistance of the circuit is 1250 ohms

• The inductance of the circuit is 0.063 mH.

Explanation:

Given;

current at resonance, I = 0.2 mA

applied voltage, V = 250 mV

resonance frequency, f₀ = 100 kHz

capacitance of the circuit, C = 0.04 μF

At resonance, capacitive reactance ($X_c$) is equal to inductive reactance ($X_l$),

$Z = \sqrt{R^2 + (X_ l - X_c)^2} \\\\But \ X_l= X_c\\\\Z = R$

Where;

R is the resistance of the circuit, calculated as;

$R = \frac{V}{I} \\\\R = \frac{250 \ \times \ 10^{-3}}{0.2 \ \times \ 10^{-3}} \\\\R = 1250 \ ohms$

The inductive reactance is calculated as;

$X_l = X_c = \frac{1}{\omega C} = \frac{1}{2\pi f_o C} = \frac{1}{2\pi (100\times 10^3)(0.04\times 10^{-6} ) } = 39.789 \ ohms$

The inductance is calculated as;

$X_l = \omega L = 2\pi f_o L\\\\L = \frac{X_l}{2\pi f_o}\\\\L = \frac{39.789}{2\pi (100 \times 10^3)} \\\\L= 6.3 \ \times \ 10^{-5} \ H\\\\L = 0.063 \times \ 10^{-3} \ H\\\\L = 0.063 \ mH$