Question

A circuit contains a 305 ohm resistor, a 1.1 micro Farad capacitor, and a 42 mH inductor. At resonance, the impedance is determined to be 336.4 ohms. What is the resistance of the inductor?

Answer 1

The resistance of the inductor at resonance is 258.76 ohms.

Step-by-step explanation:

Given;

resistance of the resistor, R = 305 ohm

capacitance of the capacitor, C = 1.1 μF = 1.1 x 10⁻⁶ F

inductance of the inductor, L = 42 mH = 42 x 10⁻³ H = 0.042 H

At resonance the inductive reactance is equal to capacitive reactance.

`\omega L = (1)/(\omega C)\\\\2\pi F_0 L = (1)/(2\pi F_0 C)\\\\F_0 = (1)/(2\pi√(LC) )`

Where;

F₀ is the resonance frequency

`F_0 = (1)/(2\pi√(LC) ) \\\\F_0 = \frac{1}{2\pi\sqrt{(0.024)(1.1*10^(-6))} }\\\\F_0 =980.4 \ Hz`

The inductive reactance is given by;

`X_l = 2\pi F_0 L\\\\X_l = 2\pi (980.4)(0.042) \\\\X_l = 258.76 \ ohms`

Therefore, the resistance of the inductor at resonance is 258.76 ohms.