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

Answer:

The resistance of the inductor at resonance is 258.76 ohms.

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.

$\omega L = \frac{1}{\omega C}\\\\2\pi F_0 L = \frac{1}{2\pi F_0 C}\\\\F_0 = \frac{1}{2\pi\sqrt{LC} }$

Where;

F₀ is the resonance frequency

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

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.