Question

An inductor has an impedance of \(30.0 \Omega\) and a resistance of $20.0 \Omega\( at a frequency of \)50.0 \mathrm{Hz}$. What is the inductance? (Model the inductor as an ideal inductor in series with a resistor.)

Short answer

Question: Given an inductor with an impedance of 30.0 Ω, a resistance of 20.0 Ω, and an operating frequency of 50.0 Hz, find the inductance of the inductor. Answer: The inductance of the inductor is approximately 0.064 H.

Step-by-step solution

Write down the impedance formula for an inductor

The impedance formula for an ideal inductor is given by \(Z_L = j \omega L\), where \(Z_L\) is the impedance of the inductor, \(j\) is the imaginary unit, \(\omega\) is the angular frequency, and \(L\) is the inductance.

Convert the frequency into angular frequency

The formula to convert frequency (\(f\)) into angular frequency (\(\omega\)) is \(\omega = 2 \pi f\). Given \(f = 50.0 \mathrm{Hz}\), we can calculate the angular frequency: \(\omega = 2 \pi (50.0) = 100 \pi \, \mathrm{rad/s}\).

Use the Pythagorean theorem to find the impedance of the ideal inductor

The impedance of the inductor (\(Z\)) is given to be \(30.0 \Omega\). Since the combined system is an ideal inductor and a resistor in series, we can use the Pythagorean theorem to find the impedance (\(Z_L\)) of the ideal inductor: \(Z = \sqrt{R^2 + Z_L^2}\). Let's solve for \(Z_L\): \(Z_L = \sqrt{Z^2 - R^2} = \sqrt{(30.0)^2 - (20.0)^2} = 20.0 \, \Omega\).

Calculate the inductance from the impedance of the ideal inductor

Using the impedance formula for an ideal inductor (\(Z_L = j \omega L\)) and the given information, we can solve for the inductance (\(L\)): \(L = \frac{Z_L}{j \omega} = \frac{20.0 \Omega}{100 \pi \, \mathrm{rad/s} \times j} = \frac{20.0}{100 \pi} H = \frac{1}{5 \pi} H \approx 0.064 \, H\). Thus, the inductance of the inductor is approximately \(0.064 \, H\).