Question

At what frequency is the reactance of a \(20.0-\mathrm{mH}\) inductor equal to \(18.8 \Omega ?\)

Short answer

Answer: The frequency is approximately 150.38 Hz.

Step-by-step solution

Write down the reactance formula for an inductor

To find the frequency, we will use the formula for the reactance of an inductor: \(X_L = 2\pi fL\), where \(X_L\) is the reactance, \(f\) is the frequency, and \(L\) is the inductance.

Substitute the given values into the formula

We know the reactance \(X_L = 18.8\Omega\) and the inductance \(L = 20.0mH = 0.020H\). Now, we can substitute these values into the formula: \(18.8 = 2\pi f (0.020)\).

Solve the equation for the frequency \(f\)

To find the frequency \(f\), we need to isolate \(f\) on one side of the equation. First, divide both sides by \(2\pi(0.020)\): \(f = \frac{18.8}{2\pi(0.020)}\).

Calculate the frequency

Now, we can calculate the frequency by plugging the numbers into the formula: \(f = \frac{18.8}{2\pi(0.020)} \approx 150.38 Hz\).

State the final answer

The frequency at which the reactance of a \(20.0mH\) inductor is equal to \(18.8\Omega\) is approximately \(150.38 Hz\).