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Chapter 21: Problem 24

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

Short Answer

Expert verified
Answer: The frequency is approximately 150.38 Hz.

Step by step solution

01

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.
02

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: \(18.8 = 2\pi f (0.020)\).
03

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 \(2\pi(0.020)\): \(f = \frac{18.8}{2\pi(0.020)}\).
04

Calculate the frequency

Now, we can calculate the frequency by plugging the numbers into the formula: \(f = \frac{18.8}{2\pi(0.020)} \approx 150.38 Hz\).
05

State the final answer

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

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