/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 29 An inductor has a reactance of \... [FREE SOLUTION] | 91Ó°ÊÓ

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An inductor has a reactance of \(56.5 \Omega\) at \(75.0 \mathrm{Hz}\). What is its reactance at \(60.0 \mathrm{Hz}\) ?

Short Answer

Expert verified
The reactance at 60.0 Hz is approximately 45.2 Ω.

Step by step solution

01

Understand Inductive Reactance Formula

Inductive reactance \( X_L \) is given by the formula \( X_L = 2\pi f L \), where \( f \) is the frequency in hertz and \( L \) is the inductance in henrys. The reactance changes with frequency.
02

Find Inductance Value

The original reactance \(56.5 \Omega\) at \(75.0 \mathrm{Hz}\) allows us to find the inductance \( L \). Rearrange the formula to \( L = \frac{X_L}{2\pi f} \) and substitute values: \( L = \frac{56.5}{2\pi \times 75.0} \approx 0.12 \mathrm{H}\).
03

Calculate New Reactance

Use the inductance found to calculate the reactance at \(60.0 \mathrm{Hz}\) using the formula \( X_L = 2\pi f L \). Substitute values: \( X_L = 2\pi \times 60.0 \times 0.12 \approx 45.2 \Omega \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Frequency Dependency
Inductive reactance is not a fixed value; it depends on the frequency of the alternating current passing through the inductor. This relationship is expressed mathematically as \( X_L = 2\pi f L \), where \( X_L \) is the inductive reactance, \( f \) is the frequency in hertz, and \( L \) is the inductance in henrys.
  • Higher frequencies lead to higher reactance.
  • Lower frequencies result in lower reactance.
This relationship shows that as the frequency changes, the reactance will adjust accordingly. In the given exercise, as the frequency decreases from 75.0 Hz to 60.0 Hz, the reactance also decreases, illustrating this dependency. Understanding this concept is crucial when calculating how different frequencies affect an inductor's behavior in AC circuits.
Inductance Calculation
Finding the inductance of an inductor is a key step in analyzing its reactance at different frequencies. The formula \( L = \frac{X_L}{2\pi f} \) is derived from rearranging the standard reactance formula. For instance, with a given reactance of 56.5 Ω at a frequency of 75.0 Hz, substitute these values into the formula:\[L = \frac{56.5}{2\pi \times 75.0} \approx 0.12 \text{ H}\]This calculated value for inductance is a vital factor used to find the new reactance at a different frequency, ensuring that calculations are accurate and internally consistent. Inductance itself remains constant regardless of frequency, serving as a fundamental property of the inductor.
Formula Manipulation
Manipulating the inductive reactance formula is essential for solving problems related to different frequencies and obtaining desired values such as inductance or reactance. Begin by understanding the basic form \( X_L = 2\pi f L \). To find the inductance given reactance and frequency, the formula can be rearranged into:\[L = \frac{X_L}{2\pi f}\]This step is crucial for calculating the inductance when initial conditions are known, as seen in the exercise. For calculating the new reactance at a different frequency:\[X_L = 2\pi f L\]Both forms demonstrate the versatility and necessity of formula manipulation for correctly analyzing circuits' responses to frequency changes. Having a firm grasp on this concept allows for solving a variety of electrical engineering problems efficiently.

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Most popular questions from this chapter

An \(R L C\) circuit with \(R=25.0 \Omega, L=325 \mathrm{mH},\) and \(C=\) \(45.2 \mu \mathrm{F}\) is connected to an ac generator with an \(\mathrm{rms}\) voltage of 24 V. Determine the average power delivered to this circuit when the frequency of the generator is (a) equal to the resonance frequency, (b) twice the resonance frequency, and (c) half the resonance frequency.

An ac generator with a frequency of \(105 \mathrm{Hz}\) and an rms voltage of \(22.5 \mathrm{V}\) is connected in series with a \(10.0-\mathrm{k} \Omega\) resistor and a \(0.250-\mu \mathrm{F}\) capacitor. What is the rms current in this circuit?

The rms current in an \(R L\) circuit is \(0.26 \mathrm{A}\) when it is connected to an ac generator with a frequency of \(60.0 \mathrm{Hz}\) and an (a) Given that the inductor has an inducrms voltage of 25 V. tance of \(145 \mathrm{mH},\) what is the resistance of the resistor? (b) Find the rms voltage across the resistor. (c) Find the rms voltage across the inductor. (d) Use your results from parts (b) and (c) to show that \(\sqrt{V_{\mathrm{rms}, R}^{2}+V_{\mathrm{rms}, l}^{2}}\) is equal to 25 V.

Find the power factor for an \(R C\) circuit connected to a \(70.0-1 \mathrm{Hz}\) generator with an rms voltage of 155 V. The values of \(R\) and \(C\). in this circuit are \(105 \Omega\) and \(82.4 \mu \mathrm{F}\), respectively.

IP Tuning a Radio A radio tuning circuit contains an RLC circuit with \(R=5.0 \Omega\) and \(L=2.8 \mu \mathrm{H.}\) (a) What capacitance is needed to produce a resonance frequency of \(85 \mathrm{MHz}\) ? (b) If the capacitance is increased above the value found in part (a), will the impedance increase, decrease, or stay the same? Explain. (c) Find the impedance of the circuit at reso- nance. (d) Find the impedance of the circuit when the capacitance is \(1 \%\) higher than the value found in part (a).

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