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Ch 31: Alternating Current
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
Chapter 31, Problem 8

(a) Compute the reactance of a 0.450-H inductor at frequencies of 60.0 Hz and 600 Hz. (b) Compute the reactance of a 2.50-μF capacitor at the same frequencies. (c) At what frequency is the reactance of a 0.450-H inductor equal to that of a 2.50-μF capacitor?

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1
To find the reactance of an inductor, use the formula for inductive reactance: \( X_L = 2\pi f L \), where \( f \) is the frequency and \( L \) is the inductance. For part (a), calculate \( X_L \) at 60.0 Hz and 600 Hz using \( L = 0.450 \) H.
For the reactance of a capacitor, use the formula for capacitive reactance: \( X_C = \frac{1}{2\pi f C} \), where \( C \) is the capacitance. For part (b), calculate \( X_C \) at 60.0 Hz and 600 Hz using \( C = 2.50 \times 10^{-6} \) F.
To find the frequency at which the reactance of the inductor equals that of the capacitor, set \( X_L = X_C \). This gives the equation \( 2\pi f L = \frac{1}{2\pi f C} \).
Solve the equation \( 2\pi f L = \frac{1}{2\pi f C} \) for \( f \). This involves rearranging the equation to find \( f = \frac{1}{2\pi \sqrt{LC}} \).
Substitute the given values of \( L = 0.450 \) H and \( C = 2.50 \times 10^{-6} \) F into the equation \( f = \frac{1}{2\pi \sqrt{LC}} \) to find the frequency at which the reactances are equal.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Inductive Reactance

Inductive reactance is the opposition that an inductor presents to alternating current, due to the magnetic field created by the current. It is calculated using the formula X_L = 2πfL, where X_L is the inductive reactance, f is the frequency, and L is the inductance. This concept is crucial for determining how an inductor behaves at different frequencies.
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Capacitive Reactance

Capacitive reactance is the opposition that a capacitor presents to alternating current, due to the electric field created between its plates. It is calculated using the formula X_C = 1/(2πfC), where X_C is the capacitive reactance, f is the frequency, and C is the capacitance. Understanding capacitive reactance is essential for analyzing how a capacitor affects AC circuits at various frequencies.
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Frequency and Reactance Equality

The frequency at which the reactance of an inductor equals that of a capacitor is determined by setting their reactance formulas equal: 2πfL = 1/(2πfC). Solving this equation gives the frequency at which the inductive and capacitive reactances are equal, highlighting the balance point in AC circuits where inductive and capacitive effects cancel each other out.
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