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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-\(\mu\)F 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-\(\mu\)F capacitor?

Short Answer

Expert verified
(a) 169.65 ohms and 1696.5 ohms. (b) 1061.03 ohms and 106.1 ohms. (c) Frequency is 149.91 Hz.

Step by step solution

01

Understand Inductive Reactance

The reactance of an inductor (denoted as \(X_L\)) is calculated using the formula \(X_L = 2\pi fL\), where \(f\) is the frequency and \(L\) is the inductance.
02

Compute Inductor Reactance for 60 Hz

Substitute \(f = 60 \text{ Hz}\) and \(L = 0.450 \text{ H}\) into the formula: \(X_L = 2\pi \times 60 \times 0.450 = 169.65 \text{ ohms}\).
03

Compute Inductor Reactance for 600 Hz

Substitute \(f = 600 \text{ Hz}\) and \(L = 0.450 \text{ H}\) into the formula: \(X_L = 2\pi \times 600 \times 0.450 = 1696.5 \text{ ohms}\).
04

Understand Capacitive Reactance

The reactance of a capacitor (denoted as \(X_C\)) is calculated by the formula \(X_C = \frac{1}{2\pi fC}\), where \(f\) is the frequency and \(C\) is the capacitance.
05

Compute Capacitor Reactance for 60 Hz

Substitute \(f = 60 \text{ Hz}\) and \(C = 2.50 \times 10^{-6} \text{ F}\) into the formula: \(X_C = \frac{1}{2\pi \times 60 \times 2.50 \times 10^{-6}} = 1061.03 \text{ ohms}\).
06

Compute Capacitor Reactance for 600 Hz

Substitute \(f = 600 \text{ Hz}\) and \(C = 2.50 \times 10^{-6} \text{ F}\) into the formula: \(X_C = \frac{1}{2\pi \times 600 \times 2.50 \times 10^{-6}} = 106.1 \text{ ohms}\).
07

Equate Reactances for Inductor and Capacitor

Set \(X_L = X_C\) to find the frequency where the reactances are equal. So, \(2\pi fL = \frac{1}{2\pi fC}\). Rearranging gives \(f^2 = \frac{1}{(2\pi)^2 LC}\).
08

Solve for Frequency

Substitute \(L = 0.450 \text{ H}\) and \(C = 2.50 \times 10^{-6} \text{ F}\) into the equation: \(f = \frac{1}{2\pi \sqrt{0.450 \times 2.50 \times 10^{-6}}} = 149.91 \text{ Hz}\).

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Key Concepts

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

Inductive Reactance
The concept of inductive reactance is crucial in understanding how inductors behave in AC circuits. Inductive reactance, denoted as \(X_L\), represents the opposition that an inductor provides to the change of current. It is calculated using the formula: \[X_L = 2\pi fL\] where \(f\) is the frequency in hertz (Hz) and \(L\) is the inductance in henrys (H).
  • The reactance increases with frequency; higher frequencies cause greater opposition.
  • This is why inductors are more effective at blocking high-frequency signals than low-frequency signals.
By substituting the given inductance and frequency values into the equation, one can easily calculate the reactance at any desired frequency, as seen in the original problem. This concept is a cornerstone of AC circuit analysis and helps to predict the inductive behavior in systems accurately.
Capacitive Reactance
Capacitive reactance, represented as \(X_C\), describes how a capacitor opposes changes in voltage in AC circuits. The formula for capacitive reactance is: \[X_C = \frac{1}{2\pi fC}\] where \(f\) is the frequency in hertz (Hz) and \(C\) is the capacitance in farads (F).
  • Unlike inductors, capacitors decrease their reactance as frequency increases. High frequencies pass through capacitors more easily.
  • This characteristic makes capacitors effective as filters in circuits that discriminate against low frequencies, a feature commonly used in electronic components like oscillators and filters.
Understanding capacitive reactance is key to predicting how capacitors will respond at varying frequencies, allowing for better circuit designs.
Frequency Calculation
Calculating the frequency at which inductive and capacitive reactances are equal is an interesting exercise. This frequency is commonly referred to as the resonant frequency in tank circuits. It can be determined by setting \(X_L = X_C\), which results in the formula: \[f^2 = \frac{1}{(2\pi)^2 LC}\] Solving for \(f\) gives: \[f = \frac{1}{2\pi \sqrt{LC}}\] By substituting the values of inductance \(L\) and capacitance \(C\) from the problem, you can find the specific frequency where both reactances match.
  • This is a fundamental principle in designing resonant circuits, such as those used in radios and other communication equipment.
  • Finding this frequency allows for the optimization of signal strengths in various applications.
Such frequency calculations are integral in ensuring efficient circuit operations and minimizing reactive losses.
Inductor and Capacitor Comparison
Inductors and capacitors are vital components in electronics, each serving distinct yet complementary roles. Understanding their differences aids in constructing effective AC circuits. Here鈥檚 a comparison:
  • Inductors resist changes in current by storing energy in a magnetic field when current flows through them.
    • They are more effective at high frequencies.
    • Commonly used in tuning circuits, filtering applications, and transformers.
  • Capacitors resist changes in voltage by storing energy in an electric field between plates.
    • They are more effective at low frequencies.
    • Typically used in power supply circuits, coupling/decoupling applications, and energy storage.
  • Working Together:
    • Both can be combined to form LC circuits that resonate at particular frequencies, making them ideal for filtering signals and tuning applications.
The synergy between inductors and capacitors is key in designing efficient electronic devices, especially in alternating current (AC) applications.

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Most popular questions from this chapter

An \(L-R-C\) series circuit is constructed using a 175-\(\Omega\) resistor, a 12.5-\(\mu\)F capacitor, and an 8.00-mH inductor, all connected across an ac source having a variable frequency and a voltage amplitude of 25.0 V. (a) At what angular frequency will the impedance be smallest, and what is the impedance at this frequency? (b) At the angular frequency in part (a), what is the maximum current through the inductor? (c) At the angular frequency in part (a), find the potential difference across the ac source, the resistor, the capacitor, and the inductor at the instant that the current is equal to one-half its greatest positive value. (d) In part (c), how are the potential differences across the resistor, inductor, and capacitor related to the potential difference across the ac source?

An \(L-R-C\) series circuit draws 220 W from a 120-V (rms), 50.0-Hz ac line. The power factor is 0.560, and the source voltage leads the current. (a) What is the net resistance \(R\) of the circuit? (b) Find the capacitance of the series capacitor that will result in a power factor of unity when it is added to the original circuit. (c) What power will then be drawn from the supply line?

An \(L-R-C\) series circuit has R = 500 \(\Omega\), L = 2.00 H, \(C\) = 0.500 \(\mu\)F, and \(V\) = 100 V. (a) For \(\omega\) = 800 rad/s, calculate \(V_R , V_L, V_C\), and \(\phi\). Using a single set of axes, graph \(v\), \(v_R , v_L\), and \(v_C\) as functions of time. Include two cycles of \(v\) on your graph. (b) Repeat part (a) for \(\omega\) = 1000 rad/s. (c) Repeat part (a) for \(\omega = 1250\space rad/s\).

A 250-\(\Omega\) resistor is connected in series with a 4.80-\(\mu\)F capacitor and an ac source. The voltage across the capacitor is v\(_C\) = (7.60 V)sin[120 rad/s)t]. (a) Determine the capacitive reactance of the capacitor. (b) Derive an expression for the voltage v\(_R\) across the resistor.

A resistance \(R\), capacitance \(C\), and inductance \(L\) are connected in series to a voltage source with amplitude \(V\) and variable angular frequency \(\omega\). If \(\omega\) = \(\omega$$_0\) , the resonance angular frequency, find (a) the maximum current in the resistor; (b) the maximum voltage across the capacitor; (c) the maximum voltage across the inductor; (d) the maximum energy stored in the capacitor; (e) the maximum energy stored in the inductor. Give your answers in terms of \(R\), \(C\), \(L\), and \(V\).
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