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

Short Answer

Expert verified
The inductive reactance is 169.65 ohms at 60 Hz and 1696.46 ohms at 600 Hz. The capacitive reactance is 1061.03 ohms at 60 Hz and 106.10 ohms at 600 Hz. The frequencies equalize at 149.44 Hz.

Step by step solution

01

Formula for Inductive Reactance

The reactance of an inductor is given by the formula: \( X_L = 2\pi f L \), where \( X_L \) is the inductive reactance, \( f \) is the frequency, and \( L \) is the inductance. To find the reactance, we'll substitute \( L = 0.450 \) H and the frequencies provided.
02

Calculate Inductive Reactance at 60 Hz

Substitute \( L = 0.450 \) H and \( f = 60 \) Hz into the formula: \( X_L = 2\pi \times 60 \times 0.450 \). After calculating, \( X_L \approx 169.65 \) ohms.
03

Calculate Inductive Reactance at 600 Hz

Substitute \( L = 0.450 \) H and \( f = 600 \) Hz into the formula: \( X_L = 2\pi \times 600 \times 0.450 \). After calculating, \( X_L \approx 1696.46 \) ohms.
04

Formula for Capacitive Reactance

The reactance of a capacitor is given by the formula: \( X_C = \frac{1}{2\pi f C} \), where \( X_C \) is the capacitive reactance, \( f \) is the frequency, and \( C \) is the capacitance. We'll use \( C = 2.50 \times 10^{-6} \) F for our calculations.
05

Calculate Capacitive Reactance at 60 Hz

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

Calculate Capacitive Reactance at 600 Hz

Substitute \( C = 2.50 \times 10^{-6} \) F and \( f = 600 \) Hz into the formula: \( X_C = \frac{1}{2\pi \times 600 \times 2.50 \times 10^{-6}} \). After calculating, \( X_C \approx 106.10 \) ohms.
07

Equate Inductive and Capacitive Reactance

Set the equations for inductive and capacitive reactance equal: \( 2\pi f L = \frac{1}{2\pi f C} \). Simplify to find \( f = \frac{1}{2\pi \sqrt{LC}} \).
08

Solve for Resonant Frequency

Substitute \( L = 0.450 \) H and \( C = 2.50 \times 10^{-6} \) F into the formula: \( f = \frac{1}{2\pi \sqrt{0.450 \times 2.50 \times 10^{-6}}} \). After calculating, \( f \approx 149.44 \) Hz, which is the frequency where the reactance of the inductor equals that of the capacitor.

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Key Concepts

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

Inductive Reactance
Inductive reactance is a property of inductors in a circuit, which opposes the change in current. It can be seen as a resistance, but only in terms of alternating current (AC). The formula for calculating the inductive reactance (\(X_L\)) is given by:\[ X_L = 2\pi f L \]Where:
  • \(X_L\) is the inductive reactance in ohms.
  • \(f\) is the frequency in hertz (Hz).
  • \(L\) is the inductance in henrys (H).

The reactance increases with frequency and inductance, meaning that higher frequencies face greater opposition when passing through an inductor.
This is important in AC circuits as it affects how the current behaves.
Inductive reactance is directly proportional to the frequency, which implies that as frequency increases, the opposition from the inductor also increases. Understanding this property is key for designing circuits that operate efficiently at specific frequencies.
Capacitive Reactance
Unlike inductive reactance, capacitive reactance arises in capacitors. It opposes the change in voltage rather than current. Capacitors store and release energy in the form of an electric field, so the reactance here depends on frequency as well. The formula is:\[ X_C = \frac{1}{2\pi f C} \]Where:
  • \(X_C\) is the capacitive reactance in ohms.
  • \(f\) is the frequency in hertz (Hz).
  • \(C\) is the capacitance in farads (F).

It's clear from the formula that the reactance of a capacitor is inversely proportional to the frequency; as frequency increases, reactance decreases.
This means capacitors allow more current to pass at higher frequencies.
Capacitive reactance is crucial when it comes to tuning circuits, as it can filter out unwanted frequencies. Knowing how capacitive reactance changes with frequency is essential for reducing noise in electronic circuits.
Resonant Frequency
Resonant frequency occurs when the reactance of the inductor (\(X_L\)) is equal to that of the capacitor (\(X_C\)) in a circuit. This creates a condition called resonance, where the impedance (total opposition to current) is minimized and only resistive elements are left.The formula to find the resonant frequency (\(f\)) is:\[ f = \frac{1}{2\pi \sqrt{LC}} \]Where:
  • \(L\) is the inductance in henrys (H).
  • \(C\) is the capacitance in farads (F).

The resonant frequency is significant in many applications such as radio transmitters and receivers, where precise frequencies are crucial for operation. At resonance, the system can oscillate with minimal external energy input, making it perfect for these technologies.
Understanding resonant frequency helps in creating efficient circuits that can maximize power usage while minimizing energy waste.
Frequency Calculation
Calculating frequency in circuits involving inductors and capacitors requires careful application of formulas relating to reactance. Typically, you would:
  • Use formulas for inductive and capacitive reactance to find reactance values at different frequencies.
  • Apply the resonant frequency formula to find where both reactances are equal.

For example, in finding where a inductor and capacitor have equal reactance, you'd set their respective formulas equal:\[ 2\pi f L = \frac{1}{2\pi f C} \]Solving gives us:\[ f = \frac{1}{2\pi \sqrt{LC}} \] This approach helps in computational problems like solving for frequencies that yield specific reactances, enhancing understanding of AC circuit behavior.
Mastering frequency calculations ensures proper circuit design and optimal functionality across different electronic applications.

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Most popular questions from this chapter

An \(L-R-C\) series circuit consists of a \(50.0-\Omega\) resistor, a \(10.0-\mu \mathrm{F}\) capacitor, a \(3.50-\mathrm{mH}\) inductor, and an ac voltage source of voltage amplitude 60.0 \(\mathrm{V}\) operating at 1250 \(\mathrm{Hz}\) . (a) Find the current amplitude and the voltage amplitudes across the inductor, the resistor, and the capacitor. Why can the voltage amplitudes add up to more than 60.0 \(\mathrm{V}\) (b) If the frequency is now doubled, but nothing else is changed, which of the quantities in part (a) will change? Find the new values for those that do change.

An \(L \cdot R-C\) series circuit consists of a \(2.50-\mu \mathrm{F}\) capacitor, a 5.00 -mH inductor, and a \(75.0-\Omega\) resistor connected across an ac source of voltage amplitude 15.0 \(\mathrm{V}\) having variable frequency. (a) Under what circumstances is the average power delivered to the circuit equal to \(\frac{1}{2} V_{\mathrm{Vms}} I_{\mathrm{rms}} ?\) (b) Under the conditions of part (a), what is the average power delivered to each circuit element and what is the maximum current through the capacitor?

The \(L \cdot R-C\) Parallel Circuit. A resistor, inductor, and capacitor are connected in parallel to an ac source with voltage amplitude \(V\) and angular frequency \(\omega\) . Let the source voltage be given by \(v=V \cos \omega t .\) (a) Show that the instantaneous voltages \(v_{R}, v_{L},\) and \(v_{C}\) at any instant are each equal to \(v\) and that \(i=i_{R}+i_{L}+i_{C},\) where \(i\) is the current through the source and \(i_{R},\) \(i_{L},\) and \(i_{C}\) are the currents through the resistor, the inductor, and the capacitor, respectively. (b) What are the phases of \(i_{R}, i_{L},\) and \(i_{C}\) with respect to \(v ?\) Use current phasors to represent \(i, i_{R}, i_{L},\) and \(i_{C}\) In a phasor diagram, show the phases of these four currents with respect to \(v\) . (c) Use the phasor diagram of part (b) to show that the current amplitude \(I\) for the current \(i\) through the source is given by \(I=\sqrt{I_{R}^{2}+\left(I_{C}-I_{L}\right)^{2}}\) (d) Show that the result of part (c) can be written as \(I=V / Z,\) with \(1 / Z=\sqrt{1 / R^{2}+(\omega C-1 / \omega L)^{2}}\)

A \(100-\Omega\) resistor, a \(0.100-\mu \mathrm{F}\) capacitor, and a \(0.300-\mathrm{H}\) inductor are connected in parallel to a voltage source with amplitude 240 \(\mathrm{V}\) (a) What is the resonance angular frequency? (b) What is the maximum current through the source at the resonance frequency? (c) Find the maximum current in the resistor at resonance. (d) What is the maximum current in the inductor at resonance? (e) What is the maximum current in the branch containing the capacitor at resonance? (f) Find the maximum energy stored in the inductor and in the capacitor at resonance.

A series circuit has an impedance of 60.0\(\Omega\) and a power factor of 0.720 at 50.0 \(\mathrm{Hz}\) . The source voltage lags the current. (a) What circuit element, an inductor or a capacitor, should be placed in series with the circuit to raise its power factor? (b) What size element will raise the power factor to unity?

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