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Chapter 31: Problem 5

(a) What is the reactance of a \(3.00 \mathrm{H}\) inductor at a frequency of \(80.0 \mathrm{~Hz} ?\) (b) What is the inductance of an inductor whose reactance is \(120 \Omega\) at \(80.0 \mathrm{~Hz} ?\) (c) What is the reactance of a \(4.00 \mu \mathrm{F}\) capacitor at a frequency of \(80.0 \mathrm{~Hz}\) ? (d) What is the capacitance of a capacitor whose reactance is \(120 \Omega\) at \(80,0 \mathrm{~Hz} ?\)

Short Answer

Expert verified
a) The reactance of a 3.00 H inductor at a frequency of 80.0 Hz is 1508 \(\Omega\). b) The inductance of an inductor whose reactance is 120 \(\Omega\) at 80.0 Hz is 0.24 H. c) The reactance of a 4.00 \(\mu F\) capacitor at a frequency of 80.0 Hz is 497.36 \(\Omega\). d) The capacitance of a capacitor whose reactance is 120 \(\Omega\) at 80.0 Hz is 16.6 nF.

Step by step solution

01

Calculate the reactance of the inductor

Using the formula \(X_L = 2\pi fL\), where f=80.0 Hz and L=3.00 H, we have \(X_L = 2\pi*80*3 = 1508 \Omega\).
02

Calculate the inductance of the inductor

To find the inductance, we rearrange the formula to be \(L = X_L/(2\pi f)\), where \(X_L=120 \Omega\) and f=80.0 Hz. Plugging in the values, we get \(L = 120/(2*\pi*80) = 0.24 H\) .
03

Calculate the reactance of the capacitor

For the capacitor, we use the formula \(X_C = 1/(2\pi fC)\) in which f=80.0 Hz and C=4.00鈥壜礔 (or 4.00*\(10^{-6}\)鈥塅 in standard SI units). The resulting value for \(X_C\) is \(X_C = 1/(2*\pi*80*4*10^{-6} = 497.36 \Omega\).
04

Calculate the capacitance of the capacitor

To find the capacitance, we can rearrange the formula to be \(C = 1/(2\pi f*X_C)\), where \(X_C=120 \Omega\) and f=80.0 Hz. Plugging in these values, we get \(C = 1/(2*\pi*80*120) = 16.6 nF\) (nano Farads) or \(16.6*10^-9 F\) in SI units.

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Key Concepts

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

Inductor
An inductor is a fascinating component in electrical circuits, often used to regulate the flow of current. It consists of a coil of wire and has the ability to store energy in its magnetic field when current flows through it. This property is called inductance, usually measured in henries (H).
You can think of an inductor as a kind of inertia for electricity鈥攋ust as a moving object resists changes to its motion, the inductor resists changes to current flow. This results in a phenomenon called reactance, which leads to the circuit opposing the change in current.
  • Inductive Reactance: It is not resistance, but rather a form of opposition due to changes in current.
  • Inductance: It measures how effectively an inductor stores energy. The larger the inductor, the more energy it can store.
  • Formulas Used: The reactance (\(X_L\)) of an inductor can be calculated using the formula \(X_L = 2\pi f L\), where \(f\) is the frequency and \(L\) is the inductance.
Understanding these concepts is crucial for analyzing how inductors influence the behavior of AC circuits.
Capacitor
Unlike inductors, capacitors play the role of storing energy in an electric field. They consist of two conductive plates separated by an insulator, known as a dielectric. Capacitors can store and release electrical energy, which makes them very useful in various electronic applications.
They cause a different kind of reactance compared to inductors, called capacitive reactance. This opposition occurs when the current attempts to move through the capacitor, which both stores and releases energy depending on the voltage across it.
  • Capacitive Reactance: This is the measure of a capacitor's ability to resist changes in voltage, and it decreases with an increase in frequency.
  • Capacitance: It measures a capacitor's ability to store charge; higher capacitance means more charge is stored. It is measured in farads (F).
  • Formulas Used: The reactance (\(X_C\)) of a capacitor is found by using the formula \(X_C = 1/(2\pi f C)\), where \(f\) is the frequency and \(C\) is the capacitance.
Capacitors are essential in tuning circuits, filtering signals, and even in power supply systems.
Frequency
Frequency is a fundamental concept in the study of AC circuits. It is defined as the rate at which current changes direction per second, measured in Hertz (Hz). The behavior of both inductors and capacitors significantly changes with frequency variation.
In simple terms, frequency dictates the oscillation rate of AC currents and voltages in a circuit.
  • Effects on Inductors: As the frequency increases, the inductive reactance also increases. This means inductors can better resist higher frequencies.
  • Effects on Capacitors: Contrary to inductors, as frequency increases, capacitive reactance decreases, which allows more current to pass through at higher frequencies.
  • Relationship with Reactance: The formulas \(X_L = 2\pi f L\) and \(X_C = 1/(2\pi f C)\) show how crucial frequency is in determining reactance.
Understanding frequency helps in designing circuits that need to work efficiently over a range of conditions.
Impedance
Impedance is a comprehensive concept in AC circuits that combines both resistance and reactance to show how much a circuit resists the flow of current. Represented in ohms (\(\Omega\)), impedance considers contributions from resistors, inductors, and capacitors.
You can view impedance as the overall opposition a circuit presents to AC current, blending the resistive and reactive effects.
  • Components: Impedance is a complex quantity, as it handles both magnitude and phase differences in AC circuits.
  • Importance of Reactance: Both inductive and capacitive reactance contribute to impedance by affecting how circuits react to changes in current and voltage.
  • Calculation: Impedance can be calculated using the formula \(Z = \sqrt{R^2 + (X_L - X_C)^2}\), where \(R\) is resistance, \(X_L\) is inductive reactance, and \(X_C\) is capacitive reactance.
Proper understanding of impedance is crucial for analyzing and designing effective AC circuits.

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Most popular questions from this chapter

An \(L-R-C\) series circuit with \(L=0.120 \mathrm{H}, R=240 \Omega,\) and \(C=7.30 \mu \mathrm{F}\) carries an \(\mathrm{rms}\) current of \(0.450 \mathrm{~A}\) with a frequency of \(400 \mathrm{~Hz}\) (a) What are the phase angle and power factor for this circuit? (b) What is the impedance of the circuit? (c) What is the rms voltage of the source? (d) What average power is delivered by the source? (e) What is the average rate at which electrical energy is converted to thermal energy in the resistor? (f) What is the average rate at which electrical energy is dissipated (converted to other forms) in the capacitor? (g) In the inductor?

You plan to take your hair dryer to Europe, where the electrical outlets put out \(240 \mathrm{~V}\) instead of the \(120 \mathrm{~V}\) seen in the United States. The dryer puts out \(1600 \mathrm{~W}\) at \(120 \mathrm{~V}\). (a) What could you do to operate your dryer via the \(240 \mathrm{~V}\) line in Europe? (b) What current will your dryer draw from a European outlet? (c) What resistance will your dryer appear to have when operated at \(240 \mathrm{~V}\) ?

A transformer connected to a \(120 \mathrm{~V}\) (rms) ac line is to supply \(12.0 \mathrm{~V}\) (rms) to a portable electronic device. The load resistance in the secondary is \(5.00 \Omega\). (a) What should the ratio of primary to secondary turns of the transformer be? (b) What rms current must the secondary supply? (c) What average power is delivered to the load? (d) What resistance connected directly across the \(120 \mathrm{~V}\) line would draw the same power as the transformer? Show that this is equal to \(5.00 \Omega\) times the square of the ratio of primary to secondary turns.

An \(L-R-C\) series circuit consists of a source with voltage amplitude \(120 \mathrm{~V}\) and angular frequency \(50.0 \mathrm{rad} / \mathrm{s},\) a resistor with \(R=400 \Omega,\) an inductor with \(L=3.00 \mathrm{H},\) and a capacitor with capacitance \(C\). (a) For what value of \(C\) will the current amplitude in the circuit be a maximum? (b) When \(C\) has the value calculated in part (a), what is the amplitude of the voltage across the inductor?

\(\operatorname{In}\) an \(L-R-C\) series circuit the source is operated at its resonant angular frequency. At this frequency, the reactance \(X_{C}\) of the capacitor is \(200 \Omega\) and the voltage amplitude across the capacitor is \(600 \mathrm{~V}\). The circuit has \(R=300 \Omega\). What is the voltage amplitude of the source?
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