/*! 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 61 An electromagnetic car alarm. Yo... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 21: Problem 61

An electromagnetic car alarm. Your latest invention is a car alarm that produces sound at a particularly annoying frequency of 3500 \(\mathrm{Hz}\) . To do this, the car-alarm circuitry must produce an alternating electric current of the same frequency. That's why your design includes an inductor and a capacitor in series. The maximum voltage across the capacitor is to be 12.0 \(\mathrm{V}\) (the same voltage as the car battery). To produce a sufficiently loud sound, the capacitor must store 0.0160 \(\mathrm{J}\) of energy. What values of capacitance and inductance should you choose for your car-alarm circuit?

Short Answer

Expert verified
Capacitance \( C \approx 222 \, \mu\text{F} \), Inductance \( L \approx 5.79 \, \text{mH} \).

Step by step solution

01

Understanding Capacitor Energy Formula

The energy stored in a capacitor is given by the formula \( E = \frac{1}{2} C V^2 \), where \( E \) is the energy, \( C \) is the capacitance, and \( V \) is the voltage across the capacitor. We need to solve for \( C \) using \( E = 0.0160 \, \text{J} \) and \( V = 12.0 \, \text{V} \).
02

Solving for Capacitance

Rearrange the formula to solve for \( C \):\[ C = \frac{2E}{V^2} \]Substituting the given values, we have:\[ C = \frac{2 imes 0.0160}{12.0^2} \approx 2.22 \times 10^{-4} \, \text{F} \] (or 222 \( \mu \text{F} \)).
03

Understanding Resonant Frequency Formula

The resonant frequency \( f \) for an LC circuit is given by \( f = \frac{1}{2\pi\sqrt{LC}} \), where \( L \) is the inductance. We need to use this formula to find the inductance \( L \) given \( f = 3500 \, \text{Hz} \) and the \( C \) we calculated.
04

Solving for Inductance

Rearrange the formula to solve for \( L \):\[ L = \frac{1}{(2\pi f)^2 C} \]Substitute \( f = 3500 \, \text{Hz} \) and \( C \approx 2.22 \times 10^{-4} \, \text{F} \):\[ L = \frac{1}{(2\pi \times 3500)^2 \times 2.22 \times 10^{-4}} \approx 5.79 \times 10^{-3} \, \text{H} \] (or 5.79 \( \text{mH} \)).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Capacitance
Capacitance is a fundamental concept in electric circuits. It refers to the ability of a capacitor to store an electric charge. The measure of capacitance is given in farads (F). In our car alarm circuit, capacitance determines how much electric charge can be stored for a given voltage. This is crucial because the energy stored in the capacitor later helps in producing the sound.Let's break it down:
  • Capacitance is determined by the size and properties of the capacitor plates and the dielectric material between them.
  • The higher the capacitance, the more charge stored at a given voltage.The formula for energy stored is: \[ E = \frac{1}{2}CV^2 \] where \(C\) is the capacitance, \(V\) the voltage, and \(E\) the energy.
  • In this exercise, rearranging to find \(C\) gives: \[ C = \frac{2E}{V^2} \]
This allows us to calculate the required capacitance based on the desired energy storage and voltage limits.
Inductance
Inductance is a key property of inductors in a circuit. It reflects the inductor's ability to resist changes in current and is measured in henrys (H). Inductance plays a vital role in determining how the LC circuit resonates at a specific frequency.Here's how it works:
  • Inductance is affected by factors such as the number of turns in the coil and the core material used.
  • The primary role of inductance in an LC circuit is to join with capacitance, enabling resonance at a specific frequency.
  • To find the inductance needed, we use the formula for resonant frequency:\[ f = \frac{1}{2\pi\sqrt{LC}} \]
By rearranging this formula, we determine \(L\):\[ L = \frac{1}{(2\pi f)^2 C} \] helps us to compute the right inductance to safely and effectively create the annoying sound in the car alarm.
Resonant Frequency
The resonant frequency is an important characteristic of LC circuits. It's the frequency at which the circuit naturally oscillates. In this car alarm, resonant frequency allows the specific sound frequency of 3500 Hz to be generated.Some essential points:
  • Resonant frequency is determined by both the capacitance \(C\) and inductance \(L\).
  • The formula that relates them is:\[ f = \frac{1}{2\pi\sqrt{LC}} \]
  • Achieving the desired resonant frequency involves precise calculations of \(C\) and \(L\).
By ensuring both components are correctly calculated, resonance allows the circuit to produce the exact sound frequency required for the car alarm.
Capacitor Energy
Capacitor energy is stored energy due to an electric field between its plates. This energy can be used to power circuits, like in our car alarm. Understanding how much energy a capacitor can store influences the loudness of the sound produced.Let's break it down further:
  • Energy storage in a capacitor is defined by the formula:\[ E = \frac{1}{2}CV^2 \]
  • This means the energy stored depends on both the capacitance and the square of the voltage.
  • The stored energy in the car alarm's capacitor translates into the power emitted by the device, impacting the output sound volume.
This concept ensures that the capacitor in the circuit not only meets the voltage demands but also provides the necessary energy to achieve the desired sound level.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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

\(\bullet\) A circular loop of wire with a radius of 12.0 \(\mathrm{cm}\) is lying flat on a tabletop. A magnetic field of 1.5 \(\mathrm{T}\) is directed vertically upward through the loop (Figure 21.49 ). (a) If the loop is removed from the field region in a time interval of 2.0 \(\mathrm{ms}\) , find the average emf that will be induced in the wire loop during the extraction process. (b) If the loop is viewed looking down on it from above, is the induced current in the loop clockwise or counterclockwise?

A toroidal solenoid has a mean radius of 10.0 \(\mathrm{cm}\) and a cross- sectional area of 4.00 \(\mathrm{cm}^{2}\) and is wound uniformly with 100 turns. A second coil with 500 turns is wound uniformly on top of the first. What is the mutual inductance of these coils?

Two coils are wound around the same cylindrical form, like the coils in Example \(21.8 .\) When the current in the first coil is decreasing at a rate of \(0.242 \mathrm{A} / \mathrm{s},\) the induced emf in the second coil has magnitude 1.65 \(\mathrm{mV}\) . (a) What is the mutual inductance of the pair of coils? (b) If the second coil has 25 turns, what is the average magnetic flux through each turn when the current in the first coil equals 1.20 \(\mathrm{A} ?(\mathrm{c})\) If the current in the second coil increases at a rate of \(0.360 \mathrm{A} / \mathrm{s},\) what is the magnitude of the induced emf in the first coil?

In a physics laboratory experiment, a coil with 200 turns enclosing an area of 12 \(\mathrm{cm}^{2}\) is rotated from a position where its plane is perpendicular to the earth's magnetic field to one where its plane is parallel to the field. The rotation takes 0.040 s. The earth's magnetic field at the location of the laboratory is \(6.0 \times 10^{-5} \mathrm{T.}\) (a) What is the total magnetic flux through the coil before it is rotated? After it is rotated? (b) What is the average emf induced in the coil?
See all solutions

Recommended explanations on Physics Textbooks

Measurements

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Electrostatics

Read Explanation

Radiation

Read Explanation

Classical Mechanics

Read Explanation

Scientific Method Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.