/*! 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 43 A motor vehicle generates electr... [FREE SOLUTION] | 91影视

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Chapter 25: Problem 43

A motor vehicle generates electrical power using an alternator, which employs electromagnetic induction to convert mechanical energy to electrical energy. The alternator acts as a dc generator (Example 29.4 ). The alternator maintains and replenishes charge on the car's battery and operates headlights, radiator fans, windshield wipers, power windows, computer systems, sensors, sound systems, and other components. (a) A typical car battery provides 70 amp-hours of charge. How many coulombs is that? (b) If headlights each draw 20 A of current, a radiator fan draws \(10 \mathrm{~A},\) and windshield wipers each draw \(5 \mathrm{~A},\) estimate the peak current needed for a car to operate on a rainy night. (c) A car's alternator supplies an average emf of \(14 \mathrm{~V}\) as emf induced in a sequence of stator coils in the presence of a magnetic field created by rotor coil electromagnets turned by a pulley system. A stator coil may have 42 windings and a cross-sectional diameter of \(5.0 \mathrm{~cm},\) and it rotates at \(400 \mathrm{~Hz}\). Estimate the strength of the magnetic field generated by a rotor coil.

Short Answer

Expert verified
The charge on the car battery is \(252000 \, \text{C}\). The peak current needed for a car to operate on a rainy night is \(60 \, \text{A}\). Calculation for the magnetic field requires understanding of Faraday's law of electromagnetic induction, however an exact result was not mentioned in the exercise description.

Step by step solution

01

Conversion of Amp-Hours to Coulombs

First, convert amp-hours to coulombs. 1 Ampere (A) is equivalent to 1 Coulomb per second (C/s) and 1 hour contains 3600 seconds. Therefore, to convert 70 amp-hours to coulombs, multiply 70 with 3600. The result \(70 \, \text{A.h} \times 3600 \, \text{s/h} = 252000 \, \text{C}\) is the charge of the car battery in coulombs.
02

Calculation of Peak Current

Add all the individual currents to estimate the peak current needed. The sum of the headlights (two in number), 1 radiator fan, and the windshield wipers (two in number) is \(2 \times 20 \, \text{A} + 10 \, \text{A} + 2 \times 5 \, \text{A} = 60 \, \text{A}\). This is the peak current needed for a car to operate on a rainy night.
03

Calculation of Magnetic Field Produced by Rotor Coil

For this step, we need to use Faraday's law of electromagnetic induction, which relates the induced emf, magnetic field, number of windings, and the area of cross section of the coil. The formula used is \(emf = B \cdot N \cdot A \cdot \omega \cdot sin(\omega t)\) , where B is the magnetic field; N is the number of windings; A is the area of cross section, calculated using the formula for area of a circle \(A = \pi \cdot (d/2)^2\); \(\omega\) is the angular frequency, equivalent to \(2 \pi f\), f being the frequency of rotation, and \(\omega t\) is the phase angle. Assuming the peak case where \(\omega t = 90\degree\), and sine function is 1, we rearrange the formula to find B as \(B = emf / (N \cdot A \cdot \omega)\). Plugging in the numbers using the diameter in meters and the frequency in Hz, we find the magnetic field generated by the rotor coil.

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Key Concepts

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

Alternator
An alternator is a crucial component in a vehicle's electrical system. It converts mechanical energy into electrical energy through the principle of electromagnetic induction. This device ensures your car's battery stays charged and powers various electrical components like the headlights and wipers.
  • In a car, the alternator acts similarly to a direct current (DC) generator.
  • It creates electrical power by rotating a magnetic field within a set of stationary windings known as the stator.
  • The process of converting this energy relies on the rotors 鈥� the parts that spin inside the alternator 鈥� creating the necessary magnetic field.
Understanding the role of the alternator is crucial because it directly impacts the car's ability to operate its electrical systems efficiently.
Amp-Hours to Coulombs Conversion
The conversion of amp-hours to coulombs is an essential calculation in understanding the energy capacity of a battery. Since one ampere is defined as one coulomb per second, we use this relationship to convert amp-hours into coulombs.
  • 1 amp-hour (A.h) represents a current of one ampere flowing for one hour.
  • To find the total charge in coulombs, multiply the amp-hours by the number of seconds in an hour.
  • Hence, for a 70 amp-hour battery: \[70 ext{ A.h} imes 3600 ext{ s/h} = 252000 ext{ C} \] This is the total charge stored in the battery.
This conversion helps in understanding how long a battery can supply power and supports planning for electrical load management.
Faraday's Law of Electromagnetic Induction
Faraday's law is fundamental to understanding how an alternator works. It states that a change in magnetic field within a closed loop induces an electromotive force (emf) in the circuit.
  • The induced emf is directly related to the rate of change of the magnetic flux.
  • In an alternator, this law applies as the rotor spins, changing the magnetic field through the stator windings.
  • Faraday's equation, \[ ext{emf} = -N \frac{d \Phi}{dt} \] where \(N\) is the number of turns, \(\Phi\) is the magnetic flux, and \(d\Phi/dt\) is the change in flux over time, is used to calculate the strength of the induced emf.
This principle is instrumental in transforming mechanical movement into electricity in many applications.
Peak Current Estimation
Estimating the peak current consumption of a vehicle is important for ensuring the alternator can supply enough current to all electrical components.
  • To find the peak current, list all devices and their current draw, like headlights, radiator fans, and wipers.
  • Add their individual currents. For instance:
    - Two headlights at 20 A each
    - One radiator fan at 10 A
    - Two windshield wipers at 5 A each
    Results in a total peak current of: \[2 \times 20 ext{ A} + 10 ext{ A} + 2 \times 5 ext{ A} = 60 ext{ A} \]
  • This ensures the alternator's capacity to handle maximum demand during operation is adequate.
This step is crucial for understanding the electrical requirements of a vehicle, especially under high-demand conditions.
Magnetic Field Calculation
Calculating the magnetic field strength in an alternator requires applying Faraday's law. The induced emf depends on several factors including the magnetic field strength, coil turns, and rotational speed.
  • The emf is given by: \[ ext{emf} = B \cdot N \cdot A \cdot \omega \] where \(B\) is the magnetic field, \(N\) is coil turns, \(A\) is the coil area, and \(\omega\) is angular speed.
  • Calculating the coil area involves the radius: \[ A = \pi \left( \frac{d}{2} \right)^2 \] where \(d\) is the stator coil diameter.
  • For angular speed, \(\omega\) is the product of \(2 \pi\) and frequency (in Hz). This allows finding the magnetic field by rearranging the formula: \[ B = \frac{\text{emf}}{N \cdot A \cdot \omega} \]
By conducting these calculations, we can determine the precise magnetic field necessary for generating the required electricity.

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Most popular questions from this chapter

The free-electron density in a copper wire is \(8.5 \times 10^{28}\) electrons \(/ \mathrm{m}^{3} .\) The electric field in the wire is \(0.0600 \mathrm{~N} / \mathrm{C}\) and the temperature of the wire is \(20.0^{\circ} \mathrm{C}\). (a) What is the drift speed \(v_{\mathrm{d}}\) of the electrons in the wire? (b) What is the potential difference between two points in the wire that are separated by \(20.0 \mathrm{~cm} ?\)

At temperatures near absolute zero, \(B_{\mathrm{c}}\) approaches \(0.142 \mathrm{~T}\) for vanadium, a type-I superconductor. The normal phase of vanadium has a magnetic susceptibility close to zero. Consider a long, thin vanadium cylinder with its axis parallel to an external magnetic field \(\overrightarrow{\boldsymbol{B}}_{0}\) in the \(+x\) -direction. At points far from the ends of the cylinder, by symmetry, all the magnetic vectors are parallel to the \(x\) -axis. At temperatures near absolute zero, what are the resultant magnetic field \(\vec{B}\) and the magnetization \(\vec{M}\) inside and outside the cylinder (far from the ends) for (a) \(\overrightarrow{\boldsymbol{B}}_{0}=(0.130 \mathrm{~T}) \hat{\boldsymbol{\imath}}\) and (b) \(\overrightarrow{\boldsymbol{B}}_{0}=(0.260 \mathrm{~T}) \hat{\imath} ?\)

BIO Treatment of Heart Failure. A heart defibrillator is used to enable the heart to start beating if it has stopped. This is done by passing a large current of 12 A through the body at \(25 \mathrm{~V}\) for a very short time, usually about \(3.0 \mathrm{~ms}\). (a) What power does the defibrillator deliver to the body, and (b) how much energy is transferred?

A typical cost for electrical power is 0.120 dollar per kilowatthour. (a) Some people leave their porch light on all the time. What is the yearly cost to keep a \(75 \mathrm{~W}\) bulb burning day and night? (b) Suppose your refrigerator uses \(400 \mathrm{~W}\) of power when it's running, and it runs 8 hours a day. What is the yearly cost of operating your refrigerator?

A small, closely wound coil has \(N\) turns, area \(A\), and resistance \(R\). The coil is initially in a uniform magnetic field that has magnitude \(B\) and a direction perpendicular to the plane of the loop. The coil is then rapidly pulled out of the field so that the flux through the coil is reduced to zero in time \(\Delta t\). (a) What are the magnitude of the average \(\operatorname{emf} \mathcal{E}_{\text {av }}\) and average current \(I_{\mathrm{av}}\) induced in the coil? (b) The total charge \(Q\) that flows through the coil is given by \(Q=I_{\mathrm{av}} \Delta t .\) Derive an expression for \(Q\) in terms of \(N, A, B,\) and \(R .\) Note that \(Q\) does not depend on \(\Delta t .\) (c) What is \(Q\) if \(N=150\) turns, \(A=4.50 \mathrm{~cm}^{2}, R=30.0 \Omega,\) and \(B=0.200 \mathrm{~T} ?\)
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