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  1. Preface
  2. Unit 1. Thermodynamics
    1. 1 Temperature and Heat
      1. Introduction
      2. 1.1 Temperature and Thermal Equilibrium
      3. 1.2 Thermometers and Temperature Scales
      4. 1.3 Thermal Expansion
      5. 1.4 Heat Transfer, Specific Heat, and Calorimetry
      6. 1.5 Phase Changes
      7. 1.6 Mechanisms of Heat Transfer
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    2. 2 The Kinetic Theory of Gases
      1. Introduction
      2. 2.1 Molecular Model of an Ideal Gas
      3. 2.2 Pressure, Temperature, and RMS Speed
      4. 2.3 Heat Capacity and Equipartition of Energy
      5. 2.4 Distribution of Molecular Speeds
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    3. 3 The First Law of Thermodynamics
      1. Introduction
      2. 3.1 Thermodynamic Systems
      3. 3.2 Work, Heat, and Internal Energy
      4. 3.3 First Law of Thermodynamics
      5. 3.4 Thermodynamic Processes
      6. 3.5 Heat Capacities of an Ideal Gas
      7. 3.6 Adiabatic Processes for an Ideal Gas
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    4. 4 The Second Law of Thermodynamics
      1. Introduction
      2. 4.1 Reversible and Irreversible Processes
      3. 4.2 Heat Engines
      4. 4.3 Refrigerators and Heat Pumps
      5. 4.4 Statements of the Second Law of Thermodynamics
      6. 4.5 The Carnot Cycle
      7. 4.6 Entropy
      8. 4.7 Entropy on a Microscopic Scale
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
  3. Unit 2. Electricity and Magnetism
    1. 5 Electric Charges and Fields
      1. Introduction
      2. 5.1 Electric Charge
      3. 5.2 Conductors, Insulators, and Charging by Induction
      4. 5.3 Coulomb's Law
      5. 5.4 Electric Field
      6. 5.5 Calculating Electric Fields of Charge Distributions
      7. 5.6 Electric Field Lines
      8. 5.7 Electric Dipoles
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
    2. 6 Gauss's Law
      1. Introduction
      2. 6.1 Electric Flux
      3. 6.2 Explaining Gauss’s Law
      4. 6.3 Applying Gauss’s Law
      5. 6.4 Conductors in Electrostatic Equilibrium
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    3. 7 Electric Potential
      1. Introduction
      2. 7.1 Electric Potential Energy
      3. 7.2 Electric Potential and Potential Difference
      4. 7.3 Calculations of Electric Potential
      5. 7.4 Determining Field from Potential
      6. 7.5 Equipotential Surfaces and Conductors
      7. 7.6 Applications of Electrostatics
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    4. 8 Capacitance
      1. Introduction
      2. 8.1 Capacitors and Capacitance
      3. 8.2 Capacitors in Series and in Parallel
      4. 8.3 Energy Stored in a Capacitor
      5. 8.4 Capacitor with a Dielectric
      6. 8.5 Molecular Model of a Dielectric
      7. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    5. 9 Current and Resistance
      1. Introduction
      2. 9.1 Electrical Current
      3. 9.2 Model of Conduction in Metals
      4. 9.3 Resistivity and Resistance
      5. 9.4 Ohm's Law
      6. 9.5 Electrical Energy and Power
      7. 9.6 Superconductors
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    6. 10 Direct-Current Circuits
      1. Introduction
      2. 10.1 Electromotive Force
      3. 10.2 Resistors in Series and Parallel
      4. 10.3 Kirchhoff's Rules
      5. 10.4 Electrical Measuring Instruments
      6. 10.5 RC Circuits
      7. 10.6 Household Wiring and Electrical Safety
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    7. 11 Magnetic Forces and Fields
      1. Introduction
      2. 11.1 Magnetism and Its Historical Discoveries
      3. 11.2 Magnetic Fields and Lines
      4. 11.3 Motion of a Charged Particle in a Magnetic Field
      5. 11.4 Magnetic Force on a Current-Carrying Conductor
      6. 11.5 Force and Torque on a Current Loop
      7. 11.6 The Hall Effect
      8. 11.7 Applications of Magnetic Forces and Fields
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    8. 12 Sources of Magnetic Fields
      1. Introduction
      2. 12.1 The Biot-Savart Law
      3. 12.2 Magnetic Field Due to a Thin Straight Wire
      4. 12.3 Magnetic Force between Two Parallel Currents
      5. 12.4 Magnetic Field of a Current Loop
      6. 12.5 Ampère’s Law
      7. 12.6 Solenoids and Toroids
      8. 12.7 Magnetism in Matter
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    9. 13 Electromagnetic Induction
      1. Introduction
      2. 13.1 Faraday’s Law
      3. 13.2 Lenz's Law
      4. 13.3 Motional Emf
      5. 13.4 Induced Electric Fields
      6. 13.5 Eddy Currents
      7. 13.6 Electric Generators and Back Emf
      8. 13.7 Applications of Electromagnetic Induction
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    10. 14 Inductance
      1. Introduction
      2. 14.1 Mutual Inductance
      3. 14.2 Self-Inductance and Inductors
      4. 14.3 Energy in a Magnetic Field
      5. 14.4 RL Circuits
      6. 14.5 Oscillations in an LC Circuit
      7. 14.6 RLC Series Circuits
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    11. 15 Alternating-Current Circuits
      1. Introduction
      2. 15.1 AC Sources
      3. 15.2 Simple AC Circuits
      4. 15.3 RLC Series Circuits with AC
      5. 15.4 Power in an AC Circuit
      6. 15.5 Resonance in an AC Circuit
      7. 15.6 Transformers
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    12. 16 Electromagnetic Waves
      1. Introduction
      2. 16.1 Maxwell’s Equations and Electromagnetic Waves
      3. 16.2 Plane Electromagnetic Waves
      4. 16.3 Energy Carried by Electromagnetic Waves
      5. 16.4 Momentum and Radiation Pressure
      6. 16.5 The Electromagnetic Spectrum
      7. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
  4. A | Units
  5. B | Conversion Factors
  6. C | Fundamental Constants
  7. D | Astronomical Data
  8. E | Mathematical Formulas
  9. F | Chemistry
  10. G | The Greek Alphabet
  11. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
    8. Chapter 8
    9. Chapter 9
    10. Chapter 10
    11. Chapter 11
    12. Chapter 12
    13. Chapter 13
    14. Chapter 14
    15. Chapter 15
    16. Chapter 16
  12. Index

Check Your Understanding

10.1

If a wire is connected across the terminals, the load resistance is close to zero, or at least considerably less than the internal resistance of the battery. Since the internal resistance is small, the current through the circuit will be large, I=εR+r=ε0+r=εr.I=εR+r=ε0+r=εr. The large current causes a high power to be dissipated by the internal resistance (P=I2r)(P=I2r). The power is dissipated as heat.

10.2

The equivalent resistance of nine bulbs connected in series is 9R. The current is I=V/9R.I=V/9R. If one bulb burns out, the equivalent resistance is 8R, and the voltage does not change, but the current increases (I=V/8R).(I=V/8R). As more bulbs burn out, the current becomes even higher. Eventually, the current becomes too high, burning out the shunt.

10.3

The equivalent of the series circuit would be Req=1.00Ω+2.00Ω+2.00Ω=5.00Ω,Req=1.00Ω+2.00Ω+2.00Ω=5.00Ω, which is higher than the equivalent resistance of the parallel circuit Req=0.50Ω.Req=0.50Ω. The equivalent resistor of any number of resistors is always higher than the equivalent resistance of the same resistors connected in parallel. The current through for the series circuit would be I=3.00V5.00Ω=0.60A,I=3.00V5.00Ω=0.60A, which is lower than the sum of the currents through each resistor in the parallel circuit, I=6.00A.I=6.00A. This is not surprising since the equivalent resistance of the series circuit is higher. The current through a series connection of any number of resistors will always be lower than the current into a parallel connection of the same resistors, since the equivalent resistance of the series circuit will be higher than the parallel circuit. The power dissipated by the resistors in series would be P=1.80W,P=1.80W, which is lower than the power dissipated in the parallel circuit P=18.00W.P=18.00W.

10.4

A river, flowing horizontally at a constant rate, splits in two and flows over two waterfalls. The water molecules are analogous to the electrons in the parallel circuits. The number of water molecules that flow in the river and falls must be equal to the number of molecules that flow over each waterfall, just like sum of the current through each resistor must be equal to the current flowing into the parallel circuit. The water molecules in the river have energy due to their motion and height. The potential energy of the water molecules in the river is constant due to their equal heights. This is analogous to the constant change in voltage across a parallel circuit. Voltage is the potential energy across each resistor.
The analogy quickly breaks down when considering the energy. In the waterfall, the potential energy is converted into kinetic energy of the water molecules. In the case of electrons flowing through a resistor, the potential drop is converted into heat and light, not into the kinetic energy of the electrons.

10.5

1. All the overhead lighting circuits are in parallel and connected to the main supply line, so when one bulb burns out, all the overhead lighting does not go dark. Each overhead light will have at least one switch in series with the light, so you can turn it on and off. 2. A refrigerator has a compressor and a light that goes on when the door opens. There is usually only one cord for the refrigerator to plug into the wall. The circuit containing the compressor and the circuit containing the lighting circuit are in parallel, but there is a switch in series with the light. A thermostat controls a switch that is in series with the compressor to control the temperature of the refrigerator.

10.6

The circuit can be analyzed using Kirchhoff’s loop rule. The first voltage source supplies power: Pin=IV1=7.20mW.Pin=IV1=7.20mW. The second voltage source consumes power: Pout=IV2+I2R1+I2R2=7.2mW.Pout=IV2+I2R1+I2R2=7.2mW.

10.7

The current calculated would be equal to I=−0.20AI=−0.20A instead of I=0.20A.I=0.20A. The sum of the power dissipated and the power consumed would still equal the power supplied.

10.8

Since digital meters require less current than analog meters, they alter the circuit less than analog meters. Their resistance as a voltmeter can be far greater than an analog meter, and their resistance as an ammeter can be far less than an analog meter. Consult Figure 10.36 and Figure 10.35 and their discussion in the text.

Conceptual Questions

1.

Some of the energy being used to recharge the battery will be dissipated as heat by the internal resistance.

3.

P=I2R=(εr+R)2R=ε2R(r+R)−2,dPdR=ε2[(r+R)−22R(r+R)−3]=0,[(r+R)2R(r+R)3]=0,r=RP=I2R=(εr+R)2R=ε2R(r+R)−2,dPdR=ε2[(r+R)−22R(r+R)−3]=0,[(r+R)2R(r+R)3]=0,r=R

5.

It would probably be better to be in series because the current will be less than if it were in parallel.

7.

two filaments, a low resistance and a high resistance, connected in parallel

9.

It can be redrawn.
Req=[1R6+1R1+1R2+(1R4+1R3+R5)−1]−1Req=[1R6+1R1+1R2+(1R4+1R3+R5)−1]−1

11.

In series the voltages add, but so do the internal resistances, because the internal resistances are in series. In parallel, the terminal voltage is the same, but the equivalent internal resistance is smaller than the smallest individual internal resistance and a higher current can be provided.

13.

The voltmeter would put a large resistance in series with the circuit, significantly changing the circuit. It would probably give a reading, but it would be meaningless.

15.

The ammeter has a small resistance; therefore, a large current will be produced and could damage the meter and/or overheat the battery.

17.

The time constant can be shortened by using a smaller resistor and/or a smaller capacitor. Care should be taken when reducing the resistance because the initial current will increase as the resistance decreases.

19.

Not only might water drip into the switch and cause a shock, but also the resistance of your body is lower when you are wet.

Problems

21.

a.

The figure shows a circuit with an emf source ε, resistor r and voltmeter V


b. 0.476W; c. 0.691 W; d. As RLRL is lowered, the power difference decreases; therefore, at higher volumes, there is no significant difference.

23.

a. 0.400Ω0.400Ω; b. No, there is only one independent equation, so only r can be found.

25.

a. 0.400Ω0.400Ω; b. 40.0 W; c. 0.0956°C/min0.0956°C/min

27.

largest, 786Ω786Ω, smallest, 20.32Ω20.32Ω

29.

29.6 W

31.

a. 0.74 A; b. 0.742 A

33.

a. 60.8 W; b. 1.56 kW

35.

a. Rs=9.00ΩRs=9.00Ω; b.I1=I2=I3=2.00AI1=I2=I3=2.00A;
c. V1=8.00V,V2=2.00V,V3=8.00VV1=8.00V,V2=2.00V,V3=8.00V; d. P1=16.00W,P2=4.00W,P3=16.00WP1=16.00W,P2=4.00W,P3=16.00W; e. P=36.00WP=36.00W

37.

a. I1=0.6mA,I2=0.4mA,I3=0.2mAI1=0.6mA,I2=0.4mA,I3=0.2mA;
b. I1=0.04mA,I2=1.52mA,I3=−1.48mAI1=0.04mA,I2=1.52mA,I3=−1.48mA; c. Pout=0.92mW,Pout=4.50mWPout=0.92mW,Pout=4.50mW;
d. Pin=0.92mW,Pin=4.50mWPin=0.92mW,Pin=4.50mW

39.

V1=42V,V2=6V,R4=18ΩV1=42V,V2=6V,R4=18Ω

41.

a. I1=1.5A,I2=2A,I3=0.5A,I4=2.5A,I5=2AI1=1.5A,I2=2A,I3=0.5A,I4=2.5A,I5=2A; b. Pin=I2V1+I5V5=34WPin=I2V1+I5V5=34W;
c. Pout=I12R1+I22R2+I32R3+I42R4=34WPout=I12R1+I22R2+I32R3+I42R4=34W

43.

I1=23VR, I2=V3R, I3=V3RI1=23VR, I2=V3R, I3=V3R

45.

a.

The resistor R subscript L is connected in series with resistor r subscript 2, voltage source ε subscript 2, resistor r subscript 1, voltage source ε subscript 1, resistor r subscript 1, voltage source ε subscript 1, resistor r subscript 1and voltage source ε subscript 1. All voltage sources have upward negative terminals.

;
b. 0.617 A; c. 3.81 W; d. 18.0Ω18.0Ω

47.

I1r1ε1+I1R4+ε4+I2r4+I4r3ε3+I2R3+I1R1=0I1r1ε1+I1R4+ε4+I2r4+I4r3ε3+I2R3+I1R1=0

49.

4.00to30.0MΩ4.00to30.0MΩ

51.

a. 2.50μF2.50μF; b. 2.00 s

53.

a. 12.3 mA; b. 7.50×10−4s;7.50×10−4s; c. 4.53 mA; d. 3.89 V

55.

a. 1.00×10−7F;1.00×10−7F; b. No, in practice it would not be difficult to limit the capacitance to less than 100 nF, since typical capacitors range from fractions of a picofarad (pF) to milifarad (mF).

57.

3.33×10−3Ω3.33×10−3Ω

59.

12.0 V

61.

400 V

63.

a. 6.00 mV; b. It would not be necessary to take extra precautions regarding the power coming from the wall. However, it is possible to generate voltages of approximately this value from static charge built up on gloves, for instance, so some precautions are necessary.

65.

a. 5.00×10−2C;5.00×10−2C; b. 10.0 kV; c. 1.001.00; d. 1.79×102°C1.79×102°C

Additional Problems

67.

a. Ceq=5.00mFCeq=5.00mF; b. τ=0.1sτ=0.1s; c. 0.069 s

69.

a. Req=20.00ΩReq=20.00Ω;
b. Ir=1.50A,I1=1.00A,I2=0.50A,I3=0.75A,I4=0.75A,I5=1.50AIr=1.50A,I1=1.00A,I2=0.50A,I3=0.75A,I4=0.75A,I5=1.50A;
c. Vr=1.50V,V1=9.00V,V2=9.00V,V3=7.50V,V4=7.50V,V5=12.00VVr=1.50V,V1=9.00V,V2=9.00V,V3=7.50V,V4=7.50V,V5=12.00V;
d. Pr=2.25W,P1=9.00W,P2=4.50W,P3=5.625W,P4=5.625W,P5=18.00WPr=2.25W,P1=9.00W,P2=4.50W,P3=5.625W,P4=5.625W,P5=18.00W;
e. P=45.00WP=45.00W

71.

a. τ=(1.38×10−5Ωm(5.00×10−2m3.14(0.05×10−32)2))10×10−3F=3.52sτ=(1.38×10−5Ωm(5.00×10−2m3.14(0.05×10−32)2))10×10−3F=3.52s; b. V=0.017A(e1.00s3.52s)351.59Ω=0.122VV=0.017A(e1.00s3.52s)351.59Ω=0.122V

73.

a. t=3A·h1.5V900Ω=1800ht=3A·h1.5V900Ω=1800h; b. t=3A·h1.5V100Ω=200ht=3A·h1.5V100Ω=200h

75.

U1=C1V12=0.72J,U2=C2V22=0.338JU1=C1V12=0.72J,U2=C2V22=0.338J

77.

a. Req=24.00ΩReq=24.00Ω; b. I1=1.00A,I2=0.67A,I3=0.33A,I4=1.00AI1=1.00A,I2=0.67A,I3=0.33A,I4=1.00A;
c. V1=14.00V,V2=6.00V,V3=6.00V,V4=4.00VV1=14.00V,V2=6.00V,V3=6.00V,V4=4.00V;
d. P1=14.00W,P2=4.04W,P3=1.96W,P4=4.00WP1=14.00W,P2=4.04W,P3=1.96W,P4=4.00W; e. P=24.00WP=24.00W

79.

a. Req=12.00Ω,I=1.00AReq=12.00Ω,I=1.00A; b. Req=12.00Ω,I=1.00AReq=12.00Ω,I=1.00A

81.

a. −400kΩ−400kΩ; b. You cannot have negative resistance. c. The assumption that Req<R1Req<R1 is unreasonable. Series resistance is always greater than any of the individual resistances.

83.

E2I2r2I2R2+I1R5+I1r1E1+I1R1=0E2I2r2I2R2+I1R5+I1r1E1+I1R1=0

85.

a. I=1.17A,I1=0.50A,I2=0.67A,I3=0.67A,I4=0.50A,I5=0.17AI=1.17A,I1=0.50A,I2=0.67A,I3=0.67A,I4=0.50A,I5=0.17A;
b. Poutput=23.4W,Pinput=23.4WPoutput=23.4W,Pinput=23.4W

87.

a. 4.99 s; b. 3.87°C3.87°C; c. 3.11×104Ω3.11×104Ω; d. No, this change does not seem significant. It probably would not be noticed.

Challenge Problems

89.

a. 0.273 A; b. VT=1.36VVT=1.36V

91.

a. Vs=VIMRM=9.99875VVs=VIMRM=9.99875V; b. RS=VPIM=199.975kΩRS=VPIM=199.975kΩ

93.

a. τ=3800sτ=3800s; b. 1.26 mA; c. t=2633.96st=2633.96s

95.

Req=(31)RReq=(31)R

97.

a. Pimheater=1cup(0.000237m3cup)(1000kgm3)(4186Jkg °C)(100°C20°C)180.00s441WPimheater=1cup(0.000237m3cup)(1000kgm3)(4186Jkg °C)(100°C20°C)180.00s441W;
b. I=441W120V+4(100W120V)+1500W120V=19.51AI=441W120V+4(100W120V)+1500W120V=19.51A; Yes, the breaker will trip.
c. I=441W120V+4(18W120V)+1500W120V=16.78AI=441W120V+4(18W120V)+1500W120V=16.78A; Yes, the breaker will trip.

99.


The figure shows a circuit with two parallel branches, one with galvanometer connected to resistors r and other with resistor R.

,
2.40×10−3Ω2.40×10−3Ω

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