Section 1 Current and Circuits: Practice Problems

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Chapter 22 Practice Problems, Review, and Assessment
Section 1 Current and Circuits: Practice Problems
1. A car battery causes a current through a lamp and produces 12 V across it as shown in Figure 4. What is the
power used by the lamp?
SOLUTION: P = IV = (2.0 A)(12 V) = 24 W
2. What is the current through a 75-W lightbulb that is connected to a 125-V outlet?
SOLUTION: P = IV
3. The current through a lightbulb connected across the terminals of a 125-V outlet is 0.50 A. At what rate does the bulb transform electrical energy to light? (Assume 100 percent efficiency.)
SOLUTION: P = IV = (0.50 A)(125 V) = 63 J/s = 63 W
4. The current through the starter motor of a car is 210 A. If the battery maintains 12 V across the motor, how much electrical energy is delivered to the starter in 10.0 s?
SOLUTION: P = IV and E = Pt
Thus, E = IVt = (210 A)(12 V)(10.0 s)
4
= 2.5×10 J
5. A 75-V generator supplies 3.0 kW of power. How much current can the generator deliver?
SOLUTION: 6. A flashlight bulb is rated at 0.90 W. If the lightbulb produces a potential drop of 3.0 V, how much current goes through it?
SOLUTION: P = IV
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5. A 75-V generator supplies 3.0 kW of power. How much current can the generator deliver?
SOLUTION: Chapter 22 Practice Problems, Review, and Assessment
6. A flashlight bulb is rated at 0.90 W. If the lightbulb produces a potential drop of 3.0 V, how much current goes through it?
SOLUTION: P = IV
7. Challenge A circuit is changed so the potential difference across a motor doubles and the current through the
lightbulb triples. How does this change the motor's power?
SOLUTION: The lightbulb’s power increases by a factor of 2×3 = 6.
8. Draw a circuit diagram to include a 60.0-V battery, an ammeter, and a resistance of 12.5 Ω in series. Draw arrows
on your diagram to indicate the direction of the current.
SOLUTION: 9. Draw a circuit diagram showing a 4.5-V battery, a resistor, and an ammeter that reads 85 mA. Show the direction of the current using conventional rules, and indicate the positive terminal of the battery.
SOLUTION: 10. Add a voltmeter to measure the potential difference across the resistors in the previous two problems. Label the
voltmeters.
SOLUTION: Both circuits will take the form:
11. Draw a circuit using a battery, a lamp, a potentiometer to adjust the lamp’s brightness, and an on-off switch.
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Chapter 22 Practice Problems, Review, and Assessment
11. Draw a circuit using a battery, a lamp, a potentiometer to adjust the lamp’s brightness, and an on-off switch.
SOLUTION: 12. Challenge Repeat the previous problem, adding an ammeter and a voltmeter across the lamp.
SOLUTION: For all problems, assume the battery voltage and the lamp resistances are constant.
13. An automobile panel lamp with a resistance of 33 Ω is placed across the battery shown in Figure 11. What is the
current through the circuit?
SOLUTION: −4 14. A sensor uses 2.0×10
of the sensor circuit?
A of current when it is operated by the battery shown in Figure 12. What is the resistance
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SOLUTION: Page 3
SOLUTION: Chapter 22 Practice Problems, Review, and Assessment
−4 14. A sensor uses 2.0×10
of the sensor circuit?
A of current when it is operated by the battery shown in Figure 12. What is the resistance
SOLUTION: 4
= 1.5×10 Ω
15. A motor with the operating resistance of 32 Ω is connected to a voltage source as shown in Figure 13. What is the
voltage of the source?
SOLUTION: V = IR = (3.8 A)(32 Ω) = 1.2×102 V
16. A lamp draws a current of 0.50 A when it is connected to a 120-V source.
a. What is the resistance of the lamp?
b. What is the power consumption of the lamp?
SOLUTION: a.
1
b. P = IV = (0.50 A)(120 V) = 6.0×10 W
17. A 75-W lamp is connected to 125 V.
a. What is the current through the lamp?
b. What is the resistance of the lamp?
SOLUTION: a.
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b.
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a.
Chapter 22 Practice Problems, Review, and Assessment
1
b. P = IV = (0.50 A)(120 V) = 6.0×10 W
17. A 75-W lamp is connected to 125 V.
a. What is the current through the lamp?
b. What is the resistance of the lamp?
SOLUTION: a.
b.
18. Challenge A resistor is added to the lamp in the previous problem to reduce the current to half its original value.
a. What is the potential difference across the lamp?
b. How much resistance was added to the circuit?
c. At what rate does the lamp transform electrical energy into radiant and thermal energy?
SOLUTION: a. The new value of the current is
V = IR = (0.30 A)(2.1×102 Ω)
1
= 6.3×10 V
b. The total resistance of the circuit is now
2
= 4.2×10 Ω
Therefore,
Rres = Rtotal − Rlamp
= 4.2×102 Ω − 2.1×102 Ω
2
= 2.1×10 Ω
c. P = IV = (0.30 A)(6.3×101 V) = 19 W
Section 1 Current and Circuits: Review
19. MAIN IDEA Explain how charged particles are related to electric current, electric circuits, and resistance.
SOLUTION: Charged particles move through an electric circuit. These moving charged particles are called an
electric current. When charged particles move through a material such as a metal wire, the flow is
impeded by the particles in the material. This flow impediment is called resistance.
20. Schematic Draw a schematic diagram of a circuit that contains a battery and a lightbulb. Make sure the lightbulb
willManual
light in
this circuit.
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SOLUTION: SOLUTION: Charged particles move through an electric circuit. These moving charged particles are called an
electric current. When charged particles move through a material such as a metal wire, the flow is
Chapter
22 Practice
and Assessment
impeded
by the Problems,
particles inReview,
the material.
This flow impediment is called resistance.
20. Schematic Draw a schematic diagram of a circuit that contains a battery and a lightbulb. Make sure the lightbulb
will light in this circuit.
SOLUTION: 21. Resistance Joe states that because R = ΔV /I, if he increases the voltage, the resistance will increase. Is Joe
correct? Explain.
SOLUTION: No, resistance depends on the device. When V increases, so will I.
22. Power A circuit has 12 Ω of resistance and is connected to a 12-V battery. Determine the change in power if the
resistance decreases to 9.0 Ω.
SOLUTION: 2
2
2
2
P1 = V /R1 = (12 V) /12 Ω = 12 W
P2 = V /R2 = (12 V) /9.0 Ω = 16 W
ΔP = P2 − P1 = 16 W − 12 W = 4.0 W
4.0 W increase
23. Resistance You want to measure the resistance of a long piece of wire. Show how you would construct a circuit
with a battery, a voltmeter, an ammeter, and the wire to be tested to make the measurement. Specify what you
would measure and how you would compute the resistance.
SOLUTION: Measure the current through the wire and the potential difference across it. Divide the potential
difference by the current to obtain the wire resistance.
3 24. Energy A circuit transforms 2.2×10 J of energy when it is operated for 3.0 min. Determine the amount of energy it will transform when it is operated for 1 h.
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Chapter
22 Practice
Problems,
Review,
andand
Assessment
Measure
the current
through
the wire
the potential difference across it. Divide the potential
difference by the current to obtain the wire resistance.
3 24. Energy A circuit transforms 2.2×10 J of energy when it is operated for 3.0 min. Determine the amount of energy it will transform when it is operated for 1 h.
SOLUTION: 25. Critical Thinking We say that power is “dissipated” in a resistor. To dissipate is to spread out or disperse. In what
sense is something being dispersed when charge flows through a resistor?
SOLUTION: The potential energy of the charges decreases as they flow through the resistor. This decrease in
potential energy is used to produce heat in the resistor.
Section 2 Using Electrical Energy: Practice Problems
26. A 15-Ω electric heater operates on a 120-V outlet.
a. What is the current through the heater?
b. How much energy is used by the heater in 30.0 s?
c. How much thermal energy is liberated in this time?
SOLUTION: a.
2
2
b. E = I Rt = (8.0 A) (15 Ω)(30.0 s)
= 2.9×104 J
4
c. 2.9×10 J, because all electric energy is transformed to thermal energy.
27. A 39-Ω resistor is connected across a 45-V battery.
a. What is the current in the circuit?
b. How much energy is used by the resistor in 5.0 min?
SOLUTION: a.
b.
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28. A 100.0-W lightbulb is 22 percent efficient. This means that 22 percent of the electrical energy is transformed to radiant energy.
Chapter 22 Practice Problems, Review, and Assessment
28. A 100.0-W lightbulb is 22 percent efficient. This means that 22 percent of the electrical energy is transformed to radiant energy.
a. How many joules does the lightbulb transform into radiant energy each minute it is in operation?
b. How many joules of thermal energy does the lightbulb output each minute?
SOLUTION: a. E = Pt = (0.22)(100.0 J/s)(1.0 min)(60 s/min) = 1.3×103 J
b. E = Pt
= (0.78)(100.0 J/s)(1.0 min)(60.0 s/min)
3
= 4.7×10 J
29. The resistance of an electric stove element at operating temperature is 11 Ω.
a. If 220 V are applied across it, what is the current through the stove element?
b. How much energy does the element transform to thermal energy in 30.0 s?
c. The element is used to heat a kettle containing 1.20 kg of water. Assume 65 percent of the thermal energy is absorbed by the water. What is the water’s increase in temperature during the 30.0 s?
SOLUTION: a. b. E = I2Rt = (2.0×101 A)2(11 Ω)(30.0 s)
5
= 1.3×10 J
c. Q = mCΔT with Q = 0.65E
= 17° C
30. Challenge A 120-V water heater takes 2.2 h to heat a given volume of water to a certain temperature. How long would a 240-V unit operating with the same current take to accomplish the same task?
SOLUTION: For a given amount of energy, doubling the voltage will divide the time by 2.
31. An electric space heater draws 15.0 A from a 120-V source. It is operated, on the average, for 5.0 h each day.
a. How much power does the heater use?
b. How much energy in kWh does it consume in 30 days?
c. At $0.12 per kWh, how much does it cost to operate the heater for 30 days?
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SOLUTION: a. P = IV = (15.0 A)(120 V) = 1800 W = 1.8 kW
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For a given amount of energy, doubling the voltage will divide the time by 2.
Chapter 22 Practice Problems, Review, and Assessment
31. An electric space heater draws 15.0 A from a 120-V source. It is operated, on the average, for 5.0 h each day.
a. How much power does the heater use?
b. How much energy in kWh does it consume in 30 days?
c. At $0.12 per kWh, how much does it cost to operate the heater for 30 days?
SOLUTION: a. P = IV = (15.0 A)(120 V) = 1800 W = 1.8 kW
b. E = Pt = (1.8 kW)(5.0 h/day)(30 days)
= 270 kWh
c. Cost = ($0.12/kWh)(270 kWh)
= $32.40
32. A digital clock has a resistance of 12,000 Ω and is plugged into a 115-V outlet.
a. How much current does it draw?
b. How much power does it use?
c. If the owner of the clock pays $0.12 per kWh, how much does it cost to operate the clock for 30 days?
SOLUTION: a. b. P = VI = (115 V)(9.6×10−3 A)
= 1.1 W
−3
c. Cost = (1.1×10 kWh)($0.12/kWh)
(30 days)(24 h/day) = $0.10
33. An automotive battery can deliver 55 A at 12 V for 1.0 h and requires 1.3 times as much energy for recharge due to its less-than-perfect efficiency. How long will it take to charge the battery using a current of 7.5 A? Assume the charging voltage is the same as the discharging voltage.
SOLUTION: Echarge = (1.3)IVt
= (1.3)(55 A)(12 V)(1.0 h)
= 858 Wh
34. Challenge Rework the previous problem by assuming the battery requires the application of 14 V when it is recharging.
SOLUTION: Echarge = (1.3)IVt
= (1.3)(55 A)(12 V)(1.0 h)
= 858 Wh
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Section 2 Using Electrical Energy: Review
Page 9
Echarge = (1.3)IVt
= (1.3)(55 A)(12 V)(1.0 h)
= 858 Wh
Chapter 22 Practice Problems, Review, and Assessment
34. Challenge Rework the previous problem by assuming the battery requires the application of 14 V when it is recharging.
SOLUTION: Echarge = (1.3)IVt
= (1.3)(55 A)(12 V)(1.0 h)
= 858 Wh
Section 2 Using Electrical Energy: Review
35. MAIN IDEA A car engine drives a generator, which transfers electrical energy to the car’s battery. The
headlamps use the energy stored in the car battery to produce light. List the forms of energy in these three
operations.
SOLUTION: Mechanical energy from the engine transforms to electric energy in the generator; electric energy
stored as chemical energy in the battery; chemical energy transformed to electric energy in the battery
and distributed to the headlamps; electric energy transformed to light and thermal energy in headlamps.
36. Resistance A hair dryer operating from 120 V has two settings, hot and warm. In which setting is the resistance likely to be smaller? Why?
SOLUTION: Hot draws more power, P = IV, so the fixed voltage current is larger. Because I = V/R the resistance is
smaller.
37. Efficiency Evaluate the impact of research to improve power transmission lines on society and the environment.
SOLUTION: Research to improve power transmission lines would benefit society in cost of electricity. Also, if less
power was lost during transmission, less coal and other power-producing resources would have to be
used, which would improve the quality of our environment.
38. Voltage Why would an electric range and an electric hot-water heater be connected to a 240-V circuit rather than
a 120-V circuit?
SOLUTION: For the same power, at twice the voltage, the current would be halved. The I2R loss in the circuit wiring
would be dramatically reduced because it is proportional to the square of the current.
39. Energy Cost A consumer uses 3098 kWh in 29 days. The utility company charges $0.077592 per kWh for the electricity plus $0.029998 per kWh for the distribution of the electricity. What is the consumer’s electric bill for the
29 days?
SOLUTION: (3098 kWh)($0.077592 kWh + $0.029998 kWH) = $333.31
40. Resistance and Power A toaster is connected to the circuit shown in Figure 18.
a. What is the resistance of the toaster?
b. At what rate does the toaster transform energy?
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electricity plus $0.029998 per kWh for the distribution of the electricity. What is the consumer’s electric bill for the
29 days?
SOLUTION: Chapter
22 Practice Problems, Review, and Assessment
(3098 kWh)($0.077592 kWh + $0.029998 kWH) = $333.31
40. Resistance and Power A toaster is connected to the circuit shown in Figure 18.
a. What is the resistance of the toaster?
b. At what rate does the toaster transform energy?
Figure 18
SOLUTION: a.
b.
41. Critical Thinking When demand for electric power is high, power companies sometimes reduce the voltage,
thereby producing a “brown out.” What is being saved?
SOLUTION: Power, not energy; most devices will have to run longer.
Chapter Assessment
Section 1 Current and Circuits: Mastering Concepts
42. Define the unit of electric current in terms terms of units of charge and time.
SOLUTION: 1 A = 1 C/s
43. How should a voltmeter be connected in Figure 19 to measure the motor’s voltage?
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Figure 19
SOLUTION: Page 11
42. Define the unit of electric current in terms terms of units of charge and time.
SOLUTION: Chapter
Problems, Review, and Assessment
1 A 22
= 1Practice
C/s
43. How should a voltmeter be connected in Figure 19 to measure the motor’s voltage?
Figure 19
SOLUTION: The positive voltmeter lead connects to the left-hand motor lead, and the negative voltmeter lead
connects to the right-hand motor lead.
44. How should an ammeter be connected in Figure 19 to measure the motor’s current?
SOLUTION: Break the circuit between the battery and the motor. Then connect the positive ammeter lead to the
positive side of the break (the side connected to the positive battery terminal) and the negative
ammeter lead to the negative side nearest the motor.
45. What is the direction of the conventional current through the motor in Figure 19?
SOLUTION: from left to right through the motor
46. BIG IDEA Refer to Figure 19 to answer the following questions.
a. Which device transforms electrical energy to mechanical energy?
b. Which device transforms chemical energy to electrical energy?
c. Which device turns the circuit on and off?
d. Which device provides a way to adjust speed?
SOLUTION: a. 4
b. 1
c. 2
d. 3
47. Which wire conducts electricity with the least resistance: one with a large cross-sectional diameter or one with a
small cross-sectional diameter? Explain.
SOLUTION: A larger-diameter wire has a smaller resistance because there are more electrons to carry the charge.
48. A simple circuit consists of a resistor, a battery, and connecting wires.
a. Draw
eSolutions
Manuala- circuit
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by Cognero of
this simple circuit.
b. How must an ammeter be connected in a circuit for the current to be correctly read?
c. How must a voltmeter be connected to a resistor for the potential difference across it to be read?
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47. Which wire conducts electricity with the least resistance: one with a large cross-sectional diameter or one with a
small cross-sectional diameter? Explain.
SOLUTION: Chapter
22 Practice Problems, Review, and Assessment
A larger-diameter wire has a smaller resistance because there are more electrons to carry the charge.
48. A simple circuit consists of a resistor, a battery, and connecting wires.
a. Draw a circuit schematic of this simple circuit.
b. How must an ammeter be connected in a circuit for the current to be correctly read?
c. How must a voltmeter be connected to a resistor for the potential difference across it to be read?
SOLUTION: a.
b. The ammeter must be connected in series.
c. The voltmeter must be connected in parallel.
49. Describe how a potentiometer controls the following devices:
a. electric motor
b. game joystick
SOLUTION: a. The potentiometer on an electric motor makes continuous changes in the speed of the motor instead
of a step-by-step change in the motor’s speed.
b. The potentiometer in a game joystick helps translate the motion of the gamer’s hand to a position on
the computer screen.
Chapter Assessment
Section 1 Currents and Circuits: Mastering Problems
50. Describe the energy transformation that occur in each of the following devices.
a. an incandescent lightbulb
b. a clothes dryer
c. a digital clock radio
d. a hand-held flashlight
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SOLUTION: a. electric energy to heat and light
b. electric energy to heat and kinetic energy
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SOLUTION: a. The potentiometer on an electric motor makes continuous changes in the speed of the motor instead
of a step-by-step change in the motor’s speed.
Chapter
22 Practice
Problems,
Review,
and Assessment
b. The
potentiometer
in a game
joystick
helps translate the motion of the gamer’s hand to a position on
the computer screen.
Chapter Assessment
Section 1 Currents and Circuits: Mastering Problems
50. Describe the energy transformation that occur in each of the following devices.
a. an incandescent lightbulb
b. a clothes dryer
c. a digital clock radio
d. a hand-held flashlight
SOLUTION: a. electric energy to heat and light
b. electric energy to heat and kinetic energy
c. electric energy to light and sound
d. chemical energy to light and thermal energy
51. A motor is connected to a 12-V battery, as shown in Figure 20. (Level 1)
a. How much power is delivered to the motor?
b. How much energy is transformed if the motor runs for 15 min?
Figure 20
SOLUTION: a. P = VI = (12 V)(1.5 A) = 18 W
b. E = Pt = (18 W)(15 min)(60 s/min)
4
= 1.6×10 J
52. Refer to Figure 21 to answer the following questions. (Level 1)
a. What should the ammeter reading be?
b. What should the voltmeter reading be?
c. How much power is delivered to the resistor?
d. How much energy is delivered to the resistor per hour?
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SOLUTION: a. P = VI = (12 V)(1.5 A) = 18 W
b. E22
= Pt
= (18 W)(15
min)(60
s/min)and Assessment
Chapter
Practice
Problems,
Review,
4
= 1.6×10 J
52. Refer to Figure 21 to answer the following questions. (Level 1)
a. What should the ammeter reading be?
b. What should the voltmeter reading be?
c. How much power is delivered to the resistor?
d. How much energy is delivered to the resistor per hour?
Figure 21
SOLUTION: a. b. 27 V
c. P = VI = (27 V)(1.5 A) = 41 W
d. E = Pt = (41 W)(3600 s) = 1.5×105 J
53. Toasters The current through a toaster that is connected to a 120-V source is 8.0 A. What power is used by the toaster? (Level 1)
SOLUTION: 2
P = IV = (8.0 A)(120 V) = 9.6×10 W
54. Lightbulbs A current of 1.2 A is measured through a lightbulb when it is connected across a 120-V source. At
what rate does the bulb transform energy? (Level 1)
SOLUTION: P = IV = (1.2 A)(120 V) = 1.4×102 W
55. Refer to Figure 22 to answer the following questions. (Level 1)
a. What should the ammeter reading be?
b. What should the voltmeter reading be?
c. How much power is delivered to the resistor?
d. How much energy is delivered to the resistor per hour?
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54. Lightbulbs A current of 1.2 A is measured through a lightbulb when it is connected across a 120-V source. At
what rate does the bulb transform energy? (Level 1)
SOLUTION: Chapter 22 Practice Problems, Review, and Assessment
P = IV = (1.2 A)(120 V) = 1.4×102 W
55. Refer to Figure 22 to answer the following questions. (Level 1)
a. What should the ammeter reading be?
b. What should the voltmeter reading be?
c. How much power is delivered to the resistor?
d. How much energy is delivered to the resistor per hour?
Figure 22
SOLUTION: a. b. 27 V
c. P = VI = (27 V)(3.0 A) = 81 W
5
d. E = Pt = (81 W)(3600 s) = 2.9×10 J
56. Refer to Figure 23 to answer the following questions. (Level 1)
a. What should the ammeter reading be?
b. What should the voltmeter reading be?
c. How much power is delivered to the resistor?
d. How much energy is delivered to the resistor per hour?
Figure 23
SOLUTION: a. eSolutions
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b. 9.0
V - Powered by Cognero
c. P = VI = (9.0 V)(0.50 A) = 4.5 W
d. E = Pt = (4.5 W)(3600 s) = 1.6×104 J
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b. 27 V
c. P22
= VI
= (27 V)(3.0
A) = Review,
81 W and Assessment
Chapter
Practice
Problems,
5
d. E = Pt = (81 W)(3600 s) = 2.9×10 J
56. Refer to Figure 23 to answer the following questions. (Level 1)
a. What should the ammeter reading be?
b. What should the voltmeter reading be?
c. How much power is delivered to the resistor?
d. How much energy is delivered to the resistor per hour?
Figure 23
SOLUTION: a. b. 9.0 V
c. P = VI = (9.0 V)(0.50 A) = 4.5 W
d. E = Pt = (4.5 W)(3600 s) = 1.6×104 J
57. A lamp draws 0.50 A from a 120-V generator. (Level 1)
a. How much power is delivered?
b. How much energy is transformed in 5.0 min?
SOLUTION: 1
a. P = IV = (0.50 A)(120 V) = 6.0×10 W
b. The definition of power is so
58. A 12-V automobile battery is connected to an electric starter motor. The current through the motor is 210 A. (Level 1)
a. How many joules of energy does the battery deliver to the motor each second?
b. What power, in watts, does the motor use?
SOLUTION: a. P = IV = (210 A)(12 V) = 2500 J/s eSolutions Manual - Powered by Cognero
or 2.5×103 J/s
3
b. P = 2.5×10 W
Page 17
Chapter 22 Practice Problems, Review, and Assessment
58. A 12-V automobile battery is connected to an electric starter motor. The current through the motor is 210 A. (Level 1)
a. How many joules of energy does the battery deliver to the motor each second?
b. What power, in watts, does the motor use?
SOLUTION: a. P = IV = (210 A)(12 V) = 2500 J/s or 2.5×103 J/s
3
b. P = 2.5×10 W
59. Dryers A 4200-W clothes dryer is connected to a 220-V circuit. How much current does the dryer draw?
(Level 1)
SOLUTION: 60. Flashlights A flashlight bulb is connected across a 3.0-V potential difference. The current through the bulb is
1.5 A. (Level 1)
a. What is the power rating of the bulb?
b. How much electrical energy does the bulb transform in 11 min?
SOLUTION: a. P = IV = (1.5 A)(3.0 V) = 4.5 W
b. The definition of power is so
61. Batteries A resistor of 60.0 V has a current of 0.40 A through it when it is connected to the terminals of a battery.
What is the voltage of the battery? (Level 1)
SOLUTION: V = IR = (0.40 A)(60.0 Ω) = 24 V
62. What voltage is applied to a 4.0-Ω resistor if the current is 1.5 A? (Level 1)
SOLUTION: V = IR = (1.5 A)(4.0 Ω) = 6.0 V
63. What voltage is placed across a motor with a 15-Ω operating resistance if there is 8.0 A of current? (Level 1)
SOLUTION: V = IR = (8.0 A)(15 Ω) = 1.2×102 V
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64. A voltage of 75 V is placed across a 150-Ω resistor. What is the current through the resistor? (Level 1)
SOLUTION: Page 18
63. What voltage is placed across a motor with a 15-Ω operating resistance if there is 8.0 A of current? (Level 1)
SOLUTION: Chapter
22 Practice Problems, Review, and Assessment
V = IR = (8.0 A)(15 Ω) = 1.2×102 V
64. A voltage of 75 V is placed across a 150-Ω resistor. What is the current through the resistor? (Level 1)
SOLUTION: 65. Some students connected a length of nichrome wire to a variable power supply to produce between 0.00 V and 10.00 V across the wire. They then measured the current through the wire for several voltages. The students recorded the data for the voltages used and the currents measured, as shown in Table 2. (Level 2)
a. For each measurement, calculate the resistance.
b. Graph I versus V.
c. Does the nichrome wire obey Ohm’s law? If not, for all the potential differences, specify the voltage range for
which Ohm’s law holds.
SOLUTION: a. R = 143 Ω, R = 148 Ω, R = 150 Ω, R = 154 Ω, R = 159 Ω, R = 143 Ω, R = 143 Ω, R = 154 Ω, R = 157
Ω, R = 161 Ω
b. eSolutions Manual - Powered by Cognero
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SOLUTION: a. R22
= 143
Ω, R Problems,
= 148 Ω, RReview,
= 150 Ω,and
R =Assessment
154 Ω, R = 159 Ω, R = 143 Ω, R = 143 Ω, R = 154 Ω, R = 157
Chapter
Practice
Ω, R = 161 Ω
b. c. Ohm’s law is obeyed when the resistance of a device is constant and independent of the potential
difference. The resistance of the nichrome wire increases somewhat as the magnitude of the voltage
increases, so the wire does not quite obey Ohm’s law.
66. Draw a series circuit diagram to include a 16-Ω resistor, a battery, and an ammeter that reads 1.75 A. Indicate the positive terminal and the voltage of the battery, the positive terminal of the ammeter, and the direction of
conventional current. (Level 2)
SOLUTION: V = IR = (1.75 A)(16 Ω) = 28 V
67. A lamp draws a 66-mA current when connected to a 6.0-V battery. When a 9.0-V battery is used, the lamp draws
75 mA. (Level 2)
a. Does the lamp obey Ohm’s law?
b. How much power does the lamp use when it is connected to the 6.0-V battery?
SOLUTION: a. No. The voltage is increased by a factor of
but the current is increased by a factor of −3
b. P = IV = (66×10 A)(6.0 V) = 0.40 W
68. Lightbulbs How much energy does a 60.0-W lightbulb use in half an hour? If the lightbulb transforms 12 percent of electrical
energybytoCognero
radiant energy, how much electrical energy is transformed to thermal energy during the Page
half 20
eSolutions
Manual - Powered
hour? (Level 2)
SOLUTION: −3
b. P22
= IV = (66×10
A)(6.0Review,
V) = 0.40
WAssessment
Chapter
Practice Problems,
and
68. Lightbulbs How much energy does a 60.0-W lightbulb use in half an hour? If the lightbulb transforms 12 percent of electrical energy to radiant energy, how much electrical energy is transformed to thermal energy during the half
hour? (Level 2)
SOLUTION: If the bulb is 12 percent efficient, 88 percent of the energy is lost to heat, so
Q = (0.88)(1.08×105 J) = 9.5×104 J
69. The current through a lamp connected across 120 V is 0.40 A when the lamp is on. (Level 2)
a. What is the lamp’s resistance when it is on?
b. When the lamp is cold, its resistance is 1/5 as great as it is when the lamp is hot. What is the lamp’s cold
resistance?
c. What is the current through the lamp as it is turned on if it is connected to a potential difference of 120 V?
SOLUTION: a. V = IR
b.
c. V = IR
70. The graph in Figure 24 shows the current through a device called a silicon diode. (Level 3)
a. A potential difference of +0.70 V is placed across the diode. What is the resistance of the diode?
b. What is the diode’s resistance when a +0.60-V potential difference is used?
Voltage (V)
Figure
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SOLUTION: a. From the graph, I = 22 m/A, and V = IR, so
Page 21
c. V = IR
Chapter 22 Practice Problems, Review, and Assessment
70. The graph in Figure 24 shows the current through a device called a silicon diode. (Level 3)
a. A potential difference of +0.70 V is placed across the diode. What is the resistance of the diode?
b. What is the diode’s resistance when a +0.60-V potential difference is used?
Voltage (V)
Figure 24
SOLUTION: a. From the graph, I = 22 m/A, and V = IR, so
b. 71. Draw a schematic diagram to show a circuit including a 90-V battery, an ammeter, and a resistance of 45 Ω connected in series. What is the ammeter reading? Draw arrows showing the direction of conventional current.
(Level 3)
SOLUTION: Section 2 Using Electrical Energy: Mastering Concepts
72. Why do incandescent lightbulbs burn out more frequently just as they are switched on rather than while they are
operating?
SOLUTION: TheManual
low resistance
of the cold filament allows a high current initially and a greater change in
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temperature, subjecting the filament to greater stress.
Page 22
Chapter 22 Practice Problems, Review, and Assessment
Section 2 Using Electrical Energy: Mastering Concepts
72. Why do incandescent lightbulbs burn out more frequently just as they are switched on rather than while they are
operating?
SOLUTION: The low resistance of the cold filament allows a high current initially and a greater change in
temperature, subjecting the filament to greater stress.
73. If a heavy copper wire is used to connect one terminal of a battery directly to the other terminal of that same
battery, the temperature of the copper wire rises rapidly. Why does this happen?
SOLUTION: The copper wire has very little resistance, which results in a high current. This causes more electrons
to collide with the atoms of the wire. This raises the atoms’ kinetic energies and the temperature of the
wire.
74. What electric quantities must be kept small to transmit electrical energy economically over long distances?
SOLUTION: the resistance of the wire and the current in the wire
75. Define the unit of power in terms of fundamental SI units.
SOLUTION: Section 2 Using Electrical Energy: Mastering Problems
76. Batteries A 9.0-V battery costs $3.00 and will deliver 0.0250 A for 26.0 h before it must be replaced. Calculate the cost per kWh. (Level 1)
SOLUTION: E = IVt = (0.0250 A)(9.0 V)(26.0 h)
= 5.9 Wh = 5.9×10
–3
kWh
= $510/kWh
77. What is the maximum current allowed in a 5.0-W, 220-Ω resistor? (Level 1)
SOLUTION: 78. A 110-V electric iron draws 3.0 A of current. How much thermal energy does it output in an hour? (Level 1)
SOLUTION: Q = E = VIt = (110 V)(3.0 A)(1.0 h)
(3600 s/h)
= 1.2×106 J
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Page 23
SOLUTION: Chapter 22 Practice Problems, Review, and Assessment
78. A 110-V electric iron draws 3.0 A of current. How much thermal energy does it output in an hour? (Level 1)
SOLUTION: Q = E = VIt = (110 V)(3.0 A)(1.0 h)
(3600 s/h)
= 1.2×106 J
1 79. For the circuit shown in Figure 25, the maximum safe power is 5.0×10 W. Use the figure to find the following:
(Level 2)
a. the maximum safe current
b. the maximum safe voltage
Figure 25
SOLUTION: a. P = I2R
2
b. P = V /R
80. Appliances A window air conditioner is estimated to have a cost of operation of $50 per 30 days. This is based on the assumption that the air conditioner will run half of the time and that electricity costs $0.090 per kWh. Determine
how much current the air conditioner will take from a 120-V outlet. (Level 2)
SOLUTION: Cost = (E)(rate)
= 556 kWh
E = IVt
= 12.9 A
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Page 24
81. Utilities Figure 26 represents an electric furnace. Calculate the monthly (30-day) heating bill if electricity costs
$0.10 per kWh and the thermostat is on one-fourth of the time. (Level 2)
2
b. P = V /R
Chapter 22 Practice Problems, Review, and Assessment
80. Appliances A window air conditioner is estimated to have a cost of operation of $50 per 30 days. This is based on the assumption that the air conditioner will run half of the time and that electricity costs $0.090 per kWh. Determine
how much current the air conditioner will take from a 120-V outlet. (Level 2)
SOLUTION: Cost = (E)(rate)
= 556 kWh
E = IVt
= 12.9 A
81. Utilities Figure 26 represents an electric furnace. Calculate the monthly (30-day) heating bill if electricity costs
$0.10 per kWh and the thermostat is on one-fourth of the time. (Level 2)
Figure 26
SOLUTION: = 2160 kWh
Cost = (2160 kWh)($0.100/kWh) = $216
82. Radios A transistor radio operates by means of a 9.0-V battery that supplies it with a 50.0-mA current. (Level 2)
a. If the cost of the battery is $2.49 and it lasts for 300.0 h, what is the cost per kWh to operate the radio in this manner?
b. The same radio, by means of a converter, is plugged into a household circuit by a homeowner who pays $0.12
per kWh. What does it now cost to operate the radio for 300.0 h?
SOLUTION: eSolutions
CogneroV)
a. PManual
= IV- =Powered
(0.050byA)(9.0
= 4.5×10
–4
kW
= 0.45 W
Page 25
= 2160 kWh
Chapter
22 Practice Problems, Review, and Assessment
Cost = (2160 kWh)($0.100/kWh) = $216
82. Radios A transistor radio operates by means of a 9.0-V battery that supplies it with a 50.0-mA current. (Level 2)
a. If the cost of the battery is $2.49 and it lasts for 300.0 h, what is the cost per kWh to operate the radio in this manner?
b. The same radio, by means of a converter, is plugged into a household circuit by a homeowner who pays $0.12
per kWh. What does it now cost to operate the radio for 300.0 h?
SOLUTION: a. P = IV = (0.050 A)(9.0 V) = 0.45 W
= 4.5×10–4 kW
= $18/kWh
b. Cost = ($0.12/kWh)(4.5 10 kW)(300 h)
= $0.02
Chapter Assessment: Applying Concepts
83. Batteries When a battery is connected to a complete circuit, charges flow in the circuit almost instantaneously.
Explain.
SOLUTION: A potential difference is felt over the entire circuit as soon as the battery is connected to the circuit.
The potential difference causes the charges to begin to flow. Note: The charges flow slowly compared to
the change in potential difference.
84. Explain why a cow experiences a mild shock when it touches an electric fence.
SOLUTION: By touching the fence and the ground, the cow encounters a difference in potential and conducts
current, thus receiving a shock.
85. Power Lines Why can birds perch on high-voltage lines without being injured?
SOLUTION: No potential difference exists along the wires, so there is no current through the birds’ bodies.
86. Describe two ways to increase the current in a circuit.
SOLUTION: Either increase the voltage or decrease the resistance.
87. Lightbulbs Two lightbulbs work on a 120-V circuit. One is 50 W and the other is 100 W. Which bulb has a higher resistance? Explain.
SOLUTION: 50-W bulb
Therefore, the lower P is caused by a higher R.
88. If the voltage across a circuit is kept constant and the resistance is doubled, what effect does this have on the
eSolutions
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circuit's
SOLUTION: Page 26
SOLUTION: 50-W bulb
Chapter 22 Practice Problems, Review, and Assessment
Therefore, the lower P is caused by a higher R.
88. If the voltage across a circuit is kept constant and the resistance is doubled, what effect does this have on the
circuit's current?
SOLUTION: If the resistance is doubled, the current is halved.
89. What is the effect on the current in a circuit if both the voltage and the resistance are doubled? Explain.
SOLUTION: No effect. V = IR, so and if the voltage and the resistance both are doubled, the current will not change.
90. If the ammeter in the simple circuit in Figure 8 were moved to the bottom of the diagram, would the ammeter have
the same reading? Explain.
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Page 27
SOLUTION: No effect. V = IR, so Chapter 22 Practice Problems, Review, and Assessment
and if the voltage and the resistance both are doubled, the current will not change.
90. If the ammeter in the simple circuit in Figure 8 were moved to the bottom of the diagram, would the ammeter have
the same reading? Explain.
Figure 8
SOLUTION: Yes, because the current is the same everywhere in this circuit.
91. Ohm’s Law Sue finds a device that looks like a resistor. When she connects it to a 1.5-V battery, she measures
−6 A, but
she
only 45×10
eSolutions
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by when
Cognero
uses a 3.0-V battery, she measures 25×10
SOLUTION: No. V = IR, so R = V/I. At 1.5 V,
−3 A. Does the device obey Ohm’s law?
Page 28
Figure 8
SOLUTION: Chapter
22 Practice Problems, Review, and Assessment
Yes, because the current is the same everywhere in this circuit.
91. Ohm’s Law Sue finds a device that looks like a resistor. When she connects it to a 1.5-V battery, she measures
only 45×10
−6 A, but when she uses a 3.0-V battery, she measures 25×10
−3 A. Does the device obey Ohm’s law?
SOLUTION: No. V = IR, so R = V/I. At 1.5 V,
A device that obeys Ohm’s law has a resistance that is independent of the applied voltage.
92. Two wires can be placed across the terminals of a 6.0-V battery. One has a high resistance, and the other has a
low resistance. Which wire will transform energy at a faster rate? Why?
SOLUTION: the wire with the smaller resistance
Smaller R results in a larger power (P), which means energy is transformed at a faster rate.
Chapter Assessment: Mixed Review
93. If a person has $5, how long could he or she play a 200 W stereo if electricity costs $0.15 per kWh? (Level 1)
SOLUTION: 94. A current of 1.2 A is measured through a 50.0-Ω resistor for 5.0 min. How much energy is transformed by the resistor? (Level 1)
SOLUTION: Q = E = I2Rt
4
= 2.2×10 J
95. A 6.0-Ω resistor is connected to a 15-V battery. (Level 1)
a. What is the current in the circuit?
b. How much energy is transformed in 10.0 min?
SOLUTION: a. VManual
= IR- Powered by Cognero
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Q = E = I2Rt
22 Practice Problems, Review, and Assessment
Chapter
4
= 2.2×10 J
95. A 6.0-Ω resistor is connected to a 15-V battery. (Level 1)
a. What is the current in the circuit?
b. How much energy is transformed in 10.0 min?
SOLUTION: a. V = IR
b. Q = E = I2Rt
4
= 2.3×10 J
96. Lightbulbs An incandescent lightbulb with a resistance of 10.0 Ω when it is not lit and a resistance of 40.0 Ω when
it is lit has 120 V placed across it. (Level 2)
a. What is the current draw when the bulb is lit?
b. What is the current draw at the instant the bulb is turned on?
c. When does the lightbulb use the most power?
SOLUTION: a. b. c. the instant it is turned on
97. A 12-V electric motor’s speed is controlled by a potentiometer. At the motor’s slowest setting, it uses 0.02 A. At its
highest setting, the motor uses 1.2 A. What is the range of the potentiometer? (Level 2)
SOLUTION: The slowest speed’s resistance is R = V/I = (12 V) / (0.02 A) = 600 Ω. The fastest speed’s resistance is
R = V/I = (12 V) / (1.2 A) = 1.0×101 Ω. The range is 1.0×101 Ω to 600 Ω.
4 98. An electric motor operates a pump that irrigates a farmer’s crop by pumping 1.0×10 L of water a vertical distance
of 8.0 m into a field each hour. The motor has an operating resistance of 22.0 Ω and is connected across a 110-V
source. (Level 3)
a. What current does the motor draw?
b. How efficient is the motor?
SOLUTION: a. V = IR
b. Ew = mgd
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Page 30
highest setting, the motor uses 1.2 A. What is the range of the potentiometer? (Level 2)
SOLUTION: The22
slowest
speed’s
resistance
is Rand
= V/I
= (12 V) / (0.02 A) = 600 Ω. The fastest speed’s resistance is
Chapter
Practice
Problems,
Review,
Assessment
1
R = V/I = (12 V) / (1.2 A) = 1.0×10 Ω. The range is 1.0×101 Ω to 600 Ω.
4 98. An electric motor operates a pump that irrigates a farmer’s crop by pumping 1.0×10 L of water a vertical distance
of 8.0 m into a field each hour. The motor has an operating resistance of 22.0 Ω and is connected across a 110-V
source. (Level 3)
a. What current does the motor draw?
b. How efficient is the motor?
SOLUTION: a. V = IR
b. Ew = mgd
5
= 8×10 J
Em = IVt = (5.0 A)(110 V)(3600 s)
6
= 2.0×10 J
= 40%
99. Appliances An electric heater is rated at 500 W. (Level 3)
a. How much energy is delivered to the heater in half an hour?
b. The heater is being used to warm a room containing 50 kg of air. If the specific heat of air is 1.10 kJ/kg °C and 50 percent of the thermal energy warms the air in the room, what is the change in air temperature in half an hour?
c. At $0.08 per kWh, how much does it cost to run the heater 6.0 h per day for 30 days?
SOLUTION: a. E = Pt = (5×102 W)(1800 s)
5
= 9×10 J
b. Q = mCΔT
= 8° C
c.
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= $7
Chapter Assessment: Thinking Critically
Page 31
Chapter 22 Practice Problems, Review, and Assessment
= 40%
99. Appliances An electric heater is rated at 500 W. (Level 3)
a. How much energy is delivered to the heater in half an hour?
b. The heater is being used to warm a room containing 50 kg of air. If the specific heat of air is 1.10 kJ/kg °C and 50 percent of the thermal energy warms the air in the room, what is the change in air temperature in half an hour?
c. At $0.08 per kWh, how much does it cost to run the heater 6.0 h per day for 30 days?
SOLUTION: a. E = Pt = (5×102 W)(1800 s)
5
= 9×10 J
b. Q = mCΔT
= 8° C
c.
= $7
Chapter Assessment: Thinking Critically
100. Apply Concepts An artist’s drawing of an electric circuit is shown in Figure 27. Draw a schematic of the
electric circuit using the correct symbols. Indicate the direction of the conventional current in your drawing.
Figure 27
SOLUTION: Models
The energy
101. Formulate
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needed to increase the potential difference of a charge (q) is represented by Page 32
E = qΔV . But in a capacitor, ΔV = q/C. Thus, as charge is added, the potential difference increases. As more
charge is added, however, it takes more energy to add the additional charge. Consider a 1.0-F “supercap” used as
Chapter 22 Practice Problems, Review, and Assessment
101. Formulate Models The energy needed to increase the potential difference of a charge (q) is represented by
E = qΔV . But in a capacitor, ΔV = q/C. Thus, as charge is added, the potential difference increases. As more
charge is added, however, it takes more energy to add the additional charge. Consider a 1.0-F “supercap” used as
an energy storage device in a personal computer. Plot a graph of ΔV as the capacitor is charged by adding 5.0 C to it. What is the voltage across the capacitor? The area under the curve is the energy stored in the capacitor. Find the
energy in joules. Is it equal to the total charge times the final potential difference? Explain.
SOLUTION: Energy E = area under curve
= 13 J
No. Graphically, total change times final potential difference is exactly twice the area under the curve.
Physically, it means that each coulomb would require the same maximum amount of energy to place it on
the capacitor. Actually, the amount of energy needed to add each charge increases as charge
accumulates on the capacitor.
102. Ranking Task Rank the following resistors according to the current through each, from greatest to least.
Specifically indicate any ties.
A: 10-MΩ resistor with a potential difference across it of 1.5 V
B: 10-MΩ resistor with a potential difference across it of 3.0 V
C: 15-MΩ resistor with a potential difference across it of 1.5 V
D: 15-MΩ resistor with a potential difference across it of 0.75 V
E: 20-kΩ resistor with a potential difference across it of 9.0 V
SOLUTION: B>C>A>D>E
103. Reverse Problem Write a physics problem with real-life objects for which the following equation would be part of
the solution:
SOLUTION: Page 33
Answers will vary, but a correct form of the answer is, “A 60-W lightbulb is connected to a 110-V outlet.
When the bulb is turned on, how much current flows through it?”
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D: 15-MΩ resistor with a potential difference across it of 0.75 V
E: 20-kΩ resistor with a potential difference across it of 9.0 V
SOLUTION: Chapter
22 Practice Problems, Review, and Assessment
B>C>A>D>E
103. Reverse Problem Write a physics problem with real-life objects for which the following equation would be part of
the solution:
SOLUTION: Answers will vary, but a correct form of the answer is, “A 60-W lightbulb is connected to a 110-V outlet.
When the bulb is turned on, how much current flows through it?”
104. Problem Posing Complete this problem so that it must be solved using Ohm’s law: “An Ohmic resistor is
connected to a 9-V battery…”
SOLUTION: Answers will vary. A possible form of the correct answer would be, “… and the current through it is 250
mA. What is its resistance?”
105. Apply Concepts The sizes of 10-Ω resistors range from a pinhead to a soup can. Why should they be different?
SOLUTION: The physical size of a resistor is determined by its power rating. Resistors rated at 100 W are much
larger than those rated at 1 W.
106. Make and Use Graphs The diode graph shown in Figure 24 is more useful than a similar graph for a resistor
that obeys Ohm’s law. Explain.
Voltage (V)
Figure 24
SOLUTION: The volt–ampere graph for a resistor obeying Ohm’s law is a straight line and is seldom necessary.
107. Make and Use Graphs Based on what you have learned in this chapter, identify and prepare two parabolic
graphs to describe a circuit's power.
SOLUTION: voltage–power and current–power
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Page 34
Voltage (V)
Figure 24
SOLUTION: Chapter
22 Practice Problems, Review, and Assessment
The volt–ampere graph for a resistor obeying Ohm’s law is a straight line and is seldom necessary.
107. Make and Use Graphs Based on what you have learned in this chapter, identify and prepare two parabolic
graphs to describe a circuit's power.
SOLUTION: voltage–power and current–power
Chapter Assessment: Writing in Physics
108. There are three kinds of equations encountered in science: (1) definitions, (2) laws, and (3) derivations. Examples of
these are: (1) an ampere is equal to one coulomb per second, (2) force is equal to mass times acceleration, (3)
power is equal to voltage squared divided by resistance. Write a one-page explanation of where “resistance is equal
to voltage divided by current” fits. Before you begin to write, first research the three categories given above.
SOLUTION: The student’s answer should include the idea (1) that, for devices obeying Ohm’s law, the voltage drop
is proportional to current through the device and (2) that the formula R = V/I, the definition of
resistance, is a derivation from Ohm’s law.
109. Recall that you learned that matter expands when it is heated. Research the relationship between thermal expansion
and high-voltage transmission lines.
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Page 35
SOLUTION: Answers will vary, but students should determine that transmission lines can become hot enough to
SOLUTION: The student’s answer should include the idea (1) that, for devices obeying Ohm’s law, the voltage drop
is proportional to current through the device and (2) that the formula R = V/I, the definition of
Chapter
22 Practice
Problems,from
Review,
andlaw.
Assessment
resistance,
is a derivation
Ohm’s
109. Recall that you learned that matter expands when it is heated. Research the relationship between thermal expansion
and high-voltage transmission lines.
SOLUTION: Answers will vary, but students should determine that transmission lines can become hot enough to
expand and sag when they have high currents. Sagging lines can be dangerous if they touch objects
beneath them, such as trees or other power lines.
Chapter Assessment: Cumulative Review
6 110. A person burns energy at the rate of about 8.4×10 J per day. How much does she increase the entropy of the
universe in that day? How does this compare to the entropy increase caused by melting 20 kg of ice?
SOLUTION: ΔS = Q/T where T is the body temperature of 310 K.
6
ΔS = (8.4×10 J)/(310 K) = 2.7×104 J/K
For melting ice
5
ΔS = (20 kg)(3.34×10 J/kg)/(273 K)
= 2.4 × 104 J/K
111. When you go up the elevator of a tall building, your ears might pop because of the rapid change in pressure. What is
the pressure change caused by riding in an elevator up a 30-story building (150 m)? The density of air is about 3
1.3 kg/m at sea level.
SOLUTION: ΔP = ρgh
= (1.3 kg/m3)(9.8 N/kg)(150 m)
= 1.9 kPa or about 2/100 of the total air pressure
112. What is the wavelength in air of a 17-kHz sound wave, which is at the upper end of the frequency range of human
hearing?
SOLUTION: = 0.020 m
= 2.0 cm
113. Light of wavelength 478 nm falls on a double slit. First-order bright bands appear 3.00 mm from the central bright band. The screen is 0.91 m from the slits. How far apart are the slits?
SOLUTION: eSolutions Manual - Powered
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–4
= 1.4×10
Page 36
m
114. A charge of +3.0×10
−6 −5 C is 2.0 m from a second charge of +6.0×10
C. What is the magnitude of the force
0.020
m Problems, Review, and Assessment
Chapter
Practice
=22
= 2.0 cm
113. Light of wavelength 478 nm falls on a double slit. First-order bright bands appear 3.00 mm from the central bright band. The screen is 0.91 m from the slits. How far apart are the slits?
SOLUTION: –4
= 1.4×10
m
114. A charge of +3.0×10
between them?
−6 −5 C is 2.0 m from a second charge of +6.0×10
C. What is the magnitude of the force
SOLUTION: = 0.41 N
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