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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 eSolutions Manual - Powered by Cognero Page 1 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. SOLUTION: eSolutions Manual - Powered by Cognero Page 2 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 eSolutions Manual - Powered by Cognero 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. eSolutions Manual - Powered by Cognero b. Page 4 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. eSolutions - Powered by Cognero Page 5 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. SOLUTION: eSolutions Manual - Powered by Cognero Page 6 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. eSolutions Manual - Powered by Cognero Page 7 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? eSolutions Manual - Powered by Cognero SOLUTION: a. P = IV = (15.0 A)(120 V) = 1800 W = 1.8 kW Page 8 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 eSolutions Manual - Powered by Cognero 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? eSolutions Manual - Powered by Cognero Page 10 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? eSolutions Manual - Powered by Cognero 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 Poweredschematic 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? Page 12 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 eSolutions Manual - Powered by Cognero SOLUTION: a. electric energy to heat and light b. electric energy to heat and kinetic energy Page 13 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? eSolutions Manual - Powered by Cognero Page 14 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? eSolutions Manual - Powered by Cognero Page 15 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 Manual 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 Page 16 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 eSolutions Manual - Powered by Cognero 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 Page 19 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 eSolutions Manual24 - Powered by Cognero 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 eSolutions - Powered by Cognero 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 eSolutions Manual - Powered by Cognero 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 eSolutions Manual - Powered by Cognero 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 Manualcurrent? - Powered by Cognero 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. eSolutions Manual - Powered by Cognero 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 Manual - Powered 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 eSolutions Page 29 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 eSolutions Manual - Powered by Cognero 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. eSolutions Manual - Powered by Cognero = $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 eSolutions Manual - Powered by Cognero 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?” eSolutions Manual - Powered by Cognero 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 eSolutions Manual - Powered by Cognero 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. eSolutions Manual - Powered by Cognero 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 by Cognero –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 eSolutions Manual - Powered by Cognero Page 37