Ch 22 Answers Physics

CHAPTER22 Current Electricity page 598 For all problems, assume that the battery voltage and lamp resistances are constant, no matter what current is present. 6. An automobile panel lamp with a resistance of 33 ⍀ is placed across a 12-V battery. What is the current through the circuit? 12 V V I ϭ ᎏ ϭ ᎏ ϭ 0.36 A R 33 ⍀ Practice Problems 22.1 Current and Circuits pages 591–600 page 594 1. The current through a lightbulb connected across the terminals of a 125-V outlet is 0.50 A. At what rate does the bulb convert electric energy to light? (Assume 100 percent efficiency.) P ϭ IV ϭ (0.50 A)(125 V) ϭ 63 J/s ϭ 63 W 2. A car battery causes a current of 2.0 A through a lamp and produces 12 V across it. What is the power used by the lamp? P ϭ IV ϭ (2.0 A)(12 V) ϭ 24 W 3. What is the current through a 75-W lightbulb that is connected to a 125-V outlet? P ϭ IV P 75 W I ϭ ᎏ ϭ ᎏ ϭ 0.60 A V 125 V Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. 7. A motor with an operating resistance of 32 ⍀ is connected to a voltage source. The current in the circuit is 3.8 A. What is the voltage of the source? V ϭ IR ϭ (3.8 A)(32 ⍀) ϭ 1.2ϫ102 V 8. A sensor uses 2.0ϫ10Ϫ4 A of current when it is operated by a 3.0-V battery. What is the resistance of the sensor circuit? V 3.0 V R ϭ ᎏ ϭ ᎏᎏ ϭ 1.5ϫ104 ⍀ Ϫ4 I 2.0ϫ10 A 4. The current through the starter motor of a car is 210 A. If the battery maintains 12 V across the motor, how much electric energy is delivered to the starter in 10.0 s? P ϭ IV and E ϭ Pt Thus, E ϭ IVt ϭ (210 A)(12 V)(10.0 s) ϭ 2.5ϫ104 J 9. 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? 120 V V R ϭ ᎏ ϭ ᎏ ϭ 2.4ϫ102 ⍀ I 0.50 A b. What is the power consumption of the lamp? P ϭ IV ϭ (0.50 A)(120 V) ϭ 6.0ϫ101 W 10. A 75-W lamp is connected to 125 V. a. What is the current through the lamp? P 75 W I ϭ ᎏ ϭ ᎏ ϭ 0.60 A V 125 V 5. A flashlight bulb is rated at 0.90 W. If the lightbulb drops 3.0 V, how much current goes through it? P ϭ IV P 0.90 W I ϭ ᎏ ϭ ᎏ ϭ 0.30 A V 3.0 V b. What is the resistance of the lamp? V 125 V R ϭ ᎏ ϭ ᎏ ϭ 2.1ϫ102 ⍀ I 0.60 A Physics: Principles and Problems Solutions Manual 445 adding an ammeter and a voltmeter across the lamp.5 ⍀ in series. and an ammeter that reads 85 mA.0 V for Practice Problem 12 and 4. 446 Solutions Manual Physics: Principles and Problems . an ammeter.Chapter 22 continued 11.2ϫ102 ⍀ I 0. Determine the resistance and label the resistor. A resistor is added to the lamp in the previous problem to reduce the current to half of its original value.5 V for Practice Problem 13.0-V battery. a division of The McGraw-Hill Companies. Battery ؊ Switch 4. A ؉ V ؊ Therefore.30 A)(2. 15. a lamp. Repeat the previous problem. Indicate the ammeter reading and the direction of the current. Choose a direction for the conventional current and indicate the positive terminal of the battery.1ϫ102 ⍀) ϭ 6.5 V R ϭ ᎏ ϭ ᎏ ϭ 53 ⍀ I 0.60 A ᎏ ϭ 0. and an on-off switch.80 A 60.2ϫ102 ⍀ Ϫ 2. the voltmeter readings will be 60.5 V 53 ⍀ I ؊ V ϭ IR ϭ (0. Draw a series-circuit diagram showing a 4. Rres ϭ Rtotal Ϫ RIamp ϭ 4. Add a voltmeter to measure the potential difference across the resistors in problems 12 and 13 and repeat the problems. Draw a circuit using a battery.30 A)(6.0 V I ϭ ᎏ ϭ ᎏ ϭ 4.1ϫ102 ⍀ ϭ 2.0 V ؊ Potentiometer 12. V V 60.30 A 14. a potentiometer to adjust the lamp’s brightness.5-V battery.085 A A ؉ 85 mA 4. Both circuits will take the following form. What is the potential difference across the lamp? The new value of the current is 0.80 A R 12.3ϫ101 V b. A ؉ I Because the ammeter resistance is assumed zero. Draw a circuit diagram to include a 60.30 A 2 V 4. Inc.1ϫ102 ⍀ c. a resistor. How much resistance was added to the circuit? The total resistance of the circuit is now V 125 V Rtotal ϭ ᎏ ϭ ᎏ ϭ 4.5 ⍀ ؉ ؊ A 13. How much power is now dissipated in the lamp? P ϭ IV ϭ (0. Lamp ؉ Copyright © Glencoe/McGraw-Hill. and a resistance of 12.5 ⍀ I 16. a.3ϫ101 V) ϭ 19 W page 600 12. 18. ؉ V ؊ b. This decrease in potential energy is used to produce heat in the resistor. or to squander. When V increases. Determine the amount of energy it will convert when it is operated for 1 h.0 A R 15 ⍀ Copyright © Glencoe/McGraw-Hill. A 39-⍀ resistor is connected across a 45-V battery.2 A R 39 ⍀ Measure the current through the wire and the potential difference across it. A 15-⍀ electric heater operates on a 120-V outlet. to waste.2ϫ103 J of energy when it is operated for 3. Schematic Draw a schematic diagram of a circuit that contains a battery and a lightbulb.2ϫ10 J E ϭ ΂ ᎏᎏ ΃(60.0 W increase 21.0 min 3 ϩ Ϫ ϭ 4.2 Using Electric Energy pages 601–605 page 603 23.0 min) 3. a division of The McGraw-Hill Companies. because all electric energy is converted to thermal energy.1 Current and Circuits pages 591–600 page 600 17. Power A circuit has 12 ⍀ of resistance and is connected to a 12-V battery. Energy A circuit converts 2. 19. How much energy is used by the heater in 30. Make sure the lightbulb will light in this circuit. and the wire to be tested to make the measurement. the resistance will increase. What is “used” when charge flows through a resistor? The potential energy of the charges decreases as they flow through the resistor. 20. 24. an ammeter.9ϫ104 J c. Resistance Joe states that because R ϭ V/I. Divide the potential difference by the current to obtain the wire resistance. No. Specify what you would measure and how you would compute the resistance.0 min. Determine the change in power if the resistance decreases to 9.4ϫ104 J 22.0 A)2(15 ⍀)(30. a.9ϫ104 J. How much thermal energy is liberated in this time? 2. a. A Practice Problems 22. Is Joe correct? Explain. What is the current through the heater? V 120 V I ϭ ᎏ ϭ ᎏ ϭ 8. resistance depends on the device. so will I. 2. What is the current in the circuit? V 45 V I ϭ ᎏ ϭ ᎏ ϭ 1.Chapter 22 continued Section Review 22. To dissipate is to use.0 s) ϭ 2. Critical Thinking We say that power is “dissipated” in a resistor. Physics: Principles and Problems Solutions Manual 447 .0 ⍀ ϭ 16 W ⌬P ϭ P2 Ϫ P1 ϭ 16 W Ϫ 12 W ϭ 4. Show how you would construct a circuit with a battery. Inc.0 ⍀. if he increases the voltage. Resistance You want to measure the resistance of a long piece of wire.0 s? E ϭ I 2Rt ϭ (8. P1 ϭ V 2/R1 ϭ (12 V)2/12 ⍀ ϭ 12 W P2 ϭ V 2/R2 ϭ (12 V)2/9. a voltmeter.0 W 4. This means that 22 percent of the electric energy is converted to light energy.0 min? V E ϭ ᎏt R (45 V) ϭ ᎏ (5. An automotive battery can deliver 55 A at 12 V for 1. 2.0 s/min) ϭ 4. At $0. It is operated. a. for 5. How many joules does the lightbulb convert into light each minute it is in operation? E ϭ Pt ϭ (0. Assume that 65 percent of the heat is absorbed by the water. How much energy in kWh does it consume in 30 days? E ϭ Pt ϭ (1.65E 0. How long would a 240-V unit operating with the same current take to accomplish the same task? E ϭ IVt ϭ I (2V )΂ ᎏ ΃ t 2 ϭ 1. How much power does the heater use? P ϭ IV ϭ (15.3ϫ103 J b.6ϫ10Ϫ3 A) ϭ 1.0 h/day)(30 days) ϭ 270 kWh c.22)(100.3ϫ105 J c.0 A)(120 V) ϭ 1800 W ϭ 1.0 J/s)(1.12 per kWh.1 h 2 page 605 28. What is the water’s increase in temperature during the 30.0 h and requires 1.000 ⍀ b.Chapter 22 continued b.12/kWh) (30 days)(24 h/day) ϭ $0.20 kg)(4180 J/kgи°C) b. How much power does it use? P ϭ VI ϭ (115 V)(9. a. what is the current through the stove element? 220 V V I ϭ ᎏ ϭ ᎏ ϭ 2. How many joules of thermal energy does the lightbulb produce each minute? E ϭ Pt ϭ (0. How much energy does the element convert to thermal energy in 30.3ϫ105 J) ᎏ ᎏ ⌬T ϭ ᎏ ϭ ᎏ mC (1. ϭ $32.0 A from a 120-V source. How much energy is used by the resistor in 5. The element is used to heat a kettle containing 1.0 J/s)(1.2 h to heat a given volume of water to a certain temperature.12 per kWh.6ϫ10Ϫ3 A R 12.7ϫ103 J 26. A 120-V water heater takes 2. If 220 V are applied across it. How long will it Physics: Principles and Problems ϭ 17°C 448 Solutions Manual .12/kWh)(270 kWh) Copyright © Glencoe/McGraw-Hill.0 min)(60 s/min) (39 ⍀) 2 2 27. on the average.0-W lightbulb is 22 percent efficient. A 100.65)(1.0 min) (60. how much does it cost to operate the clock for 30 days? Cost ϭ (1. a division of The McGraw-Hill Companies.0 h each day.2 h t ϭ ᎏ ϭ 1. How much current does it draw? V 115 V I ϭ ᎏ ϭ ᎏᎏ ϭ 9. doubling the voltage will divide the time by 2.0 min) (60 s/min) ϭ 1.1ϫ10Ϫ3 kWh)($0.0 s? E ϭ I 2Rt ϭ (2. a.20 kg of water.0ϫ101 A)2(11 ⍀)(30.000 ⍀ and is plugged into a 115-V outlet.40 29.8 kW)(5.0 s? Q ϭ mC⌬T with Q ϭ 0. Inc. A digital clock has a resistance of 12.6ϫ104 J 25.0ϫ101 A R 11 ⍀ For a given amount of energy. An electric space heater draws 15. If the owner of the clock pays $0.1 W c.0 s) ϭ 1. a.65E (0.8 kW b.10 30.3 times as much energy for recharge due to its lessthan-perfect efficiency. The resistance of an electric stove element at operating temperature is 11 ⍀.78)(100. how much does it cost to operate the heater for 30 days? Cost ϭ ($0. hot and warm. most devices will have to run longer. The headlamps use the electric charge stored in the car battery. not energy. 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? For the same power. The I 2R loss in the circuit wiring would be dramatically reduced because it is proportional to the square of the current.2 h IV (7. 34. which produces and stores electric charge in the car’s battery. Critical Thinking When demand for electric power is high. so the fixed voltage current is larger. Echarge ϭ (1. if less power was lost during transmission. 36. Energy A car engine drives a generator. 37. the current would be halved. electric energy converted to light and thermal energy in headlamps.0 h) ϭ 858 Wh E 858 Wh t ϭ ᎏ ϭ ᎏᎏ ϭ 8. Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill. chemical energy converted to electric energy in the battery and distributed to the headlamps.25 2 P2 V 1 /R V 12 35. Because I ϭ V /R the resistance is smaller. Chapter Assessment Concept Mapping page 610 38. resistance. Inc. less coal and other power-producing resources would have to be used.5 A)(14 V) Section Review 22.Chapter 22 continued take to charge the battery using a current of 7. at twice the voltage. power companies sometimes reduce the voltage.5V1)2/R P1 ᎏᎏ ϭ ᎏ ϭ ᎏᎏ ϭ 0. a division of The McGraw-Hill Companies. Rework the previous problem by assuming that the battery requires the application of 14 V when it is recharging. Echarge ϭ (1. 31. In which setting is the resistance likely to be smaller? Why? Hot draws more power. 33.0 h) ϭ 858 Wh 858 Wh E t ϭ ᎏ ϭ ᎏᎏ ϭ 9. Complete the concept map using the following terms: watt. current.3)IVt ϭ (1.3)(55 A)(12 V)(1. Electricity rate of flow rate of conversion opposition to flow current power resistance ampere watt ohm Solutions Manual 449 .5 h IV (7. P ϭ IV. List the forms of energy in these three operations.” What is being saved? Power. Research to improve power transmission lines would benefit society in cost of electricity. Power Determine the power change in a circuit if the applied voltage is decreased by one-half. Resistance A hair dryer operating from 120 V has two settings. thereby producing a “brown-out. electric energy stored as chemical energy in the battery. Also.5 A)(12 V) V 22/R (0. Mechanical energy from the engine converted to electric energy in the generator.3)(55 A)(12 V)(1.2 Using Electric Energy pages 601–605 page 605 32.3)IVt ϭ (1. which would improve the quality of our environment.5 A? Assume that the charging voltage is the same as the discharging voltage. Efficiency Evaluate the impact of research to improve power transmission lines on society and the environment. 1) 1 A ϭ 1 C/1 s 40. 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. I ؉ ؊ I b. Draw a circuit schematic of this simple circuit. and connecting wires. 450 Solutions Manual Physics: Principles and Problems . (22. Describe the energy conversions that occur in each of the following devices. Which wire conducts electricity with the least resistance: one with a large cross-sectional diameter or one with a small cross-sectional diameter? (22. Define the unit of electric current in terms of fundamental MKS units. How should an ammeter be connected in Figure 22-12 to measure the motor’s current? (22.1) from left to right through the motor 43. Figure 22-12 The positive voltmeter lead connects to the left-hand motor lead. (22. How must an ammeter be connected in a circuit for the current to be correctly read? The ammeter must be connected in series. Which device converts electric energy to mechanical energy? 4 46. Copyright © Glencoe/McGraw-Hill.1) Break the circuit between the battery and the motor. What is the direction of the conventional motor current in Figure 22-12? (22.1) a. 42. a battery.1) A larger-diameter wire has a smaller resistance because there are more electrons to carry the charge.Chapter 22 continued Mastering Concepts page 610 39. an incandescent lightbulb electric energy to heat and light 4 ؉ 1 ؊ 3 b. Which device provides a way to adjust speed? 3 44.1) a. a division of The McGraw-Hill Companies.1) a. Refer to Figure 22-12 to answer the following questions. (22. Which device converts chemical energy to electric energy? 1 c. a clothes dryer electric energy to heat and kinetic energy c. 41. (22. Inc. and the negative voltmeter lead connects to the right-hand motor lead. How should a voltmeter be connected in Figure 22-12 to measure the motor’s voltage? (22.1) b. Which device turns the circuit on and off? 2 d. A simple circuit consists of a resistor. a digital clock radio 2 ! electric energy to light and sound 45. The potential difference causes the charges to begin to flow. Lightbulbs Two lightbulbs work on a 120-V circuit. 54. 55.2) the resistance of the wire and the current in the wire 50. Which bulb has a higher resistance? Explain. subjecting the filament to greater stress. 52. Define the unit of power in terms of fundamental MKS units. c.2) The short circuit produces a high current. What electric quantities must be kept small to transmit electric energy economically over long distances? (22.2) The low resistance of the cold filament allows a high current initially and a greater change in temperature. Either increase the voltage or decrease the resistance. Power Lines Why can birds perch on highvoltage lines without being injured? No potential difference exists along the wires. Copyright © Glencoe/McGraw-Hill. Why do lightbulbs burn out more frequently just as they are switched on rather than while they are operating? (22. a division of The McGraw-Hill Companies. thus receiving a shock. If the voltage across a circuit is kept constant and the resistance is doubled. This raises the atoms’ kinetic energies and the temperature of the wire. Physics: Principles and Problems Solutions Manual 451 . so there is no current through the birds’ bodies. Explain why a cow experiences a mild shock when it touches an electric fence. How must a voltmeter be connected to a resistor for the potential difference across it to be read? The voltmeter must be connected in parallel.2) J kgиm2 C J W ϭ ᎏᎏиᎏᎏ ϭ ᎏ ϭ ᎏ ϭ ᎏ 3 s C s s s m2 kgиᎏᎏ s2 Therefore. One is 50 W and the other is 100 W. Describe two ways to increase the current in a circuit. charges flow in the circuit almost instantaneously. If a battery is short-circuited by a heavy copper wire being connected from one terminal to the other. 53. so R ϭ ᎏ R P 47. 56. 48. A potential difference is felt over the entire circuit as soon as the battery is connected to the circuit. Batteries When a battery is connected to a complete circuit. (22. Why does this happen? (22. the current is halved. 49. which causes more electrons to collide with the atoms of the wire. Inc. the temperature of the copper wire rises. the cow encounters a difference in potential and conducts current.Chapter 22 continued V Applying Concepts I A I ؉ ؊ pages 610–611 51. Note: The charges flow slowly compared to the change in potential difference. the lower P is caused by a higher R. V I ؉ ؊ I A By touching the fence and the ground. 50-W bulb V2 V2 P ϭ ᎏ . what effect does this have on the circuit’s current? If the resistance is doubled. Explain. 0-V battery.0 V.Chapter 22 continued 57. and the other has a low resistance. Two wires can be placed across the terminals of a 6.0 V At 3. What should the voltmeter reading be? 27 V c. How much power is delivered to the motor? P ϭ VI ϭ (12 V)(1. but when she uses a 3. At 1. How much power is delivered to the resistor? P ϭ VI ϭ (27 V)(1. Ohm’s Law Sue finds a device that looks like a resistor.6ϫ104 J 62. No effect.5 A) ϭ 41 W d. Inc. 452 Solutions Manual Physics: Principles and Problems . A A device that obeys Ohm’s law has a resistance that is independent of the applied voltage. would the ammeter have the same reading? Explain.5 A 18 ⍀ b.5 A ؊ 12 V ! Figure 22-13 a.5 V 4 R ϭ ᎏᎏ Ϫ6 ϭ 3.1 Current and Circuits pages 611–612 Level 1 61. R ϭ ᎏᎏ ϭ 120 ⍀ Ϫ3 25ϫ10 A ؉ Motor 1. How much energy is delivered to the resistor per hour? E ϭ Pt ϭ (41 W)(3600 s) ϭ 1.0-V battery. she measures 25ϫ10Ϫ3 A. What should the ammeter reading be? 27 V I ϭ V /R ϭ ᎏ ϭ 1. Refer to Figure 22-14 to answer the following questions. Mastering Problems 22. When she connects it to a 1. ؉ 27 V ؊ 18 ⍀ V Copyright © Glencoe/McGraw-Hill. Does the device obey Ohm’s law? No. A motor is connected to a 12-V battery. 58. Which wire will produce thermal energy at a faster rate? Why? the wire with the smaller resistance Pϭ ᎏ R V2 ! Figure 22-14 a. V ϭ IR.5 V. the current will not change. What is the effect on the current in a circuit if both the voltage and the resistance are doubled? Explain. because the current is the same everywhere in this circuit. as shown in Figure 22-13. which produces thermal energy at a faster rate. I 60. and if the voltage and the resistance both are doubled. she measures only 45ϫ10Ϫ6 A. 1. How much energy is converted if the motor runs for 15 min? E ϭ Pt ϭ (18 W)(15 min)(60 s/min) ϭ 1.5-V battery. 59.5ϫ105 J Smaller R produces larger power P dissipated in the wire.3ϫ10 ⍀ 45ϫ10 3. so R ϭ V /I. One has a high resistance. a division of The McGraw-Hill Companies. If the ammeter in Figure 22-4a on page 596 were moved to the bottom of the diagram. V ϭ IR. so I ϭ V /R.5 A) ϭ 18 W b. Yes. ϭ 18.5 W)(3600 s) ϭ 1. does the motor use? P ϭ 2.0ϫ101 W)΂ ᎏ ΃΂ ᎏ ΃ 5. a. What power.0 A.6ϫ104 J 65. How much energy is delivered to the resistor per hour? E ϭ Pt ϭ (81 W)(3600 s) ϭ 2. What power is dissipated by the bulb? P ϭ IV ϭ (1. in watts. How much current does the dryer draw? P ϭ IV P 4200 W I ϭ ᎏᎏ ϭ ᎏᎏ ϭ 19 A V 220 V Figure 22-16 a.0 A)(120 V) ϭ 9.0 ⍀ V I ! Figure 22-15 a.2 A is measured through a lightbulb when it is connected across a 120-V source. A ؉ 9.50 A)(120 V) ϭ 6.0 min 1 60 s min 64. What should the ammeter reading be? 27 V I ϭ V /R ϭ ᎏ ϭ 3. a division of The McGraw-Hill Companies. Dryers A 4200-W clothes dryer is connected to a 220-V circuit. Lightbulbs A current of 1. How many joules of energy does the battery deliver to the motor each second? P ϭ IV ϭ (210 A)(12 V) ϭ 2500 J/s or 2. Toasters The current through a toaster that is connected to a 120-V source is 8.0ϫ101 W b. How much energy is delivered to the resistor per hour? E ϭ Pt ϭ (4.0 V c. Refer to Figure 22-15 to answer the following questions.0 A 9.9ϫ105 J ϭ (6.0 ⍀ 66. How much power is delivered to the resistor? P ϭ VI ϭ (27 V)(3.8ϫ104 J 68. How much power is delivered to the resistor? P ϭ VI ϭ (9.4ϫ102 W 67. What should the ammeter reading be? 9.0 V I ϭ V /R ϭ ᎏ ϭ 0.50 A 18 ⍀ b.6ϫ102 W ؉ 27 V ؊ 9.000 J ϭ 1. What should the voltmeter reading be? 27 V c.5ϫ103 W 69. What should the voltmeter reading be? 9.50 A) ϭ 4. Copyright © Glencoe/McGraw-Hill.Chapter 22 continued 63.5 W Physics: Principles and Problems Solutions Manual 453 .0 V ؊ 18 ⍀ V a.2 A)(120 V) ϭ 1. A d. so E ϭ Pt E t b. A lamp draws 0. How much power is delivered? P ϭ IV ϭ (0.0 V)(0. The current through the motor is 210 A. Inc.0 min? The definition of power is P ϭ ᎏᎏ.5ϫ103 J/s I ! b. Refer to Figure 22-16 to answer the following questions. What power is dissipated by the toaster? P ϭ IV ϭ (8. A 12-V automobile battery is connected to an electric starter motor. How much energy is converted in 5.50 A from a 120-V generator.0 A) ϭ 81 W d. 00 Ϫ6. Inc.5 W b.00 Ϫ2.00 6.00 10.0-⍀ resistor if the current is 1.5 A)(4. What voltage is placed across a motor with a 15-⍀ operating resistance if there is 8.0510 Ϫ0.0390 Ϫ0. Batteries A resistor of 60. For each measurement. They then measured the current through the wire for several voltages.0ϫ103 J 71.00 V across the wire. ⍀ b.0280 Ϫ0.0270 0.40 A)(60.00 0.06 0.0140 Ϫ0.00 4. Does the nichrome wire obey Ohm’s law? If not.0 A of current? I (amps) a. V ϭ IR ϭ (8.00 4. A voltage of 75 V is placed across a 15-⍀ resistor.00 V and 10. Physics: Principles and Problems 454 Solutions Manual . a division of The McGraw-Hill Companies. a.40 A through it when it is connected to the terminals of a battery.00 ؊8.00 Ϫ10.0520 0. The students recorded the data for the voltages used and the currents measured.0620 ______________ ______________ ______________ ______________ ______________ ______________ ______________ ______________ ______________ ______________ E ϭ Pt ϭ (4. 0. Flashlights A flashlight bulb is connected across a 3. I Resistance. R R R R ϭ ϭ ϭ ϭ 143 154 143 161 ⍀.5 W)(11 min)΂ᎏᎏ΃ 60 s min ϭ 3.5 A.0 ⍀) ϭ 6. R ϭ 157 ⍀.0 A)(15 ⍀) ϭ 1.0 ⍀ has a current of 0.0 ⍀) ϭ 24 V 72. R ϭ 143 ⍀.0140 0. The resistance of the nichrome wire increases somewhat as the magnitude of the voltage increases. Some students connected a length of nichrome wire to a variable power supply to produce between 0.06 V (volts) Level 2 75. c. R ϭ 148 ⍀. Ohm’s law is obeyed when the resistance of a device is constant and independent of the potential difference. so t Table 22-2 Voltage.0630 Ϫ0.0 A R 15 ⍀ ؊0.00 Copyright © Glencoe/McGraw-Hill. What is the current through the resistor? V ϭ IR V 75 V I ϭ ᎏ ϭ ᎏᎏ ϭ 5.2ϫ102 V 74. R ϭ 159 ⍀. The current through the bulb is 1.0-V potential difference. R ϭ 154 ⍀. calculate the resistance. What is the power rating of the bulb? P ϭ IV ϭ (1. What is the voltage of the battery? V ϭ IR ϭ (0.0 V 73. for all the voltages.0400 0. Graph I versus V.04 ؊0. R ϭ V/I (volts) (amps) (amps) 2. What voltage is applied to a 4. V Current. so the wire does not quite obey Ohm’s law.02 ؊12.00 8.00 12.0 V) ϭ 4. ⍀.00 8.5 A? V ϭ IR ϭ (1. How much electric energy does the bulb convert in 11 min? E The definition of power is P ϭ ᎏᎏ. ⍀.04 0.Chapter 22 continued 70. as shown in Table 22-2. R ϭ 150 ⍀.00 0. specify the voltage range for which Ohm’s law holds.00 Ϫ8.5 A)(3.00 ؊4.00 Ϫ4. The current through a lamp connected across 120 V is 0. 0.0 V) ϭ 0.0-V battery. its resistance is What is the lamp’s cold resistance? ᎏ (3.6 0. a division of The McGraw-Hill Companies.0-W lightbulb use in half an hour? If the lightbulb converts 12 percent of electric energy to light energy.0ϫ102 ⍀) ϭ 6. When a 9. and V ϭ IR. What is the diode’s resistance when a ϩ0. 5 V ϭ IR ϭ (1. Draw a series circuit diagram to include a 16-⍀ resistor.02 Copyright © Glencoe/McGraw-Hill. A lamp draws a 66-mA current when connected to a 6. 88 percent of the energy is lost to heat.08ϫ105 J If the bulb is 12 percent efficient.01 0 0.2ϫ10 A b.75 A.68 W 78.0 W)(1800 s) ϭ 1. so 0.0 A 1 R 6.1 66 6.0 c.40 A when the lamp is on. and the direction of conventional current. the positive terminal of the ammeter.0ϫ102 ⍀ I 0. The voltage is increased by a 9. How much power does the lamp dissipate when it is connected to the 6.40 A A 16 ⍀ I I ‫ ؍‬1. Inc.75 A)(16 ⍀) ϭ 28 V 77.Chapter 22 continued 76. ؉ ؉ ؊ 79.5ϫ104 J a.8 Voltage (V) Figure 22-17 E ϭ Pt ϭ (60.0ϫ101 ⍀ ΂ᎏ1 5΃ 1 ᎏᎏ as great as it is when the lamp is hot. Resistance depends on voltage.2 ! 0.4 0.75 A V ‫ ؍‬28 V b.70 V V R ϭ ᎏᎏ ϭ ᎏᎏ ϭ 32 ⍀ Ϫ2 I 2. Does the lamp obey Ohm’s law? No.0 factor of ᎏᎏ ϭ 1. I ϭ 22 m/A. The graph in Figure 22-17 shows the current through a device called a silicon diode.0 V) ϭ 0.0-V battery? P ϭ IV ϭ (66ϫ10Ϫ3 A)(6. a. What is the current through the lamp as it is turned on if it is connected to a potential difference of 120 V? V ϭ IR 120 V V I ϭ ᎏ ϭ ᎏᎏ ϭ 2. What is the lamp’s resistance when it is on? V ϭ IR 120 V V R ϭ ᎏᎏ ϭ ᎏ ϭ 3. Lightbulbs How much energy does a 60. Indicate the positive terminal and the voltage of the battery.0 V? P ϭ IV ϭ (75ϫ10Ϫ3 A)(9. so Q ϭ (0. the lamp draws 75 mA. A potential difference of ϩ0. but the current 75 is increased by a factor of ᎏᎏ ϭ 1. Current in a Diode Current (A) 0.0-V battery is used.5.60-V potential difference is used? 0. How much power does it dissipate at 9.2ϫ10 A c.08ϫ105 J) ϭ 9. a battery. a.2ϫ102 ⍀ Ϫ3 I 5.40 W c. how much thermal energy does it generate during the half hour? E P ϭ ᎏᎏ t Level 3 80.0ϫ10 ⍀ b.60 V V R ϭ ᎏᎏ ϭ ᎏᎏ ϭ 1.88)(1. and an ammeter that reads 1. Physics: Principles and Problems Solutions Manual 455 . What is the resistance of the diode? From the graph. When the lamp is cold. Does the diode obey Ohm’s law? No.70 V is placed across the diode. 0 V)2 4.0 V ؊ 4.0ϫ101 W.0 h before it must be replaced. E ϭ IVt ϭ (0.0 A of current. and a resistance of 45 ⍀ connected in series.9ϫ10 kWh b. Utilities Figure 22-19 represents an electric furnace. Calculate the cost per kWh.0 ⍀ 22.15 A Ίᎏ ๶ Ί๶ 220 ⍀ R P 5. A 110-V electric iron draws 3.Chapter 22 continued 81.0ϫෆ 101 W)(40.9ϫ10Ϫ3 kWh $3.80 ⍀ 84. This is based on the assumption that the air conditioner will run half of the time Physics: Principles and Problems ΂ (240. the maximum safe current P ϭ I 2R Iϭ Ί๶ Ί๶ P ᎏ ϭ R 5.0 W Copyright © Glencoe/McGraw-Hill. What is the maximum current allowed in a 5.0ϫ101 W ᎏᎏ ϭ 1 A 40.00 cost Rate ϭ ᎏᎏ ϭ ᎏᎏ Ϫ3 E 5. Use the figure to find the following: 456 Solutions Manual . Appliances A window air conditioner is estimated to have a cost of operation of $50 per 30 days.0 h)(3600 s/h) ϭ 1. the maximum safe voltage P ϭ V 2/R (5. Calculate the monthly (30-day) heating bill if electricity costs $0. Thermostat ϭ $510/kWh 83. an ammeter. A ؉ ؉ V ؊ 40.25) ϭ 2160 kWh Cost ϭ (2160 kWh)($0.10 per kWh and the thermostat is on one-fourth of the time.0 ⍀ 2A 90 V 45 ⍀ I I ! ؊ Figure 22-18 V ϭ IR 90 V V I ϭ ᎏᎏ ϭ ᎏ ϭ 2 A 45 ⍀ R a.0 V)(26.0 ⍀) V ϭ ͙PR ෆ ϭ ͙ෆ ෆෆ ϭ 45 V 86.0 h) ϭ 5. Batteries A 9.0 A)(1.0-V battery costs $3. the maximum safe power is 5.0250 A)(9. Draw a schematic diagram to show a circuit including a 90-V battery. What is the ammeter reading? Draw arrows showing the direction of conventional current. a division of The McGraw-Hill Companies. For the circuit shown in Figure 22-18.2ϫ106 J ! Figure 22-19 E ϭ ΂ᎏᎏ΃(t ) R V2 ϭ ᎏᎏ (30 d)(24 h/d)(0.0250 A for 26. 220-⍀ resistor? P ϭ I 2R Iϭ ᎏ ϭ ᎏ ϭ 0.2 Using Electric Energy pages 612–613 Level 1 82.100/kWh) ϭ $216 87. ؉ 240.00 and will deliver 0. Inc.0-W. How much thermal energy is developed in an hour? Q ϭ E ϭ VIt ϭ (110 V)(3.80 ⍀ ΃ Level 2 85.9 Wh ϭ 5. 0 h? Cost ϭ ($0.0-⍀ resistor for 5.0 ⍀)(10.0 min.12 per kWh.0ϫ101 ⍀. Determine how much current the air conditioner will take from a 120-V outlet. What is the current draw at the instant the bulb is turned on? V 120 V I ϭ ᎏᎏ ϭ ᎏᎏ ϭ 12 A R 10. A 12-V electric motor’s speed is controlled by a potentiometer. At the motor’s slowest setting.0-⍀ resistor is connected to a 15-V battery.45 W ϭ 4. Radios A transistor radio operates by means of a 9.5) Vt ϭ 2.5 A R 6.0 ⍀ when it is lit has 120 V placed across it.0 h.0ϫ101 ⍀ to 600 ⍀.02 Copyright © Glencoe/McGraw-Hill. a.090/kWh 90.49 Cost ϭ ᎏᎏᎏ Ϫ4 (4. the motor uses 1.3ϫ104 J Level 2 92.Chapter 22 continued and that electricity costs $0. a. A 6.050 A)(9.0 ⍀ ϭ 12.0-mA current.00/kWh b.0-V battery that supplies it with a 50.0 h) b.5ϫ10Ϫ4 kW $2.5ϫ10Ϫ4 kW)(300 h) ϭ $0.5 A)2(6. Lightbulbs An incandescent lightbulb with a resistance of 10.2 A ϭ 1.2 A)2(50. Solutions Manual ϭ ᎏᎏᎏ 1 kW ($0. What does it now cost to operate the radio for 300. If the cost of the battery is $2. How much thermal energy is produced in 10.0 min? Q ϭ E ϭ I 2Rt ϭ (2.15 per kWh? E ϭ Pt ϭ ᎏᎏ Cost t ϭ ᎏᎏ (Rate)(P ) Cost Rate c. by means of a converter.9 A 88.0 ⍀)(5.0 ⍀ when it is not lit and a resistance of 40. it uses 0. a. The range is 1.0 V) ϭ 0.2ϫ104 J 91.2 A. what is the cost per kWh to operate the radio in this manner? P ϭ IV ϭ (0. What is the current in the circuit? V ϭ IR V 15 V I ϭ ᎏᎏ ϭ ᎏᎏ ϭ 2. is plugged into a household circuit by a homeowner who pays $0.2 A is measured through a 50.02 A ϭ 600 ⍀.0 min)΂ᎏᎏ΃ 60 s min ϭ 556 kWh E ϭ IVt (556 kWh)(1000 W/kW) E I ϭ ᎏᎏ ϭ ᎏᎏᎏ (120 V)(30 d)(24 h/d)(0. how long could he or she play a 200 W stereo if electricity costs $0. b. a division of The McGraw-Hill Companies. A current of 1.15/kWh)(200 W)΂ᎏ ᎏ 1000 W ΃ $5 ϭ 200 h Physics: Principles and Problems 457 .0 ⍀ ϭ $18. The fastest speed’s resistance is R ϭ V/I ϭ 12 V/1.0 A R 40.0 ⍀ Mixed Review page 613 Level 1 89. Inc.5ϫ10 kW)(300.49 and it lasts for 300.02 A. Cost ϭ (E )(rate) $50 Cost E ϭ ᎏᎏ ϭ ᎏ ᎏ rate $0. What is the range of the potentiometer? The slowest speed’s resistance is R ϭ V/I ϭ 12 V/0. What is the current draw when the bulb is lit? V 120 V I ϭ ᎏᎏ ϭ ᎏᎏ ϭ 3. When does the lightbulb use the most power? the instant it is turned on 93.12/kWh) (4. How much heat is generated by the resistor? Q ϭ E ϭ I 2Rt ϭ (1. At its highest setting.0 min)΂ᎏᎏ΃ 60 s min ϭ 2. If a person has $5. The same radio.090 per kWh. 0ϫ101 A R 4. A heating coil has a resistance of 4.80 m/s2)(8.08 (30 days)΂ ᎏᎏ ΃΂ ᎏ ΃ 6 3.0ϫ10 J 5 ϭ $4.0ϫ101 A)2(4.0 ⍀ b. Formulate Models How much energy is stored in a capacitor? The energy needed to increase the potential difference of a charge. What current does the motor draw? V ϭ IR V 110 V I ϭ ᎏ ϭ ᎏ ϭ 5. An electric motor operates a pump that irrigates a farmer’s crop by pumping 1.0 A R 22.0 A)(110 V)(3600 s) ϭ 2.0 h 3600 s Cost ϭ ΂ ᎏ ΃΂ ᎏ ΃΂ ᎏ ΃ s day h 1kWH $0.40 96.6ϫ10 J kWh ϭ 1.08 per kWh.10 kJ/kgи°C. a. and 50 percent of the thermal energy heats the air in the room. a. At $0. How much energy is delivered to the heater in half an hour? E ϭ Pt ϭ (5ϫ102 W)(1800 s) ϭ 9ϫ105 J b.08 1 kWh ᎏ ΂ ᎏᎏ 3.1ϫ10 J ϭ ᎏᎏᎏ (20. what will be the increase in the temperature of the water? Assume 100 percent of the heat is absorbed by the water. Physics: Principles and Problems 458 Solutions Manual .6ϫ106 J ΃΂ kWh ΃ 1.0 m into a field each hour.0ϫ106 J Ew Efficiency ϭ ᎏ ϫ 100 Em 8ϫ10 J ᎏ ϭ ᎏ ϫ 100 6 2. a.0 ⍀)(5. how much does it cost to operate the heating coil 30 min per day for 30 days? Cost ϭ ΂ ᎏᎏ ΃΂ ᎏ ΃(30 days) 30 min day $0.0ϫ104 L of water a vertical distance of 8.0 ⍀ and is connected across a 110-V source.08 per kWh. Q ϭ mC⌬T ϭ $7 Thinking Critically page 614 97.0 min? E ϭ I 2Rt ϭ (3. How efficient is the motor? EW ϭ mgd ϭ (1ϫ104 kg)(9. At $0.5)(9ϫ10 J) ϭ ᎏᎏᎏ (50.0 min)΂ᎏᎏ΃ 60 s min ϭ 8°C c. ϭ 40% 95.0 m) ϭ 8ϫ105 J Em ϭ IVt ϭ (5. The motor has an operating resistance of 22.0 h per day for 30 days? 500 J 6. Appliances An electric heater is rated at 500 W. What energy is supplied to the coil in 5. q.1ϫ106 J 5 min b.0 kg)(1100 J/kgиC°) 5 Copyright © Glencoe/McGraw-Hill. The heater is being used to heat a room containing 50 kg of air. Inc. If the specific heat of air is 1. If the coil is immersed in an insulated container holding 20.0 kg of water.0 kg)(4180 J/kgиC°) 6 ϭ 13°C d. a division of The McGraw-Hill Companies. What is the current in the coil while it is operating? V ϭ IR 120 V V I ϭ ᎏᎏ ϭ ᎏ ϭ 3. how much does it cost to run the heater 6.Chapter 22 continued Level 3 94.0 ⍀ Q ⌬T ϭ ᎏᎏ mC 1.0 ⍀ and operates on 120 V. what is the change in air temperature in half an hour? Q ϭ mC⌬T Q ⌬T ϭ ᎏᎏ mC (0.1ϫ106 J c. V ϭ q/C. As more charge is added. Apply Concepts A microwave oven operates at 120 V and requires 12 A of current. 1 ϭ ᎏ (5. Input 1440 J/s → Microwave converter 1440 J/s → Heat converter 1080 J/s → Useful output 810 J/s Wasted 360 J/s b.0 C Voltage V ϭ ᎏᎏ ϭ ᎏ ϭ 5. What is the voltage across the capacitor? The area under the curve is the energy stored in the capacitor.0 4.0 C to it.78°C/s d. Solve for the rate of temperature rise given the rate of energy input. Graphically. But in a capacitor. The kg unit cancels and the J unit cancels. Actually. Derive an equation for the rate of temperature increase (⌬T/s) from the information presented in Chapter 12. Physically.0 Charge (C) Figure 22-2b on page 593. it means that each coulomb would require the same maximum amount of energy to place it on the capacitor.0 V)(5.Chapter 22 continued is represented by E ϭ qV.0 3. as charge is added. Use your equation to solve for the rate of temperature rise in degrees Celsius per second when using this oven to heat 250 g of water above room temperature. however.0 4.0-F “supercap” used as an energy storage device in a personal computer. ⌬T 1 ⌬Q ᎏᎏ ϭ ᎏᎏ ᎏᎏ ⌬t mC ⌬t → Wasted 270 J/s → ΂ ΃ q 5. a. Is it equal to the total charge times the final potential difference? Explain. Discuss. it takes more energy to add the additional charge. The efficiency of conversion from microwave radiation to thermal energy in water is 75 percent. Its electric efficiency (converting AC to microwave radiation) is 75 percent.0 2. total change times final potential difference is exactly twice the area under the curve. Thus. 98. the mass. Inc.0 F C ΂ ΃ 810 J/s ϭ ᎏᎏᎏ (0.0 1. the amount of energy needed to add each charge increases as charge accumulates on the capacitor.0 3.0 5. Plot a graph of V as the capacitor is charged by adding 5.0 0. Find the energy in joules. it might be possible to find a new Solutions Manual ϭ 13 J No. leaving °C/s.0 C) 2 ϭ 0.0 V 1. e. It might be possible to use a different frequency of electromagnetic radiation to improve this rating. Or. and its conversion efficiency from microwave radiation to heating water is also 75 percent. Potential difference (V) 5. the potential difference increases. Consider a 1. and the specific heat of a substance. Review your calculations carefully for the units used and discuss why your answer is in the correct form.0 2. 1 ⌬Q ⌬T ᎏᎏ ϭ ᎏᎏ ᎏᎏ mC ⌬t ⌬t c. It might be possible to find a way to convert electric energy to radiation using a different approach that would be more efficient. different ways in which you could increase the efficiency of microwave heating. The efficiency of conversion from electric energy to microwave energy is 75 percent.0 1. a division of The McGraw-Hill Companies. Label the function of each block according to total joules per second. Draw a block power diagram similar to the energy diagram shown in Physics: Principles and Problems 459 .25 kg)(4180 J/kgи°C) Energy E ϭ area under curve Copyright © Glencoe/McGraw-Hill. in general terms. In the case of heating a cup of water. This can lead to overheating of the oven components and to their failure. Formulate a diplomatic reply to the clerk. The empty oven means that the microwave energy has to be dissipated in the oven. P ϭ I 2R 10 0 5 10 15 20 Voltage across a 10-⍀ resistor 2 PϭV R 40 30 20 10 b. Hint: Think about heating a specific object. Explain. why microwave ovens are not useful for heating everything. but it depends on the specific application. g. why it is not a good idea to run microwave ovens when they are empty.” 460 Solutions Manual 0 1 2 Amperes through a 10-⍀ resistor Physics: Principles and Problems . In the case of heating a potato. Discuss. a microwave oven heats mostly the potato and is more efficient than an electric oven or skillet. a. Make and Use Graphs The diode graph shown in Figure 22-17 on page 612 is more useful than a similar graph for a resistor that obeys Ohm’s law. Formulate an argument to refute the clerk’s claim. A microwave oven uses two energy conversions (electricity to microwave radiation to heat) and is typically around 50 percent efficient. It’s not as good for other materials. identify and prepare two parabolic graphs. Hint: Think about heating a specific object. Watts dissipated 100. The volt–ampere graph for a resistor obeying Ohm’s law is a straight line and is seldom necessary. The conversion efficiency from microwave energy to thermal energy is good for water. 102. a division of The McGraw-Hill Companies. racks. in general terms. The physical size of a resistor is determined by its power rating.Chapter 22 continued geometry of radiating objects to be heated to improve the efficiency. in efficiency terms. etc. 101. voltage–power and current–power 40 Watts dissipated 30 20 Copyright © Glencoe/McGraw-Hill. c. 99. Inc. Analyze and Conclude A salesclerk in an appliance store states that microwave ovens are the most electrically efficient means of heating objects. Explain. an immersion heater uses only resistance for energy conversion and is nearly 100 percent efficient. Resistors rated at 100 W are much larger than those rated at 1 W. The containers and dishes designed for use with microwave ovens convert little of the energy. Make and Use Graphs Based on what you have learned in this chapter. f. Discuss. cabinets. which also heats the air. “It can be true. Formulate an argument to support the clerk’s claim. Apply Concepts The sizes of 10-⍀ resistors range from a pinhead to a soup can. your ears might pop because of the rapid change in pressure. first research the three categories given above.0 m from a second charge of ϩ6.4ϫ10Ϫ4 m 109. (3) power is equal to voltage squared divided by resistance.Chapter 22 continued Writing in Physics page 614 103. which is at the upper end of the frequency range of human hearing? (Chapter 15) v ϭ ␭f 343 m/s v ␭ ϭ ᎏ ϭ ᎏᎏ ϭ 0.3 kg/m3 at sea level. Light of wavelength 478 nm falls on a double slit. The screen is 0. Before you begin to write. When you go up the elevator of a tall building.020 m ϭ 2. but students should determine that transmission lines can become hot enough to expand and sag when they have high currents. for devices obeying Ohm’s law. First-order bright bands appear 3. A person burns energy at the rate of about 8.9 kPa or about 2/100 of the total air pressure 107.3 kg/m3)(9.34ϫ105 J/kg)/(273 K) ϭ 2.41 N Solutions Manual 461 . Sagging lines can be dangerous if they touch objects beneath them.4ϫ106 J per day. 104. (2) force is equal to mass times acceleration.91 m) (3.00ϫ10 m) q q d ϭ (9.0 cm f 17. The student’s answer should include the idea (1) that. What is the wavelength in air of a 17-kHz sound wave.4ϫ104 J/K 106. ⌬S ϭ (8. such as trees or other power lines.0 m)2 ϭ 0.7ϫ104 J/K For melting ice Physics: Principles and Problems (478ϫ10Ϫ9 m)(0. Write a one-page explanation of where “resistance is equal to voltage divided by current” fits. Research the relationship between thermal expansion and high-voltage transmission lines. you learned that matter expands when it is heated.00 mm from the central bright band.80 m/s2)(150 m) ϭ 1. How far apart are the slits? (Chapter 19) ␭ ϭ ᎏᎏ d ϭ ᎏᎏ ϭ ᎏᎏᎏ Ϫ3 ϭ 1. the definition of resistance.0ϫ109 Nиm2/C2) (3. (Chapter 13) ⌬P ϭ ␳gh ϭ (1.91 m from the slits.0ϫ10Ϫ6 C is 2. There are three kinds of equations encountered in science: (1) definitions. 108. What is the magnitude of the force between them? (Chapter 20) A B F ϭ Kᎏ 2 xd L ␭L x Cumulative Review page 614 105. 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? (Chapter 12) ⌬S ϭ Q/T where T is the body temperature of 310 K. Examples of these are: (1) an ampere is equal to one coulomb per second. ⌬S ϭ (20 kg)(3.0ϫ10Ϫ5 C) ᎏ ᎏ ᎏ ᎏ (2. a division of The McGraw-Hill Companies. and (3) derivations.000 Hz Copyright © Glencoe/McGraw-Hill.4ϫ106 J)/(310 K) ϭ 2. (2) laws. Inc. In Chapter 13.0ϫ10Ϫ5 C.0ϫ10Ϫ6 C)(6. is a derivation from Ohm’s law. What is the pressure change caused by riding in an elevator up a 30-story building (150 m)? The density of air is about 1. A charge of ϩ3. Answers will vary. the voltage drop is proportional to current through the device and (2) that the formula R ϭ V/I. Switch 1 is closed.Chapter 22 continued Challenge Problem page 604 Use the figure to the right to help you answer the questions below. Switch 1 is now opened. What is the voltage across the capacitor? 15 V 2. the current remains at 13 mA because the battery voltage is constant at 15 V. 15 V Switch 2 Switch 1 1. Switch 2 is closed. What is the voltage across the capacitor and the current through the resistor immediately after Switch 2 is closed? 15 V and 13 mA 4. and the current eventually would become zero due to battery depletion. Initially. and Switch 2 remains open. the capacitor is uncharged. 462 Solutions Manual Physics: Principles and Problems . what happens to the voltage across the capacitor and the current through the resistor? The capacitor voltage remains at 15 V because there is no path to discharge the capacitor.5 ␮F 1200 ⍀ Copyright © Glencoe/McGraw-Hill. What is the voltage across the capacitor? Why? It remains 15 V because there is no path for the charge to be removed. and Switch 2 remains open. Next. 1. Inc. However. As time goes on. a division of The McGraw-Hill Companies. if the battery and capacitor were real components instead of ideal circuit components. while Switch 1 remains open. the capacitor voltage eventually would become zero due to leakage. 3.