CHAPTER
Current Electricity
22
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 ! IV ! (0.50 A)(125 V) ! 63 J/s ! 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 ! IV ! (2.0 A)(12 V) ! 24 W 3. What is the current through a 75-W lightbulb that is connected to a 125-V outlet? P ! IV
6. An automobile lamp with a resistance of 33 ! is placed across a 12-V battery. What is the current through the circuit? 12 V V I ! $ ! $ ! 0.36 A 33 !
R
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 ! IR ! (3.8 A)(32 !) ! 1.2"102 V 8. A sensor uses 2.0"10#4 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 ! $ ! $$ ! 1.5"104 ! #4 I
P 75 W I ! $ ! $ ! 0.60 A V 125 V Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
page 598 For all problems, assume that the battery voltage and lamp resistances are constant, no matter what current is present.
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 ! IV and E ! Pt Thus, E ! IVt ! (210 A)(12 V)(10.0 s) !
2.5"104
J
5. A flashlight bulb is rated at 0.90 W. If the lightbulb drops 3.0 V, how much current goes through it? P ! IV P 0.90 W I ! $ ! $ ! 0.30 A V 3.0 V
Physics: Principles and Problems
2.0"10
A
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 ! $ ! $ ! 2.4"102 ! I
0.50 A
b. What is the power consumption of the lamp? P ! IV ! (0.50 A)(120 V) ! 6.0"101 W 10. A 75-W lamp is connected to 125 V. a. What is the current through the lamp? P 75 W I ! $ ! $ ! 0.60 A V
125 V
b. What is the resistance of the lamp? V 125 V R ! $ ! $ ! 2.1"102 ! I
0.60 A
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Chapter 22 continued 11. A resistor is added to the lamp in the previous problem to reduce the current to half of its original value.
A 85 mA
#
4.5 V
a. What is the potential difference across the lamp?
53 !
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The new value of the current is V 4.5 V R ! $ ! $ ! 53 !
0.60 A $ ! 0.30 A 2
I
V ! IR ! (0.30 A)(2.1"102 !) !
6.3"101 V
b. How much resistance was added to the circuit?
0.085 A
14. Add a voltmeter to measure the potential difference across the resistors in problems 12 and 13 and repeat the problems. Both circuits will take the following form.
The total resistance of the circuit is now V 125 V Rtotal ! $ ! $ ! 4.2"102 ! I 0.30 A
A #
V "
I
Therefore, Rres ! Rtotal # RIamp ! 4.2"102 ! # 2.1"102 ! ! 2.1"102 ! c. How much power is now dissipated in the lamp? P ! IV ! (0.30 A)(6.3"101 V) ! 19 W
15. Draw a circuit using a battery, a lamp, a potentiometer to adjust the lamp’s brightness, and an on-off switch. Lamp #
Battery "
Switch
Potentiometer
A 4.80 A
#
60.0 V
12.5 !
"
16. Repeat the previous problem, adding an ammeter and a voltmeter across the lamp.
I
V 60.0 V I ! $ ! $ ! 4.80 A R
12.5 !
13. Draw a series-circuit diagram showing a 4.5-V battery, a resistor, and an ammeter that reads 85 mA. Determine the resistance and label the resistor. Choose a direction for the conventional current and indicate the positive terminal of the battery.
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Physics: Principles and Problems
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
page 600 12. Draw a circuit diagram to include a 60.0-V battery, an ammeter, and a resistance of 12.5 ! in series. Indicate the ammeter reading and the direction of the current.
Because the ammeter resistance is assumed zero, the voltmeter readings will be 60.0 V for Practice Problem 12 and 4.5 V for Practice Problem 13.
Chapter 22 continued
Section Review
P1 ! V 2/R1 ! (12 V)2/12 ! ! 12 W
22.1 Current and Circuits pages 591–600
P2 ! V 2/R2 ! (12 V)2/9.0 ! ! 16 W
page 600 17. Schematic Draw a schematic diagram of a circuit that contains a battery and a lightbulb. Make sure the lightbulb will light in this circuit.
4.0 W increase
$P ! P2 # P1 ! 16 W # 12 W ! 4.0 W
21. Energy A circuit converts 2.2"103 J of energy when it is operated for 3.0 min. Determine the amount of energy it will convert when it is operated for 1 h. 3
2.2"10 J E ! ! $$ "(60.0 min) 3.0 min
! 4.4"104 J
$ #
22. Critical Thinking We say that power is “dissipated” in a resistor. To dissipate is to use, to waste, or to squander. What is “used” when charge flows through a resistor?
18. Resistance Joe states that because R % V/I, if he increases the voltage, the resistance will increase. Is Joe correct? Explain.
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
No, resistance depends on the device. When V increases, so will I. 19. 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.
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.
Practice Problems 22.2 Using Electric Energy pages 601–605 page 603 23. A 15-! electric heater operates on a 120-V outlet. a. What is the current through the heater?
A
V 120 V I ! $ ! $ ! 8.0 A R
#
V "
15 !
b. How much energy is used by the heater in 30.0 s? E ! I 2Rt ! (8.0 A)2(15 !)(30.0 s) ! 2.9"104 J
Measure the current through the wire and the potential difference across it. Divide the potential difference by the current to obtain the wire resistance. 20. Power A circuit has 12 ! of resistance and is connected to a 12-V battery. Determine the change in power if the resistance decreases to 9.0 !.
c. How much thermal energy is liberated in this time? 2.9"104 J, because all electric energy is converted to thermal energy. 24. A 39-! resistor is connected across a 45-V battery. a. What is the current in the circuit? V 45 V I ! $ ! $ ! 1.2 A R
Physics: Principles and Problems
39 !
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447
Chapter 22 continued b. How much energy is used by the resistor in 5.0 min? 2
V E ! $t R
2
(45 V) ! $ (5.0 min)(60 s/min) (39 !)
! 1.6"104 J 25. A 100.0-W lightbulb is 22 percent efficient. This means that 22 percent of the electric energy is converted to light energy. a. How many joules does the lightbulb convert into light each minute it is in operation? E ! Pt ! (0.22)(100.0 J/s)(1.0 min) (60 s/min) !
1.3"103
J
b. How many joules of thermal energy does the lightbulb produce each minute? E ! Pt
! 4.7"103 J
t 2
For a given amount of energy, doubling the voltage will divide the time by 2. 2.2 h t ! $ ! 1.1 h 2
page 605 28. 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? P ! IV ! (15.0 A)(120 V) ! 1800 W ! 1.8 kW b. How much energy in kWh does it consume in 30 days?
a. If 220 V are applied across it, what is the current through the stove element? 220 V V I ! $ ! $ ! 2.0"101 A 11 !
b. How much energy does the element convert to thermal energy in 30.0 s? E ! I 2Rt ! (2.0"101 A)2(11 !)(30.0 s) ! 1.3"105 J c. The element is used to heat a kettle containing 1.20 kg of water. Assume that 65 percent of the heat is absorbed by the water. What is the water’s increase in temperature during the 30.0 s? Q ! mC$T with Q ! 0.65E 0.65E (0.65)(1.3"105 J) $T ! $ ! $$$ mC (1.20 kg)(4180 J/kg%°C)
! 17°C Solutions Manual
! 270 kWh c. At $0.12 per kWh, how much does it cost to operate the heater for 30 days? Cost ! ($0.12/kWh)(270 kWh) ! $32.40 29. A digital clock has a resistance of 12,000 ! and is plugged into a 115-V outlet. a. How much current does it draw? V 115 V I ! $ ! $$ ! 9.6"10#3 A R
12,000 !
b. How much power does it use? P ! VI ! (115 V)(9.6"10#3 A) ! 1.1 W c. If the owner of the clock pays $0.12 per kWh, how much does it cost to operate the clock for 30 days? Cost ! (1.1"10#3 kWh)($0.12/kWh) (30 days)(24 h/day) ! $0.10 30. 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 lessthan-perfect efficiency. How long will it Physics: Principles and Problems
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
26. The resistance of an electric stove element at operating temperature is 11 !.
448
E ! IVt ! I(2V )! $ "
E ! Pt ! (1.8 kW)(5.0 h/day)(30 days)
! (0.78)(100.0 J/s)(1.0 min) (60.0 s/min)
R
27. 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?
Chapter 22 continued take to charge the battery using a current of 7.5 A? Assume that the charging voltage is the same as the discharging voltage. Echarge ! (1.3)IVt ! (1.3)(55 A)(12 V)(1.0 h) ! 858 Wh 858 Wh E t ! $ ! $$ ! 9.5 h IV
(7.5 A)(12 V)
31. Rework the previous problem by assuming that the battery requires the application of 14 V when it is recharging. Echarge ! (1.3)IVt ! (1.3)(55 A)(12 V)(1.0 h) ! 858 Wh E 858 Wh t ! $ ! $$ ! 8.2 h IV
(7.5 A)(14 V)
Section Review
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
22.2 Using Electric Energy pages 601–605 page 605 32. Energy A car engine drives a generator, which produces and stores electric charge in the car’s battery. The headlamps use the electric charge stored in the car battery. List the forms of energy in these three operations. Mechanical energy from the engine converted to electric energy in the generator; electric energy stored as chemical energy in the battery; chemical energy converted to electric energy in the battery and distributed to the headlamps; electric energy converted to light and thermal energy in headlamps. 33. 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? Hot draws more power, P ! IV, so the fixed voltage current is larger. Because I ! V/R the resistance is smaller. 34. Power Determine the power change in a circuit if the applied voltage is decreased by one-half. Physics: Principles and Problems
V 22/R (0.5V1)2/R P1 $$ ! $ ! $$ ! 0.25 2 P2 V 1 /R V 12
35. Efficiency Evaluate the impact of research to improve power transmission lines on society and the environment. 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. 36. 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, at twice the voltage, the current would be halved. The I 2R loss in the circuit wiring would be dramatically reduced because it is proportional to the square of the current. 37. Critical Thinking When demand for electric power is high, power companies sometimes reduce the voltage, thereby producing a “brown-out.” What is being saved? Power, not energy; most devices will have to run longer.
Chapter Assessment Concept Mapping page 610 38. Complete the concept map using the following : watt, current, resistance. Electricity
rate of flow
rate of conversion
opposition to flow
current
power
resistance
ampere
watt
ohm
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Chapter 22 continued
Mastering Concepts page 610 39. Define the unit of electric current in of fundamental MKS units. (22.1)
b. Which device converts chemical energy to electric energy? 1 c. Which device turns the circuit on and off? 2
1 A ! 1 C/1 s 40. How should a voltmeter be connected in Figure 22-12 to measure the motor’s voltage? (22.1)
d. Which device provides a way to adjust speed? 3 44. Describe the energy conversions that occur in each of the following devices. (22.1)
4
a. an incandescent lightbulb electric energy to heat and light
#
3
1 "
b. a clothes dryer electric energy to heat and kinetic energy c. a digital clock radio
2 ■
Figure 22-12
The positive voltmeter lead connects to the left-hand motor lead, and the negative voltmeter lead connects to the right-hand motor lead.
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. 42. What is the direction of the conventional motor current in Figure 22-12? (22.1) from left to right through the motor 43. Refer to Figure 22-12 to answer the following questions. (22.1) a. Which device converts electric energy to mechanical energy?
45. Which wire conducts electricity with the least resistance: one with a large cross-sectional diameter or one with a small cross-sectional diameter? (22.1) A larger-diameter wire has a smaller resistance because there are more electrons to carry the charge. 46. A simple circuit consists of a resistor, a battery, and connecting wires. (22.1) a. Draw a circuit schematic of this simple circuit.
I
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b. How must an ammeter be connected in a circuit for the current to be correctly read? The ammeter must be connected in series.
4
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Physics: Principles and Problems
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
41. How should an ammeter be connected in Figure 22-12 to measure the motor’s current? (22.1)
electric energy to light and sound
Chapter 22 continued
Applying Concepts
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pages 610–611 51. Batteries When a battery is connected to a complete circuit, charges flow in the circuit almost instantaneously. Explain.
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c. 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. V
52. Explain why a cow experiences a mild shock when it touches an electric fence.
I
By touching the fence and the ground, the cow encounters a difference in potential and conducts current, thus receiving a shock.
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47. Why do lightbulbs burn out more frequently just as they are switched on rather than while they are operating? (22.2)
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
The low resistance of the cold filament allows a high current initially and a greater change in temperature, subjecting the filament to greater stress. 48. If a battery is short-circuited by a heavy copper wire being connected from one terminal to the other, the temperature of the copper wire rises. Why does this happen? (22.2) The short circuit produces a high current, which causes more electrons to collide with the atoms of the wire. This raises the atoms’ kinetic energies and the temperature of the wire. 49. What electric quantities must be kept small to transmit electric energy economically over long distances? (22.2) the resistance of the wire and the current in the wire 50. Define the unit of power in of fundamental MKS units. (22.2) m2 kg%$$ s2
J kg%m2 C J W ! $$%$$ ! $ ! & ! $ 3 s C
s
A potential difference is felt over the entire circuit as soon as the battery is connected to the circuit. The potential difference causes the charges to begin to flow. Note: The charges flow slowly compared to the change in potential difference.
s
Physics: Principles and Problems
53. Power Lines Why can birds perch on highvoltage lines without being injured? No potential difference exists along the wires, so there is no current through the birds’ bodies. 54. Describe two ways to increase the current in a circuit. Either increase the voltage or decrease the resistance. 55. Lightbulbs Two lightbulbs work on a 120-V circuit. One is 50 W and the other is 100 W. Which bulb has a higher resistance? Explain. 50-W bulb V2 V2 P ! & , so R ! & R
P
Therefore, the lower P is caused by a higher R. 56. If the voltage across a circuit is kept constant and the resistance is doubled, what effect does this have on the circuit’s current? If the resistance is doubled, the current is halved.
s
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Chapter 22 continued 57. What is the effect on the current in a circuit if both the voltage and the resistance are doubled? Explain.
Motor 1.5 A
No effect. V ! IR, so I ! V/R, and if the voltage and the resistance both are doubled, the current will not change.
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12 V
58. Ohm’s Law Sue finds a device that looks like a resistor. When she connects it to a 1.5-V battery, she measures only 45"10#6 A, but when she uses a 3.0-V battery, she measures 25"10#3 A. Does the device obey Ohm’s law? No. V ! IR, so R ! V/I. At 1.5 V, 1.5 V 4 R ! $$ #6 ! 3.3"10 ! 45"10
3.0 V At 3.0 V, R ! $$ ! 120 ! #3 25"10
■
Figure 22-13
a. How much power is delivered to the motor? P ! VI ! (12 V)(1.5 A) ! 18 W b. How much energy is converted if the motor runs for 15 min? E ! Pt ! (18 W)(15 min)(60 s/min) ! 1.6"104 J
A
A device that obeys Ohm’s law has a resistance that is independent of the applied voltage.
62. Refer to Figure 22-14 to answer the following questions. A
59. If the ammeter in Figure 22-4a on page 596 were moved to the bottom of the diagram, would the ammeter have the same reading? Explain.
# 27 V
18 !
"
V
I
60. 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 produce thermal energy at a faster rate? Why?
a. What should the ammeter reading be?
the wire with the smaller resistance
b. What should the voltmeter reading be?
V2
P! $ R
Smaller R produces larger power P dissipated in the wire, which produces thermal energy at a faster rate.
Mastering Problems 22.1 Current and Circuits pages 611–612 Level 1 61. A motor is connected to a 12-V battery, as shown in Figure 22-13.
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Figure 22-14
27 V I ! V /R ! $ ! 1.5 A 18 !
27 V c. How much power is delivered to the resistor? P ! VI ! (27 V)(1.5 A) ! 41 W d. How much energy is delivered to the resistor per hour? E ! Pt ! (41 W)(3600 s) ! 1.5"105 J
Physics: Principles and Problems
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Yes, because the current is the same everywhere in this circuit.
Chapter 22 continued 63. Refer to Figure 22-15 to answer the following questions.
E ! Pt ! (4.5 W)(3600 s) ! 1.6"104 J
A
65. Toasters The current through a toaster that is connected to a 120-V source is 8.0 A. What power is dissipated by the toaster?
# 27 V
9.0 !
"
V
P ! IV ! (8.0 A)(120 V) ! 9.6"102 W I
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d. How much energy is delivered to the resistor per hour?
Figure 22-15
a. What should the ammeter reading be? 27 V I ! V/R ! $ ! 3.0 A 9.0 !
b. What should the voltmeter reading be? 27 V
67. A lamp draws 0.50 A from a 120-V generator. a. How much power is delivered? b. How much energy is converted in 5.0 min?
P ! VI ! (27 V)(3.0 A) ! 81 W
E t
d. How much energy is delivered to the resistor per hour? E ! Pt ! (81 W)(3600 s) !
P ! IV ! (1.2 A)(120 V) ! 1.4"102 W
P ! IV ! (0.50 A)(120 V) ! 6.0"101 W
c. How much power is delivered to the resistor?
2.9"105
66. Lightbulbs A current of 1.2 A is measured through a lightbulb when it is connected across a 120-V source. What power is dissipated by the bulb?
The definition of power is P ! $$, so E ! Pt ! (6.0"101 W)! $ "! $ "
J
5.0 min 1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
64. Refer to Figure 22-16 to answer the following questions.
# 18 !
"
V
a. How many joules of energy does the battery deliver to the motor each second? P ! IV ! (210 A)(12 V) ! 2500 J/s or 2.5"103 J/s
I ■
! 18,000 J ! 1.8"104 J 68. A 12-V automobile battery is connected to an electric starter motor. The current through the motor is 210 A.
A
9.0 V
60 s min
Figure 22-16
a. What should the ammeter reading be? 9.0 V I ! V/R ! $ ! 0.50 A 18 !
b. What should the voltmeter reading be? 9.0 V c. How much power is delivered to the resistor?
b. What power, in watts, does the motor use? P ! 2.5"103 W 69. Dryers A 4200-W clothes dryer is connected to a 220-V circuit. How much current does the dryer draw? P ! IV P 4200 W I ! $$ ! $$ ! 19 A V
220 V
P ! VI ! (9.0 V)(0.50 A) ! 4.5 W
Physics: Principles and Problems
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Chapter 22 continued 70. Flashlights A flashlight bulb is connected across a 3.0-V potential difference. The current through the bulb is 1.5 A.
Table 22-2 Voltage, V Current, I Resistance, R ! V/I
a. What is the power rating of the bulb?
(volts)
(amps)
2.00
0.0140
______________
4.00
0.0270
______________
P ! IV ! (1.5 A)(3.0 V) ! 4.5 W b. How much electric energy does the bulb convert in 11 min? E The definition of power is P ! $$, so t
E ! Pt ! (4.5 W)(11 min)!$$" 60 s min
! 3.0"103 J 71. Batteries A resistor of 60.0 ! 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?
10.00
0.0630
______________
#2.00 #4.00
#0.0140 #0.0280
______________ ______________
#6.00
#0.0390
______________
#8.00 #10.00
#0.0510 #0.0620
______________ ______________
! ! ! !
143 154 143 161
!, R ! 148 !, R ! 150 !, !, R ! 159 !, R ! 143 !, !, R ! 154 !, R ! 157 !, !
0.04
I (amps)
0.02 "12.00 "8.00 "4.00 0.00
4.00
8.00 12.00
"0.04 "0.06
15 !
V (volts)
Level 2 75. 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 22-2.
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c. Does the nichrome wire obey Ohm’s law? If not, for all the voltages, specify the voltage range for which Ohm’s law holds. 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. Physics: Principles and Problems
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
R
______________
0.06
73. What voltage is placed across a motor with a 15-! operating resistance if there is 8.0 A of current?
V 75 V I ! $ ! $$ ! 5.0 A
______________
0.0520
b. Graph I versus V.
V ! IR ! (1.5 A)(4.0 !) ! 6.0 V
V ! IR
0.0400
8.00
R R R R
72. What voltage is applied to a 4.0-! resistor if the current is 1.5 A?
74. A voltage of 75 V is placed across a 15-! resistor. What is the current through the resistor?
6.00
a. For each measurement, calculate the resistance.
V ! IR ! (0.40 A)(60.0 !) ! 24 V
V ! IR ! (8.0 A)(15 !) ! 1.2"102 V
(amps)
Chapter 22 continued 76. Draw a series circuit diagram to include a 16-! resistor, a battery, and an ammeter that reads 1.75 A. Indicate the positive terminal and the voltage of the battery, the positive terminal of the ammeter, and the direction of conventional current. #
V % 28 V
a. What is the lamp’s resistance when it is on? V ! IR 120 V V R ! $$ ! $ ! 3.0"102 ! I
A
0.40 A
b. When the lamp is cold, its resistance is
I % 1.75 A
#
79. The current through a lamp connected across 120 V is 0.40 A when the lamp is on.
16 !
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V ! IR ! (1.75 A)(16 !) ! 28 V 77. 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. a. Does the lamp obey Ohm’s law? No. The voltage is increased by a 9.0 factor of $$ ! 1.5, but the current 6.0
75 is increased by a factor of $$ ! 1.1 66
b. How much power does the lamp dissipate when it is connected to the 6.0-V battery?
1 $$ as great as it is when the lamp is hot. 5
What is the lamp’s cold resistance?
!$15$"(3.0"102 !) ! 6.0"101 ! c. What is the current through the lamp as it is turned on if it is connected to a potential difference of 120 V? V ! IR 120 V V I ! $ ! $$ ! 2.0 A 1 R
6.0"10 !
Level 3 80. The graph in Figure 22-17 shows the current through a device called a silicon diode. Current in a Diode
c. How much power does it dissipate at 9.0 V? P ! IV ! (75"10#3 A)(9.0 V) ! 0.68 W 78. Lightbulbs How much energy does a 60.0-W lightbulb use in half an hour? If the lightbulb converts 12 percent of electric energy to light energy, how much thermal energy does it generate during the half hour? E P ! $$ t
E ! Pt ! (60.0 W)(1800 s) ! 1.08"105 J If the bulb is 12 percent efficient, 88 percent of the energy is lost to heat, so Q ! (0.88)(1.08"105 J) ! 9.5"104 J
Current (A)
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
P ! IV ! (66"10#3 A)(6.0 V) ! 0.40 W 0.02
0.01
0
0.2
0.4
0.6
0.8
Voltage (V) ■
Figure 22-17
a. A potential difference of &0.70 V is placed across the diode. What is the resistance of the diode? From the graph, I ! 22 m/A, and V ! IR, so 0.70 V V R ! $$ ! $$ ! 32 ! #2 I
2.2"10
A
b. What is the diode’s resistance when a &0.60-V potential difference is used? 0.60 V V R ! $$ ! $$ ! 1.2"102 ! #3 I
5.2"10
A
c. Does the diode obey Ohm’s law? No. Resistance depends on voltage. Physics: Principles and Problems
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Chapter 22 continued 81. Draw a schematic diagram to show a circuit including a 90-V battery, an ammeter, and a resistance of 45 ! connected in series. What is the ammeter reading? Draw arrows showing the direction of conventional current.
# V
40.0 !
"
A 2A
#
90 V
I
45 !
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a. the maximum safe current
V ! IR
P ! I 2R
90 V V I ! $$ ! $ ! 2 A 45 ! R
I!
22.2 Using Electric Energy pages 612–613 Level 1 82. 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. E ! IVt ! (0.0250 A)(9.0 V)(26.0 h) ! 5.9 Wh ! 5.9"10#3 kWh $3.00 cost Rate ! $$ ! $$ #3 E
5.9"10
kWh
!" !"
b. the maximum safe voltage P ! V 2/R 101 W)(40.0 !) V ! #PR $ ! #(5.0" $$ $$ ! 45 V 86. Utilities Figure 22-19 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.
#
P ! I 2R
240.0 V
$ ! 0.15 A !$"R$ ! !" 220 ! 5.0 W
84. A 110-V electric iron draws 3.0 A of current. How much thermal energy is developed in an hour? Q ! E ! VIt ! (110 V)(3.0 A)(1.0 h)(3600 s/h) !
1.2"106
J
Level 2 85. For the circuit shown in Figure 22-18, the maximum safe power is 5.0"101 W. Use the figure to find the following:
456
Solutions Manual
4.80 !
"
■
Figure 22-19
V2
E ! !$$"(t ) R
!
(240.0 V)2 4.80 !
"
! $$ (30 d)(24 h/d)(0.25) ! 2160 kWh Cost ! (2160 kWh)($0.100/kWh) ! $216 87. 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 Physics: Principles and Problems
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Thermostat
83. What is the maximum current allowed in a 5.0-W, 220-! resistor? P
5.0"101 W && ! 1 A 40.0 !
P $ ! R
! $510/kWh
I!
Figure 22-18
Chapter 22 continued and that electricity costs $0.090 per kWh. Determine how much current the air conditioner will take from a 120-V outlet.
Q ! E ! I 2Rt
Cost ! (E )(rate) $50 Cost E ! $$ ! $ $ $0.090/kWh
rate
E ! IVt (556 kWh)(1000 W/kW) E I ! $$ ! $$$ (120 V)(30 d)(24 h/d)(0.5) Vt
! 12.9 A
60 s min
91. A 6.0-! resistor is connected to a 15-V battery. a. What is the current in the circuit? V ! IR
88. Radios A transistor radio operates by means of a 9.0-V battery that supplies it with a 50.0-mA current. a. If the cost of the battery is $2.49 and it lasts for 300.0 h, what is the cost per kWh to operate the radio in this manner? P ! IV ! (0.050 A)(9.0 V) ! 0.45 W ! 4.5"10#4 kW $2.49 Cost ! $$$ #4 (4.5"10
kW)(300.0 h)
! $18.00/kWh
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
! (1.2 A)2(50.0 !)(5.0 min)!$$" ! 2.2"104 J
! 556 kWh
b. The same radio, by means of a converter, is plugged into a household circuit by a homeowner who pays $0.12 per kWh. What does it now cost to operate the radio for 300.0 h? Cost ! ($0.12/kWh) (4.5"10#4 kW)(300 h) ! $0.02
Mixed Review page 613 Level 1 89. If a person has $5, how long could he or she play a 200 W stereo if electricity costs $0.15 per kWh? Cost Rate
E ! Pt ! $$ Cost t ! $$ (Rate)(P)
$5
!
90. A current of 1.2 A is measured through a 50.0-! resistor for 5.0 min. How much heat is generated by the resistor?
&&& 1 kW ($0.15/kWh)(200 W)!$ $ 1000 W "
! 200 h Physics: Principles and Problems
V 15 V I ! $$ ! $$ ! 2.5 A R
6.0 !
b. How much thermal energy is produced in 10.0 min? Q ! E ! I 2Rt ! (2.5 A)2(6.0 !)(10.0 min)!$$" 60 s min
! 2.3"104 J Level 2 92. Lightbulbs An incandescent lightbulb with a resistance of 10.0 ! when it is not lit and a resistance of 40.0 ! when it is lit has 120 V placed across it. a. What is the current draw when the bulb is lit? V 120 V I ! $$ ! $$ ! 3.0 A R
40.0 !
b. What is the current draw at the instant the bulb is turned on? V 120 V I ! $$ ! $$ ! 12 A R
10.0 !
c. When does the lightbulb use the most power? the instant it is turned on 93. 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? The slowest speed’s resistance is R ! V/I ! 12 V/0.02 A ! 600 !. The fastest speed’s resistance is R ! V/I ! 12 V/1.2 A ! 1.0"101 !. The range is 1.0"101 ! to 600 !. Solutions Manual
457
Chapter 22 continued Level 3 94. An electric motor operates a pump that irrigates a farmer’s crop by pumping 1.0"104 L of water a vertical distance of 8.0 m into a field each hour. The motor has an operating resistance of 22.0 ! and is connected across a 110-V source. a. What current does the motor draw? V ! IR
Q $T ! $$ mC
6
1.1"10 J ! $$$ (20.0 kg)(4180 J/kg%C°)
! 13°C d. At $0.08 per kWh, how much does it cost to operate the heating coil 30 min per day for 30 days? 1.1"106 J 5 min
Cost ! ! $$ "! $ "(30 days)
V 110 V I ! $ ! $ ! 5.0 A R 22.0 !
$0.08 1 kWh $ ! $$ 3.6"106 J "! kWh "
b. How efficient is the motor? EW ! mgd
! $4.40
! (1"104 kg)(9.80 m/s2)(8.0 m) ! 8"105 J Em ! IVt ! (5.0 A)(110 V)(3600 s) ! 2.0"106 J
5
2.0"10 J
! 40% 95. A heating coil has a resistance of 4.0 ! and operates on 120 V.
b. The heater is being used to heat a room containing 50 kg of air. If the specific heat of air is 1.10 kJ/kg'°C, and 50 percent of the thermal energy heats the air in the room, what is the change in air temperature in half an hour? Q ! mC$T Q $T ! $$ mC
V ! IR
5
(0.5)(9"10 J) ! $$$
120 V V I ! $$ ! $ ! 3.0"101 A
(50.0 kg)(1100 J/kg%C°)
4.0 !
b. What energy is supplied to the coil in 5.0 min? E ! I 2Rt ! (3.0"101 A)2(4.0 !)(5.0 min)!$$" 60 s min
1.1"106
J
c. If the coil is immersed in an insulated container holding 20.0 kg of water, what will be the increase in the temperature of the water? Assume 100 percent of the heat is absorbed by the water. Q ! mC$T
Solutions Manual
! 8°C c. At $0.08 per kWh, how much does it cost to run the heater 6.0 h per day for 30 days? 500 J 6.0 h 3600 s Cost ! ! $ "! $ "! $ " s
day
h
1kWH $0.08 (30 days)! $$ "! $ " 6 3.6"10 J
kWh
! $7
Thinking Critically page 614 97. Formulate Models How much energy is stored in a capacitor? The energy needed to increase the potential difference of a charge, q, Physics: Principles and Problems
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
a. What is the current in the coil while it is operating?
458
a. How much energy is delivered to the heater in half an hour? ! 9"105 J
8"10 J ! $$ " 100 6
!
96. Appliances An electric heater is rated at 500 W.
E ! Pt ! (5"102 W)(1800 s)
Ew Efficiency ! $ " 100 Em
R
30 min day
Chapter 22 continued
Potential difference (V)
4.0 3.0
→
Microwave converter 1440 J/s
→
Heat converter 1080 J/s
Wasted 360 J/s
Wasted 270 J/s
→
Useful output 810 J/s
b. Derive an equation for the rate of temperature increase ((T/s) from the information presented in Chapter 12. Solve for the rate of temperature rise given the rate of energy input, the mass, and the specific heat of a substance.
!
"
1 $Q $T $$ ! $$ $$ mC $t $t
2.0 1.0 0.0
1.0 2.0 3.0 4.0 5.0 Charge (C)
q 5.0 C Voltage V ! $$ ! $ ! 5.0 V 1.0 F C
Energy E ! area under curve Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Input 1440 J/s
→
5.0
Figure 22-2b on page 593. Label the function of each block according to total joules per second.
→
is represented by E % qV. But in a capacitor, V % 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 V 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.
1 ! $ (5.0 V)(5.0 C) 2
! 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. 98. Apply Concepts A microwave oven operates at 120 V and requires 12 A of current. Its electric efficiency (converting AC to microwave radiation) is 75 percent, and its conversion efficiency from microwave radiation to heating water is also 75 percent. a. Draw a block power diagram similar to the energy diagram shown in Physics: Principles and Problems
c. 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.
!
"
$T 1 $Q $$ ! $$ $$ $t mC $t
810 J/s ! $$$ (0.25 kg)(4180 J/kg%°C)
! 0.78°C/s d. Review your calculations carefully for the units used and discuss why your answer is in the correct form. The kg unit cancels and the J unit cancels, leaving °C/s. e. Discuss, in general , 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. It might be possible to find a way to convert electric energy to radiation using a different approach that would be more efficient. The efficiency of conversion from microwave radiation to thermal energy in water is 75 percent. It might be possible to use a different frequency of electromagnetic radiation to improve this rating. Or, it might be possible to find a new Solutions Manual
459
Chapter 22 continued
f.
Discuss, in efficiency , why microwave ovens are not useful for heating everything. The conversion efficiency from microwave energy to thermal energy is good for water. It’s not as good for other materials. The containers and dishes designed for use with microwave ovens convert little of the energy.
g. Discuss, in general , why it is not a good idea to run microwave ovens when they are empty. The empty oven means that the microwave energy has to be dissipated in the oven. This can lead to overheating of the oven components and to their failure. 99. Analyze and Conclude A salesclerk in an appliance store states that microwave ovens are the most electrically efficient means of heating objects.
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. 101. 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. Explain. The volt–ampere graph for a resistor obeying Ohm’s law is a straight line and is seldom necessary. 102. Make and Use Graphs Based on what you have learned in this chapter, identify and prepare two parabolic graphs. voltage–power and current–power
In the case of heating a cup of water, an immersion heater uses only resistance for energy conversion and is nearly 100 percent efficient. A microwave oven uses two energy conversions (electricity to microwave radiation to heat) and is typically around 50 percent efficient.
In the case of heating a potato, a microwave oven heats mostly the potato and is more efficient than an electric oven or skillet, which also heats the air, cabinets, racks, etc. c. Formulate a diplomatic reply to the clerk. “It can be true, but it depends on the specific application.” 460
Solutions Manual
30 20 10
0
5
10
15
20
Voltage across a 10-! resistor 2 P!V R
40 Watts dissipated
b. Formulate an argument to the clerk’s claim. Hint: Think about heating a specific object.
P ! I 2R
40
30 20 10
0
1
2
Amperes through a 10-! resistor
Physics: Principles and Problems
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
a. Formulate an argument to refute the clerk’s claim. Hint: Think about heating a specific object.
100. Apply Concepts The sizes of 10-! resistors range from a pinhead to a soup can. Explain.
Watts dissipated
geometry of radiating objects to be heated to improve the efficiency.
Chapter 22 continued
Writing in Physics page 614 103. 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. 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.
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
104. In Chapter 13, you learned that matter expands when it is heated. Research the relationship between thermal expansion and high-voltage transmission lines. 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.
$S ! (20 kg)(3.34"105 J/kg)/(273 K) ! 2.4"104 J/K 106. 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 1.3 kg/m3 at sea level. (Chapter 13) $P ! !gh ! (1.3 kg/m3)(9.80 m/s2)(150 m) ! 1.9 kPa or about 2/100 of the total air pressure 107. 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? (Chapter 15) v ! "f 343 m/s v " ! $ ! $$ ! 0.020 m ! 2.0 cm f
17,000 Hz
108. Light of wavelength 478 nm falls on a double slit. First-order bright bands appear 3.00 mm from the central bright band. The screen is 0.91 m from the slits. How far apart are the slits? (Chapter 19) xd L
" ! $$ "L x
d ! $$ (478"10#9 m)(0.91 m) (3.00"10 m)
Cumulative Review
! $$$ #3
page 614 105. A person burns energy at the rate of about 8.4"106 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? (Chapter 12)
! 1.4"10#4 m
$S ! Q/T where T is the body temperature of 310 K. $S ! (8.4"106 J)/(310 K) ! 2.7"104 J/K For melting ice Physics: Principles and Problems
109. A charge of &3.0"10#6 C is 2.0 m from a second charge of &6.0"10#5 C. What is the magnitude of the force between them? (Chapter 20) q q d
A B F ! K$ 2
! (9.0"109 N%m2/C2) (3.0"10#6 C)(6.0"10#5 C) $$$ $ (2.0 m)2
! 0.41 N Solutions Manual
461
Chapter 22 continued
Challenge Problem page 604 Use the figure to the right to help you answer the questions below. 1. Initially, the capacitor is uncharged. Switch 1 is closed, and Switch 2 remains open. What is the voltage across the capacitor?
Switch 2 Switch 1 1.5 !F 15 V
1200 !
15 V 2. Switch 1 is now opened, and Switch 2 remains open. What is the voltage across the capacitor? Why? It remains 15 V because there is no path for the charge to be removed. 3. Next, Switch 2 is closed, while Switch 1 remains open. 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. As time goes on, what happens to the voltage across the capacitor and the current through the resistor?
462
Solutions Manual
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
The capacitor voltage remains at 15 V because there is no path to discharge the capacitor; the current remains at 13 mA because the battery voltage is constant at 15 V. However, if the battery and capacitor were real components instead of ideal circuit components, the capacitor voltage eventually would become zero due to leakage, and the current eventually would become zero due to battery depletion.
Physics: Principles and Problems