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Chapter 22-23 Assignment Solutions Table of Contents Page 610 #39, 44-45, 61-75 .......................................................................................................................... 1 Page 610 #38, 53-57, 82-85, 89-92 ............................................................................................................... 5 Page 636 #36-47, 57-63, 77-81 ..................................................................................................................... 8 Chapter 22 Review ...................................................................................................................................... 12 Chapter 23 Review ...................................................................................................................................... 12 Combined Circuit Practice........................................................................................................................... 12 Page 610 #39, 44-45, 61-75 39) Define the unit of electric current in terms of fundamental MKS units. (22.1) 1A= 1C 1s 44) Describe the energy conversions that occur in each of the following devices. (22.1) a) An incandescent lightbulb Electric energy to heat and light b) A clothes dryer Electric energy to heat and kinetic energy c) A digital clock radio Electric energy to light and sound 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. 61) A motor is connected to a 12-V battery, as shown in Figure 22-13. a) How much power is delivered to the motor? π = πΌπ = (1.5 A)(12 V) = 18 W b) How much energy is converted if the motor runs for 15 min? 60 s ππΈ = ππ‘ = (18 W)(15 min) ( ) = 1.6 × 104 J 1 min 62) Refer to Figure 22-14 to answer the following questions. a) What should the ammeter reading be? πΌ= π 27 V = = 1.5 A π 18 Ξ© b) What should the voltmeter reading be? 27 V c) How much power is delivered to the resistor? π = πΌπ = (1.5 A)(27 V) = 41 W d) How much energy is delivered to the resistor per hour? 60 min 60 s ππΈ = ππ‘ = (41 W)(1 hour) ( )( ) = 1.5 × 105 J 1 hour 1 min 63) Refer to Figure 22-15 to answer the following questions a) What should the ammeter reading be? πΌ= π 27 V = = 3.0 A π 9.0 Ξ© b) What should the voltmeter reading be? 27 V c) How much power is delivered to the resistor? π = πΌπ = (3.0 A)(27 V) = 81 W d) How much energy is delivered to the resistor per hour? 60 min 60 s ππΈ = ππ‘ = (81 W)(1 hour) ( )( ) = 2.9 × 105 J 1 hour 1 min 64) Refer to Figure 22-16 to answer the following questions a) What should the ammeter reading be? πΌ= π 9.0 V = = 0.50 A π 18 Ξ© b) What should the voltmeter reading be? 9.0 V c) How much power is delivered to the resistor? π = πΌπ = (0.50 A)(9.0 V) = 4.5 W d) How much energy is delivered to the resistor per hour? 60 min 60 s ππΈ = ππ‘ = (4.5 W)(1 hour) ( )( ) = 1.6 × 104 J 1 hour 1 min 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? π = πΌπ = (8.0 A)(120 V) = 9.6 × 102 W 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? π = πΌπ = (1.2 A)(120 V) = 1.4 × 102 W 67) A lamp draws 0.50 A from a 120-V generator. a) How much power is delivered? π = πΌπ = (0.50 A)(120 V) = 6.0 × 101 W b) How much energy is converted in 5.0 min? 60 s ππΈ = ππ‘ = (60. W)(5.0 min) ( ) = 1.8 × 104 J 1 min 68) A 12-V automobile battery is connected to an electric starter motor. The current through the motor is 210 A. a) How many joules of energy does the battery deliver to the motor each second? ππΈ = ππ‘ = πΌππ‘ = (210 A)(12 V)(1 s) = 2500 J b) What power, in watts, does the motor use? π = 2500 W 69) Dryers: A 4200-W clothes dryer is connected to a 220-V circuit. How much current does the dryer draw? π = πΌπ πΌ= π 4200 W = = 19 A π 220 V 70) Flashlights: A flashlight bulb is connected across a 3.0-V potential difference. The current through the bulb is 1.5 A. a) What is the power rating of the bulb? π = πΌπ = (1.5 A)(3.0 V) = 4.5 W b) How much electric energy does the bulb convert in 11 min? 60 s ππΈ = ππ‘ = (4.5 W)(11 min) ( ) = 3.0 × 103 J 1 min 71) Batteries: A resistor of 60.0 Ξ© 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? π = πΌπ = (0.40 A)(60.0 Ξ©) = 24 V 72) What voltage is applied to a 4.0-Ξ© resistor if the current is 1.5 A? π = πΌπ = (1.5 A)(4.0 Ξ©) = 6.0 V 73) What voltage is placed across a motor with a 15-Ξ© operating resistance if there is 8.0 A of current? π = πΌπ = (8.0 A)(15 Ξ©) = 120 V 74) A voltage of 75 V is placed across a 15-Ξ© resistor. What is the current through the resistor? π = πΌπ πΌ= π 75 V = = 5.0 A π 15 Ξ© 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. a) For each measurement, calculate the resistance. Voltage, π (volts) Current, πΌ Resistance, π = π/πΌ (ohms) (amps) 2.00 0.0140 143 Ξ© 4.00 0.0270 148 Ξ© 6.00 0.0400 150 Ξ© 8.00 0.0520 154 Ξ© 10.00 0.0630 159 Ξ© -2.00 -0.0140 143 Ξ© -4.00 -0.0280 143 Ξ© -6.00 -0.0390 154 Ξ© -8.00 -0.0510 157 Ξ© -10.00 -0.0620 161 Ξ© b) Graph πΌ versus π. c) Does the nichrome wire obey Ohmβs law? If not for all the voltages, specify the voltage range for which Ohmβs law holds. Ohmβs law is obeyed when the resistance of a device is constant, independent of π. The wire does not obey Ohmβs law for all π, but does from about β4 V to about 2 V. Page 610 #38, 53-57, 82-85, 89-92 38) Complete the concept map using the following terms: watt, current, resistance. 53) Power Lines: Why can birds perch on high-voltage lines without being injured? The birds do not provide a conduit for current to flow anywhere but back into the wire. Since the birdsβ bodies have much higher resistance than the wire, the current takes the path of least resistance β through the wire β rather than through the bird. 54) Describe two ways to increase the current in a circuit. ο· ο· Increase the electric potential difference of the power source. Reduce the resistance of the circuit. o Add more paths in parallel. o Use a material with lower resistivity. o Use wire with a larger cross-section. o Remove resistors/lamps/etc. from the circuit. 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. The 50-W bulb. π= π2 π2 β π = π π Therefore, the lower π is caused by the higher π . 56) If the voltage across a circuit is kept constant and the resistance is doubled, what effect does this have on the circuitβs current? π = πΌπ β πΌ = π π If the resistance is doubled, the current is cut in half. 57) What is the effect on the current in a circuit if both the voltage and the resistance are doubled? Explain. π No effect. π = πΌπ , so πΌ = π , and if the electric potential difference and the resistance are both doubled, 2π π πΌ2 = 2π = π = πΌ, the current will not change. 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. ππΈ = ππ‘ = πΌππ‘ = (0.0250 A)(9.0 V)(26.0 h) = 5.9 Wh = 5.9 × 10β3 kWh π ππ‘π = πΆππ π‘ $3.00 = = $510/kWh ππΈ 5.9 × 10β3 kWh 83) What is the maximum current allowed in a 5.0-W, 220-Ξ© resistor? π = πΌ2 π π 5.0 W πΌ=β =β = 0.15 A π 220 Ξ© 84) A 110-V electric iron draws 3.0 A of current. How much thermal energy is developed in an hour? 60 min 60 s π = ππΈ = ππ‘ = πΌππ‘ = (3.0 A)(110 V)(1 hour) ( )( ) = 1.2 × 106 J 1 hour 1 min 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: a) the maximum safe current π = πΌ2 π π 5.0 × 101 W πΌ=β =β = 1.1 A π 40.0 Ξ© b) the maximum safe voltage π2 π= π 1 π = βππ = β(5.0 × 10 W)(40.0 Ξ©) = 45 V 89) If a person has $5, how long could he or she play a 200 W stereo if electricity costs $0.15 per kWh? ππΈ = ππ‘ = π‘= πΆππ π‘ π ππ‘π πΆππ π‘ $5 = (π ππ‘π)(π) ($0.15/kWh)(200 1 kW W) (1000 W) = 200 hours 90) A current of 1.2 A is measured through a 50.0-Ξ© resistor for 5.0 min. How much heat is generated by the resistor? π = ππΈ = ππ‘ = πΌ 2 π π‘ = (1.2 A)2 (50.0 Ξ©)(5.0 min) ( 60 s ) = 2.2 × 104 J 1 min 91) A 6.0-Ξ© resistor is connected to a 15-V battery. a) What is the current in the circuit? π = πΌπ π 15 V πΌ= = = 2.5 A π 6.0 Ξ© b) How much thermal energy is produced in 10.0 min? π = ππΈ = ππ‘ = πΌ 2 π π‘ = (2.5 A)2 (6.0 Ξ©)(10.0 min) ( 60 s ) = 2.3 × 104 J 1 min 92) Lightbulbs: An incandescent lightbulb with a resistance of 10.0 Ξ© when it is not lit and a resistance of 40.0 Ξ© when it is lit has 120 V placed across it. a) What is the current draw when the bulb is lit? π = πΌπ π 120 V πΌ= = = 3.0 A π 40.0 Ξ© b) What is the current draw at the instant the bulb is turned on? At the instant the bulb is turned on, its resistance is still 10.0 Ξ©. It takes time (a very small amount of time, but still some time) for its resistance to increase to 40.0 Ξ©. π = πΌπ π 120 V πΌ= = = 12 A π 10.0 Ξ© c) When does the lightbulb use the most power? The instant the bulb is turned on Page 636 #36-47, 57-63, 77-81 36) Complete the concept map using the following terms: series circuit, π = π 1 + π 2 + π 3 , constant current, parallel circuit, constant potential. 37) Why is it frustrating when one bulb burns out on a string of holiday tree lights connected in series? (23.1) When one bulb burns out, the circuit is open and all the bulbs go out. 38) Why does the equivalent resistance decrease as more resistors are added to a parallel circuit? (23.1) Each new resistor provides an additional path for the current. 39) Several resistors with different values are connected in parallel. How do the values of the individual resistors compare with the equivalent resistance? (23.1) The equivalent resistance will be less than the resistance of each of the resistors. 40) Why is household wiring constructed in parallel instead of in series? (23.1) Appliances/lights/etc. connected in parallel can be run independently of one another. If they were connected in series, they would have to all be on or all be off at any given time. 41) Why is there a difference in equivalent resistance between three 60 Ξ© resistors connected in series and three 60 Ξ© resistors connected in parallel? (23.1) In a series circuit, the current is opposed by each resistor in turn. The total resistance is the sum of the individual resistances (180 Ξ©). In a parallel circuit, each resistor provides an additional path for current. The result is a decrease in total resistance, to 20 Ξ©. 42) Compare the amount of current entering a junction in a parallel circuit with that leaving the junction. (A junction is a point where three or more conductors are joined.) (23.1) The amount of current entering a junction is equal to the amount of current leaving the junction. The is known as Kirchoffβs junction rule. 43) Explain how a fuse functions to protect an electric circuit. (23.2) The purpose of a fuse is to prevent conductors from being overloaded with current, causing fires due to overheating. A fuse is simply a short length of wire that will melt from the heating effect if the current exceeds a certain maximum. This opens the circuit and stops the current from flowing. 44) What is a short circuit? Why is a short circuit dangerous? (23.2) A short circuit is a circuit that has extremely low resistance. A short circuit is dangerous because any potential difference will produce a large current. The heating effect of the current can cause a fire. 45) Why is an ammeter designed to have a very low resistance? (23.2) An ammeter must have low resistance because it is placed in series in the circuit. If its resistance were high, it would significantly change the total resistance of the circuit and thus serve to reduce the current in the circuit, thereby changing the current it is meant to measure. 46) Why is a voltmeter designed to have a very high resistance? (23.2) A voltmeter is placed in parallel with the portion of the circuit whose electric potential difference is to be measured. A voltmeter must have very high resistance for the same reason that an ammeter must have very low resistance. If the voltmeter had low resistance, it would lower the resistance of the portion of the circuit it is connected across and increase the current in the circuit. This would produce a larger decrease in electric potential across the part of the circuit where the voltmeter is connected, changing the electric potential difference it is measuring. 47) How does the way in which an ammeter is connected in a circuit differ from the way in which a voltmeter is connected? (23.2) An ammeter is connected in series; a voltmeter is connected in parallel. 57) Ammeter 1 in Figure 23-14 reads 0.20 A. a) What should ammeter 2 indicate? 0.20 A, because current is constant in a series circuit b) What should ammeter 3 indicate? 0.20 A, because current is constant in a series circuit 58) Calculate the equivalent resistance of these series-connected resistors: 680 Ξ©, 1.1 kΞ©, and 10 kΞ©. π = 0.68 kΞ© + 1.1 kΞ© + 10 kΞ© = 12 kΞ© 59) ) Calculate the equivalent resistance of these parallel-connected resistors: 680 Ξ©, 1.1 kΞ©, and 10.2 kΞ©. 1 1 1 1 = + + π π 1 π 2 π 3 π = 1 1 = = 0.40 kΞ© 1 1 1 1 1 1 + + + + π 1 π 2 π 3 0.68 kΞ© 1.1 kΞ© 10.2 kΞ© 60) A series circuit has two voltage drops: 5.50 V and 6.90 V. What is the supply voltage? According to Kirchoffβs loop rule, the supplied Ξπ must equal the drops in Ξπ. ππ ππ’πππ = π1 + π2 = 5.50 V + 6.90 V = 12.40 V 61) A parallel circuit has two branch currents: 3.45 A and 1.00 A. What is the current in the energy source. πΌπ ππ’πππ = πΌ1 + πΌ2 = 3.45 A + 1.00 A = 4.45 A 62) Ammeter 1 in Figure 23-14 reads 0.20 A. a) What is the total resistance in the circuit? π = π 1 + π 2 = 15 Ξ© + 22 Ξ© = 37 Ξ© b) What is the battery voltage? π = πΌπ = (0.20 A)(37 Ξ©) = 7.4 V c) How much power is delivered to the 22 Ξ© resistor? π = πΌ 2 π = (0.20 A)2 (22 Ξ©) = 0.88 W d) How much power is supplied by the batter? π = πΌπ = (0.20 A)(7.4 V) = 1.5 W 63) Ammeter 2 in Figure 23-14 reads 0.50 A. a) Find the voltage across the 22-Ξ© resistor. π = πΌπ = (0.50 A)(22 Ξ©) = 11 V b) Find the voltage across the 15-Ξ© resistor. π = πΌπ = (0.50 A)(15 Ξ©) = 7.5 V c) What is the battery voltage? π = π1 + π2 = 11 V + 7.5 V = 19 V 77) Refer to Figure 23-19 and assume that all the resistors are 30.0 Ξ©. Find the equivalent resistance. 1 1 1 = + β π π = 15.0 Ξ© π π 30.0 Ξ© 30.0 Ξ© π = 30.0 Ξ© + 15.0 Ξ© = 45.0 Ξ© 78) Refer to Figure 23-19 and assume that each resistor dissipates 120 mW. Find the total dissipation. π = 3(120 mW) = 360 mW 79) Refer to Figure 23-19 and assume that πΌπ΄ = 13 mA and πΌπ΅ = 1.7 mA. Find πΌπΆ . πΌπ΄ = πΌπ΅ + πΌπΆ πΌπΆ = πΌπ΄ β πΌπ΅ = 13 mA β 1.7 mA = 11 mA 80) Refer to Figure 23-19 and assume that πΌπ΅ = 13 mA and πΌπΆ = 1.7 mA. Find πΌπ΄ . πΌπ΄ = πΌπ΅ + πΌπΆ = 13 mA + 1.7 mA = 15 mA 81) Refer to Figure 23-20 to answer the following questions. a) Determine the total resistance. The 30.0-Ξ© and 20.0-Ξ© resistors are in series. π 1 = 30.0 Ξ© + 20.0 Ξ© = 50.0 Ξ© The 30.0-Ξ© and 20.0-Ξ© resistors are in series. π 2 = 10.0 Ξ© + 40.0 Ξ© = 50.0 Ξ© π 1 and π 2 are in parallel. 1 1 1 = + π π 1 π 2 1 1 π = = = 25.0 Ξ© 1 1 1 1 + + π 1 π 2 50.0 Ξ© 50.0 Ξ© π is in series with the top 25.0-Ξ© resistor. π πππ‘ππ = 25.0 Ξ© + 25.0 Ξ© = 50.0 Ξ© b) Determine the current through the 25-Ξ© resistor. All of the current must flow through the 25-Ξ© resistor. We can use Ohmβs Law and the total resistance to calculate the current. π = πΌπ πππ‘ππ π 25 π πΌ= = = 0.50 A π πππ‘ππ 50.0 Ξ© c) Which resistor is the hottest? Coolest? The temperature of the resistor is directly proportional to the power consumed by the resistor. π = πΌ2 π π25Ξ© = (0.50 A)2 (25.0 Ξ©) = 6.25 W The 0.50 A is split evenly between the two parallel branches because the sum of resistances in each branch are equal. π30Ξ© = (0.25 A)2 (30.0 Ξ©) = 1.9 W π20Ξ© = (0.25 A)2 (20.0 Ξ©) = 1.2 W π10Ξ© = (0.25 A)2 (10.0 Ξ©) = 0.62 W π40Ξ© = (0.25 A)2 (40.0 Ξ©) = 2.5 W The 25.0-Ξ© resistor is the hottest. The 10.0-Ξ© resistor is the coolest. Chapter 22 Review (coming soon) Chapter 23 Review (coming soon) Combined Circuit Practice (coming soon)