Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Homework #1: Definition of Current and Resistance. 1. In a particular television picture tube, the measured beam current is 60.0 μA. How many electrons strike the screen every second? 2. A lightbulb has a resistance of 240 Ω when operating at a voltage of 120 V. What is the current in the bulb? 3. A 1.00-V potential difference is maintained across a 10.0-Ω resistor for a period of 20.0 s. What total charge passes through the wire in this time interval? 4. What is the resistance of a toaster if 120 V produces a current of 4.2 A? 5. What voltage will produce 0.25 A of current through a 3800- resistor? 6. A bird stands on a dc electric transmission line carrying 2800 A. The line has 2.5 10 5 resistance per meter, and the bird’s feet are 4.0 cm apart. What is the potential difference between the bird’s feet? Homework #2: Resistivity. 7. Calculate the diameter of a 2.0-cm length of tungsten filament in a small lightbulb if its resistance is 0.050 Ω. 8. Eighteen-gauge wire has a diameter of 1.024 mm. Calculate the resistance of 15 m of 18-gauge copper wire at 20°C. 9. A potential difference of 12 V is found to produce a current of 0.40 A in a 3.2-m length of wire with a uniform radius of 0.40 cm. What is (a) the resistance of the wire? (b) the resistivity of the wire? 10. A wire 50.0 m long and 2.00 mm in diameter is connected to a source with a potential difference of 9.11 V, and the current is found to be 36.0 A. Assume a temperature of 20°C, and, using Table 17.1, identify the metal out of which the wire is made. 11. A wire of initial length L0 and radius r0 has a measured resistance of 1.0 Ω. The wire is drawn under tensile stress to a new uniform radius of r = 0.25r0. What is the new resistance of the wire? 12. Suppose that you wish to fabricate a uniform wire out of 1.00 g of copper. If the wire is to have a resistance R = 0.500 Ω, and if all of the copper is to be used, what will be (a) the length and (b) the diameter of the wire? Hint: The density of copper is 8.92 g/cm3, and the volume of the copper does not change. Homework #3: Temperature Dependence. 13. A certain lightbulb has a tungsten filament with a resistance of 19 Ω when cold and 140 Ω when hot. Assume that R Ro 1 T To can be used over the large temperature range involved here, and find the temperature of the filament when it is hot. Assume an initial temperature of 20°C. 14. At 20°C, the carbon resistor in an electric circuit connected to a 5.0-V battery has a resistance of 200 Ω. What is the current in the circuit when the temperature of the carbon rises to 80°C? 15. A 100-cm-long copper wire of radius 0.50 cm has a potential difference across it sufficient to produce a current of 3.0 A at 20°C. (a) What is the potential difference? (b) If the temperature of the wire is increased to 200°C, what potential difference is now required to produce a current of 3.0 A? Homework #4: Power. 16. A toaster is rated at 600 W when connected to a 120-V source. What current does the toaster carry, and what is its resistance? 17. If electrical energy costs 12 cents, or $0.12, per kilowatt-hour, how much does it cost to (a) burn a 100-W lightbulb for 24 h? (b) operate an electric oven for 5.0 h if it carries a current of 20.0 A at 220 V? 18. A high-voltage transmission line with a resistance of 0.31 Ω/km carries a current of 1,000 A. The line is at a potential of 700 kV at the power station and carries the current to a city located 160 km from the station. (a) What is the power loss due to resistance in the line? 2 {Hint: Use I R for the power loss to the wire.} (b) What fraction of the transmitted power does this loss represent? {Hint: Use V I for the power delivered to the city.} 19. The heating element of a coffeemaker operates at 120 V and carries a current of 2.00 A. Assuming that the water absorbs all of the energy converted by the resistor, calculate how long it takes to heat 0.500 kg of water from room temperature (23.0°C) to the boiling point. 20. What is the required resistance of an immersion heater that will increase the temperature of 1.50 kg of water from 10.0°C to 50.0°C in 10.0 min while operating at 120 V? 21. A small motor draws a current of 1.75 A from a 120-V line. The output power of the motor is 0.20 hp. (a) At a rate of $0.060/kWh, what is the cost of operating the motor for 4.0 h? (b) What is the efficiency of the motor? 1 horsepower = 746 W 22. It has been estimated that there are 270 million plug-in electric clocks in the United States, approximately one clock for each person. The clocks convert energy at the average rate of 2.50 W. To supply this energy, how many metric tons of coal are burned per hour in coal-fired electric generating plants that are, on average, 25.0% efficient? The heat of combustion for coal is 33.0 MJ/kg. 23. A steam iron draws 6.0 A from a 120-V line. (a) How many joules of internal energy are produced in 20 min? (b) How much does it cost, at $0.080/kWh, to run the steam iron for 20 min? Solutions: 1. Q I t and the number of electrons is 6 Q I t 60.0 10 C s 1.00 s n 3.75 1014 electrons -19 e e 1.60 10 C V 120 V 0.500 A 500 m A R 240 V Q V 3. The current is I . Thus, the change that passes is Q t , giving t R R 1.00 V Q t 0.100 A 20.0 s 2.00 C 10.0 V 120 V 29 4. R I 4.2 A 2. I 5. V IR 0.25 A 3800 950 V 6. Find the potential difference from the resistance and the current. R 2.5 105 m 4.0 10 2 m 1.0 10 6 V IR 2800 A 1.0 106 2.8 10 3 V 7. From R d 4 L R 8. R 9. (a) L A R (b) L A , we obtain A 4 5.6 108 m d 2 4 d 4 2 R 2.0 10 2 0.050 L L 41.7 108 m 1.024 10 3 , or m 1.7 10 15 m m 2 4 m 0.17 m m 0.31 V 12 V 30 I 0.40 A From, R L A , 2 2 R A 30 0.40 10 m 4.7 10 4 m L 3.2 m V 9.11 V 0.253 , so the resistivity of the metal is 10. The resistance is R I 36.0 A 2 3 R A R d 4 0.253 2.00 10 m L L 4 50.0 m Thus, the metal is seen to be silver . 2 1.59 108 m 11. The volume of material, V AL0 r02 L0 , in the wire is constant. Thus, as the wire is stretched to decrease its radius, the length increases such that rf2 Lf r02 L0 giving 2 2 r r 2 Lf 0 L0 0 L0 4.0 L0 16L0 rf 0.25r0 The new resistance is then Rf Lf Af Lf r 2 f 16L0 r0 4 2 L 2 16 4 02 256R 0 256 1.00 256 r0 12. The volume of the copper is V 1.00 103 kg m density 8.92 103 kg m 3 1.12 107 m Since, V A L , this gives A L 1.12 107 m (a) From R L A 3 3 (1) , we find that 1.70 108 m 8 A L L 3.40 10 m L 0.500 R Inserting this expression for A into equation (1) gives 3.40 10 8 From equation (1), A (b) d m L2 1.12 107 m 3 , which yields 4 1.12 107 m L 3 d2 4 1.12 107 m L 41.12 107 m 1.82 m 3 L 1.82 m 3 , or 2.80 104 m 0.280 m m 13. Solving R R0 1 T T0 for the final temperature gives T T0 14. At 80°C, R R0 140 19 20C 1.4 103 C 1 3 R0 4.5 10 C 19 I or V V 5.0 V R R0 1 T T0 200 1 0.5 103 C 1 80C 20C I 2.6 102 A 26 m A 15. (a) At 20°C, the resistance of this copper wire is R0 L A L r2 and the potential difference required to produce a current of I 3.0 A is 1.00 m L V0 IR0 I 2 3.0 A 1.7 108 m r 0.50 102 m or 2 V0 6.5 104 V 0.65 m V (b) At T 200C , the potential difference required to maintain the same current in the copper wire is V IR IR0 1 T T0 V0 1 T T0 or V 0.65 m V 1 3.9 103 °C -1 200C 20C 1.1 m V 16. I P 600 W 5.00 A V 120 V V 120 V 24.0 I 5.00 A 17. (a) The energy used by a 100-W bulb in 24 h is and R E Pt 100 W 24 h 0.100 kW 24 h 2.4 kW h and the cost of this energy, at a rate of $0.12 per kilowatt-hour is cost E rate 2.4 kW h $0.12 kW h $0.29 (b) The energy used by the oven in 5.0 h is 1 kW E Pt I V t 20.0 C s 220 J C 5.0 h 22 kW h 3 10 J s and the cost of this energy, at a rate of $0.12 per kilowatt-hour is cost E rate 22 kW h $0.12 kW h $2.6 18. (a) The power loss in the line is Ploss I2R 1000 A 0.31 km 2 (b) 160 km 5.0 107 W 50 M W The total power transmitted is Pinput V I 700 103 V 1000 A 7.0 108 W 700 M W Thus, the fraction of the total transmitted power represented by the line losses is fraction loss Ploss 50 M W 0.071 or 7.1% Pinput 700 M W 19. The energy required to bring the water to the boiling point is E m c T 0.500 kg 4186 J kg C 100C 23.0C 1.61 105 J The power input by the heating element is Pinput V I 120 V 2.00 A 240 W 240 J s Therefore, the time required is t E Pinput 1.61 105 J 1 m in 672 s 11.2 m in 240 J s 60 s 20. The energy input required is E m c T 1.50 kg 4186 J kg C 50.0C 10.0C 2.51 105 J and, if this is to be added in t 10.0 m in 600 s, the power input needed is P E 2.51 105 J 419 W t 600 s The power input to the heater may be expressed as P V R , so the needed resistance is 2 R V P 2 120 V 419 W 2 34.4 21. (a) The power input to the motor is Pinput V I 120 V 1.75 A 210 W 0.210 kW At a rate of $0.06/kWh, the cost of operating this motor for 4.0 h is cost Energy used rate Pinputt rate 0.210 kW (b) 4.0 h 6.0 cents kW h 5.0 cents The efficiency is Eff Poutput Pinput 0.20 hp 0.746 kW hp 0.210 kW 0.71 or 71% 22. The total power converted by the clocks is P 2.50 W 270 106 6.75 108 W and the energy used in one hour is E Pt 6.75 108 W 3600 s 2.43 10 12 J The energy input required from the coal is Ecoal 2.43 1012 J E 9.72 1012 J efficiency 0.250 The required mass of coal is thus m Ecoal 9.72 1012 J 2.95 105 kg 6 heatof com bustion 33.0 10 J kg 1 m etric ton m 2.95 105 kg 295 m etric tons 3 10 kg 23. (a) The power required by the iron is or à V I 120 V 6.0 A 7.2 102 W and the energy transformed in 20 minutes is J 60 s 5 E à t 7.2 102 20 m in 8.6 10 J s 1 m in (b) The cost of operating the iron for 20 minutes is cost E rate 1 kW h 8.6 105 J $0.080 kW h $0.019 1.9 cents 6 3.60 10 J