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Physics 100 Energy in Today’s World Homework Ch. 10 1. Prof. Menningen p. 1 of 4 The electric potential energy of a charged object at point A is known to be 48 J. If it requires 12 J of work to move the object from A to B, what is the potential energy at point B? PE PE B PE A work PE B PE A work 48 J 12 J 60 J 2. How much work does a 8.0-Volt battery do in pushing 2.9 milliCoulombs of charge through a circuit containing one light bulb? V work work charge V 0.0029 C 8.0 V 0.0232 J charge 3. When you stack three flashlight batteries in the same direction, you get a voltage of 3 × 1.5 volts = 4.5 volts. What voltage do you get if one of the batteries is turned around end-forend? a. 10.0 V V 1.5 1.5 1.5 V 1.5 V b. 3.0 V c. 1.5 V d. −1.5 V 4. What are the analogs to the following terms in the water model of electricity? a. wire pipe ____________ friction b. resistance ____________ c. voltage pressure or GPE ____________ d. charge water ____________ water wheel e. light bulb ____________ f. current flow rate ____________ pump ____________ h. switch valve ____________ g. battery 5. If the voltage in a circuit is doubled, what happens to the current? a. It increases by a factor of four. b. It decreases by a factor of four. c. It is doubled. d. It is cut in half. 6. Label as "series" or "parallel" the following circuit characteristics: series a. Has a large total resistance: ____________ parallel b. Used in household circuits: ____________ 7. c. Same voltage throughout: parallel ____________ d. Same current throughout: series ____________ In what unit is a household's consumption of electricity measured? a. watt/hour b. kilowatt/hour c. watt·sec d. kilowatt·hour e. kilowatt·sec Physics 100 Energy in Today’s World Homework Ch. 10 8. Prof. Menningen p. 2 of 4 Why do power companies use high voltages for their long-distance transmission lines? Write a brief but complete explanation. Using a high voltage reduces the current required to produce the same amount of power. Reducing the current greatly reduces the power loss Ploss = I2R, where R is the resistance of the wires. 9. A 750-MW power plant sends its power out on a 530,000 V high-voltage line. If the total resistance of the line is 0.70 Ω, what percentage of the power is lost due to resistive heating? P 750 106 W P VI I 1415 A V 530,000 V Ploss I 2 R 1415 A 0.70 1.40 106 W 1.40 MW 2 % loss Ploss 1.40 MW 100 0.00187 0.187% P 750 MW 10. For the following questions assume household electricity is provided at 120 V. (a) What is the resistance of a light bulb that draws 0.570 A of current? V IR R V 120 V 211 I 0.570 A (b) What is the power used by a toaster that draws a current of 9.95 A? P VI 120 V 9.95 A 1194 W 1.19 kW (c) A 1250-W heater for a sauna requires 65 minutes to heat the sauna to 120 °F. What does this cost if electricity sells for 9¢/kWh? P E 1 hr E Pt 1.250 kW 65 min 1.354 kWh t 60 min cost 1.354 kWh 9¢/kWh 12.2¢ Physics 100 Energy in Today’s World Homework Ch. 10 Prof. Menningen p. 3 of 4 11. (a) Three resistors with values 1.0 Ω, 4.0 Ω, and 7.0 Ω are connected in series with a 9.00 V battery. How much current is drawn from the battery? Rtotal 1.0 4.0 7.0 12.0 I V 9.00 V 0.750 A Rtotal 12.0 (b) If the same three resistors are each connected in parallel to the same battery, how much current is drawn from the battery? I1 V 9.00 V 9.00 V 9.00 V 9.0 A, I 2 2.25 A, I 3 1.29 A R1 1.0 4.0 7.0 I total 9.00 2.25 1.29 A 12.5 A 12. Two V = 1.65-V batteries are connected in parallel to a 7.00-Ω resistor. How much current does each battery supply? V 1.65 V 0.236 A R 7.00 That's the total current through the resistor. Each battery V IR I supplies half the current: 1 2 0.236 A 0.118 A 13. The common exit sign in a public building uses two 15 W incandescent bulbs. These could be replaced with light emitting diodes (LEDs) at 1.9 W per sign. If there are 45 million signs in use in the United States, what savings in electricity use for an entire year (8,760 hours) could be secured with this changeover? Use $0.10/kWh as the cost of electricity. P 2 15 W 1.9 W 28.1 W 0.0281 kW per sign E Pt 0.0281 kW 8760 h 246 kWh per sign savings 45 106 signs 246 kWh/sign $0.10 kWh 1 $1.11 109 1.11 billion dollars Physics 100 Energy in Today’s World Homework Ch. 10 Prof. Menningen p. 4 of 4 14. (a) Suppose you have an electric water heater with efficiency of 86%, and you take a shower using 54 kg of water (~15 gallons). At a price of $0.10 per kWh, how much did you pay for the shower? Assume it heated the water from 16 °C (~61 °F) to 42 °C (~108 °F). Q mcT 54 kg 4186 J/kg/ C 42 16C 5.877 106 J That's the heat added to the water. The energy input to the heater is: Q 5.877 106 J E 6.834 106 J e 0.86 1 kWh $0.10 E 6.834 106 J $0.190 6 3.6 10 J kWh (b) Consider the above scenario. If the power plant is 34% efficient and the transmission lines are 84% efficient, how many pounds of coal were burned for your shower? Each pound of coal releases about 1.2×107 J of thermal energy. The energy sent into the transmission lines by the power plant: W E etransmission 6.834 106 J 8.136 106 J 0.84 The fuel energy supplied to the power plant is: QH W epower plant 8.136 106 J 2.393 107 J 0.34 QH 2.393 107 J 1 lb coal 1.99 lb of coal 1.2 107 J (c) How many pounds of coal will be burned each year to provide showers for a family of four, assuming one shower per day for each family member? 1.99 lb 4 shower 7.98 lb/day 365 day/year shower day 2911 lb of coal/year 1.46 ton of coal/year Note that this calculation does not include the cost of keeping the water heater hot during the rest of the day!

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