Electrodynamics. Review 8


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1 Unit 8 eview: Electrodynamics eview 8 Electrodynamics 1. A 9.0 V battery is connected to a lightbulb which has a current of 0.5 A flowing through it. a. How much power is delivered to the b. How much energy will the bulb use in? c. How long would the bulb have to stay on to use 1 kwh of energy? 2. A 60 W lightbulb is connected to a 115 V power source. a. What is the current through the b. What is the resistance of the 3. A circuit is set up as shown in the diagram at the top right. a. What should the reading on the ammeter be? b. What should the reading on the voltmeter be? c. How much power is delivered to the resistor? d. How much energy is delivered to the resistor per hour? 4. A microwave draws a 15.0 A current from a 120 V power source. a. How much power is delivered to the microwave? b. How much energy does the microwave use each minute? 5. A 100 W lightbulb is turned on for two fifths of the time for 30 days. At a price of $0.29 per kwh, how much does the lightbulb cost to run during the 30 days? 6. A current of 0.85 A is measured through a 27 Ω resistor for. How much heat does the resistor generate? 7. A 60 W lightbulb has a resistance of 240 Ω. What is the current in the 8. What is the current through an 80.0 Ω resistor if the voltage drop across the resistor is 20 V? 9. A 60 W lightbulb and a 100 W lightbulb both are connected to a 120 V power source. Which bulb has the greater resistance? 10. A 12 V battery is used to power a 300 ma video camera. a. How much power does the battery provide to the camera? b. How much energy does the camera use in 1 s? c. How much energy does the camera use in? d. How long would it take the video camera to use 1 kwh of energy? 11. A lightbulb powered by a 120 V source has a resistance of 5.0 Ω at room temperature and Ω at operating temperature. a. What is the current when a 60 W bulb is turned on at room temperature? b. What is the power of this room temperature bulb? c. What is the current when the 60 W bulb is at operating temperature? d. What is the power of the bulb at operat ing temperature? 1
2 Unit 8 eview: Electrodynamics 12. A 240 V water heater has a resistance of 15 Ω. a. What is the power of the heater? b. What thermal energy is supplied by the heater in 15 min? c. Compare the power of the same 240 V water heater to a 120 V water heater, also with a resistance of 15 Ω. d. Compare the thermal energy of the 240 V water heater to a 120 V heater in a 15 min period? 13. How much power is dissipated in wires that have a current of 50 A and a resistance of Ω? 14. A longdistance hightension wire uses 500,000 V. a. What is the power output of these wires in watts, kilowatts, and megawatts if the current is 20 A? b. What is the power dissipated in the wires if the resistance is Ω? 15. Compare the energy a 120 W bulb uses per hour with the energy use in our of a 60 W bulb. 16. How much money would be saved by turning off one 100 W lightbulb 3 h/day for 365 days if the cost of electricity is $0.12 per kwh? 17. How much money can be saved by turning off a 500 W television set for /day for 365 days at $0.14 per kwh? 18. How much does it cost to operate a 1000 W blow dryer for 10 minutes at $0.23 per kwh? 19. A common household electric lightbulb produces of power in the form of light and heat from a filament powered by a standard electric socket of 120 V. a. What is the current through this bulb s filament? b. The lightbulb s efficiency is typically only 10%, which means only 10% of the power used is converted into light energy. How much energy in the form of heat does this bulb produce during of use? How much energy in the form of light does it produce during this same time interval? c. How much electric charge will pass through this bulb during? d. What is the resistance of the circuit? e. Explain why the filament in the circuit glows but the wires going to and from the lightbulb do not glow. 20. An electric stove uses a heating element that has several settings: warm, low, medium, and high. The owner s manual tells you that the control knob for the heating element is a potentiometer, a device that produces a variable resistance. a. If you increase the resistance of the heating element circuit, what happens to the current? b. How will increasing the resistance affect the power used by the circuit? c. Explain why a circuit with a high power rating gives off more heat than a circuit with a low power rating. d. redict which setting has the highest resistance. e. The warm setting uses 180 W, while the high setting uses 1200 W. Calculate the resistance of the circuit for each setting, assuming the voltage is 240 V. Does this calculation support your prediction? Explain. 2
3 Unit 8 eview: Electrodynamics 1a 1b = (9 V)(0.5 A) = 4.5 W = E t E = t E = (4.5 W)() E = J 3d 4a = E t E = t E = (54 W)() E = J = (120 V)(15 A) = 1800 W 1c 2a J = t J t = 222 h 2b 3a 3b 3c I = V 4b = E t E = t E = (1800 W)(60 s) E = J I = 115 V I = 0.52 A I = V = V I = 115 V 0.52 A = 220 Ω I = V I = 36 V 24 Ω I = 1.5 A 36 V = (36 V)(1.5 A) = 54 W J 1 s x 60 s 60 min x x 24 h 1 min 1 day x days 1 kwh x J x $ kwh = $ E = t E = I 2 t ( ) 2 ( 27 Ω) ( ) E = 0.85 A E = J 7 = I 2 8 I = I = 240 Ω I = 0.5 A I = V I = 20 V 80 Ω I = 0.25 A 3
4 Unit 8 eview: Electrodynamics 9 10a = V 2 = 120 V = 240 Ω = V 2 = 120 V 100 W = 144 Ω The has a greater resistance. 10b 3.6 J 10c 10d 11a 11b 11c 11d ( )( 0.3 A) = 12 V = 3.6 W 3.6 J 1 s = E E = J = t t = 278 h I = V I = 120 V 5 Ω I = 24 A ( )( 24 A) = 120 V = 2880 W I = V I = 120 V 100 Ω I = 1.2 A ( )( 1.2 A) = 120 V = 144 W 12a 12b 12c = 240 V 15 Ω = 3840 W E = t ( )( 900 s) E = 3840 W E = J = 120 V 15 Ω = 9 This is ¼ the power than it was at 240 V. 12d The energy will also be ¼ of what it was at 240 V. 13 = I 2 14a 14b ( ) 2 ( Ω) = 50 A = 37.5 W ( )( 20 A) = V = W = kw =10 MW = I 2 ( ) 2 ( Ω) = 20 A = 6 W 15 Half the power will use half the energy. 4
5 Unit 8 eview: Electrodynamics J 1 s 1 kwh J $ h 365 days 1 day 1 year $ e 20a The filament has a much higher resistance than the lead wires. All wires have the same current. = I 2 Higher resistance results in lower current. I = V/ J 1 kwh 1 s J $ days 1 day 1 year $ J 1 kwh 1 s J $ s 10 min 1 min 1 $ a I = V I = 120 V I = 0.5 A 20b Higher resistance results in lower power. / 20c ower is defined as the rate at which energy is transferred. Higher power means more energy transferred each second. 20d Warm is the lowest setting so it must have the least current and therefore the highest resistance. 20e = V 2 = 240 V 180 W = 320 Ω = V 2 = 240 V 1200 W = 48 Ω 19b 19c 19d 1 s 90% = J 1 s 10% = J 1 s = J = V 2 = 120 V = 240 Ω The higher resistance does result in lower power as predicted. 5
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