Chapter 23 continued Answer Key - Henry County Schools ... ?· Chapter 23 Study Guide ... Chapter 23…

Download Chapter 23 continued Answer Key - Henry County Schools ... ?· Chapter 23 Study Guide ... Chapter 23…

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<ul><li><p>Chapter 23 Study GuideSeries and Parallel CircuitsVocabulary Review</p><p>1. h2. e3. i4. g5. b6. c7. d8. a9. k</p><p>10. f11. j</p><p>Section 23.1 Simple Circuits</p><p>1. b2. c3. d4. a5. a6. b7. b 8. The circuits electric energy source raises the</p><p>potential by an amount equal to the poten-tial drop produced when the current passesthrough the resistors.</p><p>9. Multiply the current in the circuit by theresistance of the individual resistor.</p><p>10. A series circuit is used as a voltage divider.11. A voltage divider produces a voltage source</p><p>of desired magnitude from a higher-voltagebattery.</p><p>12. The resistance of a photoresistor depends onthe amount of light that strikes it. A pho-toresistor is made a material that naturallyhas a higher resistance in the dark.</p><p>13. A circuit that includes a photoresistor isused in a light meter. The electronic circuitdetects the potential difference and convertsit to a measurement of illuminance.</p><p>14. b15. b16. c</p><p>17. d18. a </p><p>Section 23.2 Applications of Circuits</p><p>1. true2. thickness3. closes4. true 5. parallel6. large7. First draw a schematic of the circuit. Then</p><p>reduce the problem to a set of series circuitsand a set of parallel circuits. Combine theresistances of the parallel circuits into onecircuit, and calculate the single equivalentresistance that can replace them. That leavesonly a series circuit. Add the resistors inseries to calculate the equivalent resistance.</p><p>8. a9.</p><p>125 V 20 !</p><p>10 !</p><p>60 !</p><p>V</p><p>C</p><p>125 V 20 !</p><p>10 !</p><p>60 !</p><p>A</p><p>B</p><p>125 V 20 !</p><p>10 !</p><p>60 !</p><p>A</p><p>A</p><p>Physics: Principles and Problems Chapters 2125 Resources 187</p><p>Chapter 23 continued Answer KeyC</p><p>opyr</p><p>ight</p><p> G</p><p>lenc</p><p>oe/M</p><p>cGra</p><p>w-H</p><p>ill, a</p><p> div</p><p>isio</p><p>n of</p><p> The</p><p> McG</p><p>raw</p><p>-Hill</p><p> Com</p><p>pani</p><p>es, I</p><p>nc.</p></li><li><p>10. voltmeter11. ammeter12. high13. low14. series15. parallel16. ammeter17. high, parallel18. voltmeter19. resistance20. ammeter</p><p>Section 23-1 Quiz1. The equivalent resistance is the sum of all</p><p>the individual resistances along the seriescircuit. To calculate equivalent resistance in aseries circuit, add all the individual resis-tances of the resistors (R. " R</p><p>A# R</p><p>B).</p><p>2. A voltage divider is a series circuit used toproduce a voltage source of desired magni-tude from a higher-voltage battery. To createa voltage divider, a circuit designer wouldconnect two resistors in a series circuit withthe battery. The desired voltage is the voltagedrop that can be measured across the secondresistor. The values of the two resistorswould be chosen based on the voltage of thebattery and the desired voltage.</p><p>3. R " RA # RB # RCR " 25 ! # 30 ! # 40 ! " 95 !</p><p>I " " " 0.063 A</p><p>4. " # #</p><p>" # #</p><p>R " !</p><p>I "</p><p>I " " 0.6 A</p><p>Section 23-2 Quiz1. When a short circuit occurs, the electricity is</p><p>traveling along a shorter path because a siz-able part of the circuit has been cut out.There is a higher-current intensity in the partof the circuit that remains. A fuse is a shortpiece of metal that melts when too muchcurrent passes through it. The fuse acts as asafety device by preventing this high currentflow from continuing and possibly causingan electrical fire.</p><p>2. An ammeter measures current in any part ofa circuit. To use an ammeter, place it inseries in the circuit in the place where youwant to measure the current. A voltmetermeasures the voltage drop across a resistor.Connect the voltmeter in parallel to theresistor to measure the voltage drop.</p><p>3.</p><p>R " RA # RBR " 20.0 ! # 30.0 ! " 50.0 !</p><p>I " " " 0.18 A</p><p>4.</p><p>" #</p><p>" #</p><p>R " 12.0 !</p><p>I " " " 0.75 A9.0 V$12.0 !V$R</p><p>1$30.0 !</p><p>1$20.0 !</p><p>1$R</p><p>1$RB</p><p>1$RA</p><p>1$R</p><p>9.0 V 20.0 30.0 </p><p>9.0 V$50.0 !</p><p>V$R</p><p>9.0 V</p><p>20.0 </p><p>30.0 </p><p>6.0 A!!</p><p>!$65090$ !"</p><p>V$R</p><p>600$59</p><p>1$40 !</p><p>1$30 !</p><p>1$25 !</p><p>1$R</p><p>1$RC</p><p>1$RB</p><p>1$RA</p><p>1$R</p><p>6.0 V$95 !</p><p>V$R</p><p>Chapter 23 continuedAnswer Key</p><p>188 Chapters 2125 Resources Physics: Principles and Problems</p><p>Copyright G</p><p>lencoe/McG</p><p>raw-H</p><p>ill, a division of The McG</p><p>raw-H</p><p>ill Com</p><p>panies, Inc.</p></li></ul>