Quick QuizzesAnswers to Even Numbered Conceptual QuestionsAnswers to Even Numbered ProblemsProblem SolutionsChapter 18 Direct-Current Circuits Quick Quizzes 1. (a), (d). Bulb R1 becomes brighter. Connecting a wire from b to c provides a nearly zero resistance path from b to c and decreases the total resistance of the circuit from R1 + R2 to just R1. Ignoring internal resistance, the potential difference maintained by the battery is unchanged while the resistance of the circuit has decreased. The current passing through bulb R1 increases, causing this bulb to glow brighter. Bulb R2 goes out because essentially all of the current now passes through the wire connecting b and c and bypasses the filament of Bulb R2. 2. (b). When the switch is opened, resistors R1 and R2 are in series, so that the total circuit resistance is larger than when the switch was closed. As a result, the current decreases. 3. (a). When the switch is closed, resistors R1 and R2 are in parallel, so that the total circuit resistance is smaller than when the switch was open. As a result, the total current increases. 4. (a) unchanged; (b) unchanged; (c) increase; (d) decrease. Neglecting internal resistance, the terminal potential difference of the battery remains constant (equal to the emf) as bulbs are added. Since each bulb is added in parallel with the battery, the current in each bulb, and hence the brightness of the bulb, is unchanged as new bulbs are added. The total current supplied by the battery increases as bulbs are added in parallel. Thus, the power delivered by the battery increases and its lifetime decreases. 5. ((a) decrease; (b) decrease; (c) decrease; (d) increase. Adding bulbs in series increases the total resistance of the circuit and results in a decrease in the current in the bulbs. Thus, the brightness of individual bulbs deceases as bulbs are added. Neglecting internal resistance, the terminal potential difference of the battery remains constant (equal to the emf) and the power supplied by the battery decreases as bulbs are added. This means that the battery will supply energy at a lower rate, and its lifetime will be increased. 6. (c). After the capacitor is fully charged, current flows only around the outer loop of the circuit. This path has a total resistance of 3 Ω, so the 6-V battery will supply a current of 2 Amperes. 99100 CHAPTER 18 Answers to Even Numbered Conceptual Questions 2. RRRRRRRRRRRRRRRRR 4. (a) ii (b) i 6. (a) ii (b) i 8. A short circuit can develop when the last bit of insulation frays away between the two conductors in a lamp cord. Then the two conductors touch each other, opening a low resistance branch in parallel with the lamp. The lamp will immediately go out, carrying no current and presenting no danger. A very large current will be produced in the power source, the house wiring, and the wire in the lamp cord up to and through the short. The circuit breaker will interrupt the circuit quickly but not before considerable heating and sparking is produced in the short-circuit path. 10. A wire or cable in a transmission line is thick and made of material with very low resistivity. Only when its length is very large does its resistance become significant. To transmit power over a long distance it is most efficient to use low current at high voltage, minimizing the power loss in the transmission line. 2IR12. The bulbs of set A are wired in parallel. The bulbs of set B are wired in series, so removing one bulb produces an open circuit with infinite resistance and zero current. 14. (a) ii. The power delivered may be expressed as P , and while resistors connected in series have the same current in each, they may have different values of resistance. (b) ii. The power delivered may also be expressed as 2IR=()2VR=∆P , and while resistors connected in parallel have the same potential difference across them, they may have different values of resistance. 16. Compare two runs in series to two resistors connected in series. Compare three runs in parallel to three resistors connected in parallel. Compare one chairlift followed by two runs in parallel to a battery followed immediately by two resistors in parallel. The junction rule for ski resorts says that the number of skiers coming into a junction must be equal to the number of skiers leaving. The loop rule would be stated as the total change in altitude must be zero for any skier completing a closed path. 18. Because water is a good conductor, if you should become part of a short circuit when fumbling with any electrical circuit while in a bathtub, the current would follow a pathway through you, the water, and to ground. Electrocution would be the obvious result.Direct-Current Circuits 101 20. Even if the fuse were to burn out or the circuit breaker to trip, there would still be a current path through the device, and it would not be protected. 22. When connected in series, all bulbs carry the same current. Thus, the one with the lowest resistance dissipates the least power 2IR=P and glows the dimmest. This is seen to be the bulb that is labeled “200 W”. If they were connected in parallel, all bulbs would have the same potential difference across them. Then, the one with the lowest resistance will dissipate the most power ()2V=∆P R and glow the brightest. 24. The break is closer to A.102 CHAPTER 18 Answers to Even Numbered Problems 2. (a) (b) 1.0 A (c) 24 Ω48122.18 , 6.0 A, 3.0 A, 2.0 AIIIΩ== = 4. (a) 30 V (b) 2.3 V 6. 15 Ω 8. (a) (b) 4.53 V 5.13 Ω10. 6.0 , 3.0ABRR=Ω= ΩI12. (a) 75.0 V (b) 25.0 W, 6.25 W, 6.25 W; 37.5 W 14. 14.2 W, 28.4 W, 1.33 W, 4.00 W 16. 0.714 A, 1.29 A, 12.6 V 18. 5.4 V with a at a higher potential than b 20. 212.0 A (flows from toward ), 1.0 AIba=− =22. 0.50 W 24. (a) (b) 0.0816 or 8.16% 4.59 Ω26. 0.28 A (dead battery), 1. (starter) 27 10 A×28. (a) 9.0 V with b at a higher potential than a (b) 0.42 A directed from b to a 32. (a) 2.00 ms (b) 180 Cµ (c) 114 Cµ 34. 47 Fµ 36. (a) 10.0 Fµ (b) 415 Cµ 38. (a) 8.0 A (b) 120 V (c) 0.80 A (d) 25.8 10 W×40. No. Their combined power requirements exceed the 1 800 W available. 42. (a) (b) 114.1 10 J−× 0.56 Aµ 44. 0.865 or 86.5% 46. (a) (b) 56 W (c) 2.0 A 14 Ω48. (a) 40.0 W (b) 80.0 V, 40.0 V, 40.0 VDirect-Current Circuits 103 50. (a) 0 in the resistor and 3-kΩ 333