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solutions_chapter19

solutions_chapter19 - 19 CURRENT RESISTANCE AND...

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19-1 C URRENT , R ESISTANCE , AND D IRECT -C URRENT C IRCUITS 19 Answers to Multiple-Choice Problems 1. B 2. B 3. D 4. B, D 5. A 6. A 7. C 8. A 9. C 10. C 11. C 12. C 13. C 14. A 15. E Solutions to Problems 19.1. Set Up: An electron has charge of magnitude The current tells us how much charge flows in a given time. We can find the number of electrons that correspond to a certain amount of charge. Solve: (a) (b) (c) Reflect: The amount of charge associated with typical currents is quite large. In Chapter 17 we found that the force between objects that have net charge 1.0 C is immense. The same amount of charge enters and leaves any part of the circuit each second and each circuit element remains neutral. 19.2. Set Up: Solve: 19.3. Set Up: The number of ions that enter gives the charge that enters the axon in the specified time. Solve: 19.4. Set Up: Negative charge moving from A to B is equivalent to an equal magnitude of positive charge going from B to A. The current direction is the direction of flow of positive charge. Solve: The total positive charge moving from B to A is Positive charge flows from B to A so the current is in this direction. 19.5. Set Up: gives the amount of charge that passes a point in the wire in time From we can find the number of electrons, since an electron has charge Solve: Reflect: Typical currents correspond to the flow of a large number of electrons. 1 5.50 3 10 2 3 C / s 2 1 1 1.60 3 10 2 19 C / electron 2 5 3.44 3 10 16 electrons / s 2 e . D Q D t . D Q 5 I D t I 5 D Q D t 5 1.85 C 30 s 5 62 mA. 1.85 C. D Q 5 1 5.11 3 10 18 1 2 3 3.24 3 10 18 421 1.60 3 10 2 19 C 2 5 I 5 D Q D t . I 5 D Q D t 5 9.0 3 10 2 8 C 10 3 10 2 3 s 5 9.0 m A D Q 5 1 5.6 3 10 11 ions 21 1.60 3 10 2 19 C / ion 2 5 9.0 3 10 2 8 C I 5 D Q D t . D Q 5 I D t 5 1 25,000 A 21 40 3 10 2 6 s 2 5 1.0 C I 5 D Q D t D t 5 D Q I 5 7.50 C 1.50 A 5 5.00 s D Q 5 I D t 5 1 1.50 A 21 300 s 2 5 450 C 1.50 A 5 1 1.50 C / s 2 1 1 1.60 3 10 2 19 C / electron 2 5 9.38 3 10 18 electrons / s e 5 1.60 3 10 2 19 C. 1A 5 1 C / s.
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19.6. Set Up: For copper, Table 19.1 gives that and for silver, Solve: (a) (b) so 19.7. Set Up: From Table 19.1, copper has Solve: 19.8. Set Up: For copper, Solve: 19.9. Set Up: The length of the wire in the spring is the circumference of each coil times the number of coils. Solve: Reflect: The value of we calculated is about a factor of 100 times larger than for copper. The metal of the spring is not a very good conductor. 19.10. Set Up: For aluminum, For copper, Solve: so 19.11. Set Up: For aluminum, For copper, Solve: (a) (b) 19.12. Set Up: Solve: R new 5 r L new p 1 d new / 2 2 2 5 r 3 L p 1 2 d / 2 2 2 5 3 4 r L p 1 d / 2 2 2 5 3 4 R d new 5 2 d L new 5 3 L ; R 5 r L A 5 r L p 1 d / 2 2 2 . L 5 R p r 2 r c 5 1 2.00 3 10 2 4 V 2 p 1 0.750 3 10 2 3 m 2 2 1.72 3 10 2 8 V 5 0.0205 m 5 2.05 cm R 5 r c L p r 2 . R 5 r al L A 5 1 2.63 3 10 2 8 V # m 21 3.80 m 2 1 1.00 3 10 2 2 m 21 5.00 3 10 2 2 m 2 5 2.00 3 10 2 4 V r c 5 1.72 3 10 2 8 V # m. r al 5 2.63 3 10 2 8 V # m. R 5 r L A . d c 5 d al Å r c r al 5 1 3.26 mm 2 Å 1.72 3 10 2 8 V # m 2.63 3 10 2 8 V # m 5 2.64 mm.
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