15 HMK #4 Chp 25

# 15 HMK #4 Chp 25 - Ch. 25 PHYSICS 2B Spring 2011 Homework...

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Ch. 25 PHYSICS 2B Spring 2011 Homework Solutions – Chapter 25 Author: Evan Grohs - [email protected] Ch. 25 Problems: 55, 58, 62, 64, 66, 68, 71, 74, 83, 86 Page 1/6

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Ch. 25 page 2/6 C URRENT , R ESISTANCE , AND E LECTROMOTIVE F ORCE 25.55. (a) I DENTIFY and S ET U P : Use . L R A E XECUTE :    2 3 8 0.104 1.25 10 m 3.65 10 m 14.0 m RA L    E VALUATE : This value is similar to that for good metallic conductors in Table 25.1. (b) I DENTIFY and S ET U P : Use V = EL to calculate E and then Ohm's law gives I. E XECUTE :   1.28 V/m 14.0 m 17.9 V VE L 17.9 V 172 A 0.104 V I R E VALUATE : We could do the calculation another way: 72 8 1.28 V/m so 3.51 10 A/m 3.65 10 m E EJ J 2 3 172 A, IJ A   which checks (c) I DENTIFY and S ET U P : Calculate / o r / JIA JE and then use Eq. (25.3) for the target variable d . v E XECUTE : dd J nqv nev 3 d 28 3 19 2.58 10 m/s 2.58 mm/s 8.5 10 m 1.602 10 C J v ne     E VALUATE : Even for this very large current the drift speed is small. 25.58. I DENTIFY : Conservation of charge requires that the current is the same in both sections. The voltage drops across each section add, so Cu Ag . R RR  The total resistance is the sum of the resistances of each section. I A , so , IR E L where R is the resistance of a section and L is its length. S ET U P : For copper, 8 Cu 1.72 10 m . For silver, 8 Ag 1.47 10 m. E XECUTE : (a) Cu Ag . VV I R 8 Cu Cu Cu 42 Cu (1.72 10 m)(0.8 m) 0.049 ( /4)(6.0 10 m) L R A and 8 Ag Ag Ag Ag (1.47 10 m)(1.2 m) 0.062 . (/ 4 ) ( 6 . 01 0 m ) L R A This gives 5.0 V 45 A. 0.049 0.062 I  The current in the copper wire is 45 A. (b) The current in the silver wire is 45 A, the same as that in the copper wire or else charge would build up at their interface.
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## This note was uploaded on 12/18/2011 for the course PHYS 2B 2b taught by Professor Hirsch during the Spring '10 term at UCSD.

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15 HMK #4 Chp 25 - Ch. 25 PHYSICS 2B Spring 2011 Homework...

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