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Ch25Word - Chapter 25 p 1 CHAPTER 25 Electric Currents and...

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Chapter  25   p. 1 CHAPTER 25 – Electric Currents and Resistance 1. The rate at which electrons pass any point in the wire is the current: = 1.50 A = (1.50 C/s)/(1.60 × 10 –19  C/electron) =         9.38 × 10 18  electron/s . 2. The charge that passes through  the battery is ? Q  =  ? t  = (5.7 A)(7.0 h)(3600 s/h)  =         1.4 × 10 5  C . 3. We find the current from = ? Q /? t  = (1000 ions)(1.60 × 10 –19  C/ion)/(7.5 × 10 –6  s) =        2.1 × 10 –11  A . 4. We find the resistance from  V  =  IR 110 V = (4.2 A) R , which gives  R  =        26  . 5. We find the voltage from  V  =  IR  = (0.25 A)(3000  ) =        7.5 × 10 2  V . 6. For the device we have  V  =  IR .   ( a ) If we assume  constant resistance and  divide expressions for the two conditions, we get V 2 / V 1  =  I 2 / I 1 ; 0.90 =  I 2 /(5.50 A), which gives  I 2  =        4.95 A . ( b ) With the voltage constant, if we divide  expressions for the two conditions, we get I 2 / I 1  =  R 1 / R 2 ; I 2 /(5.50 A) = 1/0.90, which gives  I 2  =        6.11 A . 7. The rate at which electrons leave the battery is the current: V / R  = [(9.0 V)/(1.6  )](60 s/min)/(1.60 × 10 –19  C/electron) =         2.1 × 10 21  electron/min . 8. We find the resistance from  V  =  IR 12 V = (0.50 A) R , which gives  R  =        24  . The energy taken out of the battery is Energy =  Pt  =  IVt  = (0.50 A)(12 V)(1 min)(60 s/min)  =        3.6 × 10 2  J . 9. ( a ) We find the resistance from  V  =  IR 120 V = (7.5 A) R , which gives  R  =        16  . ( b ) The charge that passes through  the hair dryer is ? Q  =  ? t  = (7.5 A)(15 min)(60 s/min)  =         6.8 × 10 3  C . 10. We find the potential difference across the bird’s feet from  V  =  IR  = (2500 A)(2.5 × 10 –5   /m)(4.0 × 10 –2  m) =        2.5 × 10 –3  V . 11. We find the radius from  R  =  ρ L / A  =  ρ L /p r 2 ; 0.22    = (5.6 × 10 –8       m)(1.00 m)/p r 2 , which gives  = 2.85 × 10 –4  m, so the diameter is 5.7 × 10 –4  m =        0.57 mm . 12. We find the resistance from R     ρ L / A  =  ρ L / # p d 2  
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Chapter  25   p. 2 = (1.68 × 10 –8       m)(3.5 m)/ # π (1.5 × 10 –3  m) 2  =         0.033  . 13. From the expression for the resistance,  R  =  ρ L / A , we form the ratio R Al / R Cu = ( ρ Al / ρ Cu )( L Al / L Cu )( A Cu / A Al ) = ( ρ Al / ρ Cu )( L Al / L Cu )( d Cu / d Al ) 2 = [(2.65 × 10 –8       m)/(1.68 × 10 –8       m)][(10.0 m)/(20.0 m)][(2.5 mm)/(2.0 mm)] 2 = 1.2,   or       R Al  = 1.2 R Cu 14. Yes ,      if we select the appropriate diameter.  From the expression for the resistance,  R  =  ρ L / A we form the ratio R T / R Cu  = ( ρ T / ρ Cu )( L T / L Cu )( A Cu / A T ) = ( ρ T / ρ Cu )( d Cu / d T ) 2 ; 1 = [(5.6 × 10 –8       m)/(1.68 × 10 –8       m)][(2.5 mm)/ d T ] 2
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