# Ch25Word - Chapter 25 p. 1 CHAPTER 25 – Electric Currents...

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Unformatted text preview: 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: I = 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 = I ? t = (5.7 A)(7.0 h)(3600 s/h) = 1.4 × 10 5 C . 3. We find the current from I = ? 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 Ω . 5. We find the voltage from V = IR = (0.25 A)(3000 Ω ) = 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: I = V / R = [(9.0 V)/(1.6 Ω )](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 Ω . 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 Ω . ( b ) The charge that passes through the hair dryer is ? Q = I ? 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 Ω /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 Ω = (5.6 × 10 –8 Ω · m)(1.00 m)/p r 2 , which gives r = 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 Chapter 25 p. 2 = ( 1 . 6 8 × 10 –8 Ω · m)(3.5 m)/ # π (1.5 × 10 –3 m) 2 = 0.033 Ω . 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 . 6 5 × 10 –8 Ω · m)/(1.68 × 10 –8 Ω · 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 Ω · m)/(1.68 × 10 –8 Ω · m)][(2.5 mm)/ d T ] 2 , which gives d T = 4.6 mm . 15. Because the material and area of the two pieces are the same, from the expression for the resistance, R = ρ L / A , we see that the resistance is proportional to the length: R 1 / R 2 = L 1 / L 2 = 5.0....
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## This note was uploaded on 02/05/2010 for the course PHYSICS 37,38,39 taught by Professor Fielding during the Spring '09 term at Los Angeles Valley College.

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

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