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

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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 . 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: I = 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 = I ? t = (7.5 A)(15 min)(60 s/min) = 6.8 10 3 C . 10. We find the potential difference across the birds 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 = L / A = L /p r 2 ; 0.22 = (5.6 10 8 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 = L / A = L / # p d 2 Chapter 25 p. 2 = (1.68 10 8 m)(3.5 m)/ # (1.5 10 3 m) 2 = 0.033 . 13. From the expression for the resistance, R = L / A , we form the ratio R Al / R Cu = ( Al / Cu )( L Al / L Cu )( A Cu / A Al ) = ( Al / 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)]m)][(10....
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Ch25Word - Chapter 25 p. 1 CHAPTER 25 Electric Currents and...

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