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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

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.

Page1 / 12

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online