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Electric current
Electric current
Electric current
Electric current
Electric current
Electric current
Electric current
Electric current
Electric current
Electric current
Electric current
Electric current
Electric current
Electric current
Electric current
Electric current
Electric current
Electric current
Electric current
Electric current
Electric current
Electric current
Electric current
Electric current
Electric current
Electric current
Electric current
Electric current
Electric current
Electric current
Electric current
Electric current
Electric current
Electric current
Electric current
Electric current
Electric current
Electric current
Electric current
Electric current
Electric current
Electric current
Electric current
Electric current
Electric current
Electric current
Electric current
Electric current
Electric current
Electric current
Electric current
Electric current
Electric current
Electric current
Electric current
Electric current
Electric current
Electric current
Electric current
Electric current
Electric current
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Electric current

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  • 1. Current and ResistanceFriday, July 22, 2011
  • 2. Current Convention : Current depicts flow of positive (+) chargesFriday, July 22, 2011
  • 3. Current Convention : Current depicts flow of positive (+) charges Area +Friday, July 22, 2011
  • 4. Current Convention : Current depicts flow of positive (+) charges Area + Ammeter (measures current)Friday, July 22, 2011
  • 5. Current Convention : Current depicts flow of positive (+) charges Area + + + Ammeter (measures current)Friday, July 22, 2011
  • 6. Current Convention : Current depicts flow of positive (+) charges Area + + + Ammeter (measures current)Friday, July 22, 2011
  • 7. Current A measure of how much charge passes through an amount of time + + + Ammeter (measures current)Friday, July 22, 2011
  • 8. Current Count how many charges flow through + + +Friday, July 22, 2011
  • 9. Current Count how many charges flow through Expand surface to a volume + + +Friday, July 22, 2011
  • 10. Current Count how many charges flow through Expand surface to a volume + + + Area = AFriday, July 22, 2011
  • 11. Current Count how many charges flow through Expand surface to a volume + + + Area = A length = !xFriday, July 22, 2011
  • 12. Current Count how many charges flow through Expand surface to a volume + + Total volume V = (A)(!x) + Area = A length = !xFriday, July 22, 2011
  • 13. Current Count how many charges flow through Expand surface to a volume + + Total volume V = (A)(!x) + Area = A length = !x Number of charges = (charge density or charge per volume)*(volume)Number of charges = (n) * (A!x)Friday, July 22, 2011
  • 14. Current Count how many charges flow through Expand surface to a volume + + Total volume V = (A)(!x) + Area = A length = !x Number of charges = (charge density or charge per volume)*(volume)Number of charges = (n) * (A!x) Total amount of charge = (number of charges)*(charge) !Q = (n A !x)*(q)Friday, July 22, 2011
  • 15. Current !Q = (n A !x)*(q) + + Total volume V = (A)(!x) + Area = A length = !xFriday, July 22, 2011
  • 16. Current !Q = (n A !x)*(q) but charges have drift velocity vd = !x/!t + + Total volume V = (A)(!x) + Area = A length = !x = vd !tFriday, July 22, 2011
  • 17. Current !Q = (n A !x)*(q) but charges have drift velocity vd = !x/!t + + Total volume V = (A)(!x) + Area = A length = !x = vd !t !Q = (n A vd !t)*(q)Friday, July 22, 2011
  • 18. Current !Q = (n A !x)*(q) but charges have drift velocity vd = !x/!t + + Total volume V = (A)(!x) + Area = A length = !x = vd !t !Q = (n A vd !t)*(q) !Q/!t = (n A vd)*(q) I = n q vd AFriday, July 22, 2011
  • 19. Current This is the reason why large wires are needed to support large currentsFriday, July 22, 2011
  • 20. Current This is the reason why large wires are needed to support large currentsFriday, July 22, 2011
  • 21. Resistance Current density (J) current per areaFriday, July 22, 2011
  • 22. Resistance Current density (J) current per area Direction of current (flow of positive charges) is same with direction of electric fieldFriday, July 22, 2011
  • 23. Resistance Current density (J) current per area Direction of current (flow of positive charges) is same with direction of electric field conductivityFriday, July 22, 2011
  • 24. Resistance Current density (J) current per area Direction of current (flow of positive charges) is same with direction of electric field conductivity (material property) resistivity (material property)Friday, July 22, 2011
  • 25. Resistance Current density (J) current per area Direction of current (flow of positive charges) is same with direction of electric field conductivity resistivity Current is proportional to conductivity but inversely proportional to resistivity!Friday, July 22, 2011
  • 26. Resistance Current is proportional to conductivity but inversely proportional to resistivity!Friday, July 22, 2011
  • 27. Resistance Current is proportional to conductivity but inversely proportional to resistivity! Current is proportional to the electric potential (specifically potential difference)Friday, July 22, 2011
  • 28. Resistance Current is proportional to conductivity but inversely proportional to resistivity! Current is proportional to the electric potential (specifically potential difference) Ohm’s Law Potential difference Resistance currentFriday, July 22, 2011
  • 29. Resistance Current is proportional to conductivity but inversely proportional to resistivity! Current is proportional to the electric potential (specifically potential difference) Ohm’s Law Potential difference Resistance current a much better form than ΔV = I RFriday, July 22, 2011
  • 30. Resistance Current is proportional to conductivity but inversely proportional to resistivity! Current is proportional to the electric potential (specifically potential difference) Ohm’s Law Potential difference Resistance current a much better form Increasing !V increases I than ΔV = I R Increasing R decreases IFriday, July 22, 2011
  • 31. Resistance Current is proportional to conductivity but inversely proportional to resistivity! Current is proportional to the electric potential (specifically potential difference) Ohm’s Law Potential difference Resistance current a much better form Increasing !V increases I than ΔV = I R Increasing R decreases I !V = I R Increasing R does not increase !V Current (I) is increased because !V is increasedFriday, July 22, 2011
  • 32. ResistanceFriday, July 22, 2011
  • 33. Resistance Important points: same with capacitance, resistance does not depend on !V and I Resistance depends on material property resistivity ", length of wire l and cross sectional area A conventional current is flowing positive (+) charges though in reality electrons flow direction of the current I is same as direction of electric fieldFriday, July 22, 2011
  • 34. Recent Equations → → →E J = σE = ρ → → J = nq v d A → → I J = A ∆V I= R ρl R= AFriday, July 22, 2011
  • 35. Exercise Rank from lowest to highest amount of current Derive the equation R = "L/A from V = IR, J = E/" = I/A, V = ELFriday, July 22, 2011
  • 36. Resistance and Temperature ρl R= A ρ = ρ0 (1 + α∆T ) ∆T = T − T0 T0 is usually taken to be 25 °C T ↑ ρ↑Friday, July 22, 2011
  • 37. Power ∆U P = ∆t ∆(q∆V ) P = ∆t (∆q)(∆V ) P = ∆t ∆q P = ∆V ∆t P = I∆VFriday, July 22, 2011
  • 38. Power P = I∆V ∆V I= R V2 P = P = I 2R RFriday, July 22, 2011
  • 39. Exercises The electron beam emerging from a certain high-energy electron accelerator has a circular cross section of radius 1.00 mm. (a) The beam current is 8.00 µA. Find the current density in the beam, assuming that it is uniform throughout. (b) The speed of the electrons is so close to the speed of light that their speed can be taken as c = 3.00 x 108 m/s with negligible error. Find the electron density in the beam. (c) How long does it take for Avogadroʼs number of electrons to emerge from the accelerator? An aluminum wire having a cross-sectional area of 4.00 x 10-6 m2 carries a current of 5.00 A. Find the drift speed of the electrons in the wire. The density of aluminum is 2700 kg/m3. Assume that one conduction electron is supplied by each atom. Molar mass of Al is 27 g/mol. Four wires A, B, C and D are made of the same material but of different lengths and radii. Wire A has length L but has radius R. Wire B has length 2L but with radius ½R. Wire C has length ½L but with radius 2R. Wire D has length ½L but with radius ½R. Rank with increasing resistance A 0.900-V potential difference is maintained across a 1.50-m length of tungsten wire that has a cross-sectional area of 0.600 mm2. What is the current in the wire? resistivity of tungsten is 5.6 x 10-8 Ω-mFriday, July 22, 2011
  • 40. Exercises An electric heater is constructed by applying a potential difference of 120 V to a Nichrome wire that has a total resistance of 8.00 Ω. Find the current carried by the wire and the power rating of the heater. A 500-W heating coil designed to operate from 110 V is made of Nichrome wire 0.500 mm in diameter. (a) Assuming that the resistivity of the Nichrome remains constant at its 20.0°C value, find the length of wire used. (b) What If? Now consider the variation of resistivity with temperature. What power will the coil of part (a) actually deliver when it is heated to 1200°C? ρ = 1.50 x 10-6 Ω-mFriday, July 22, 2011
  • 41. More exercises If the magnitude of the drift velocity of free electrons in a copper wire is 7.84 x 10-4 m/s, what is the electric field in the conductor? A certain lightbulb has a tungsten filament with a resistance of 19.0 Ω when cold and 140 Ω when hot. Assume that the resistivity of tungsten varies linearly with temperature even over the large temperature range involved here, and find the temperature of the hot filament. Assume the initial temperature is 20.0°C. 4.5 x 10-3 C-1 The cost of electricity varies widely through the United States; $0.120/kWh is one typical value. At this unit price, calculate the cost of (a) leaving a 40.0-W porch light on for two weeks while you are on vacation, (b) making a piece of dark toast in 3.00 min with a 970-W toaster, and (c) drying a load of clothes in 40.0 min in a 5 200-W dryer.Friday, July 22, 2011
  • 42. Electromotive Force The electromotive force is denoted as “ε” A force that moves charges The emf ε is the maximum possible voltage that the battery can provide. ε = ∆V in batteriesDirect current - current that is constant in direction and magnitudeFriday, July 22, 2011
  • 43. Resistors in Series ∆V Recall: I= R use the equation to calculate the equivalent resistance ReqFriday, July 22, 2011
  • 44. Resistors in Series Convert to simple equivalent circuitFriday, July 22, 2011
  • 45. Resistors in Series I1 I2 ∆V1 ∆V2 Conservation of matter = Current is conserved I = I1 = I2Friday, July 22, 2011
  • 46. Resistors in Series I1 I2 ∆V1 ∆V2 Conservation of matter = Current is conserved I = I1 = I2 Conservation of energy ∆V = ∆V1 + ∆V2Friday, July 22, 2011
  • 47. Resistors in Series I1 I2 ∆V1 ∆V2 Conservation of matter = Current is conserved Ohms Law I = I1 = I2 ∆V Conservation of energy I= R ∆V = ∆V1 + ∆V2Friday, July 22, 2011
  • 48. Resistors in Series I1 I2 ∆V = I1 R1 + I2 R2 ∆V1 ∆V2 ∆V = IR1 + IR2 ∆V = I(R1 + R2 ) ∆V = IReq Req = R1 + R2 Conservation of matter = Current is conserved Ohms Law I = I1 = I2 ∆V Conservation of energy I= R ∆V = ∆V1 + ∆V2Friday, July 22, 2011
  • 49. Resistors in Series I1 I2 ∆V1 ∆V2 Conservation of matter = Current is conserved Ohms Law I = I1 = I2 ∆V Conservation of energy I= R ∆V = ∆V1 + ∆V2Friday, July 22, 2011
  • 50. Resistors in Parallel 1. Imagine positive charges pass first I1 I2 through R1 and then through%R2. Compared to the current in R1, the current in R2 is ∆V1 ∆V2 (a) smaller (b) larger (c) the same. 2. With the switch in the circuit of closed (left), there is no current in R2, because the current has an alternate zero-resistance path through the switch. There is current in R1 and this current is measured with the ammeter (a device for measuring current) at the right side of the circuit. If the switch is opened (right), there is current in R2. What happens to the reading on the ammeter when the switch is opened? (a) the reading goes up (b) the reading goes down (c) the reading does not change.Friday, July 22, 2011
  • 51. Resistors in Parallel ∆V Recall: I= R use the equation to calculate the equivalent resistance ReqFriday, July 22, 2011
  • 52. Resistors in Parallel Convert to simple equivalent circuitFriday, July 22, 2011
  • 53. Resistors in Parallel I1 ∆V1 I2 ∆V2 Conservation of matter = Current is conserved I = I1 + I2Friday, July 22, 2011
  • 54. Resistors in Parallel I1 ∆V1 I2 ∆V2 Conservation of matter = Current is conserved I = I1 + I2 Conservation of energy ∆V = ∆V1 = ∆V2Friday, July 22, 2011
  • 55. Resistors in Parallel I1 ∆V1 I2 ∆V2 Conservation of matter = Current is conserved Ohms Law I = I1 + I2 ∆V Conservation of energy I= R ∆V = ∆V1 = ∆V2Friday, July 22, 2011
  • 56. Resistors in Parallel I1 ∆V1 I = I1 + I2 ∆V ∆V1 ∆V2 I2 = + ∆V2 R R1 R2 ∆V ∆V ∆V = + R R1 R2 1 1 1 = + R R1 R2 Conservation of matter = Current is conserved Ohms Law I = I1 + I2 ∆V Conservation of energy I= R ∆V = ∆V1 = ∆V2Friday, July 22, 2011
  • 57. Resistors in Parallel I1 ∆V1 I2 ∆V2 Conservation of matter = Current is conserved Ohms Law I = I1 + I2 ∆V Conservation of energy I= R ∆V = ∆V1 = ∆V2Friday, July 22, 2011
  • 58. Recall: Ohms Law Capacitance ∆V I= Q = C∆V R Series ParallelFriday, July 22, 2011
  • 59. Exercise Find the current passing through each resistor Find the voltage drop (potential difference) through each resistorFriday, July 22, 2011
  • 60. Kirchhoff’s Rules Junction Rule “conservation of matter” Loop Rule “conservation of energy” Σ ∆V = 0 closed loopFriday, July 22, 2011
  • 61. Exercise In solving complicated circuit problems apply Junction rule first (conservation of current) You may assign any direction of current as long as it is reasonable (does not violate common sense!) A Then apply the loop rule B Write down the equations for loop rules concerning loop A, B, C and the outer loop of the circuit following C clockwise direction. (there must be four equations!)Friday, July 22, 2011

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