2. Current Convention : Current depicts flow of positive (+) chargesSunday, July 24, 2011
3. Current Convention : Current depicts flow of positive (+) charges Area +Sunday, July 24, 2011
4. Current Convention : Current depicts flow of positive (+) charges Area + Ammeter (measures current)Sunday, July 24, 2011
5. Current Convention : Current depicts flow of positive (+) charges Area + + + Ammeter (measures current)Sunday, July 24, 2011
6. Current Convention : Current depicts flow of positive (+) charges Area + + + Ammeter (measures current)Sunday, July 24, 2011
7. Current A measure of how much charge passes through an amount of time + + + Ammeter (measures current)Sunday, July 24, 2011
8. Current Count how many charges flow through + + +Sunday, July 24, 2011
9. Current Count how many charges flow through Expand surface to a volume + + +Sunday, July 24, 2011
10. Current Count how many charges flow through Expand surface to a volume + + + Area = ASunday, July 24, 2011
11. Current Count how many charges flow through Expand surface to a volume + + + Area = A length = !xSunday, July 24, 2011
12. Current Count how many charges flow through Expand surface to a volume + + Total volume V = (A)(!x) + Area = A length = !xSunday, July 24, 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)Sunday, July 24, 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)Sunday, July 24, 2011
15. Current !Q = (n A !x)*(q) + + Total volume V = (A)(!x) + Area = A length = !xSunday, July 24, 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 !tSunday, July 24, 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)Sunday, July 24, 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 ASunday, July 24, 2011
19. Current This is the reason why large wires are needed to support large currentsSunday, July 24, 2011
20. Current This is the reason why large wires are needed to support large currentsSunday, July 24, 2011
21. Resistance Current density (J) current per areaSunday, July 24, 2011
22. Resistance Current density (J) current per area Direction of current (flow of positive charges) is same with direction of electric fieldSunday, July 24, 2011
23. Resistance Current density (J) current per area Direction of current (flow of positive charges) is same with direction of electric field conductivitySunday, July 24, 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)Sunday, July 24, 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!Sunday, July 24, 2011
26. Resistance Current is proportional to conductivity but inversely proportional to resistivity!Sunday, July 24, 2011
27. Resistance Current is proportional to conductivity but inversely proportional to resistivity! Current is proportional to the electric potential (specifically potential difference)Sunday, July 24, 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 currentSunday, July 24, 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 RSunday, July 24, 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 ISunday, July 24, 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 increasedSunday, July 24, 2011
32. ResistanceSunday, July 24, 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 fieldSunday, July 24, 2011
34. Recent Equations → → →E J = σE = ρ → → J = nq v d A → → I J = A ∆V I= R ρl R= ASunday, July 24, 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 = ELSunday, July 24, 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↑Sunday, July 24, 2011
37. Power ∆U P = ∆t ∆(q∆V ) P = ∆t (∆q)(∆V ) P = ∆t ∆q P = ∆V ∆t P = I∆VSunday, July 24, 2011
38. Power P = I∆V ∆V I= R V2 P = P = I 2R RSunday, July 24, 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-mSunday, July 24, 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-mSunday, July 24, 2011
41. Sunday, July 24, 2011
42. 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-cmSunday, July 24, 2011
43. More exercises 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.Sunday, July 24, 2011
44. 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-1Sunday, July 24, 2011
45. 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. $0.120 $0.120 1kW 1hour $3.33 × 10−8 = = 1kWh 1kWh 1000W 3600secs 1Joule ∆U ∆U 1week 1day 1hour ∆U(a) P = ∆t = 2weeks 7days 24hours 3600secs = 1209600secs ∆U 40.0W = 1209600s $3.33 × 10−8 ∆U = 48384kJ 4.84 × 107 J = $1.61 1Joule(b) $5.82 × 10− 3(c) $0.416Sunday, July 24, 2011
46. 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 magnitudeSunday, July 24, 2011
47. Resistors in Series ∆V Recall: I= R use the equation to calculate the equivalent resistance ReqSunday, July 24, 2011
48. Resistors in Series Convert to simple equivalent circuitSunday, July 24, 2011
49. Resistors in Series I1 I2 ∆V1 ∆V2 Conservation of matter = Current is conserved I = I1 = I2Sunday, July 24, 2011
50. Resistors in Series I1 I2 ∆V1 ∆V2 Conservation of matter = Current is conserved I = I1 = I2 Conservation of energy ∆V = ∆V1 + ∆V2Sunday, July 24, 2011
51. 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 + ∆V2Sunday, July 24, 2011
52. 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 + ∆V2Sunday, July 24, 2011
53. 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 + ∆V2Sunday, July 24, 2011
54. 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.Sunday, July 24, 2011
55. Resistors in Parallel ∆V Recall: I= R use the equation to calculate the equivalent resistance ReqSunday, July 24, 2011
56. Resistors in Parallel Convert to simple equivalent circuitSunday, July 24, 2011
57. Resistors in Parallel I1 ∆V1 I2 ∆V2 Conservation of matter = Current is conserved I = I1 + I2Sunday, July 24, 2011
58. Resistors in Parallel I1 ∆V1 I2 ∆V2 Conservation of matter = Current is conserved I = I1 + I2 Conservation of energy ∆V = ∆V1 = ∆V2Sunday, July 24, 2011
59. 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 = ∆V2Sunday, July 24, 2011
60. 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 = ∆V2Sunday, July 24, 2011
61. 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 = ∆V2Sunday, July 24, 2011
62. Recall: Ohms Law Capacitance ∆V I= Q = C∆V R Series ParallelSunday, July 24, 2011
63. Exercise Find the current passing through each resistor Find the voltage drop (potential difference) through each resistorSunday, July 24, 2011
64. Kirchhoff’s Rules Junction Rule “conservation of matter” Loop Rule “conservation of energy” Σ ∆V = 0 closed loopSunday, July 24, 2011
65. 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!)Sunday, July 24, 2011
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