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Current and Resistance

Current and Resistance. Chapter 31. Batteries. Battery. Batteries provide Chemical Electricity Electrons “bunch up” or have the potential to flow from the negative end Electrons can’t flow in an isolated battery. -. +. e e. e e. Chemical Reaction that produces electrons.

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Current and Resistance

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  1. Current and Resistance Chapter 31

  2. Batteries

  3. Battery • Batteries provide Chemical Electricity • Electrons “bunch up” or have the potential to flow from the negative end • Electrons can’t flow in an isolated battery - + e e e e Chemical Reaction that produces electrons Chemical Reaction that absorbs electrons

  4. Circuits

  5. Drift Speed • Electrons do not flow through wires like pipes • Electric field gives direction to the random motion of electrons. (vD = drift speed)

  6. 0.05 mm/s • About 5 ½ hours to travel one meter (coin waterfall at Chuck-E-Cheese) • About a year to go 1 mile Electron Current (ie) ie = neAvD ne = electron density

  7. Calculate the electron current in a 2.0 mm diamter copper wire if the electron drift speed is 1.0 X 10-4 m/s. (2.7 X 1019 s-1)

  8. Conventional Current • Flows positive to negative • Opposite of electron flow (electron current)

  9. Current (I) • Current – Net amount of charge per unit time • 1 coulomb/second = 1Ampere I = DQ Dt I = dQ I = eie electron current dt

  10. The electron current through a wire is 1.2 X 1019 electrons/s. • Calculate the current, I (1.9 A) • Calculate the amount of charge that flows each hour (6800 C)

  11. Current Density (J)

  12. A 1.0 A current passes through a 1.0 mm diameter wire. • Calculate the current density. (1.3 X 106 A/m2) • Calculate the drift speed of the electron. (0.13 mm/s)

  13. A 5.0 A current passes through a 3.2 mm diameter wire. • Calculate the current density. (6.2 X 105 A/m2) • Calculate the drift speed of the electron. (0.05 mm/s)

  14. Current: Ex. 1 A steady current of 2.5 A flows through a wire for 4.0 min. How much charge passed through any point in the circuit? I = DQ Dt DQ = IDt DQ = (2.5 A)(240 s) = 600 C

  15. How many electrons would this be? 1 electron = 1.60 X 10-19 C 600 C 1 electron = 3.8 X 1021electrons 1.60 X 10-19 C

  16. Current Density (J) Conductivity (s) Resistivitiy (r) Current density

  17. A 2.0 mm diameter aluminum wire carries a current of 800 mA. • Calculate the current density using J = I/A (2.55 X 105 A/m2) • Calculate the electric field inside the wire (0.0072 V/m)

  18. A copper wire has a diameter of 3.2 mm. The current is 5.0 A. • Calculate the current density of the wire (6.2 X 105 A/m2) • Calculate the electric field inside the wire (0.01 V/m)

  19. Ohm’s Law V = IR DV = IR V = Voltage (V) I = Current (A) R = Resistance (Ohms, W) (only works for metal conductors, not semiconductors (nonohmic))

  20. Resistors • Color coded to determine resistance • Devices that heat have high resistance (light bulbs, electric stoves, toasters)

  21. A small flashlight bulb draws 300 mA from a 1.5 V battery. • Calculate the resistance of the bulb (5.0 W) • If the voltage dropped to 1.2 V and the resistance stayed at 5.0 W, what current would flow. (0.24 A)

  22. Resistivity R = rL A R = Resistance L = Length (longer wire, greater resistance) A = Area (wider wire, less resistance) r = Resistivity of the material http://www.earthsci.unimelb.edu.au/ES304/MODULES/RES/NOTES/resistivity.html

  23. What is the resistance of a 2.00mm diameter, 10.0 meter copper wire? A = pr2 = (3.14)(0.001 m)2 = 3.14 X 10-6 m2 R = rL = (1.68 X 10-8Wm)(10.0 m) A (3.14 X 10-6 m2) R = 0.0535 W of 53.5 m W

  24. A speaker wire must be 20.0 m long and have a resistance of less than 0.100 W per wire. • What diameter copper wire should be used? (2.06 mm) • What is the voltage drop across each wire at a current of 4.00 A? (0.40 V)

  25. A wire of length L is stretched to twice its normal length. • Calculate the new cross sectional area (assume the volume does not charge (Anew =1/2A) • Calculate the new resistance (Rnew= R)

  26. R = rL A A = rL R A = [(1.68 X 10-8Wm)(20.0 m)]/0.100 W A = 3.36 X 10-6 m2 A = pr2 r = (A/p)1/2 r = (3.36 X 10-6 m2 /3.14)1/2 = 1.03 X 10-3 m D = 2r = 2.06 X 10-3 m or 2.06 mm

  27. Resistance and Temperature • Metals • Resistance increases with temp. • Atoms more disorderly • Interferes with flow of electrons • Semiconductors • Resistance sometimes decreases with temperature • Some electrons become excited and able to flow

  28. Superconductivity • Superconductivity – resistance of a material becomes zero • No loss of current over a wire • Generally near absolute zero • Record as of 2007 is 138 K • Maglev trains

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