Chapter 27 Current and esistance 27.1 Electric Current Electric current: dq dt, unit: ampere 1A = 1C s The rate at which charge flows through a surface. No longer have static equilibrium. E and Q can 0 inside a conductor! The direction of the current is defined as the direction of the flow of positive charge, which is opposite the direction of flow of electrons. Microscopic View of Electric Current Within a time t, the number of charge canflow through a surface is: Volume charge density = A(v d t) n v d : the drift speed (average speed) of charge carriers. n: the density of charge carriers (Carriers move in a zigzag fashion due to collision with atoms.) Total charge: Q = A(v d t) n q q: the charge on each charge carrier. Q Thus, = lim t0 t = nqv d A Example: Drift Speed in a Copper Wire A copper wire of cross-sectional area 3 10-6 m 2, carries a current of 10A. The density of copper is 8.95 g/cm 3, and atomic mass is 63.5 g/mol. Find v d Solution: v d = nqa Need to find n =? charge/ m 3 1 m 3 copper weights 1 m 3 8.95 g/cm 3 =8.95 10 6 g t contains 8.95 10 6 g/63.5 (g/mol) = 1.41 10 5 mol of copper. t contains (1.41 10 5 mol)(6.02 10 23 ) = 8.48 10 28 atoms Assume each copper atom contributes one free electron. Then: n = 8.48 10 28 electrons/ m 3 v d = nqa = 10C / s (8.48 10 28 m 3 )(1.6 10 19 C)(3 10 6 m 2 ) = 2.46 10-4 m/s = 0.246 mm/s t is slow! For electron to drift 1 meter long copper wire: 1000 mm/ 0.246 (mm/s) = 4065 s = 67 minutes Q: Why a light turn on so fast when a switch is thrown?
27.2 esistance and Ohm s Law Define current density : current per unit area J / A = nqv d (A/m 2 ) The current density is proportional to the electric field: J = E, Where, is called the conductivity of the conductor. Also: = 1/ is the esistivity of a conductor Ohm s Law: many materials follows: J = E, and is a constant. Ohmic: Materials obey Ohm s law, such as most metals. Nonohmic: Materials do not obey Ohm s law Since J = E and V =EL, V = EL = J L L = ( JA) A = L A = V = = L A is called resistance of a conductor Unit for : ohm (), 1 1V 1A Unit for : ( m) Material esistivity Temperature Coefficient ( m) [( C) -1 ] Silver 1.59 10-8 3.8 10-3 Gold 2.44 10-8 3.4 10-3 Copper 1.7 10-8 3.9 10-3 Carbon 3.5 10-5 -0.5 10-3 Silicon 640-75 10-3 Glass 10 10 ~ 10 14 Example: A copper wire, 10 m long, A=1.0 10-7 m 2. (a) Calculate the resistance: = L A = (1.7 10m 108 m) 110 7 m 2 =1.7 (b) f a potential difference of 1 V is maintained across the wire, what is the current in the wire? = V = 1V = 0.59A = 590mA 1.7 27.3 A Classical Model for electrical conduction Electrons are treated as molecules of gas. When E = 0, average speed of electron = 0. When E 0, the electrons move with a drift velocity v d. Acceleration of the electron: a = qe m Between collisions, the speed of electron:
v = v0 + at = v0 + qe m t Take an average: v d = qe m Where, is the average time interval between collisions. = 0 [1+(T T 0 )] 0 is the resistivity at the reference temperature T 0 (usually 20 C). is the tempereature coefficient of resistivity. (Unit 1/ C). Current density: J = nqv d = nq 2 E = E m Therefore, = nq2 m, and = 1 = m nq 2 However, calculated v d is about 10x slower than the true value, and predicted (T) is also not correct. The correct description can only be given using quantum mechanics. 27.4 esistance and Temperature metal T silicon T > 0 < 0 Since, similarly: = 0 [1+ (T T 0 )] Example: Nichrome wire has = 0.4 10-3 / C, it is initially at 20 C. Calculate the temperature at which its resistance is doubled. Solution: Since > 0, increases with temperature, Let = 2 0, then 2 0 = 0 [1+ (T T 0 )] (T T 0 ) = 1 = 1 3 C = 2500 C 0.4 10 T = 2500 C + T 0 = 2500 C + 20 C = 2520 C esistivity can be approximated by:
27.5 Superconductors A class of material whose resistance goes to zero below a certain temperature, T c. (critical tempereature). () Hg 4.2 T(K) (b) Steady currents can persist in a superconducting loop for years without any applied voltage. (b) Applications: superconducting magnets,.. (b) Current active research: high-temperature superconductors 27.6 Electrical Energy and Power (c) When a charge Q moves through a resistor with potential difference of V, the electrical potential energy decreases by VQ. (d) f the charge is in vacuum, VQ will be converted to the kinetic energy. (quiz 2 problem) (e) n a resistor, VQ become thermal energy, not kinetic energy, due to collision of electrons with atoms. This energy heats up the resistor. The rate at which the charge Q loses potential energy: U = VQ = V t t The power P dissipated in a resistor: P = V Unit: watt. 1W = 1V1A + _ Using, V =, P can also be written as: p = V = 2 = V 2
Example: A potential difference of 110 V is applied to a Nichrome-wire electric heater of = 8. (a)find the current. = V = 110V 8 = 13.8A (b) Find the power rating of the heater. + _ P = 2 = (13.8 A) 2 (8 )= 1.52 kw Example: A lightbulb is rated at 120V/75W. (a) Find the current in the bulb = P V = 75W 120V = 0.625A (b) The resistance of the bulb Ohm s law, = V = 120V 0.625 = 192 Example: Calculate the cost to burn a 100-W lightbulb for 24 h if electricity costs 8 cents per kilowatt hour (kwh). Energy consumed = (0.1 kw)(24 h) = 2.4 kwh Cost = (2.4 kwh)($ 0.08 / kwh) = $ 0.19 Example: A charge of 100,000 C flows through a 60W lightbulb. f the price of electricity is 8 /kwh, what is the cost? Energy consumed = W = qv =110 5 C 120V = 1.210 7 J = 1.210 7 (kwh)/3.610 6 = 3.3 kwh Cost = 3.3 kwh 8 /kwh = 26.4