Current, Resistance Electric current and current density

Transcription

1 General Physics Current, Resistance We will now look at the situation where charges are in otion - electrodynaics. The ajor difference between the static and dynaic cases is that E = 0 inside conductors for the static case, but E 0 (i.e when a potential applied to the conductor) inside conductors for the dynaic case. nd, if E 0, then charges in the conductor feel a force (F = qe) and ove in response to that force. n the static case E=0, there is no charge inside the conductor. ll charges are distributed over the surface of the conductor. There exists a theral otion, and the average position of the charges does not change (even though electrons will be oving at approxiately 0 5 /s between collisions when the teperature is 300 K.) n the dynaic case: E 0: When the electric field is present, the oveent of the charges caused by the electric field is superiposed on the theral otion, and there is a net otion of the charges (in the direction of the force exerted on the charges due to the electric field in the sae direction as the E-field for positive charges and in the opposite direction of V the E-field for negative charges). The average speed at which the charges ove is called the drift velocity, v d. t is this flow of charge that we will study. The otion of electrons (negative charge) in a particular direction can be replaced equivalently by the otion of positive charges in the opposite direction. n our discussion of charge otion we will deal priarily with otion of positive charges. ny device that supplies the energy to cause the charges to separate is referred to as an EMF (electrootive force). The EMF produces the necessary electric field to cause the charges to ove, e.g., battery (cheical energy), generator (echanical energy), etc. The electric field produced by the battery causes the charges to ove, and this otion of charges is called an electric current. Electric current and current density agine a section of the wire above, with a cross-sectional area and with charges flowing with a velocity v d. The direction of current flow is taken as the direction in which positive charges flow (even though in wires the negative charges, electrons, are the ones flowing). The electric current is defined as the aount of charge crossing an iaginary boundary in the wire per unit tie: aount of charge crossing a boundary electric current = tie t is atheatically defines as dq = dt [ The units of current are: []= Q ] [] t = Coulob pere = p = second Departent of Engineering Physics

2 General Physics The value of the electric current in a wire is the sae no atter how the cross-sectional area of the wire ight change. nother quantity that is closely related to current and does depend on the cross sectional area of the wire is current density J. The current density is the electric current per cross-sectional area, that is, J = d d When the current density is constant: J=/. Now let's ask the seeingly harless question, how fast are the electrons going? d d l l v d E J Positive charges v d E J Negative charges The speed of the electrons can be written as, υ = t. The tie can be found using the definition of current as in exaple, dq Ne t dq = dt dt 0 Ne = t t = Ne. 0 J Now the speed becoes, vd = = = Ne N e ne Define the free electron density and the current density, n N The Definition of Free Electron Density vol Exaple What is the drift velocity of the electrons in a 2 diaeter copper wire carrying a current of? Take n = 8.5 x 0 28 electrons per cubic eter for copper. Before aking the calculation, what is your guess? J = / = /( π 2 0 ) = / ; vd = J ne= = / /(8.85*0 *.6*0 ) (.8*0 )/s This velocity is so sall that it is hard to understand why a light bulb goes on alost instantly when the switch is flipped. There is soething we haven't accounted for in our odel of charge flow in conductors. Departent of Engineering Physics 2

3 General Physics Resistance and Resistivity ccording to this odel of current flow the charge will have a constant acceleration. Using Newton's Second Law, F = a a = F = qe = ev. f the acceleration is constant, the speed will increase indefinitely. This contradicts the concept of a drift velocity. Because collisions take place as the electrons ove through the wires, we can say that the wires are producing an ipeding effect on the flow of those charges. This ipeding effect is called electrical resistance. We ust include collisions between the electrons and atos. f we call τ the average tie between collisions, the average velocity of the electrons will roughly be υ aτ. Using the acceleration above and the equation for drift velocity, j ne = ev τ ne = ev τ V = τne 2. Define the resistivity as, ρ τne 2 The Definition of Resistivity Notice that ρ is dependent on icroscopic properties of the conducting aterial. The resistivity is difficult to calculate fro these nubers, but it is not hard to easure. You will find tables of resistivity values. Soeties the conductivity is tabulated. Conductivity, σ, is the reciprocal of the resistivity, σ ρ. Now, V = ρ. Define resistance as, R ρ The Definition of Resistance Resistance takes into account the geoetrical properties of the aterial. Finally we have Oh's Rule: V = R Oh's Rule The units of resistance are: [ [ R]= V ] [] = Volt p Oh = Ω Soe values of resistivity, ρ, and conductivity, σ = /ρ. copper: ρ =.7 x 0-8 oh σ = 5.9 x 0 7 (oh ) - carbon: ρ = 3.5 x 0-5 oh σ = 2.9 x 0 4 (oh ) - glass: ρ 0 2 oh σ 0-2 (oh ) - We notice that the resistivity depends on teperature. How does the resistivity of various types of aterials change when the teperature is changed? Why? When the teperature of the aterial is increased then the charges in the aterial gains the energy. For the teperature dependence of the resistivity an approxiate epirical expression can be written as: ρ = ρ + α T T ( ( )) 0 0 here ρ 0 is resistivity of the aterial at the reference teperature T 0. The quantity α is the teperature coefficient of the resistivity. Exaple Departent of Engineering Physics 3

4 General Physics Two conductors of the sae aterial and length have different resistances. Conductor is a solid.00 diaeter wire. Conductor B is a tube of inner diaeter.00 and outer diaeter Find the ratio of the resistances of conductor to conductor B. Fro the definition of resistance, R ρ and R B ρ Conductor. B The ratio is ρ R R B ρ = B = π(d D 2 ) πd = 22 2 = B Conductor B D D D 2 Oh s Law Our ai here to understand why etals obey Oh s Law. We know that acroscopic relation is given by V = R without further discussion we use the relations V=E.l and the relations between resistance and resistivity and definition of the current density we obtain E = ρj We can say that at constant teperature the resistivity of the aterial is independent to the applied electric field E.(Oh s law, icroscopic view) We can calculate drift speed inters of the applied field E. Fron Newton s second law: ee a = Consider an electron that has just undergone a typical effective collision. n the average tie interval τ to the next collision, the electron will change its velocity in the direction of E by an aount of aτ : ee vd = aτ = τ Then we can obtain another relation ee J v = d τ = n e Reeber E = ρ J, leads to ρ = 2 ne τ This is the icroscopic Oh s law. Note that τ is independent fro electric filed. Exercises n electrical circuits, E and J are not easured, but V (voltage) and (current) are. Look at the relationship between these quantities. Consider a length of the wire where an electric field E and the resulting current density J are present. Find the potential difference between points and B, that is, find V B = V V B. E B Departent of Engineering Physics 4

5 General Physics Define resistance. What is the unit of the resistance. Take a length of copper wire with a diaeter of 2 carrying a current of 2 at a teperature of 20 o C. What is the resistance of the wire, and what is the voltage between the ends of the wire? Suppose the teperature of the above wire is increased to 220 o C. What is the new resistance? Can you find the resistance of a truncated cone ade fro a aterial whose resistivity is ρ, whose radii are a and b, and whose length is L? a b L Energy Transfer in Circuits The work required to ove a charge dq through a potential difference V is dw = (dq)v. Then, the power needed to accoplish this is Power = P = dw dq dq = V, since =, Power P = V. dt dt dt This is an expression that can be used for any circuit eleent. The units are aps.volts ( v) = Watt (W). Using Oh's Rule, power can be written without V or without : Exaple P = V = V 2 R = 2 R Electric Power The extension cord of exaple 3 is connected to a 0V source. Find the (a)power supplied by the source, (b)power lost in the cord and (c)power supplied to the load. (a)the electrical power supplied is P = V = (25.0)(0V) = 2750W. (b)the power lost in the cord can be found fro the voltage drop, P = V = (25.0)(8.34V) = 209W. The resistance of the cord could be used instead, P = 2 R = (25.0) 2 (0.334Ω) = 209W. (c)by the Law of Conservation of Energy, P sup ply = P cord + P load P load = P sup ply P cord = = 2540W. Do you know why extension cords are rated for axiu length? Departent of Engineering Physics 5

6 General Physics Explain why power lines use high voltage instead of high current. Suary The Definition of Current dq dt The Definition of Free Electron Density n N vol The Definition of Current Density J = Drift Velocity J = nev d The Definition of Resistivity ρ τne 2 The definition of conductivity σ = ρ The Definition of Resistance R ρ Oh's Rule V = R Electric Power P = V = V 2 R = 2 R Departent of Engineering Physics 6