Presentation on theme: "Usually a diluted salt solution chemical decomposition"— Presentation transcript:
1 Usually a diluted salt solution chemical decomposition Chapter 5. Steady Electric CurrentsTypes of electric currentsConvection currents result from motion of electrons and/or holes in a vacuum or rarefied gas. (electron beams in a CRT, violent motion of charged particles in a thunder storm). Convection current , the result of hydrodynamic motion involving a mass transport, are not governed by Ohm’s law.Electrolytic current are the result of migration of positive and negative ions.ElectrolyteUsually a diluted salt solutionElectrolysischemical decomposition+ion-ioncurrent- Conduction currents result from drift motion of electrons and/or holes in conductors and semiconductors.Atoms of the conducting medium occupy regular positions in a crystalline structure and do not move.Electrons in the inner shells are tightly bound to the nuclei are not free to move away.Electrons in the outermost shells do not completely fill the shells; they are valence or conduction electrons and are only very loosely bound to the nuclei.When an external E field is applied, an organized motion of the conduction electrons will result (e.g. electron current in a metal wire).The average drift velocity of the electrons is very low (10-4~10-3 m/s) because of collision with the atoms, dissipating part of their kinetic energy as a heat.
2 The total current flowing through an arbitrary surface S is Steady current densityElectric currents : motion of free chargesCurrent density : current per unit areaConsider a tube of charge with volume charge density vmoving with a mean velocity u along the axis of the tube.Over a period t, the charges move a distance l = u t.The amount of charge that crosses the tube's cross-sectionalsurface s' in time t is thereforeIf the charges are flowing through a surface s whose surfacenormal is not necessarily parallel to u,: (volume) current densityThe total current flowing through an arbitrary surface S is
3 For semiconductors : cf) resistivity : In vacuum,Example 5-1.In conductors and semiconductors, electrons and/or holes can not be accelerated due to the collision.The drift velocity is proportional to the applied E field.For metal,For semiconductors : cf) resistivity : Ohmic media : material following Ohm's law: point form of Ohm's law σ: conductivity (S/m)The resistance of a material having a straight length , uniformcross section area S, and conductivity :
4 Electromotive force (emf) Static (conservative) electric field :This equation tells us that a steady current cannot be maintained in the same direction in a closed circuit by an electrostatic field (Charge carriers collide with atoms and therefore dissipate energy in the circuit). This energy must come from a nonconservative field source for continuous current flow (e.g. battery, generator, thermocouples, photovoltaic cells, fuel cells, etc.). These energy sources, when connected in an electric circuit, provide a driving force to push a current in a circuit : impressed electric field intensity Ei .EMF of a battery : the line integral of the impressed field intensity Ei from the negative to the positive electrode inside the battery.For an ohmic material :Insidethe sourceOutsidethe sourceInsidethe sourceCurrent flows from (-) to (+) inside source!
5 Kirchhoff's voltage law When a resistor is connected between terminal 1 and 2 of the battery to complete the circuit : the total electric field intensity (E + Ei) must be used in the point form of Ohm's law.If the resistor has a conductivity , length , and uniform cross section S, J = I / S.Kirchhoff's voltage law : Around a closed path in an electric circuit, the algebraic sum of the emf’s (voltage rises) is equal to the algebraic sum of the voltage drops across the resistances.R
6 Charge relaxation : Equation of continuity Equation of continuity and Kirchhoff's current lawPrinciple of conservation of charge : If a net current I flows across the surface out of (into) the region, the charge in the volume must decrease (increase) at a rate that equals the current.Kirchhoff's current law : Algebraic sum of all the dc currents flowing out of (into) a junction in an electric circuit is zero.: Equation of continuityFor steady currents, and thereforeFor ac currents, and thereforeReally?Quasi-static case(at low frequency =0)Charge relaxation For a good conductor(e.g. copper), = 1.5210-19 [s]: relaxation timedecay to 1/e (36.8% value)
7 Power dissipation and Joule's law Power dissipated in a conducting medium in the presence of an electrostatic field EMicroscopically, electrons in the conducting medium moving under the influence of an electric field collide with atoms or lattice sites Energy is thus transmitted from the electric field to the atoms in thermal vibration.The work W done by an electric field E in moving a charge q a distance isFor a given volume V, the total electric power converted into heat isIn a conductor of a constant cross section, , with measured in the direction J.Since V = RI,Power density under steady-current conditionsJoule’s law
8 Boundary conditions for current density For steady current density J in the absence of nonconservative energy sources(1) Normal component : the normal component of a divergenceless vector field is continuous.(2) Tangential component : the tangential component of a curl-free vector field is continuous across an interface.[HW] Solve Example 5-3.Differential formIntegral formThe ratio of the tangential components of J at two sidesof an interface is equal to the ratio of the conductivities.
9 Resistance calculations We have calculated the resistance of a conducting medium having a straight length, uniform cross section area S, and conductivity .This equation can not be used if the S of the conductor is not uniform How can we calculate the resistance?Procedures for resistance calculation(1) Choose an appropriate coordinate system for the given geometry.(2) Assume a potential difference V0 between the conductor material.(3) Find E within the conductor (by solving Laplace's equation and taking ).(5) Find resistance R by taking the ratio V0 / I.(4) Find the total current I from
10 Example 5-6 : Resistance of a conducting flat circular washer Sol.(1) Choose a coordinate system : CCS(2) Assume a potential difference V0.(3) Find E .Boundary conditions are :(4) Find the total current I.(5) Find R.+V0-(at = /2 surface)