同济大学：《大学物理》课程教学资源（电子教案，英文版）第十七章 交流电路

AC circuit HAFONHA EAHKAJYTOCAABHJS 100 MARODHA BANKA JUGOSLAVIE 100

AC circuit

Agenda Today 1.Self-Inductance and Multi Inductance 2.Energy stored in Magnetic Field 3.RL Circuit 4.AC Circuit

Agenda Today 1. Self-Inductance and Multi Inductance 2. Energy stored in Magnetic Field 3. RL Circuit 4. AC Circuit

Self-inductance(自感系数) L=Ψ/I Unit:H According to Faraday's Law: Ψ d(LI) dI 8L= dt dt dt

Self-inductance(自感系数) L  / I Unit: H According to Faraday’s Law: dt dI L dt d LI dt d L        ( ) 

Self-inductance of ideal solenoid Cross section area A length:1 current I number of turns:N permeability:u B=μ1

Self-inductance of ideal solenoid l Cross section area : A length: l current : I number of turns: N permeability:  I l N B  

N2 Ψ=NBS= IS 半专4S=Wherenis画 per unit length Multi induction(互感) Emf caused by the change in current carried by another loop

IS l N NBS 2     S n V l N I L 2 2       Where n is turns per unit length Multi induction（互感） Emf caused by the change in current carried by another loop

Multi inductance(互感系数) 平，=N,Φ21=M2 Y2=NΦ12=M122 M12=M21=M 821=-M dt 612=-M dt

Multi inductance（互感系数）1 2 I1 I2 12 21 21 2 21 21 1   N   M I 12 1 12 12 2   N   M I M12  M21  M dt dI M dt dI M 2 12 1 21      

互感概念演示仪

Tesla coil

Tesla coil

Example:Inside the magnetic material with permeability u is a straight long wire and rectangular loop as the figure shows.find the multi inductance solution B= 2πr dΦ=BdS =41 ldr dr 2元r a+b Φ L= n 2元r 2元 a Φ M= =w1 a+b 2π a

Example: Inside the magnetic material with permeability  is a straight long wire and rectangular loop as the figure shows.find the multi inductance solution r dr a b l I r I B o   2  ldr r I d BdS o   2    a Il a b dr r a b Il a o o       ln 2 2    a l a b I M o     ln 2 