资料课件讲义物理课件ch26-27

1、1,Current, Resistance, and Electromotive Force 电流，电阻和电动力 Direct-Current Circuits 直流电路,几个问题,什么是电流？电流有什么作用？ 电源的作用是什么?,2,导体中的电子运动模型,理想气体分子的运动模型 电子的运动速度106m/s 电子的漂移速度 （理想气体分子的自由程）,Drift Velocity 漂移速度,一 电流,电流为通过截面S 的电荷随时间的变化率,为电子的漂移速度大小,单位: 1A,二 电流密度,大小规定：等于在单位时间内过该点附近垂直于正电荷运动方向的单位面积的电荷,一 电阻率,电阻率,一段电路的欧姆

2、定律,电阻定律,电阻率（电导率）不但与材料的种类有关，而且还和温度有关 . 一般金属在温度不太低时,二 超导体,有些金属和化合物在降到接近绝对零度时，它们的电阻率突然减小到零，这种现象叫超导.,汞在4.2K附近电阻突然降为零,欧姆定律的 微分形式,三 欧姆定律的微分形式,一般金属或电解液，欧姆定律在相当大的电压范围内是成立的， 但对于许多导体或半导体，欧姆定律不成立，这种非欧姆导电特性有很大的实际意义，在电子技术，电子计算机技术等现代技术中有重要作用.,欧姆定律的 微分形式,表明任一点的电流密度 与电场强度 方向相同，大小成正比,例1 一内、外半径分别为 和 的金属圆筒，长度 , 其电阻率 ，

3、若筒内外电势差为 ，且筒内缘电势高，圆柱体中径向的电流强度为多少 ？,解法二,一 电源 电动势（electromotive force，emf）,恒 定 电 流,三 稳恒电流,若闭合曲面 S 内的电荷不随时间而变化，有,单位时间内通过闭合曲面向外流出的电荷，等于此时间内闭合曲面里电荷的减少量 .,1）在恒定电流情况下，导体中电荷分布不随时间变化形成恒定电场； 2）恒定电场与静电场具有相似性质（高斯定理和环路定理），恒定电场可引入电势的概念； 3）恒定电场的存在伴随能量的转换.,恒 定 电 流,非静电力: 能不断分离正负电荷使正电荷逆静电场力方向运动.,电源：提供非静电力的装置.,非静电电场强度

4、 : 为单位正电荷所受的非静电力.,电动势的定义：单位正电荷绕闭合回路运动一周，非静电力所做的功.,电动势,电源电动势大小等于将单位正电荷从负极经电源内部移至正极时非静电力所作的功.,电源电动势,从点A出发 , 顺时针 绕行一周各部分电势降 落总和为零 , 即,全电路的欧姆定律,例 已知 =2V =4V 2 = 6 . 求 (1) I ？（2）A，B，C 相邻两点电势降？并作图表示.,解（1）回路电势降落之和为零,得,（2）,21,电流的稳恒条件和导电规律,必有 I = 0,稳恒情况必有 I入= I出,基尔霍夫第一定律,规定从节点流出： I 0 ，,流入节点：I r = 2 W,Chargin

5、g a battery Example 26.4,Junction rule at “a”: 2A + 1A = I or +2 + 1 I = 0 I = 3 Amps Loop rule starting at “a” around (2) -E + 1A(1W) 2A(3W) = 0 = EMF (E ) = -5V Negative value for EMF = Battery should be “flipped”,Charging a battery Example 26.4,Junction rule at “a”: 2A + 1A = I or +2 + 1 I = 0 I

6、= 3 Amps Loop rule starting at “a” around (3) +12 V 3A(2W) 1A(1) +E = 0 = E = -5V (again!) Check your values with third loop!,Charging a battery (cont.) Example 26.5,What is the power delivered by the 12V power supply, and by the battery being recharged? What is power dissipated in each resistor?,Ch

7、arging a battery (cont.) Example 26.5,What is the power delivered by the 12V power supply, and by the battery being recharged? Psupplied = EMF x Current = 12 V x 3 Amps = 36 Watts Pdissipated in supply = i2r = (3Amps)2 x 2W = 18W Net Power = 36 18 = 18 Watts,Charging a battery (cont.) Example 26.5,W

8、hat is the power delivered by the 12V power supply, and by the battery being recharged? PEMF = E x Current = -5 V x 1 Amps = -5 Watts Negative = power not provided power is being stored!,Charging a battery (cont.) Example 26.5,What is power dissipated in each resistor? Pdissipated in battery = i2r =

9、 (1Amps)2 x 1W = 1W Pdissipated in bulb = i2r = (2Amps)2 x 3W = 12W,Charging a battery (cont.) Example 26.5,Total Power: +36W from supply - 18 W to its internal resistance r - 5 W to charge dead battery - 1 W to dead batterys internal resistance - 12 W to indicator light.,Find Current in each resist

10、or! Find equivalent R!,1W,1W,1W,2W,1W,13 V,+,Step 1: Junction Rule! Define current directions and labels,1W,1W,1W,2W,1W,13 V,+,Step 1: Junction Rule! Define current directions and labels,1W,1W,1W,2W,1W,13 V,+,NOTE for Junction Rule! Actual directions of current may differ, but value of current deriv

11、ed is correct!,Step 1: Junction Rule! Define current directions and labels,I2,13 V,+,I3,I4,I5,I6,I1,Step 1: Junction Rule! Define current directions and labels,I2,13 V,+,I3,I4,I5,I6,I1,I1 - I2 - I3 = 0 or I1 = I2 + I3,Step 1: Junction Rule! Define current directions and labels,I2,13 V,+,I3,I4,I5,I6,

12、I1,I2 I5 I4 = 0 or I2 = I4 + I5,Step 1: Junction Rule! Define current directions and labels,I2,13 V,+,I3,I4,I5,I6,I1,I4 + I3 I6 = 0 or I6 = I4 + I3,A complex network Example 26.6,Step 1: Junction Rule! Define current directions and labels,I2,13 V,+,I3,I4,I5,I6,I1,I5 + I6 I1 = 0 or I1 = I5 + I6,I1,St

13、ep 1: Junction Rule! Define current directions and labels,I1,13 V,+,I2,I3,I1 I3,I2 + I3,I1 + I2,NOTE for Junction Rule! How you divide current doesnt matter, but it can simplify solution steps,Step 1: Junction Rule! Define current directions and labels,13 V,Lots of ways to do this none is necessaril

14、y better than another. Direction WILL affect final signs in your answer.,Step 2: Loop Rule! Define loop directions and labels,1W,1W,1W,2W,1W,13 V,+,Step 2: Loop Rule! Define loop directions and labels,1W,1W,1W,2W,1W,13 V,+,Note: Loop Rule! Loop directions do NOT have to be in any particular directio

15、n nor order!,Step 2: Loop Rule! Define loop directions and labels,1W,1W,1W,2W,1W,13 V,+,Loop 1,Step 2: Loop Rule! Define loop directions and labels,1W,1W,1W,2W,1W,13 V,+,Loop 2,Step 2: Loop Rule! Define loop directions and labels,1W,1W,1W,2W,1W,13 V,+,Loop 3,Step 2: Loop Rule! Additional loops avail

16、able!,1W,1W,1W,2W,1W,13 V,+,Loop 4,Step 2: Loop Rule! Additional loops available!,1W,1W,1W,2W,1W,13 V,+,Loop 5,Any closed path will work. Extra loops good for checking,Step 3: Make Loop Equations!,Step 4: Solve equations (substitution or matrix),+13 1I1 - 1(I1-I3) = 0 +13 1I2 2(I2 + I3) = 0 -1I1 - 1

17、I3 + 1I2 = 0,Step 4: Solve equations (substitution or matrix),I1 = +6 A I2 = 5 A I3 = -1 A,So I3 is really going from b to c,Step 5: Check with extra loop equations!,1W,1W,1W,2W,1W,13 V,+,Loop 5,DArsonval galvanometer,A dArsonval galvanometer measures the current through it (see Figures 26.13 and 26

18、.14 below). Many electrical instruments, such as ammeters and voltmeters, use a galvanometer in their design.,Ohmmeters and potentiometers,An ohmmeter is designed to measure resistance. A potentiometer measures the emf of a source without drawing any current from the source.,Adding Capacitors to DC

19、circuits!,RC circuits include Batteries (Voltage sources!) Resistors Capacitors and switches! RC circuits will involve TIMING considerations Time to fill up a capacitor with charge Time to drain a capacitor that is already charged,Charging a capacitor,Start with uncharged capacitor! What happens?,Ad

20、ding Capacitors to DC circuits!,In charging RC circuits the time constant is t t increases with R Larger resistors decrease current Less charge/time arrives at capacitor It takes longer to fill up capacitor t increases with C Larger capacitors have more capacity! They take longer to fill up!,Chargin

21、g a capacitor,The time constant is = RC. In ONE time constant: Current drops to 1/e of initial value (about 36%) Charge on capacitor plates rises to 64% of maximum value,Charging a Capacitor,Vab = iR C = q/Vbc so Vbc = q/C E iR q/C = 0 Current is a function of time i = dq/dt E (dq/dt)R q(t)/C = 0 A

22、differential equation involving charge q,Charging a Capacitor,E (dq/dt)R q/C = 0 Boundary conditions relate q and t at key times: q(t) on capacitor = 0 t=0 q= Qmax, i(t) = 0 t = (when capacitor is full),Charging a Capacitor,E iR q/C = 0 General Solution q(t) = Qmax (1 - e-t/RC) Check? q(t) on capaci

23、tor = 0 t=0 i(t) = dq/dt = (Qmax/RC) e-t/RC i(0) is maximum current, Qmax/RC = E /R = I0 i(t) = 0 when capacitor is full, t = ,Discharging a capacitor,Disconnect Battery let Capacitor “drain”,Adding Capacitors to DC circuits!,In discharging RC circuits the time constant is still t t increases with R

24、 Larger resistors decrease current Less charge/time leaves capacitor It takes longer to drain capacitor t increases with C Larger capacitors have more capacity! They take longer to drain!,Discharging a capacitor,Time constant is still RC! In one time time constant: Charge on plates DROPS by 64% Curr

25、ent through resistor DROPS by 64% (“negative”),Charging a Capacitor,Vab = iR Vbc = q/C iR + q/C = 0 Another differential equation involving charge i = dq/dt (dq/dt)R + q(t)/C = 0,Charging a Capacitor,iR + q/C = 0 (dq/dt)R + q(t)/C = 0 Boundary conditions: q(t) on capacitor = Qmax t=0 q= 0, i(t) = 0 t = when capacitor is empty,Charging a Capacitor,iR + q/C = 0 General Solution q(t