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1、专业好文档 恒定电流一、概念荐入1.电流电流的定义式:,适用于任何电荷的定向移动形成的电流。 对于金属导体有i=nqvs(q是每个电荷的电荷量,n为单位体积内的自由电子个数,s为导线的横截面积,v为自由电子的定向移动速率),这个公式只适用于金属导体,千万不要到处套用。方向的规定:和正电荷的定向移动的方向相同,和负电荷的定向移动相反常用单位有:a安,ma毫安,微安,注意相互之间的进级。形成电流的条件:内因是有自由移动的电荷,外因是导体两端有电势差。例1:在2s内共有12.501018个电子通过导体某横截面,则通过导体的电流强度大小是 a1.0a b6.251018a c6.251017a d0.

2、10a2.电阻定律导体的电阻r跟它的长度l成正比,跟它的横截面积s成反比。是反映材料导电性能的物理量,叫材料的电阻率(反映该材料的性质,不是每根具体的导线的性质)。纯金属的电阻率小,合金的电阻率大。材料的电阻率与温度有关系:金属的电阻率随温度的升高而增大。半导体的电阻率随温度的升高而减小。有些物质当温度接近0 k时,电阻率突然减小到零这种现象叫超导现象。能够发生超导现象的物体叫超导体。材料由正常状态转变为超导状态的温度叫超导材料的转变温度tc。3. 欧姆定律: 公式:(适用于金属导体和电解液,不适用于气体导电)。 电阻的伏安特性曲线:注意i-u曲线和u-i曲线的区别()。还要注意:当考虑到电阻

3、率随温度的变化时,电阻的伏安特性曲线不再是过原点的直线(此时,也就是此时的电阻为图线上的点与坐标原点连线的斜率而不是切线的斜率) 例2 实验室用的小灯泡灯丝的i-u特性曲线可用以下哪个图象来表示:a. b. c. d.i i i io u o u o u o u4. 电功和电功率 (1)电功和电功率:电流做功的实质是电场力对电荷做功。电场力对电荷做功,电荷的电势能降低,电势能转化为其他形式的的能量,因此电功,这是计算电功普遍适用的公式。(2) 电功率:单位时间内电流做的功,这是计算电功率普遍适用的公式。(3) 电热和焦耳定律:焦耳定律,电热q=i2rt对纯电阻而言,电功等于电热:w=q=uit

4、=i 2r t=对非纯电阻电路(如电动机和电解槽),由于电能除了转化为电热以外还同时转化为机械能或化学能等其它能,所以电功必然大于电热:wq,这时电功只能用w=uit计算,电热只能用q=i 2rt计算,两式不能通用。例3 某一电动机,当电压u1=10v时带不动负载,因此不转动,这时电流为i1=2a。当电压为u2=36v时能带动负载正常运转,这时电流为i2=1a。求这时电动机的机械功率是多大?例题 直流电动机线圈的电阻为r,当电动机工作时通过线圈的电流是i,此时它两端的电压为u,则电动机的输出功率为( ) aui;biui2r ciu一i2r; du2r5. 电阻的连接(简单说明串联和并联电路的

5、电流电压电阻电压分配功率分配的关系) 此处单独出题目直接考查以上知识的可能性不大,一般都是结合复杂电路跟电功率最大最小啊,额定功率啊来出题,做这类题目的最根本还是电路分析,画等效电路图一般用等势点法。已知如图,两只灯泡l1、l2分别标有“110v,60w”和“110v,100w”,另外有一只滑动例题:变阻器r,将它们连接后接入220v的电路中,要求两灯泡都正常发光,并使整个电路消耗的总功率最小,应使用下面哪个电路?l1 l2 l1 l2 l1 l2 l1r r r l2ra. b. c. d. 例题: 图中的 a为理想电流表,v1和v2为理想电压表,r1为定值电阻,r2为可变电阻,电池e内阻不

6、计,下列说法中不正确的是 ar2不变时,v2读数与a读数之比等于r1 br2不变时,v1读数与a读数之比等于r1 cr2改变一定量时,v2读数的变化量与v1读数变化量之和一定为零 dr2改变一定量时,v2读数的变化量与v1读数变化量之和不一定为零6. 闭合电路的欧姆定律 ueor内容:闭合电路的电流跟电源的电动势成正比,与内、外电路的电阻之和成反比,即i=e/(r+r)由euir可知,电源电势等于内外电压之和;适用条件:纯电阻电路(3)路端电压跟负载的关系路端电压:根据可知,当r增大时,u增大;当r减小时,u减小。u一i关系图由闭合电路欧姆定律知:ueir , 路端电压随着电路中电流的增大而减

7、小;当电路断路即i0时,纵坐标的截距为电动势e;当外电路电压为u0时,横坐标的截距i短=e/r为短路电流;图线的斜率的绝对值为电源的内电阻(做题的时候需要注意的是纵坐标数值在坐标原点是否是从零开始的)(4).闭合电路的输出功率功率关系:电源的输出功率与外电路电阻的关系:当rr时也即i=e/2r时,电源的输出功率最大, 由图象可知,对应于电源的非最大输出功率p可以有两个不同的外电阻rl和r2,不难证明由图象还可以看出:当rr时,若r增大,则p出减小应注意:对于内外电路上的固定电阻,其消耗的功率仅取决于电路中的电流大小电源的供电效率(说明:电源的效率总小于1,不能过分追求效率大)e rr2r1例题

8、:已知如图,e =6v,r =4,r1=2,r2的变化范围是010。求:电源的最大输出功率;r1上消耗的最大功率;r2上消耗的最大功率。r1r2r3r4e r闭合电路中只要有一只电阻的阻值发生变化,就会影响整个电路,使总电路和每一部分的电流、电压都发生变化。讨论依据是:闭合电路欧姆定律、部分电路欧姆定律、串联电路的电压关系、并联电路的电流关系。以右图电路为例:设r1增大,总电阻一定增大;由,i一定减小;由u=e-ir,u一定增大;因此u4、i4一定增大;由i3= i-i4,i3、u3一定减小;由u2=u-u3,u2、i2一定增大;由i1=i3 -i2,i1一定减小。总结规律如下:总电路上r增大

9、时总电流i减小,路端电压u增大;变化电阻本身和总电路变化规律相同;和变化电阻有串联关系(通过变化电阻的电流也通过该电阻)的看电流(即总电流减小时,该电阻的电流、电压都减小);和变化电阻有并联关系的(通过变化电阻的电流不通过该电阻)看电压(即路端电压增大时,该电阻的电流、电压都增大)。 例题 如图所示,电源电动势为e,内电阻为r当滑动变阻器的触片p从右端滑到左端时,发现电压表v1、v2示数变化的绝对值分别为u1和u2,下列说法中正确的是 v2v1l1l2l3pa.小灯泡l1、l3变暗,l2变亮 b.小灯泡l3变暗,l1、l2变亮c.u1u2winger tuivasa-sheck, who sc

10、ored two tries in the kiwis 20-18 semi-final win over england, has been passed fit after a lower-leg injury, while slater has been named at full-back but is still recovering from a knee injury aggravated against usa.both sides boast 100% records heading into the encounter but australia have not conc

11、eded a try since josh charnleys effort in their first pool match against england on the opening day.aussie winger jarryd hayne is the competitions top try scorer with nine, closely followed by tuivasa-sheck with eight.but it is recently named rugby league international federation player of the year

12、sonny bill williams who has attracted the most interest in the tournament so far.the kiwi - with a tournament high 17 offloads - has the chance of becoming the first player to win the world cup in both rugby league and rugby union after triumphing with the all blacks in 2011.id give every award back

13、 in a heartbeat just to get across the line this weekend, said williams.the (lack of) air up there watch mcayman islands-based webb, the head of fifas anti-racism taskforce, is in london for the football associations 150th anniversary celebrations and will attend citys premier league match at chelse

14、a on sunday.i am going to be at the match tomorrow and i have asked to meet yaya toure, he told bbc sport.for me its about how he felt and i would like to speak to him first to find out what his experience was.uefa hasopened disciplinary proceedings against cskafor the racist behaviour of their fans

15、 duringcitys 2-1 win.michel platini, president of european footballs governing body, has also ordered an immediate investigation into the referees actions.cska said they were surprised and disappointed by toures complaint. in a statement the russian side added: we found no racist insults from fans o

16、f cska. baumgartner the disappointing news: mission aborted.the supersonic descent could happen as early as sunda.the weather plays an important role in this mission. starting at the ground, conditions have to be very calm - winds less than 2 mph, with no precipitation or humidity and limited cloud

17、cover. the balloon, with capsule attached, will move through the lower level of the atmosphere (the troposphere) where our day-to-day weather lives. it will climb higher than the tip of mount everest (5.5 miles/8.85 kilometers), drifting even higher than the cruising altitude of commercial airliners

18、 (5.6 miles/9.17 kilometers) and into the stratosphere. as he crosses the boundary layer (called the tropopause),e can expect a lot of turbulence.the balloon will slowly drift to the edge of space at 120,000 feet ( then, i would assume, he will slowly step out onto something resembling an olympic di

19、ving platform.they blew it in 2008 when they got caught cold in the final and they will not make the same mistake against the kiwis in manchester.five years ago they cruised through to the final and so far history has repeated itself here - the last try they conceded was scored by englands josh char

20、nley in the opening game of the tournament.that could be classed as a weakness, a team under-cooked - but i have been impressed by the kangaroos focus in their games since then.they have been concentrating on the sort of stuff that wins you tough, even contests - strong defence, especially on their own goal-line, completing sets and a good kick-chase. theyve been great at all

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