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九年级电流和电路中考真题汇编[解析版]

九年级电流和电路中考真题汇编[解析版]

一、初三物理电流和电路易错压轴题(难)

1.归纳式探究——研究电子在电场中的偏转:

如图1,给两块等大、正对、靠近的平行金属加上电压,两板之间就有了电场。若将电子沿着平行于两板的中线方向入射到电场中,电子就会发生偏转。若两板间距为d,板长为L,所加的电压为U,电子入射初速度为v0,离开电场时偏移的距离为y,则经研究得到如下数据:

次数d/m L/m U/V v0/(m·s-1)y/m

14×10-20.2401×107 3.6×10-2

28×10-20.2401×107 1.8×10-2

34×10-20.1401×1070.9×10-2

48×10-20.21601×1077.2×10-2

58×10-20.22402×107 2.7×10-2

(1)y=k__________,其中k=_________(填上数值和单位)。本实验在探究影响电子离开电场时偏移的距离时,运用了_________法;

(2)相同情况下,电子的入射速度越大,偏移距离越________。它们间的关系可以用图像2中的图线________表示;

(3)现有两块平行相对的长度为5cm,间距为1cm的金属板,为了让初始速度为3×107m/s 的电子从一端沿两板间中线方向入射后,刚好能从另一端的金属板边缘处射出,需要加_____V的电压。

【答案】2

20

UL dv ()1022

910m /V s ?? 控制变量 小 b 200

【解析】 【分析】 【详解】

(1)[1]分析表中数据可知,1与2相比L 、U 、0v 均相同,而d 增大一倍,y 减小为原来的

1

2

,可知y 与d 成反比;同理,1与3相比,y 与2L 成正比;2与4相比,y 与U 成正比;将第2次实验电压U 增大至6倍,则y 增大至6倍,此时

240V U =

210.810m y -=?

将此时的数据与第5次实验相比,y 与2

0v 成反比,综上所述可得

2

20

UL y k dv =

[2]将表格中第3次数据(其他组数据也可)代入2

20

UL y k dv =计算可得 ()1022910m /V s k =??

[3]本实验在探究影响电子离开电场时偏移的距离时,运用了控制变量法。

(2)[4]在其他情况相同时,y 与2

0v 成反比,所以电子入射的初速度越大,偏移距离越小。 [5]它们之间的关系可以用图乙中的图线b 来表示。

(3)[6]将已知数据中的y 、d 、L 、0v 和前面算出的k 分别代入220

UL y k dv =并计算可得 200V U =

2.小敏和小英通过实验探究“并联电路中干路电流与各支路电流的关系”.她们连接的电路如图所示.

(1)当小敏用开关“试触”时,发现电流表A 1无示数、电流表A 2的指针快速右偏,两灯均不发光.由此可知,她们连接的电路发生了________故障;若闭合开关S ,可能造成的后果是________.

(2)小英只改接了电路中的一根导线,电路便连接正确了,请你在图中画出她的改法.(在错接的导线上画×,然后画出正确的接线) (___________)

(3)她们利用改正后的电路进行实验:闭合开关S后,读出两块电流表的示数,得出的初步结论是:________.

【答案】短路电流表A2被烧坏在并联电路中,

干路电流大于支路电流

【解析】

【分析】

【详解】

(1)电流表A1无示数、电流表A2的指针快速右偏,两灯均不发光,由此可知电路发生了短路故障,若闭合开关S,由于电路发生短路,电路中电流过大,可能造成的后果是电流表A2被烧坏.

(2)改正电路如图:

(3)利用改正后的电路进行实验,闭合开关S后,读出两块电流表的示数,得出的初步结论是:在并联电路中,干路电流大于支路电流.

【点睛】

电路中发生短路的危害,可能烧毁电流表或电源,为避免短路发生,可采用试触法.

3.归纳式探究—.研究电磁感应现象中的感应电流:

磁场的强弱用磁感应强度描述,用符号B表示,单位是特斯拉,符号是T.强弱和方向处处相同的磁场叫做匀强磁场.

如图甲所示,电阻R1与圆形金属线圈R2连接成闭合回路,R1和R2的阻值均为R0,导线的电阻不计,

在线圈中半径为r的圆形区域内存在垂直于线圈平面向里的匀强磁场,磁感应强度B随时间t变化的关系图象如图乙所示,图线与横、纵坐标的截距分别为t0和B0.则0至t1时间内通过R1的电流I与阻值R0、匀强磁场的半径r、磁感应强度B0和时间t0的关系数据如下表:

次数R0/Ωr/m B0/T T0/s I/A

1100.1 1.00.15π×l0-2

2200.1 1.00.1 2.5π×l0-2 3200.2 1.00.110π×l0-2 4100.10.30.1 1.5π×l0-2 5200.10.10.050.5π×l0-2

(1)I=_____k,其中k=________(填上数值和单位)

(2)上述装置中,改变R0的大小,其他条件保持不变,则0至t1时间内通过R1的电流I 与R0的关系可以用图象中的图线____表示.

【答案】

2000B r R t 2

5A s

T m π?Ω?? c 【解析】 【分析】 【详解】

(1)[1]由图像分析可得

2E

I R

=……………1 E t

=

?...............2 =?S B ?Φ? (3)

120+2R R R R == (4)

由1234联立得:

220000

22B B r r I t R t R ππ?=?=??

由于

2

π

为定值,故 2

000

B r I k R t =

[2]将第一组数据带入上式得:

k =

2

5A s

T m π?Ω??

(2)[3]若R 0变,其他为定值,则2

00

2B r t π均为定值,可看作0'k I R =,此为反比例函数,

故可用图线c 表示.

4.如图是一个模拟交通路口红绿灯工作的实验电路,请你用笔画线代替导线,只添加两根导线,实物电路图补充完整.要求:红灯亮时,黄灯和绿灯都不亮; 当红灯灭时,黄灯和绿灯可以分别亮一盏.

【答案】

【解析】

由题意知红灯亮时,黄灯和绿灯都不亮;当红灯灭时,黄灯和绿灯可以分别亮一盏.说明三盏灯互不影响,也就是说三盏灯是并联;结合实物图,S1是控制红灯和黄灯;S2控制绿灯和黄灯,故连接电路图如图.

5.(一)用下图所示的装置探究摩擦力跟接触面粗糙程度的关系。

(1)实验时,用弹簧测力计水平拉动木块,使它沿长木板做运动,根据知识,从而测出木块与长木板之间的滑动摩擦力。

(2)第一次实验中弹簧测力计的示数如图所示为 N,]分析表中数据可以得到的结论是。

实验次数123

接触面情况木块和长木板木块和棉布木块和毛巾

摩擦力/N1.92.4

(3)实验结束后,小丽同学想探究摩擦力是否与接触面的大小有关,她用弹簧测力计测出木块在水平面上的摩擦力,然后将木块沿竖直方向锯掉一半,测得摩擦力的大小也变为原来的一半.她由此得出:当接触面的粗糙程度一定时,接触面越小,摩擦力越小。你认为她的结论正确吗?,理由是。

(二)襄襄和樊樊在“探究并联电路中干路电流与各支路电流有什么关系”时,利用一个开关、一个电流表、一个学生电源(有多个电压档位)、四个阻值不等的电阻以及若干条导线,进行了大胆地探究。如图所示是他们的实验电路图。

(1)他们的猜想是:(只写一种猜想)。

[进行实验]

(2)襄襄按照电路图正确地进行实验,在连接电路时开关处于;得到了表1中的实验数据。襄襄在实验中,是靠只改变而得到实验数据的:测量三次的目的是:。

(3)樊樊也按照上述同样的器材和同样的电路进行了实验,却得到了表2中的数据。樊樊在实验中是靠只改变而得到实验数据的。

(4)在襄襄同学另外在测量串联电路电压规律中测L2两端的电压时,为了节省实验时间,采用以下方法:电压表所接的B接点不动,只断开A接点,并改接到C接点上;瑞瑞同学用上面的方法能否测出L2两端的电压? 为什么?__ .

【答案】(1)匀速直线二力平衡 1.2 压力一定时,接触面越粗糙,受到的摩擦力越大

不正确没有控制压力不变;

(2)并联电路中,干路电流等于各支路电流之和断开 R1电阻大小多次测量,避免实验偶然性,寻找普遍规律;电源电压不可以,电压表正负接线柱接反了

【解析】

试题分析:(一)(1)实验时,我们利用二力平衡的知识,使水平方向的摩擦力等于弹簧测力计的示数,因此必须水平匀速直线拉动木块;

(2)弹簧测力计的量程是5N,分度值是0.2N,示数为1.2N;

分析表中数据可以得到,压力一定时,接触面越粗糙,受到的摩擦力越大;

(3)因为有压力大小、接触面的粗糙程度影响摩擦力的大小,因此如果想探究摩擦力是否与接触面的大小有关,她应该控制压力大小和接触面的粗糙程度一定,但她沿竖直方向锯掉一半时,压力大小改变了,因此没有控制压力一定,因此实验是错误的;

(二)(1)他们的猜想是:并联电路中,干路电流等于各支路电流之和;

(2)在连接实验电路时开关处于断开,避免接通最后一根导线时,电路中有电流;

表1中的实验数据,第二条支路电流没有变化,说明襄襄在实验中,是靠只改变R1电阻大小,而得到实验数据的:测量三次的目的是:多次测量,避免实验偶;

(3)樊樊得到了表2中的数据,三次实验中,R1和R2的电流成倍数增大,说明樊樊在实验中是靠只改变电源电压得到实验数据的;

(4)在测量串联电路电压规律中测L2两端的电压时,将电压表所接的B接点不动,只断开A接点,并改接到C接点上,是不对的,电压表的正负极接线柱接反了。

考点:探究摩擦力跟接触面粗糙程度的实验探究干路电流与支路电流的关系实验

6.为了防止电路中电流过大,发生危险,电路中常常需要安装保险丝.保险丝安装在玻璃管中,称为保险管.实验室有熔断电流分别为1A和2A的两种保险管,保险管上印刷的文字已经模糊不清,但小星知道:熔断电流较大的保险丝,其电阻较小,因此,小星设计了如图甲所示实验电路图,想通过比较它们的电阻大小来区别这两种保险管.

(1)根据图甲电路图,在图乙的实物图中,用笔代善导线把电路连接完整。

(________)

(2)开关闭合前,滑动变阻器的滑片应该移至最_______(选填“左”或“右”端)。

(3)在图甲中,若电压表V2的读数为2.1V,电压表V1读数如图丙所示,则保险管B两端的电压为_______V.

(4)保险管A的熔断电流为_______A

(5)将这两个保险管并联接入某一电路中,当电路中的电流达到_______A时,保险管将熔断.

【答案】左 1.5 2 2.8A

【解析】

【详解】

(1)根据图甲电路图可知,电压表V1测保险管A的电压,电压表V2测两保险管总电压。再由第(3)小题中“电压表V2的读数为2.1V”可知,两电压表都接小量程。连接电压表

V1“-”接线柱到保险管A左端接线柱、“3”接线柱到保险管A右端接线柱;再连接电压表V2“-”接线柱到保险管B左端接线柱、“3”接线柱到滑动变阻器右下端(或保险管A右端)接线柱。如图:

(2)为保护电路,开关闭合前,滑动变阻器的滑片应该移至阻值最大端。因滑动变阻器接右下端接线柱,所以滑片移至最左端阻值最大。

(3)电压表V1测保险管A的电压,其分度值为0.1V,则U A=0.6V。电压表V2测两保险管总电压,即U A+B=2.1V。则保险管B两端的电压为U B=U A+B-U A=2.1V-0.6V=1.5V。

(4)因U A=0.6V、U B= 1.5V,即U A

(5)由上题可知,R A:R B= U A:U B=0.6V: 1.5V=2:5。将这两个保险管并联接入某一电路中,电流之比为I A:I B= R B:R A=5:2。若当保险管A电流达到熔断电流2A时,保险管B电流为

I B=2

5

I A=

2

5

×2A=0.8A,此时电路总电流达到I=2A+0.8A=2.8A,保险管A刚熔断,保险管B不

能熔断;若当保险管B电流达到熔断电流1A时,保险管A电流为

I A=5

2

I B=

5

2

×1A=2.5A>2A,说明A早已熔断。综合以上分析可知,将这两个保险管并联接入

某一电路中,当电路中的电流达到2.8A时,保险管将熔断。

7.小明在探究并联电路电流规律的实验中,如图甲是实验的电路图。

(1)在连接电路时发现,刚接好最后一根导线,表的指针就发生了偏转,由此可知在连接电路时,他忘了_____。

(2)他先将电流表接A处,闭合开关后,观察到灯L2发光,但灯L1不发光,电流表的示数为零,电路可能存在的故障是:_____。

(3)他在测量B处的电流时,发现电流表的指针偏转如图乙所示,原因是_____;在排除故障后,电流表的示数如图丙所示,则电流表的示数为_____A。

(4)在解决了以上问题后,将电流表分别接入A、B、C三点处,闭合开关,测出了一组电流并记录在表格中,立即得出了并联电路的电流规律。请你指出他们实验应改进方法是_____。

(5)实验结束后,小明又利用器材连接了如图丁所示的电路图,当开关S由断开到闭合时,电流表A2的示数_____(选填“变大”“变小”或“不变”)。

【答案】断开开关; L1断路;电流表正负接线柱接反了; 0.24;换用不同的灯泡测量多组数据;变大。

【解析】

【分析】

【详解】

(1)[1]在连接电路时发现,刚接好最后一根导线,表的指针就发生了偏转,由此可知在连接电路时,他忘了断开开关;

(2)[2]将电流表接A处,闭合开关后,观察到灯L2发光,但灯L1不发光,电流表的示数为零,说明L1所在支路断路;

(3)[3]电流表的指针反向偏转原因是:电流表正负接线柱接反了;

[4]图中电流表选用小量程,分度值为0.02A,则电流表的示数为0.24A;

(4)[5]只测量了一组数据得出的结论有偶然性,应改进方法是:换用不同的灯泡测量多组数据;

(5)[6]实验结束后,小明又利用器材连接了如图丁所示的电路图,当开关S断开时,电路中只有L1,电流表A2测电路中电流,S闭合时,两灯并联,电流表A2测干路的电路中,因L1的电压和电流不变,根据并联电路电流的规律,电流表A2的示数变大。

8.小明用如图甲所示的电路图来探究并联电路中干路电流与各支路电流的关系.

(1)请根据电路图,用笔画线代替导线把图乙的电路连接完整. (____)

(2)连接电路后,小明把电流表接入图甲的A 处,闭合开关S 后,发现小灯泡1L 不亮,

2L 亮,电流表无示数,产生这种现象的原因可能是______.排除电路故障再做实验,电流

表在A 处的示数如图丙所示,请你帮小明把该示数填入下表空格处.然后小明把电流表分别接入电路中的B 、C 两处测电流,并把电流表的示数记录在下表中.

A 处的电流/A A I

B 处的电流/A B I

C 处的电流/A C I

______

0.15

0.25

小明分析表格中的数据认为:并联电路中干路电流等于______.

(3)小明想,要使上述结论更具普通性,还要用不同的方法进行多次实验,于是小明和同学们讨论了以下三种方案:

方案一:在图甲的基础上,反复断开、闭合开关,测出A 、B 、C 三处的电流. 方案二:在图甲的基础上,只改变电源电压,测出A 、B 、C 三处的电流.

方案三:在图甲的基础上,在其中一条支路换上规格不同的灯泡,测出A 、B 、C 三处的电流.

以上三种方案,你认为不可行的方案______(填“一”“二”或“三”). 小明选择上述可行方案之一,做了三次实验,并把实验数据填入下表中. 实验次序 A 处的电流/A A I

B 处的电流/A B I

C 处的电流/A C I

1 0.1

2 0.18 0.30 2 0.14 0.21 0.35 3

0.20

0.30

0.50

请你根据小明的实验步骤和有关数据回答下列问题: ①在拆电路改装时,开关必须______;

②小明选择的方案是方案______(填“一”“二”或“三”).

L支路有断路 0.10 各支路电流之和一断【答案】1

开二

【解析】

【分析】

【详解】

(1)[1]由电路图可知,两灯泡并联,因此可得实物图如下:

(2)[2]电流表无示数说明电路不通,即1L支路有断路;

[3]由图丙可知,A处电流为0.1A;

[4]由表格数据可知,干路C处电流,等于A支路,B支路电流之和,因此可得并联电路中干路电流等于各支路电流之和;

(3)[5] 方案二和方案三:换用不同的规格的小灯泡,或者增加电池的节数改变电源电压,测出三处的电流,可以得出多组不同的实验数据,故可以得出普遍规律;而方案一中反复断开,闭合开关,测得到的实验数据都是相同的,故不可行;

[6]为了保护电路,在拆电路改装时,开关必须断开;

[7] 分析A、B两处的电流的变化关系可知,通过A处电流的变化倍数和通过B电流的变化倍数相等,说明选择的实验方案是二.

9.宝宝和玲玲同学想探究并联电路电流规律,

猜想与假设:

(1)宝宝同学猜想:并联电路中各支路电流相等;

(2)玲玲同学猜想:并联电路干路中的电流等于各支路电流之和。

设计实验与制定计划:宝宝和玲玲同学分别从实验室选取电流表3只,灯泡2只,开关1个,滑动变阻器1个,干电池,导线若干。实验电路图如图所示。

进行实验与收集证据:

(1)宝宝同学根据电路图连接好实验电路,连接过程中,开关应该_______;检查电路无误后,开始实验,正确读出电流表示数如下表:

(2)玲玲同学根据电路图正确连接好实验电路,开始实验,正确读出电流表示数如下表:

分析与论证:

分析记录的实验数据,宝宝同学得出:并联电路中各支路电流相等;玲玲同学得出:并联电路总电流有时等于各支路电流之和,有时不等于各支路电流值和。

评估:

(1)宝宝同学得出错误的实验结论,主要原因是______;

(2)玲玲同学的实验记录表格中,实验次数______读数错误,原因是______。

(3)为了使实验结论更具科学性,请你提出合理化建议(1条即可):______。

【答案】断开选用的小灯泡规格是相同的 3 看错了量程应该再选用不同规格的灯泡再进行多次测量得出结论

【解析】

【分析】

【详解】

[1]宝宝同学根据电路图连接好实验电路,连接过程中,开关应该断开;

[2]检查电路无误后,开始实验,实验得到的数据具有特殊性,每一次两支路电流都相等,由于并联电压相等,所以可知他所选用的小灯泡规格是相同的,因此得出的实验结论不具有普遍性和科学性;

[3]分析小红同学的实验记录表格中数据可以看出,第三次的数据明显差别很大,所以这一

次是错误的,干路电流表示数比支路电流表示数小,原因只有一种就是读数时看错量程了;

[4]为了使实验结论更具科学性,普遍性,应该换用不同规格的灯泡再进行多次测量得出实验结论。

10.按要求完成填空:

(1)如图甲所示,用天平测得一正方体物块的质量为_______________g ,若用刻度尺测得该物块的边长为2. 00cm ,则它的密度为_______________g/cm 3.

(2)如图乙所示,在长木板上速拉动木块时弹簧测力计示数如图,则木块受到的摩擦力为_____________N ;图示实验探究的是_______________对滑动摩擦力大小的影响. (3)如图丙所示,该冲程是_______________冲程;该冲程中能量转化方式为_______________.

(4)如图丁所示,两个气球分别与毛皮摩擦后相互靠近,图示现象表明了_______________.

【答案】61.6 7.7 1.8 压力大小 压缩 机械能(或动能)转化为内能 同种电荷相互排斥 【解析】 【详解】

(1)图中标尺的分度值为0.2g ,物块的质量m =50g+10g+1.6g=61.6g ,该物块的体积V =a 3=(2cm )3=8cm 3;它的密度ρ=

361.6g 8cm

m V =7.7g/cm 3; (2)图乙中弹簧测力计的示数为1.8N ,由二力平衡可知,当木块被匀速拉动时,这时滑动摩擦力的大小等于弹簧测力计拉力的大小1.8N ;实验中接触面的粗糙程度相同,压力不同,探究压力对滑动摩擦力大小的影响;

(3)由图可知,此时汽油机的两个气门都关闭,活塞由下向上运动,压缩内部的混合物,使其内能增大、温度升高,故该冲程是压缩冲程,即是把机械能转化为内能的过程; (4)两个气球分别与毛皮摩擦后相互靠近,由图可知,两只气球相互远离,说明两只气球带上了同种电荷而相互排斥。

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