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电学实验是高考实验考查的重点,考查的实验 ( http: / / www.21cnjy.com )有测定金属丝的电阻率、描绘小灯泡的伏安特性曲线、测定电源的电动势和内阻、多用电表的使用等.高考对电学实验的考查侧重于考查电路设计、实物图连接、数据处理和误差分析,并且高考对电学实验的考查向着拓展实验、创新实验的方向发展.要求考生掌握常规实验的数据处理方法,能将课本中分组实验和演示实验的实验原理、实验方法迁移到新的背景中,深刻理解物理概念和规律,并能灵活运用,要求考生有较强的创新能力.
考题1 电表读数和多用电表的使用
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(1)下列关于用多用电表欧姆挡测电阻的说法中正确的是( )
A.测量电阻时,如果红、黑表笔分别插在负、正插孔,则不会影响测量结果
B.双手捏住两表笔金属杆,测量值将偏大
C.测量电路中的电阻时,应该把该电阻与电路断开
D.欧姆表使用一段时间后,电池电动势变小,内阻变大,但仍能调零,其测量结果与原来相比不变
(2)用多用电表探测二极管的极性,用欧姆挡 ( http: / / www.21cnjy.com )测量,黑表笔接a端,红表笔接b端时,指针向右偏转角较大,然后黑、红表笔反接指针偏转角较小,说明________(填“a”或“b”)端是二极管正极.
(3)某同学想通过多用电表A中的欧姆档,直 ( http: / / www.21cnjy.com )接去测量相同规格的多用电表B中量程为2.5 V的电压档的内阻,如图1甲连线,选择倍率为×1 k欧姆挡,测量前应对多用电表A进行______________,按正确的操作步骤测量,两表指针位置相同,如图乙所示,从A表读出B表内阻约为________________kΩ,从B表读出电压约为________ V.
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图1
计算出多用电表A的电源电动势约为________ V(已知表盘正中电阻刻度值为15,计算结果保留两位有效数字).
解析 (2)当用多用电表的 ( http: / / www.21cnjy.com )红表笔接触二极管a端、黑表笔接触二极管b端时,b端电势高、a端电势低,此时发现多用电表的指针偏转量非常小,说明此时电阻很大,说明此时二极管反向偏压,则a端是二极管的正极.
(3)选挡后要先进行欧姆调零,欧姆表读 ( http: / / www.21cnjy.com )数=表盘读数×倍率=13×1 kΩ=13 kΩ;电压表读数为1.32 V;欧姆表内电阻等于中值电阻,为r=15 kΩ;根据闭合电路欧姆定律,有:E=Ir+U≈2.8 V.
答案 (1)AC (2)a (3)欧姆调零 13 1.32 2.8
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1.(2014·重庆·6(1))某照明电路出现故障,其电路如图2所示,该电路用标称值12 V的蓄电池为电源,导线及其接触完好.
维修人员使用已调好的多用表直流50 V挡检测故障.他将黑表笔接在c点,用红表笔分别探测电路的a、b点.
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图2
①断开开关,红表笔接a点时多用表指 ( http: / / www.21cnjy.com )示如图3所示,读数为________ V,说明________正常(选填“蓄电池”、“保险丝”、“开关”或“小灯”).
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图3
②红表笔接b点,断开开关时,表针不 ( http: / / www.21cnjy.com )偏转,闭合开关后,多用表指示仍然和图3相同,可判断发生故障的器件是________.(选填“蓄电池”、“保险丝”、“开关”或“小灯”)
答案 ①11.5(11.2~11.8的值均可) 蓄电池 ②小灯
解析 ①因用直流50 V挡检测,故满偏时电压 ( http: / / www.21cnjy.com )为50 V,刻度均匀,在图示位置时,读数为11.5(11.2~11.8之间均可)V,说明蓄电池正常.
②红表笔接b点,断开开关,指针不偏转,闭合开关后多用表读数与原来相同,说明保险丝、开关均正常,发生故障的器件只能为小灯.
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1.欧姆表的原理
(1)欧姆表内有电源,红表笔与内部电源负极相连,黑表笔与内部电源的正极相连,故其电流方向为“红表笔流进,黑表笔流出”.
(2)测电阻的原理是闭合电 ( http: / / www.21cnjy.com )路欧姆定律.当红、黑表笔短接时,调节滑动变阻器R0(即欧姆调零),使灵敏电流计满偏,Ig=,此时中值电阻R中=Rg+R0+r,当两表笔接入电阻Rx时I=,电阻Rx与电路中的电流相对应,但不是线性关系,故欧姆表刻度不均匀.
2.欧姆表使用六注意
(1)选挡接着调零;(2 ( http: / / www.21cnjy.com ))换挡重新调零;(3)待测电阻与电路、电源要断开;(4)尽量使指针指在表盘中间位置附近;(5)读数之后要乘以倍率得阻值;(6)用完后,选择开关要置于“OFF”挡或交流电压的最高挡.
考题2 测量电阻的实验
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图4
在保证安全和满足需求的情况下,使测量范围尽可能大.实验具体步骤如下:
(ⅰ)连接好电路,将滑动变阻器R调到最大;
(ⅱ)闭合S,从最大值开始调节电阻箱R ( http: / / www.21cnjy.com )1,先调R1为适当值,再调节滑动变阻器R,使A1示数I1=0.15 A,记下此时电阻箱的阻值R1和A2的示数I2;
(ⅲ)重复步骤(ⅱ),再测量6组R1和I2的值;
(ⅳ)将实验测得的7组数据在如图5所示坐标纸上描点.
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图5
根据实验回答以下问题:
①现有四只供选用的电流表
A.电流表(0~3 mA,内阻为2.0 Ω)
B.电流表(0~3 mA,内阻未知)
C.电流表(0~0.3 A,内阻为5.0 Ω)
D.电流表(0~0.3 A,内阻未知)
A1应选用________,A2应选用________.
②测得一组R1和I2值后,调整电阻箱R ( http: / / www.21cnjy.com )1,使其阻值变小,要使A1示数I1=0.15 A,应让滑动变阻器R接入电路的阻值________(选填“不变”、“变大”或“变小”).
③在坐标纸上画出R1与I2的关系图.
④根据以上实验得出Rx=________ Ω.
答案 ①D C ②变大 ③关系图线如图所示 ④31
解析 ①该实验测Rx电阻的原理为并 ( http: / / www.21cnjy.com )联电路两支路电压相等,即(Rx+r2)I2=(R0+R1+r1)I1,其中r1、r2分别为A1、A2两表的内阻.又因为Rx的阻值约几十欧,与R0+R1在同一数量级,A1示数为I1=0.15 A,表示电流表A1只能从C、D中选择,A2电流值的数量级与A1相同.由测量原理可知:(Rx+r2)I2=(R0+R1+r1)I1,因Rx未知,故r2阻值应已知,故A1选D,A2选C.
②因A1示数I1=0.15 A不变,R1变小,(R0+R1+r1)I1变小,滑动变阻器所分电压变大,R应变大.
③该图线为一条直线,离线较远的一组数据为错误数据,应舍去,其他点应对称分布在直线两侧.
④由实验原理得:
(R0+R1+r1)I1=(Rx+r2)I2
其中R0=10 Ω,r2=5.0 Ω,I1=0.15 A,从图线上选择两组R1、I2代入不同数值可解得Rx约为31 Ω.
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2.(2014·新课标Ⅱ·22)在伏安法测电阻的实验中,待测电阻Rx约为200 Ω,电压表INCLUDEPICTURE "../../../../../../../2014/幻灯片/考前三个月/物理/通用/全书完整的Word版文档/圈V.TIF" \* MERGEFORMAT INCLUDEPICTURE "../../../../../../../2014/幻灯片/考前三个月/物理/通用/全书完整的Word版文档/圈V.TIF" \* MERGEFORMAT INCLUDEPICTURE "../../../../../../../2014/幻灯片/考前三个月/物理/通用/全书完整的Word版文档/圈V.TIF" \* MERGEFORMAT INCLUDEPICTURE "\\\\0慧慧\\e\\2014\\幻灯片\\考前三个月\\物理\\通用\\全书完整的Word版文档\\圈V.TIF" \* MERGEFORMATINET INCLUDEPICTURE "\\\\0慧慧\\e\\2014\\幻灯片\\考前三个月\\物理\\通用\\全书完整的Word版文档\\圈V.TIF" \* MERGEFORMATINET 的内阻约为2 kΩ,电流表INCLUDEPICTURE "../../../../../../../2014/幻灯片/考前三个月/物理/通用/全书完整的Word版文档/圈A.TIF" \* MERGEFORMAT INCLUDEPICTURE "../../../../../../../2014/幻灯片/考前三个月/物理/通用/全书完整的Word版文档/圈A.TIF" \* MERGEFORMAT INCLUDEPICTURE "../../../../../../../2014/幻灯片/考前三个月/物理/通用/全书完整的Word版文档/圈A.TIF" \* MERGEFORMAT INCLUDEPICTURE "\\\\0慧慧\\e\\2014\\幻灯片\\考前三个月\\物理\\通用\\全书完整的Word版文档\\圈A.TIF" \* MERGEFORMATINET 的内阻约为10 Ω,测量电路中电流表的连接方式如图6(a)或(b)所示,结果由公式Rx=计算得出,式中U与I分别为电压表和电流表的示数.若将图(a)和图(b)中电路测得的电阻值分别记为Rx1和Rx2,则________(填“Rx1”或“Rx2”)更接近待测电阻的真实值,且测量值Rx1________(填“大于”、“等于”或“小于”)真实值,测量值Rx2________(填“大于”、“等于”或“小于”)真实值.
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图6
答案 Rx1 大于 小于
解析 根据题意知>,电压表的分流作用 ( http: / / www.21cnjy.com )较显著,故Rx1更接近待测电阻的真实值.题图(a)的测量值是Rx与RA串联的电阻阻值,故Rx1>Rx真;题图(b)的测量值是Rx与RV并联的电阻阻值,故Rx23.(2014·浙江·22)小明对2B铅笔芯的导电性能感兴趣,于是用伏安法测量其电阻值.
(1)图7是部分连接好的实物电路图,请用电流表外接法完成接线并在图中画出.
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图7
(2)小明用电流表内接法和外接法分别 ( http: / / www.21cnjy.com )测量了一段2B铅笔芯的伏安特性,并将得到的电流、电压数据描到U-I图上,如图8所示.在图中,由电流表外接法得到的数据点是用________(填“○”或“×”)表示的.
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图8
(3)请你选择一组数据点,在图上用作图法作图,并求出这段铅笔芯的电阻为________ Ω.
答案 (1)见解析图 (2)× (3) ( http: / / www.21cnjy.com )见解析图 用“×”表示的数据连线时,1.2(1.1~1.3均可),用“○”表示的数据连线时,1.6(1.5~1.7均可)
解析 (1)连线如图所示.
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(2)U-I图像如图所示.
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U-I图像的斜率反映了电阻的大小,而用电流表内接法时测得的电阻偏大,外接法时测得的电阻偏小,所以外接法的数据点是用“×”表示的.
(3)在U-I图像上,选用外接法所得的“×”连线,则R==1.2 Ω,选用内接法所得的“○”连线,则R==1.6 Ω.
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1.电路结构:实验电路至少包括三个部分:(1)测量电路;(2)控制电路(滑动变阻器、开关);(3)电源.
2.电路设计:(1)电表选择注 ( http: / / www.21cnjy.com )意三方面:①安全性,要求量程不能太小,电表不反接;②准确性,要求量程不能太大,读数一般要超过量程的;③若电表量程不合适,要利用有准确内阻的电表进行改装.
(2)测量电路设计:若>,则应把电流表内接;若<,则应把电流表外接.
(3)控制电路:以小控大用分压,相差无 ( http: / / www.21cnjy.com )几用限流,即当滑动变阻器的阻值较小时,常采用分压式接法;当滑动变阻器的阻值与负载相差不大时,常采用限流式接法.另外滑动变阻器必须接为分压式的情况有三种:①电压要求从零开始变化;②滑动变阻器太小,不能起到限流的作用;③限流式不能获取有区分度的多组数据.若两种接法均可,则采用限流式,因为限流式损耗功率小.
考题3 描绘小灯泡(或者其他元件)的伏安特性曲线
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A.直流电源3 V(内阻可不计)
B.直流电流表0~600 mA(内阻约0.5 Ω)
C.直流电压表0~3 V(内阻约3 kΩ)
D.滑动变阻器(10 Ω,1 A)
E.滑动变阻器(1 kΩ,300 mA)
F.开关、导线若干
(1)本实验中滑动变阻器选用________.(填“D”或“E”)
(2)某同学用导线a、b、c、d、e、f、 ( http: / / www.21cnjy.com )g和h连接的电路如图9所示,电路中所有元器件都是完好的,且电压表和电流表已调零.闭合开关后发现电压表的示数为2 V,电流表的示数为零,小灯泡不亮,则可确定断路的导线是________;若电压表示数为零,电流表的示数为0.3 A,小灯泡亮,则断路的导线是________;若反复调节滑动变阻器,小灯泡亮度发生变化,但电压表、电流表示数不能调为零,则断路的导线是________.
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图9
(3)下表中的各组数据是该同学在实验中测得的,根据表格中的数据在如图10所示的方格纸上作出该灯泡的伏安特性曲线.
U/V 0 0.5 1.0 1.5 2.0 2.5
I/A 0 0.17 0.30 0.39 0.45 0.49
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图10
④如图11所示,将两个这样的 ( http: / / www.21cnjy.com )灯泡并联后再与5 Ω的定值电阻R0串联,接在电压恒为4 V的电路上,每个灯泡的实际功率为________ W(结果保留两位有效数字).
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图11
解析 (1)因描绘小灯泡的伏安特性曲线,需测量多组数据,滑动变阻器需用分压式接法,为了调节方便,滑动变阻器用阻值较小的,故选择D.
(2)电压表有示数,说明通过电压表的电 ( http: / / www.21cnjy.com )路为通路,而小灯泡不亮,通过与电压表并联的小灯泡的电路为断路,所以导线d断开;若电压表示数为零,而电流表有示数,说明电路接通,电压表没有接入电路,所以h断开;若电压表、电流表示数不能调为零,是因为滑动变阻器是限流式接法,故是g断开.
(3)描点画图,如图所示.
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(4)设灯泡的电压为U,电流为I,根据 ( http: / / www.21cnjy.com )闭合电路欧姆定律得:U0=U+2IR0,所以U=-2R0I+U0,代入数据U=-10I+4,在灯泡的伏安特性曲线坐标图上画出该直线,其交点(U,I)对应的电压电流值即为灯泡的实际电压、电流值,可得灯泡消耗的实际功率P=UI=0.30 W.
答案 (1)D (2)d h g (3)见解析图 (4)0.30
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4.某实验小组的同学在学校实验室中发现一 ( http: / / www.21cnjy.com )电学元件,该电学元件上标有“最大电流不超过6 mA,最大电压不超过7 V”,同学们想通过实验描绘出该电学元件的伏安特性曲线,他们设计的一部分电路如图12所示,图中定值电阻R=1 kΩ,用于限流;电流表量程为10 mA,内阻约为5 Ω;电压表(未画出)量程为10 V,内阻约为10 kΩ;电源电动势E为12 V,内阻不计.
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图12
(1)实验时有两个滑动变阻器可供选择:
A.阻值0~200 Ω,额定电流0.3 A
B.阻值0~20 Ω,额定电流0.5 A
应选的滑动变阻器是________(填“A”或“B”).
正确接线后,测得数据如下表:
次数 1 2 3 4 5 6 7 8 9 10
U/V 0.00 3.00 6.00 6.16 6.28 6.32 6.36 6.38 6.39 6.40
I/mA 0.00 0.00 0.00 0.06 0.50 1.00 2.00 3.00 4.00 5.50
(2)由以上数据分析可知,电压表应并联在M与________(填“O”或“P”)之间;
(3)在图12中将电路图补充完整;
(4)从表中数据可知,该电学元件的电阻特点是:______________________________
________________________________________________________________________.
答案 (1)A (2)P (3)电路图见解析图
(4)当元件两端的电压小于6 V时,元件电阻非常大,不导电;当元件两端电压大于6 V时,随着电压的升高电阻逐渐变小
解析 (1)滑动变阻器A允许的最大电 ( http: / / www.21cnjy.com )压为0.3×200 V=60 V>12 V,滑动变阻器B两端所能加的最大电压为0.5×20V=10 V<12 V,为保证安全,滑动变阻器应选A.
(2)由表中实验数据可知,电学元件电阻 ( http: / / www.21cnjy.com )最小测量值约为:R= Ω≈1 163.6 Ω,电流表内阻约为5 Ω,电压表内阻约为10 kΩ,相对来说,元件电阻远大于电流表内阻,电流表应采用内接法,因此电压表应并联在M与P之间.
(3)描绘伏安特性曲线,电压与电流应从零开始变化,滑动变阻器应采用分压式接法,由(2)可知,电流表采用内接法,实验电路图如图所示:
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(4)由表中实验数据可知,当元件两端的电 ( http: / / www.21cnjy.com )压小于6 V时,电路电流很小,几乎为零,由欧姆定律可知,元件电阻非常大,不导电;当元件两端电压大于6 V时,随着电压的升高电流迅速增大,电压与电流的比值减小,电阻变小.
知识专题练 训练13
题组1 电表读数和多用电表的使用
1.(1)用多用电表的欧姆挡测量阻 ( http: / / www.21cnjy.com )值约为几十千欧的电阻Rx,以下给出的是可能的操作步骤,其中S为选择开关,P为欧姆挡调零旋钮,把你认为正确的步骤前的字母按合理的顺序填写在下面的横线上________.
a.将两表笔短接,调节P使指针对准刻度盘上欧姆挡的零刻度,断开两表笔
b.将两表笔分别连接到被测电阻的两端,读出Rx的阻值后,断开两表笔
c.旋转S使其尖端对准欧姆挡×1 k
d.旋转S使其尖端对准欧姆挡×100
e.旋转S使其尖端对准交流500 V挡,并拔出两表笔
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图1
根据如图1所示指针位置,此被测电阻的阻值约为________ Ω.
(2)下述关于用多用电表欧姆挡测电阻的说法中正确的是( )
A.测量电阻时如果指针偏角过大,应将选择开关S拨至倍率较小的挡位,重新调零后测量
B.测量电阻时,如果红、黑表笔分别插在负、正插孔,则会影响测量结果
C.测量电路中的某个电阻,应该把该电阻与电路断开
D.测量阻值不同的电阻时都必须重新调零
答案 (1)cabe 30 k (2)AC
解析 (1)测量几十千欧的电阻Rx我 ( http: / / www.21cnjy.com )们一般选择较大的挡位先粗测,使用前应先进行欧姆调零,然后依据欧姆表的示数,再更换挡位,重新欧姆调零,再进行测量;使用完毕应将选择开关置于OFF位置或者交流电压最大挡,拔出表笔.欧姆表的示数乘以相应挡位的倍率即为待测电阻的阻值30 kΩ.
(2)欧姆挡更换规律“大小,小大”,即 ( http: / / www.21cnjy.com )当指针偏角较大时,表明待测电阻较小,应换较小的挡位;反之应换较大的挡位.电流总是从红表笔流入从黑表笔流出多用电表,每次换挡一定要进行欧姆调零,测量电阻一定要断电作业.
题组2 测量电阻的实验
2.(2014·安徽·21Ⅱ)某同学为了测量一个量程为3 V的电压表的内阻,进行了如下实验.
(1)他先用多用表进行了正确 ( http: / / www.21cnjy.com )的测量,测量时指针位置如图2所示,得出电压表的内阻为3.00×103 Ω,此时电压表的指针也偏转了.已知多用表欧姆挡表盘中央刻度值为“15”,表内电池电动势为1.5 V,则电压表的示数应为________ V(结果保留两位有效数字).
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图2
(2)为了更准确地测量该电压表的内阻RV,该同学设计了如图3所示的电路图,实验步骤如下:
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图3
A.断开开关S,按图3连接好电路;
B.把滑动变阻器R的滑片P滑到b端;
C.将电阻箱R0的阻值调到零;
D.闭合开关S;
E.移动滑动变阻器R的滑片P的位置,使电压表的指针指到3 V的位置;
F.保持滑动变阻器R的滑片 ( http: / / www.21cnjy.com )P的位置不变,调节电阻箱R0的阻值使电压表指针指到1.5 V的位置,读出此时电阻箱R0的阻值,此值即为电压表内阻RV的测量值;
G.断开开关S.
实验中可供选择的实验器材有:
a.待测电压表
b.滑动变阻器:最大阻值2 000 Ω
c.滑动变阻器:最大阻值10 Ω
d.电阻箱:最大阻值9 999.9 Ω,阻值最小改变量为0.1 Ω
e.电阻箱:最大阻值999.9 Ω,阻值最小改变量为0.1 Ω
f.电池组:电动势约6 V,内阻可忽略
g.开关、导线若干
按照这位同学设计的实验方法,回答下列问题:
①要使测量更精确,除了选用电池组、导线、开 ( http: / / www.21cnjy.com )关和待测电压表外,还应从提供的滑动变阻器中选用________(填“b”或“c”),电阻箱中选用________(填“d”或“e”).
②电压表内阻RV的测量值R测和真实值R真相比 ( http: / / www.21cnjy.com ),R测________R真(填“>”或“<”);若RV越大,则越________(填“大”或“小”).
答案 (1)1.0 (2)①c d ②> 小
解析 (1)欧姆表的中值电阻即为其内阻值.由 ( http: / / www.21cnjy.com )欧姆挡中央刻度值为“15”可知欧姆表内阻为1.5×103 Ω,根据电阻分压原理可知,电压表的示数U=·RV=×3×103 V=1.0 V.
(2)①明确半偏法测电阻的实 ( http: / / www.21cnjy.com )验原理.由半偏法测电阻和滑动变阻器的分压接法可知,滑动变阻器的阻值应远小于电表内阻,故滑动变阻器选c.电阻箱的最大阻值应与电压表内阻相近或大于电压表内阻,故电阻箱选d.
②接入电阻箱后,电路总电阻阻值变 ( http: / / www.21cnjy.com )大,干路电流变小,变阻箱R0和电压表分担的电压变大,则UR0>URV,故R测>R真.RV越大,R测越接近真实值,故越小.
题组3 描绘小灯泡(或者其他元件)的伏安特性曲线
3.某探究小组要描绘一个标有“4 V,1 W”的小灯泡的R-U曲线,所供选择的器材除了导线和开关外,还有以下一些器材可供选择:
A.电压表V(量程5 V,内阻约为5 kΩ)
B.直流电源E(电动势4.5 V,内阻不计)
C.电流表A1(量程150 mA,内阻约为2 Ω)
D.电流表A2(量程300 mA,内阻约为1 Ω)
E.滑动变阻器R1(阻值0~10 Ω)
F.滑动变阻器R2(阻值0~200 Ω)
(1)实验中要求小灯泡电压从零逐渐增大到额 ( http: / / www.21cnjy.com )定电压,测量误差尽可能小且方便调节,电流表应选用________,滑动变阻器应选用________(填写仪器符号).请在虚线框中画出该实验的电路图.
(2)根据实验数据,计算并描绘出了R— ( http: / / www.21cnjy.com )U图象,如图4所示.由图象可知,当所加电压为3.0 V时,小灯泡实际消耗的电功率为________ W(计算结果保留两位有效数字).
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图4
(3)小灯泡的P-U图象及I-U图 ( http: / / www.21cnjy.com )象如图5甲、乙所示.则小灯泡的电功率P随电压U变化的曲线是________;小灯泡的伏安特性曲线是________.(填“a”、“b”、“c”、“d”)
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图5
答案 (1)A2 R1 见解析图 (2)0.75 (3)b c
解析 (1)待测小灯泡为“4 V ( http: / / www.21cnjy.com ),1 W”,故其额定电流I==0.25 A=250 mA,故电流表只能选择A2;要描绘小灯泡RL的R-U曲线,电压要从0开始连续可调,故滑动变阻器采用分压式接法,为方便实验操作滑动变阻器应选R1.灯泡正常发光时的电阻R== Ω=16 Ω,电压表内阻约为5 kΩ,电流表内阻约为1 Ω,电压表内阻远大于灯泡电阻,电流表应采用外接法;描绘灯泡伏安特性曲线,电压与电流应从零开始变化,滑动变阻器应采用分压式接法,实验电路图如图所示.
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(2)当电压为3 V时,由题图可知小灯泡的电阻为12 Ω,故此时小灯泡消耗的实际功率P== W≈0.75 W.
(3)灯泡电功率P=,由于灯泡电阻随 ( http: / / www.21cnjy.com )温度的升高而增大,由小灯泡的R-U图象可知,U变大,I变大,而P=IU,因此P-U图象的斜率应随电压的增大而增大,故b正确;由小灯泡的R-U图象可知随电压的增大灯泡的电阻增大,故逐渐减小,故I-U图象的斜率随电压的增大而减小,故c正确.
4.要描绘一个标有“3 V,0.8 W”小灯泡的伏安特性曲线,已选用的器材有:
电源(电动势为4.5 V,内阻约1 Ω)
电流表(量程为0~300 mA,内阻约5 Ω)
电压表(量程为0~3 V,内阻约3 kΩ)
滑动变阻器(最大阻值10 Ω,额定电流1 A)
开关一个和导线若干.
(1)为便于实验操作,并确保实验有尽可能高的精度,则实验的电路图应选用下图中的________(填字母代号).
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(2)图6是实验器材实物图,图中已连接了部分导线.请根据在(1)问中所选的电路图补充完成图中实物间的连线(用笔画线代替导线).
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图6
(3)测量结束后,先把滑动变阻器滑片移到________(填“左端”或“右端”),然后断开开关,接着拆除导线,整理好器材.
(4)实验得到小灯泡的伏安特性曲线 ( http: / / www.21cnjy.com )如图7所示.由图象可知小灯泡的电阻值随工作电压的增大而________ (填“不变”、“增大”或“减小”).
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图7
答案 (1)C
(2)如图所示
INCLUDEPICTURE "../../../../../../../2014/幻灯片/考前三个月/物理/通用/全书完整的Word版文档/398.TIF" \* MERGEFORMAT INCLUDEPICTURE "../../../../../../../2014/幻灯片/考前三个月/物理/通用/全书完整的Word版文档/398.TIF" \* MERGEFORMAT INCLUDEPICTURE "../../../../../../../2014/幻灯片/考前三个月/物理/通用/全书完整的Word版文档/398.TIF" \* MERGEFORMAT INCLUDEPICTURE "\\\\0慧慧\\e\\2014\\幻灯片\\考前三个月\\物理\\通用\\全书完整的Word版文档\\398.TIF" \* MERGEFORMATINET INCLUDEPICTURE "\\\\0慧慧\\e\\2014\\幻灯片\\考前三个月\\物理\\通用\\全书完整的Word版文档\\398.TIF" \* MERGEFORMATINET
(3)左端 (4)增大