专题2.17 电器铭牌问题(解析版)
2021年高考物理100考点最新模拟题千题精练（选修3-1）
第二部分恒定电流
专题2.17 电器铭牌问题
一．选择题
1．（2020浙江七彩阳光名校联盟模拟）平衡车仅仅依靠人体重心的改变，便可以实现车辆的启动、加速、减速、停止等动作。因为其炫酷的操作，平衡车已经从年轻人的玩具，变成了日常通勤的交通工具。某款电动平衡车的部分参数如表所示，则下列说法中不正确的是（）
额定功率250W额定电压36V
锂电池容量4400mAh充电器输出44V，2A
百公里耗电量 1.1kwh质量12kg
A. 电池最多储存的电能约为5.7×105J
B. 电池从完全没电到充满电所需的时间为2.2h
C. 充满电的平衡车以额定功率行驶的最长时间约为83min
D. 该平衡车能骑行的最大里程约为14.4km
【参考答案】.C
【命题意图】本题以电动平衡车铭牌参数为信息，考查对铭牌参数的理解汽及其相关知识点，考查的核心素养是科学思维与推理。
【解题思路】锂电池容量q=4400mAh=4.4×3600C=1.584×104C，电池最多储存的电能W=qU=1.584×104×36J=5.7×105J，选项A正确；充电器充电电流I=2A，由q=It解得电池从完全没有电到充满电所需要的时间为t=q/I=2.2h，选项B正确；由W=Pt解得充满电的平衡车以额定功率行驶的最长时间t=W/P=2280s=38min，选项C不正确；锂电池容量q=4400mAh=4.4Ah，电池最多储存的电能W=qU=4.4Ah×36V=15.8Wh=1.58×10-2kWh，由百公里耗电量E100=1.1kWh可知1km耗电量E1=0.011kWh，由W=E0x解得该平衡车能够骑行的最大里程约为x=14.4km，选项D正确。
【易错警示】解答此题常见错误主要有：一是对铭牌参数理解有误；二是相关推理计算错误。
2．（2020江苏高考仿真模拟2）新能源电动汽车是当代生活中重要的交通工具。某新能源电动观光车的主要技术参数如下表： 空车质量 1180kg 动力电源 96V ，210Ah 满载人员 12人 输出电压 96V 车身长度 3.8m 额定功率 5.04 kW 最大车速
36km/h
效率
80%
根据表格有关数据，下列说法正确的是 （ ） A ．该车电动机的内阻约为1.82 Ω
B ．该车在水平路面匀速行驶时受到的阻力为1.4×102 N
C ．该车保持额定功率行驶的最长时间为4 h
D ．该车以额定功率工作半小时消耗电能约为7.3×106 J ， 【参考答案】C
【名师解析】由题意可知，额定输出功率5.04k W ，发热功率为P r ＝P×20%＝1008W ，额定工作电流为I ＝P
U
＝52.5 A ，由公式P r ＝I 2r ，解得r ≈0.37Ω，选项A 错误；由公式m fv P =，故该车在水平路面匀速行驶时受到的阻力为5.04×102N ，选项B 错误；根据电池容量Q ＝210 A·h ，电流为52.5 A ，则可得t ＝q
I ＝4 h ，选
项C 正确；该车以额定功率工作半小时消耗电能为==Pt W 9.072×106 J ，选项D 错误。
3.（2019江苏泰州12月联考）智能手机耗电量大，移动充电宝应运而生，它是能直接给移动设备充电的储能装置。充电宝的转化率是指电源放电总量占电源容量的比值，一般在0.60-0.70之间（包括移动电源和被充电池的 线路板、接头和连线的损耗)。如图为某一款移动充电宝，其参数见下表，下列说法正确的是（ ）
容量 20000mAh 兼容性 所有智能手机 边充边放
否
保护电路 是
输入
DC5V2AMAX
输出
DC5V0.1A-2.5A
尺寸156*82*22mm 转换率0.60
产品名称索扬SY10-200 重量约430g
A．充电宝充电时将电能转化为内能
B．该充电宝最多能储存能量为3.6×l06J
C．该充电宝电量从零到完全充满电的时间约为2h
D．该充电宝给电量为零、容量为3000mAh的手机充电，则理论上能充满4次
【参考答案】D
【名师解析】充电宝充电时将电能转化为化学能，故A错误；容量
最多能储存能量为，故B错误；，故C错误；由于其转换率为0.6所以释放的电荷量为，对电池充电的次数为，故D正确。
4. (2019河北保定期末)下表是某种型号的扫地机器人产品的铭牌，已知当扫地机器人剩余电池容量为总容量的20%时．机器人将停止扫地，寻觅充电基座自动回去充电。如果该机器人扫地时一直处于额定工作状态，则
额定工作电压12V
额定功率30W
电池容量2400mAh
A．扫地时电流为2400mA
B．扫地时电流为3.33A
C．扫地时机器人电阻为3.5Ω
D．机器人可以连续扫地60min
【参考答案】A
【名师解析】由P=UI可得扫地时电流I=P/U=40/12A=3.33A，选项A错误B正确；扫地机器人为非纯电阻元件，由扫地机器人铭牌数据不能得出扫地时机器人电阻，选项C错误；设机器人可以连续扫地时间为t，由
2400mAh=2.4×3600C=3.33t可得t=43.2×60s=43.2min，选项D错误。
5.（2018西城二模）．电动汽车由于节能环保的重要优势，越来越被大家认可。电动汽车储能部件是由多个
蓄电池串联叠置组成的电池组，如图所示。某品牌电动小轿车蓄电池的数据如下表所示。下列说法正确的是
电池只数输入电压充电参数放电时平均电压/只电池容量/只
100只交流220 420V，20A 3.3V 120Ah
A．将电池组的两极直接接在交流电上进行充电
B．电池容量的单位Ah就是能量单位
C．该电池组充电时的功率为4.4kW
D．该电池组充满电所储存的能量为1.4╳108J
【参考答案】D
【名师解析】电池组需要通过专用充电器接在220V交流电上进行充电，选项A错误；电池容量的单位Ah 是电流与时间乘积，根据q=It可知，是电荷量单位，选项B错误；根据表中充电参数420V，20A，可知该电池组充电时的功率为P=UI=420×20W=8400W=8.4kW，选项C错误；q=It=120×3600C，该电池组充满电所储存的能量为E=100Uq=100×3.3×120×3600J=1.4╳108J，选项D正确。
6．（2017北京东城二模）移动电源（俗称充电宝）解决了众多移动设备的“缺电之苦”，受到越来越多人的青睐。目前市场上大多数充电宝的核心部件是锂离子电池（电动势3.7 V）及其充放电保护电路、充放电管理电路、升压电路等。其中的升压电路可以将锂离子电池的输出电压提升到手机、平板电脑等移动设备所要求的输入电压(5 V)。
由于锂离子电池的材料特性，在电池短路、过高或过低温度、过度充电或放电等情况下都有可能引起电池漏液、起火或爆炸。为安全起见，中国民航总局做出了相关规定，如图1所示。
为了给智能手机充电，小明购买了一款移动电源，其铭牌如图2所示。给手机充电时该移动电源的效率按80%计算。
根据以上材料，请你判断
A．这款移动电源能为手机提供的最大充电量为8000 mAh
B．这款移动电源充满电后所储存的总化学能为37 Wh
C．乘飞机出行时，这款移动电源可以托运
D．Wh与mAh均为能量单位
【参考答案】B
【名师解析】这款移动电源能为手机提供的最大充电量为10 000 mAh，选项A错误；这款移动电源充满电后所储存的总电能为10 000 mAh×3.7V=37Wh，由能量守恒定律可知总化学能为37 Wh，选项B正确。乘飞机出行时，这款移动电源可以随身携带，不可以托运，选项C错误。Wh为能量单位，mAh为电量单位，选项D错误。此题正确选项为B。
7. （2018浙江名校联考）如图是某品牌吊扇的相关参数，假设吊扇的吊杆下方的转盘与扇叶的总质量为7 kg, 则吊扇（g=10m/s2）（）
A.吊扇的内电阻为880 Ω
B.以额定功率工作时通过吊扇的电流为0.25 A
C.正常工作时吊扇机械功率等于55 W
D.正常工作时吊杆对转盘的拉力大于70 N
【参考答案】B
【命题意图】本题考查对吊扇铭牌数据的理解，意在考查运用电功率、能量守恒定律、牛顿运动定律等相关知识分析解决实际问题的能力。
【解题思路】由于吊扇不是纯电阻元件，根据题述和铭牌数据，不能计算出吊扇的内电阻，选项A 错误；由P=UI 可得：以额定功率工作时通过吊扇的电流为I=P/U=0.25 A ，选项B 正确；铭牌中给出的是吊扇额定电功率55W ，由于电流热效应和机械损耗，正常工作时吊扇机械功率一定小于55 W ，选项C 错误；吊扇正常工作时，对空气有向下的作用力，根据牛顿第三定律，空气对吊扇有向上的作用力，所以正常工作时吊杆对转盘的拉力一定小于70 N ，选项D 错误。
8.(多选)下表列出了某品牌电动自行车及所用电动机的主要技术参数，不计其自身机械损耗。若该车在额定状态下以最大运行速度行驶，则( )
自重 40 kg 额定电压 48 V 载重
75 kg 额定电流 12 A 最大行驶速度
20 km/h
额定输出功率
350 W
A.电动机的输入功率为576 W
B.电动机的内电阻为4 Ω
C.该车获得的牵引力为104 N
D.该车受到的阻力为63 N 【参考答案】AD
【名师解析】 由于U ＝48 V ，I ＝12 A ，则P ＝IU ＝576 W ，故选项A 正确；因P 入＝P 出＋I 2r ，r ＝
P 入－P 出
I 2
＝
576－350
122
Ω＝1.57 Ω，故选项B 错误；由P 出＝Fv ＝F f v ，F ＝F f ＝63 N ，故选项C 错误，D 正确。
9、(多选)下表列出了某品牌电动自行车及所用电动机的主要技术参数，不计其自身机械损耗。
若该车在额定状态下以最大运行速度行驶，则()
B．电动机的内电阻为4 Ω
C．该车获得的牵引力为104 N
D．该车受到的阻力为63 N
【参考答案】AD
【名师解析】由于U＝48 V，I＝12 A，则P＝IU＝576 W，故选项A正确；因P入＝P出＋I2r，r
＝P入－P出
I2＝
576－350
122Ω＝1.57 Ω，故选项B错误；由P出＝Fv＝F f v，F＝F f＝63 N，故选项C错
误，D正确。
10.某电力公司曾举办“计量日进您家”活动，免费上门为市民进行家庭用电耗能诊断分析，针对每户家庭提出个性化的节能建议。在上门实测过程中，电力技术人员发现，家电待机耗电成为最容易被市民忽略的问题。以电视机为例，待机一天的耗电量在0.2度左右，小小机顶盒一天待机耗电量更是高达0.4度。根据专家统计：每使用1度(千瓦时)电，就相应消耗了0.4 kg标准煤，同时产生0.272 kg碳粉尘、0.997 kg二氧化碳、0.03 kg二氧化硫、0.015 kg氮氧化物，根据下表提供的数据，若按照1度电电费为0.5元，估算一户普通家庭待机一年
A.多消耗电能14度
B.多交电费7元
C.相应多消耗了标准煤56 kg
D.相应产生的二氧化碳为1 400 kg
【参考答案】C
【名师解析】一户普通家庭待机一年消耗电能为E＝(4＋2×1＋2×4＋1×2)×24×365÷1 000度≈140度，
相应消耗了标准煤m1＝140×0.4 kg＝56 kg，选项A错误，C正确；多交电费140×0.5元＝70元，选项B错误；相应产生的二氧化碳为m2＝140×0.997 kg≈140 kg，选项D错误。
11.（2019浙江大学附中模拟）如图所示的电动自行车既可以电动骑行，也可以脚踏骑行。当电动骑行
时，蓄电池对车上的电动机供电，电动机为车提供动力。下表是某型号电动自行车的主要技术参数，根据学过的物理知识，判断以下估算结果中合理的是（）
整车整车质量40kg
最高车速30km/h 最大噪声62dB
蓄电池
电压48V
容量12Ah
电动机
额定电压48V
额定功率240W
A. 电动骑行过程中，电动机的工作电流保持5A不变
B. 蓄电池一次充足电，放电时可输出的电能最多约为
C. 若蓄电池储存能量的80%用于驱动电动自行车在平均阻力为40N的水平公路上匀速行驶，蓄电池一
次充满电后，最多可供电动骑行约41m
D. 蓄电池一次充电后，该电动自行车最多可供电动骑行2.4h
【参考答案】C
【名师解析】由P=UI可得，电动机的额定电流：I电动机===5A，因电动机行驶过程中不一定以额定功率行驶，所以，电动机的工作电流不一定是5A，选项A错误；蓄电池一次充足电后放电时可输出的最多电能：W电=U电I电t电=48V×12A×3600s=2.0736×106J，故B错误；每次充电可以给电动机提供的电能：
W机械=W电×80%=2.0736×106J×80%=1.65888×106J，因电动车匀速行驶，所以，受到的牵引力F=f=40N，由W=Fs 可得，最远行驶的距离：s===41472m≈41km，选项C正确；由I=可得，该电动自行车最多可供电动车骑行的时间：t==═2.4h，因电动车行驶时，蓄电池不能完全放电，
所以，该电动自行车最多可供电动车骑行的时间小于2.4h，选项D错误。
【关键点拨】知道电动机的额定电压和额定功率，根据P=UI求出额定电流，电动机行驶过程中不一定以额定功率行驶，通过电动机的电流不一定等于额定电流；知道蓄电池的电压和容量，根据W=UIt求出蓄电池
一次充足电后放电时可输出的最多电能；根据效率公式求出蓄电池提供的电能，电动车匀速行驶时受到的牵引力和阻力是一对平衡力，二力的大小相等，根据W=Fs 求出蓄电池一次充电最多能连续电动车行驶的距离；根据I=
求出蓄电池一次充电后，完全放电时该电动自行车最多可供电动车骑行的时间，但蓄电池不
能完全放电。本题考查了学生对电功率公式、电功公式、做功公式、效率公式的掌握和运用，要注意正常行驶时的功率和额定功率相等，否则两个功率不相等，要学会从题目所给信息中找到有用的数据。
12．如图所示为某家庭使用的智能电饭锅的铭牌。假定该家庭用此电饭锅每天煮饭两次，每次在额定功率下工作30分钟，电价1 kW·h 为0.54元。则该电饭锅(
)
A ．工作时的额定电流为0.41 A
B ．保温时的电流为0.2 A
C ．每次煮饭需消耗电能4.5 kW·h
D ．一年因煮饭而产生的电费约为177元 【参考答案】D
【名师解析】根据额定电压和额定功率可知工作时的额定电流为4.1 A ，选项A 错误；根据保温能耗30W ，可知保温时的电流为0.14A ，选项B 错误；该家庭用此电饭锅每天煮饭两次，每次在额定功率下工作30分钟，每次煮饭需消耗电能W=Pt=0.9kW×0.5h=0.45 kW·h ，选项C 错误；一年因煮饭而消耗电能W=365×2×Pt=365×2×0.45 kW·h=320 kW·h ，产生的电费约为320×0.54元=177元，选项D 正确。
二．计算题
1.下表是一辆电动自行车的部分技术指标，其中额定车速是指电动自行车满载情况下在水平平直道路上以额定功率匀速行驶的速度。
请参考表中数据，完成下列问题 (g 取10 m/s 2)：
(1) 此电动机的电阻是多少？正常工作时，电动机的效率是多少？
(2) 在水平平直道路上行驶过程中电动自行车受阻力是车重（包括载重）的k 倍，试计算k 的大小。 (3) 仍在上述道路上行驶，若电动自行车满载时以额定功率行驶，当车速为2m/s 时的加速度为多少？
额定车速 整车质
量 载重
额定输出功
率 电动机额定工作电压和
电流 18km/h
40kg
80kg
180W
36V/6A
【名师解析】
(1) 从表中可知，输出功率P 出= 180W ，输入功率P 入=UI =36×6W=216W
P 损= P 入-P 出 =36W ，，
由 22=636W P I r r =?=损 解得此电动机的电阻 r =1 Ω 。 正常工作时，电动机的效率180
=
=83.3%216
P P η=
出入
(2)根据题述，阻力f= k (M+m )g ，额定车速v m =18km/h=5m/s ， 在水平平直道路上行驶过程中，F=f 。 由P 出=Fv m =k (M+m )gv m 解得：k =
()m
P M m gv +出
=1800.03(4080)105=+?? (3)由 P 出=Fv 解得当车速为2m/s 时的牵引力F=90N
由牛顿第二定律，F -k （M+m ）g =（M+m ）a 解得加速度为a =0.45 m/s 2
2.(14分)表中是一辆电动自行车说明书上的一些技术参数，根据表中提供的信息，探究以下问题：
(1)(2)假设行驶过程中所受阻力是车和人总重的0.02，在最大载重量的情况下，人骑车行驶的最大速度为多大.(g 取10 m/s 2)
【名师解析】：(1)由题目提供的表可知，电机的输出功率为P 出＝120 W ，额定电压为U 0＝40 V ，额定电流为I 0＝3.5 A
电机正常工作时输入功率为 P 入＝U 0I 0＝40×3.5 W ＝140 W 所以电机的效率为
η＝P 出P 入×100%＝120
140×100%≈85.7%
设电机的内阻为r ，
则由欧姆定律知P入－P出＝I20r
解得r≈1.63 Ω.
(2)设车的质量为m，人的质量为M，由题意知行驶时所受阻力为F阻＝k(M＋m)g，当达到最大速度v max 时，应用P出＝F阻v max解得v max＝4 m/s.
答案：(1)1.63 Ω85.7%(2)4 m/s
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2015高考新课标卷II 短文改错 解析
2015高考新课标卷II 短文改错解析 山东郯城第一中学高级教师徐保国假定英语课上老师要求同桌之间交换修改作文，请你修改你同桌写的以下短文。短文中共有10处错误，每句中最多有两处。错误涉及一个单词的增加、删除或修改。 增加：在缺词处加一个漏字符号(∧), 并在此符号下面写出该加的词。 删除：把多余的词用斜线（＼）划掉。 修改：在错的词下划一横线, 并在该词下面写出修改后的词。 注意：1、每处错误及其修改均仅限一词； 2、只允许修改10处, 多者(从第11处起) 不计分。 One day, little Tony went to a shopping center with his parent. It was very crowded. Tony saw a toy on a shop window. He liked it so very much that he quickly walked into the shop. After looks at the toy for some time, he turned around and found where his parents were missing. Tony was scared and begun to cry. A woman saw him crying and telling him to wait outside a shop. Five minutes later. Tony saw parents. Mom said, ”How nice to see you again! Dad and I were terrible worried.” Tony promised her that this would never happen again. 第一处：parent改为parents 考查名词。根据下文，Five minutes later. Tony saw parents.可知此句中parent应该用复数形式。 第二处：on改为in考查介词。在商店橱窗里，用介词in。 第三处：删去very。前面有so，不用very，so和that构成固定句式。 第四处：looks改为looking。考查非谓语动词。after是介词，后面应该用动名词作宾语。第五处：where改为that考查宾语从句引导词。that引导宾语从句，在从句中不做成分，也没有意义，而where引导宾语从句时，在从句中作地点状语，有“在哪里”的意思。 第六处：begun改为began考查动词。begin是不及物的，没有被动语态，此处直接用过去式形式和前面的was scared一起做并列谓语。 第七处：telling改为told考查谓语动词。and后tell和saw一起做并列谓语，而并非和crying 并列做宾补。 第八处：a改为the考查冠词。上面已经提到shop，再次提到用定冠词。 第九处：saw后加his。名词parents前缺少限定词。 第十处：terrible改为terribly考查副词。修饰形容词worried，用副词形式。
专题09 备战高中高考英语短文改错-2021高考英语短文改错专项练习(解析版)
改错专项09 1.假定英语课上老师要求同桌之间交换修改作文，请你修改你同桌写的以下作文。文中共有10处语言错误，每句中最多有两处。每处错误仅涉及一个单词的增加、删除或修改。 增加：在缺词处加一个漏字符号(∧)，并在其下面写出该加的词。 删除：把多余的词用斜线(＼)划掉。 修改：在错的词下划一横线，并在该词下面写出修改后的词。 注意：⒈每处错误及其修改均仅限一词；⒉只允许修改10处，多者(从第11处起)不计分。 The summer holiday is coming. My classmates and I are talking about what to spend our holiday. We can chose between staying at home and take a trip. If we stayed at home, it will be comfortably and there will be no need to spend money. Therefore, in that case, we will learn little about the world. If we go on a trip abroad, we can broaden our view and gain knowledges we cannot get from books. Some classmates suggest we can go to places of interest nearby. I think that it is good idea. It does not cost many, but we can still learn a lot. 【答案】①what改为how或者where ②chose改为choose ③take改为taking ④stayed改为stay ⑤comfortably 改为comfortable ⑥Therefore改为However ⑦knowledges改为knowledge ⑧删掉can或can改为should ⑨增加a ⑩many改为much 【解析】 （1）考查连词。“疑问词+不定式”结构中。疑问代词who, what, which等和疑问副词when, where, how等后面跟不定式，以及连接副词whether与不定式连用构成不定式短语（这个不定时短语相当于一个名词）。这种结构在句中可作主语、宾语、表语、宾语的补语，状语，同位语等句子成分。注意：疑问副词why后不能跟动词不定式。句意：同学们和我在谈论怎样/在哪度暑假。。“特殊疑问词+to do”在本句中作为短语talk about的宾语，用how/where，故what改为how/where 。 （2）考查动词。情态动词后面跟动词原形，故chose改为choose。 （3）考查非谓语动词。本句中动名词短语staying at home与taking a trip构成并列关系，都作为介词between的宾语，因此take改为taking 。 （4）考查连词。句意：如果我们待在家里，很舒服也没有必要花钱。这是一个条件状语从句，从句用一般现在时表将来，故stayed改为stay。 （5）考查形容词。句意：如果我们待在家里，很舒服也没有必要花钱。be动词后接形容词作表语，故comfortably改为comfortable。 （6）考查副词。句意：但是，那样的话，我们对这个世界了解不多。前后是转折关系，故Therefore改为However。