Edexcel IGCSE物理 电流 电压 电阻 电功率
电学是 Edexcel International GCSE Physics（4PH1）specification 里的第 2 个主题，分成四个子块：单位、市电、电路中的能量与电压、电荷。试卷设置值得先看一眼：卷一 2 小时 110 分，占 61.1%，只考不带 P 标记的核心内容；卷二 1 小时 15 分钟 70 分，占 38.9%，核心内容加带 P 标记的加深内容一起考。两卷分数直接相加，180 分不折算，最后按 1 到 9 级评定。电学在两卷都出现，卷一偏基本量关系，卷二才会碰到静电那几条 2.22P 到 2.28P。下面的十三个小节按 2.1 到 2.21 的顺序走一遍公式、图像与失分点。
Electricity is topic 2 of the Pearson Edexcel International GCSE Physics specification, and it splits into four strands: units, mains electricity, energy and voltage in circuits, and electric charge. Paper 1 runs two hours for 110 marks at 61.1 per cent and covers only the core content without a P reference, while Paper 2 lasts 75 minutes for 70 marks at 38.9 per cent and tests everything, including the P material. The two papers add up to 180 marks that are never scaled, and the grade runs from 9 down to 1.
七个单位：安培、库仑、焦耳、欧姆、秒、伏特、瓦特
4PH1 的 2.1 把电学要用的单位一次列清：电流用安培（A）、电荷用库仑（C）、能量用焦耳（J）、电阻用欧姆（Ω）、时间用秒（s）、电压用伏特（V）、功率用瓦特（W）。这七个符号会在卷二的计算题里反复出现，漏写一个通常就是白白丢掉一分。很多学生以为单位只是数字后面的装饰，实际上先落笔写单位能反过来检查公式有没有用错。算出 0.5 A 乘 6 V 得到 3 的时候，如果心里认定结果是焦耳每秒，就该写成 3 W 而不是 3 J。常见的错误是把瓦特与焦耳混着用，功率的答案标成 J，能量的答案标成 W。欧姆符号 Ω 手写时容易被看成 2 或者小写 w，考前把写法固定下来省事。
The specification lists seven units in clause 2.1: ampere, coulomb, joule, ohm, second, volt and watt. Writing the unit before the number is a habit worth building, because the unit alone tells you whether the relationship you used can possibly be right. Handwritten ohms cause genuine trouble in scripts, where a careless omega reads as a two or a small w to the marker.
电流是电荷的流动：固体金属里真正移动的是电子
2.14 的原文是电流等于电荷流动的速率，2.16 再补一句，固体金属导体里的电流是一群带负电的电子在移动。这两条放在一起常被读成矛盾，因为题目里的电流方向是从正极经外电路回到负极，电子却从负极跑向正极。阅卷时看到学生写电子沿电流方向移动，基本可以判断概念没有接上。约定的电流方向是正电荷会走的方向，属于历史留下的规定，与真实载流子是谁无关。算题的时候不需要处理这个差别，解释题就必须写清楚。电解液里移动的是正负离子，这也正是 2.16 只限定固体金属导体的原因。
Clause 2.14 defines current as the rate of flow of charge, and 2.16 adds that the current in a solid metallic conductor is a flow of negatively charged electrons. Conventional current points from positive to negative around the external circuit, which is a historical convention rather than a description of what actually moves. Answering an explanation question with electrons drifting along the conventional direction is a very common slip.
Q 等于 I 乘 t：把时间换成秒再动手
2.15 给出的关系是电荷等于电流乘时间，记作 Q = I × t。一盏灯通过 0.25 A 的电流持续 4 分钟，电荷量就是 0.25 乘 240，等于 60 C。这里唯一容易翻车的地方是时间单位，试卷给分钟、给小时都正常，公式里的 t 必须是秒。这一步最容易出错，很多人在 4 分钟那里直接乘 4，答案差了六十倍。反过来用也常见，给总电荷与时间求平均电流，只要记得把结果标上 A。充电宝、手机电池的容量标的是毫安时，本质还是电荷量，换算时先变成安培乘秒再看。
Clause 2.15 states that charge equals current multiplied by time, or Q = I × t. A lamp drawing 0.25 A for four minutes therefore moves 60 coulombs, because four minutes is 240 seconds. Nearly every error in this calculation comes from leaving the time in minutes, which shrinks the answer by a factor of sixty.
电压是单位电荷获得的能量：一伏特等于一焦耳每库仑
2.20 把电压说得很直白：电压是每单位电荷通过时转移的能量，一伏特就是一焦耳每库仑。由这个定义直接得到 2.21 的关系，能量等于电荷乘电压，写作 E = Q × V。12 C 的电荷通过 6 V 的电源，转移的能量是 72 J。很多学生以为电压是电路里某种被消耗掉的东西，其实它描述的是每一库仑电荷能拿到多少能量。串联电路里各元件分到的电压加起来等于电源电压，原因就在这里，能量按元件分配掉了。写解释题时把每单位电荷这几个字写出来，比只写电压是推力更容易拿到分。
Clause 2.20 defines voltage as the energy transferred per unit charge passed, which makes one volt exactly one joule per coulomb. The relationship E = Q × V follows straight from that definition, so 12 coulombs driven through 6 volts carries 72 joules. Voltage is not something used up along a wire; it is the energy each coulomb can collect, which is why the voltages across components in series add up to the supply voltage.
V 等于 I 乘 R：欧姆定律与它的两个变形
2.13 的关系是电压等于电流乘电阻，写作 V = I × R，变形后得到 I = V ÷ R 与 R = V ÷ I。一个 12 V 电源接 4 Ω 的电阻，电流是 3 A；若换成 12 Ω，电流降到 1 A。2.10 要求能定性说明电阻变大时电流怎么变，在电源电压不变的前提下，电阻变大电流变小，两者成反比。常见的错误是把电阻当成元件的固有属性就完全与电流无关，实际上灯丝这种元件的电阻会随温度变化，所以它不服从这条直线的规律。做题先判断元件是不是欧姆元件，再决定用哪一套说法。
Clause 2.13 gives voltage as current multiplied by resistance, from which I equals V over R and R equals V over I. A 12 volt supply across 4 ohms drives 3 amps, and swapping in 12 ohms drops that to one amp at the same supply voltage. Clause 2.10 asks only for the qualitative direction, so state that current falls when resistance rises with the supply unchanged, then say what happens to the total current.
I-V 图像：导线、电阻、灯丝灯与二极管为什么形状不同
2.9 要能描述电流随电压变化的情况，对象是导线、电阻、金属灯丝灯与二极管，并且知道怎么用实验测出来。导线与固定电阻在温度基本不变时给出一条过原点的直线，斜率恒定意味着电阻恒定。灯丝灯的曲线会弯，电压升高时灯丝温度升高，电阻随之变大，所以同样再升一伏，电流增加得越来越少。二极管只在一个方向导通，反向几乎不导电，正向要过一个小门槛才开始有电流。作图时横轴放电压、纵轴放电流，读斜率得到的是电阻的倒数，单位是安培每伏特。把横纵轴放反了斜率就变成电阻，单位是欧姆，两种画法都能得分，前提是轴标清楚。
Clause 2.9 covers how current varies with voltage for wires, resistors, metal filament lamps and diodes, plus the experiment used to collect the data. A wire or fixed resistor at steady temperature gives a straight line through the origin, while the filament lamp bends because a hotter filament has a higher resistance. A diode conducts in one direction only and needs a small threshold before current rises at all.
串联电路：电流处处相等，电压按电阻分配
2.17 说明电流在节点处守恒，所以串联回路里每个位置的电流都相同。2.19 要求能算两个电阻元件串联时的电流、电压与电阻，总电阻就是两个电阻相加。6 V 电源接 2 Ω 与 4 Ω，总电阻 6 Ω，电流 1 A，两个电阻上的电压分别是 2 V 与 4 V，加回来正好 6 V。2.8 要求理解串联电路里的电流如何取决于所加电压以及其他元件的数量与性质，加一个元件总电阻变大，电流变小。很多学生以为电流流过第一个电阻会被用掉一部分，其实电流不变，被分配的是电压。串联的一个隐患是任一位置断开整条回路都不通，圣诞树灯串常这么接。
Clause 2.17 explains why current is conserved at a junction, so every point in a series loop carries the same current. Clause 2.19 asks you to calculate currents, voltages and resistances for two resistive components in series, where the resistances simply add. With 6 volts across 2 ohms and 4 ohms the current is one amp, and the two voltages are 2 volts and 4 volts.
并联电路：支路电压相同，电流按电阻分
2.18 明确写出并联的两个元件电压相同，这一点与串联正好相反。并联时总电流等于各支路电流之和，所以并联的支路越多，电源要输出的电流越大。2.7 要求解释为什么某些场合更适合串联或并联，家用照明是常考的对照：灯并联，各自的电压都是市电电压，坏一只不影响其他灯；若串联，一只坏了整条回路都不亮，而且每只灯只分到一部分电压因而偏暗。常见的错误是把并联的总电阻当成两个电阻相加，实际上并联后的总电阻比任何一个支路电阻都小。算并联电阻可以用 1 除以各支路倒数之和，也可以用总电流除以电源电压来验证。
Clause 2.18 states that the voltage across two components in parallel is the same, which is the mirror image of the series rule. In parallel the supply current splits between branches, so adding a branch increases the total current the supply must deliver. Domestic lighting is the standard comparison in clause 2.7: lamps in parallel each receive the full mains voltage and one failure leaves the rest working.
电功率 P 等于 I 乘 V：额定值与保险丝的选择
2.4 的关系是功率等于电流乘电压，写作 P = I × V，并要求把这个关系用在保险丝的选择上。一只标着 2000 W 的电水壶接在 230 V 的市电上，工作电流约 8.7 A，选 13 A 的保险丝合适，选 5 A 的就会在正常工作时熔断。选保险丝的规则是额定电流略高于设备正常工作电流，既要容得下正常电流，又要在故障电流出现时及时熔断。很多学生以为保险丝越大越安全，方向正好反了，额定值过大等于让小故障继续烧下去。题目有时候给功率与电压让你反推电流，也会给电流与电阻让你用 P 等于 I 平方乘 R 求功率，两条路径结果一致。
Clause 2.4 links power, current and voltage as P = I × V, and the specification expects that relationship to be used when choosing a fuse. A 2000 watt kettle on a 230 volt supply draws roughly 8.7 amps, so a 13 amp fuse is sensible and a 5 amp fuse would blow in normal use. Fuses are chosen just above the working current: high enough for everyday operation, low enough to fail quickly under fault conditions.
电能 E 等于 I 乘 V 乘 t：焦耳与千瓦时
2.5 的关系是能量等于电流乘电压乘时间，写作 E = I × V × t，与 E = P × t 是一回事。2000 W 的电水壶烧 3 分钟，能量是 2000 乘 180 秒，等于 360000 J；同一件事写成千瓦时就是 2 kW 乘 0.05 h，等于 0.1 kWh，电费单用的正是后一种。2.3 要求理解电流在电阻中为什么会带来能量转移与温度升高，并把这一点联系到家用电器上。电水壶、电熨斗、白炽灯都是靠这个原理工作的，能量从电能变成内能，发热是设计目的而不是副作用。灯丝灯只有一小部分能量变成光，其余变成热，所以它的效率在几种照明里偏低。算电费时先统一单位，瓦要换成千瓦，分钟要换成小时。
Clause 2.5 gives energy transferred as current times voltage times time, which is the same statement as E = P × t. A 2000 watt kettle boiling for three minutes transfers 360,000 joules, or 0.1 kilowatt hours, and household bills use the second form. Clause 2.3 explains that a current in a resistor transfers energy and raises its temperature, the principle behind kettles, irons and filament lamps.
市电、交流与直流：230 V 与 50 Hz 背后的三根线
2.6 要求分清市电是交流电、电池供电是直流电，两者的区别在于电流方向是否周期性反转。英国市电大致是 230 V、50 Hz，方向每秒来回 50 次，这解释了三根线的分工。火线带电压，零线构成回路，地线与金属外壳相连，外壳意外带电时电流走地线而不是穿过人体。2.2 要求理解绝缘、双重绝缘、接地、保险丝与断路器如何保护设备与使用者，其中双重绝缘的电器不需要地线，因为它靠两层绝缘把人与带电部件隔开。断路器与保险丝作用相同，断路器可以复位，保险丝熔断后要更换。火线与零线都碰不得，插头接线时火线接 L、零线接 N、地线接 E，接反了外壳就可能带电。
Clause 2.6 separates mains electricity, which is alternating current, from the direct current supplied by cells and batteries, where the direction does not reverse. UK mains sits at roughly 230 volts and 50 hertz, so the current changes direction fifty times a second, which is why the three wires in a cable have different jobs. Clause 2.2 asks how insulation, double insulation, earthing, fuses and circuit breakers each protect a device or its user.
LDR 与热敏电阻：电阻随光照和温度变化
2.11 要求定性描述光敏电阻的电阻随光照强度如何变化、热敏电阻的电阻随温度如何变化。光敏电阻在光照变强时电阻下降，所以它可以串在分压电路里当光控开关，天亮时让路灯断电。常用的热敏电阻在温度升高时电阻下降，用它做的探头能把温度变化变成电压变化，烤箱与恒温器都用得上。很多学生以为这类元件也服从 V 等于 I 乘 R，其实它们依然服从这条关系，只是 R 不再是个常数。题目常要求解释为什么某个位置的电压会升或降，判断思路是先定电阻变化方向，再看它在串联回路里分到的电压怎么变。
Clause 2.11 covers how the resistance of a light-dependent resistor varies with illumination and how a thermistor varies with temperature. An LDR drops in resistance as light increases, which makes it a natural light-operated switch inside a potential divider. A thermistor behaves the other way round as it warms, and both components still obey V = I × R; only the resistance is no longer constant.
常见失分点：公式不在卷面给出，轴的斜率要看清楚
specification 的附录 7 写明一件事：这些关系式不会在试卷上提供，无论原式还是变形都不给，所以 Q = I × t、V = I × R、P = I × V、E = Q × V、E = I × V × t 这几条必须背到能直接变形。第二个高频失分点是图线的斜率，横轴电压纵轴电流时斜率是电阻的倒数，横轴电流纵轴电压时斜率才是电阻。第三个是把题目给的 230 V 当成已经对了的中间结果，其实额定电压下算出的电流只是工作电流，故障电流要靠电阻或功率另算。阅卷时最常看到的是对完答案却漏掉单位，或者把千瓦时与焦耳混在一起写。备考时把 2.1 到 2.21 的编号对着考试局官网的 specification 逐条过一遍，比刷十套题有用。
Appendix 7 of the specification says plainly that these relationships will not be provided in the paper in either their given or their rearranged form, so Q = I × t, V = I × R, P = I × V, E = Q × V and E = I × V × t have to be recalled cold. The second trap is the gradient of an I-V graph, which gives the reciprocal of resistance when current sits on the vertical axis. Reading the specification clause by clause on the Pearson Edexcel website is a faster way to close gaps than grinding through past papers.
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