OCR GCSE物理 电学 电流 电压 电阻 串并联
电学在 OCR GCSE Physics 里的分量不小，Topic P3 就叫 Electricity and magnetism，spec 把它拆成 Static and charge 与 Simple circuits 两小节，Practical Activity Group 里还专门安排了 PAG6 测元件的 I-V 曲线、PAG7 比较串并联灯泡的亮度。OCR 官方 spec（GCSE (9-1) Physics A Gateway Science，代码 J249，2016 年 9 月起施教，首考 2018 年 6 月）把这一块的公式分成两栏，一栏印在卷子上，一栏要求背下来。这一块的题型很稳定：给一张电路图让你算电流，或者给一组读数让你判读图线，或者问并联之后总电阻为什么变小。下面按 spec 的顺序拆成八个知识块，每块落到具体公式和数字上，看完拿一张空白电路图自己标一遍电流方向，标不顺的地方就是下次复习的入口。
Electricity carries a large share of the marks in OCR GCSE Physics, where Topic P3 is called Electricity and magnetism and splits into Static and charge plus Simple circuits; the Practical Activity Groups add PAG6 for I-V characteristics and PAG7 for comparing the brightness of bulbs in series and parallel. The specification for Physics A (Gateway Science), code J249, first teaching September 2016 with first assessment in June 2018, separates the equations printed on the paper from those students are expected to recall. Question style is repetitive: a circuit diagram to work through, a set of readings to turn into a graph, or an explanation of why adding a parallel branch lowers the total resistance. The blocks below follow the order of the specification and stay with numbers, units and mark-scheme wording rather than general advice.
电荷量与电流的计算: Q = It
电流的定义是每秒通过导体某一截面的电荷量，写成公式是 Q = I x t，其中电荷量 Q 用库仑（C），电流 I 用安培（A），时间 t 用秒（s）。1 A 等于每秒 1 C，所以 2 A 的电流持续 30 s，通过的总电荷是 60 C，换算成电子个数用元电荷量 1.6 x 10⁻¹⁹ C，得到大约 3.75 x 10²⁰ 个电子。很多学生以为电流会被灯泡用掉一部分，回路里电荷其实处处守恒，灯泡只是把电能转成光和热，灯泡前后的电流读数完全相同。静电那一节讲的是电荷不流动时的行为，摩擦起电靠电子从一个物体转移到另一个物体，得到电子的带负电，失去电子的带正电。同种电荷相互排斥，异种电荷相互吸引，带电气球靠近细水流会让水流偏转，原因是水流被静电场极化后受到吸引。做题时先看清题目给的是电流与时间，还是电荷量与时间，两者用的是同一个公式的不同变形。
Current is defined as the charge passing a point in a conductor each second, written Q = I x t with charge in coulombs, current in amperes and time in seconds; 2 A for 30 seconds therefore moves 60 C, which at an elementary charge of 1.6 x 10⁻¹⁹ C corresponds to about 3.75 x 10²⁰ electrons. Charge is conserved around a circuit, so a lamp converting energy into light and heat does not consume the current itself, and the reading before the lamp equals the reading after it. Static electricity deals with the same charges when they are trapped, and rubbing transfers electrons so that one object ends up negative and the other positive. Like charges repel and unlike charges attract, which is why a charged balloon polarises and bends a thin stream of water.
电路图符号与闭合回路
电路图里每个元件都有固定符号：一长一短两条线表示电池，带十字的圆圈表示灯泡，矩形表示电阻，斜线表示开关。读图的第一件事是确认回路是否闭合，开关断开时整条回路没有电流，电源的电势差再大也一样。电源提供电势差（potential difference），单位伏特（V），它把能量交给电荷，电荷绕一圈经过各个元件时再把能量交出去。图上标的电流方向是常规电流（conventional current），从正极出发经外电路回到负极，电子实际流动方向与它相反，笔试只认常规电流这一套。安培表测电流要串联在待测支路里，伏特表测电势差要并联在元件两端，很多学生以为两块表接法一样，接错的结果是零读数或者烧表。题目要求同时测灯泡的电压和这条支路的电流时，正确画法是伏特表跨在灯泡两边、安培表串在回路里。
A circuit diagram uses fixed symbols: a long and a short line for a cell, a circle with a cross for a lamp, a rectangle for a resistor and a diagonal for a switch. The first thing to establish is whether the loop is closed, because an open switch stops the current no matter how large the supply potential difference happens to be. The supply transfers energy to the charges, which hand it on to the components so that little is left by the time they return to the cell; current direction is drawn as conventional current from the positive terminal round the outside to the negative one, with electron flow the other way. Ammeters go in series with the component under test and voltmeters in parallel across it, and swapping those two connections gives a zero reading or a damaged meter instead of a measurement.
串联电路的三条关系: 电流 电势差 电阻
串联只有一条路径，所以电流处处相同，任何位置的读数都一样，这条关系通常是串联题的入口。电势差按元件分摊，电源的电势差等于各元件电势差之和，两个电阻串在 6 V 电源上，两端读数加起来一定回到 6 V。总电阻等于各电阻之和，R = R1 + R2，串联的元件越多总电阻越大，电流越小，旧式灯串多接一个灯泡就整体变暗，原因就在这里。算例：6 V 电池串接 2 Ω 与 4 Ω 两个电阻，先求总电阻 6 Ω，再求电流 I = V / R = 6 / 6 = 1 A，回代得两个电阻上的电势差分别是 2 V 与 4 V，加起来正好是 6 V，这一步可以当验算用。学生最常把串联的电流当成各元件分摊，写成电流被灯泡用掉一部分，这类答案在阅卷时拿不到过程分。串联电路里任何一处断开，整条回路都没有电流，串联开关因此能一次控制所有元件。
A series circuit offers one path only, so the current is the same at every point, which is usually the quickest way into the question; potential differences add up so that the supply value equals the sum across the components, and two resistors on a 6 V supply must share exactly 6 V. Total resistance is the sum of the individual values, which explains why adding one more lamp to an old string of lights dims all of them. For a 6 V battery with 2 Ω and 4 Ω in series the total is 6 Ω, the current works out at 1 A, and the two potential differences come back as 2 V and 4 V, a useful self-check because they have to return to 6 V. Breaking the loop anywhere stops the current everywhere, which is how a single switch controls every part of the same series path.
并联电路的支路规律与总电阻
并联的每个支路都直接接在电源两端，各支路两端的电势差相同，都等于电源的电势差，这一条与串联刚好相反。电流则分流，总电流等于各支路电流之和，接的支路越多，从电源取出的电流越大。总电阻的变化方向常被问：并联等于多开了一条通路，总电阻一定小于所有支路中最小的那个电阻，4 Ω 与 6 Ω 并联得到 2.4 Ω，而不是 10 Ω。算例：12 V 电源上并联 4 Ω 与 6 Ω，支路电流分别是 3 A 与 2 A，总电流 5 A，用 R = V / I 反推总电阻 12 / 5 = 2.4 Ω。常见的错误是把并联的总电阻直接相加，或者把各支路电流加完就算完，忘记电源供给的是这个总电流。家里的电器全部并联，接线也简单，每件电器都在额定电势差下独立工作，关掉一件不影响其他。
In a parallel circuit every branch connects straight across the supply, so the potential difference across each branch equals that of the supply, the reverse of the series case. The current divides instead: the total drawn from the supply is the sum of the branch currents, so more branches mean a larger supply current. Total resistance always falls below the smallest branch value and never rises, and 4 Ω with 6 Ω in parallel gives 2.4 Ω rather than 10 Ω. With 12 V across 4 Ω and 6 Ω the branch currents are 3 A and 2 A, the supply current is 5 A and the total resistance comes back as 2.4 Ω. Household appliances are wired in parallel so that each works at its rated potential difference and switching one off leaves the rest running.
欧姆定律与三种 I-V 图线的判读
欧姆定律写成 V = I x R，三个量知道两个就能求第三个，电阻的定义式是 R = V / I，单位欧姆（Ω）。OCR 官方 spec 要求通过 PAG6 测出某个元件的 I-V 曲线，做法是改变电源电压，逐点记录电流与电势差，再把数据画成坐标图。固定电阻的图线是过原点的直线，斜率 ΔI / ΔV 等于 1 / R，斜率越大电阻越小，直线的形状说明电阻不随电压改变。灯丝的图线会弯，电压升高时电流上升得越来越慢，原因是灯丝温度升高、电阻变大，答题时必须写成电阻随温度升高而增大，写成电阻变大而不提温度通常只给一半分。二极管在正向电压达到约 0.6 V 之前几乎没有电流，之后迅速上升，反向电压下电流为零，只允许一个方向通过。判读图线时分三步：看是否过原点，看是否弯曲，看有没有单向截止，三个特征足够把三种元件分完。
Ohm's law is written V = I x R, with resistance defined as R = V / I in ohms, and the specification expects an I-V characteristic to be obtained through PAG6 by varying the supply and recording current and potential difference point by point. A fixed resistor gives a straight line through the origin whose gradient is proportional to 1 / R, and a flatter line means a larger resistance that does not change with potential difference. A filament lamp curves instead, because the wire heats up and its resistance rises, so the answer has to say that resistance increases with temperature rather than treating it as a constant. A diode conducts almost nothing until the forward potential difference reaches about 0.6 V, then rises steeply, and passes no current in reverse, so the three shapes can always be separated by checking the origin, the curvature and the cut-off.
电功率与电能: P = VI 与 E = Pt
电功率衡量电能转成其他形式的快慢，公式 P = V x I，单位瓦特（W），家电的额定功率就是它正常工作时每秒消耗的焦耳数。电能写成 E = P x t，单位焦耳（J），家庭用电量习惯用千瓦时（kWh），1 kWh 等于 3.6 x 10⁶ J，也就是常说的 1 度电。算例：电水壶接在 230 V 电源上，工作电流 4 A，功率是 920 W，连续工作 5 分钟即 300 s，消耗的电能是 920 x 300 = 276000 J，换算成千瓦时约 0.077 kWh。反过来求电流也常考：2300 W 的电器在 230 V 下工作，I = P / V = 10 A，选保险丝要选额定电流略大于 10 A 的那一档，太小会频繁熔断，太大就失去保护作用。保险丝靠电流过大时发热熔断来断开电路，所以它串在火线上，不串在零线上。很多学生以为功率大的电器一定耗电多，实际还要看使用时间，短时间大功率的耗电可能低于长时间小功率。
Electrical power measures how quickly energy is transferred, P = V x I in watts, so a rated appliance converts that many joules each second, and the energy transferred is E = P x t in joules, with domestic consumption normally quoted in kilowatt hours where 1 kWh equals 3.6 x 10⁶ J. A kettle drawing 4 A from a 230 V supply has a power of 920 W and transfers 276000 J over 300 seconds, which is roughly 0.077 kWh. Working the other way, a 2300 W appliance on 230 V draws 10 A, so a fuse has to be rated just above that value, since one that is too small keeps blowing and one that is too large stops protecting anything. A fuse melts when the current overheats the wire and breaks the circuit, which is why it sits in the live wire; power alone does not settle energy use, because the time it runs matters just as much.
家庭用电: 交流 火线零线 接地与安全
市电是交流电，方向每秒改变 50 次，也就是频率 50 Hz，电压在正负之间来回摆动，有效值约 230 V，OCR 官方 spec 在电学这一块写出的正是 230 V 与 50 Hz 这两个数字。电池和太阳能电池板输出的是直流，方向不变，示波器上交流是一条正弦曲线，直流是一条水平直线，判断类型只看形状就够。三孔插座里火线（live）把电流送进电器，零线（neutral）把电流送回，地线（earth）平时没有电流，只在金属外壳漏电时把电流导走。带金属外壳的电器要接地，一旦内部导线碰到外壳，电流走地线而不是穿过人的身体，同时大电流把保险丝熔断，电路被切断。双重绝缘的电器塑料外壳把带电部分完全包住，不做接地也安全，插头只有两根线。把危险和保护措施对上号是这类题的常规要求：绝缘层破损可能触电，接保险丝与地线属于保护，插座过载会让导线过热起火。
Mains electricity is alternating, changing direction 50 times a second at 50 Hz with an effective value of about 230 V, the two figures the specification states for this topic, whereas a battery or a solar cell supplies direct current that never reverses; on an oscilloscope the a.c. trace is a sine wave and the d.c. trace a flat line. In a three-pin plug the live wire carries current in, the neutral carries it back and the earth wire normally carries nothing until a fault sends current through a metal case instead of through a person. Double-insulated appliances keep their live parts inside plastic and need no earth connection, so a two-core cable is enough. Hazards and their fixes pair up neatly: damaged insulation risks a shock, earthing and fusing protect against it, and an overloaded socket can overheat the cable enough to start a fire.
电学实验与结构化题的答题要点
电学实验题要求把步骤写全：测 I-V 曲线先画好电路，安培表串联、伏特表并联，改变电源电压后逐点读数并列表记录，坐标轴标出量与单位，最后说明图线的形状说明了什么。PAG7 比较串并联灯泡亮度时，两次测量之间保持电源电压不变，只改连接方式，变量不唯一的话结论无效。结构化题常把三条线索放在一起：一张电路图、一组读数、一个解释要求，先算数再解释，解释句里写出因果关联，只写数字拿不到解释分。读数题要看量程与最小分度，0-1 A 量程的安培表每一小格 0.02 A，读数保留两位小数最稳，单位漏写是另一处常见的扣分点。单位换算集中在几处，mA 与 A 差 1000 倍，kΩ 与 Ω 差 1000 倍，小时与秒差 3600 倍，算电能时最容易混。阅卷时能明显看出两类答案：一类写了公式、代入、结果和单位，另一类只写结果，分数差距往往就在前半段的推导上。
Practical work on this topic needs a complete method: build the circuit with the ammeter in series and the voltmeter in parallel, vary the supply, record each pair of readings in a table, label both axes with quantity and unit, and then state what the shape of the graph shows. When comparing bulb brightness in series and parallel for PAG7, keep the supply potential difference fixed and change only the connection, since moving two variables at once makes the conclusion worthless. Structured questions normally combine a diagram, a set of readings and an explanation task, so do the arithmetic first and then write the reason as a sentence that links cause to effect, because a bare number scores nothing on an explanation line. Meter ranges repay attention as well, since a 0-1 A ammeter reads to 0.02 A per division, and missing units lose marks that were otherwise earned.
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