WJEC物理 GCSE 电路 波 比热容 半衰期 红移
WJEC GCSE Physics 的学生最常做的事是把课本重读一遍，然后带着对三张卷结构的模糊印象进考场。这门课只有三个单元：Unit 1 是电学、能量与波，Unit 2 是力、宇宙与放射性，两张笔试各占 45%、各 80 分、时长都是 1 小时 45 分钟；第三单元占 10%，在校内完成、由 WJEC 外部评分。WJEC 官方 spec（代码 3420，2016 年 9 月起施教、2018 年首次认证）对数学技能的权重写得很明确：至少 30% 的分值落在计算与图表上，科学探究技能至少占 15%。下面的内容按三个单元的顺序，把最常考的计算、图形与失分动作逐条写清。
Revision for WJEC GCSE Physics usually means rereading a textbook, yet the paper rewards three specific skills; the qualification has only three units, with Unit 1 covering electricity, energy and waves and Unit 2 covering forces, space and radioactivity at 45% each and 80 marks per paper. Unit 3 carries 10% and is assessed in centre but marked externally; the WJEC specification (code 3420, first teaching September 2016, first award 2018) puts mathematical skills at a minimum of 30% of the marks. Scientific enquiry skills carry at least 15%, and every section below follows that official unit order with the calculations that appear most often.
三张卷的分值结构与 2026 新版 Unit 3
Unit 1 与 Unit 2 各 1 小时 45 分钟、80 分、占 45%，题型混合了短答、结构化问答、延展写作与数据题，其中一部分情境题直接给实验数据。Unit 3 占 10%，现行 spec 里它叫 Practical Assessment，30 分，在 1 月到 2 月由校内完成再寄回 WJEC 外部评分。WJEC 在 2026 年 2 月发布的 spec 第 3 版把它改名为 Scientific Enquiry，二选一探究，含 1 小时实践任务与 1 小时笔答任务，总分 28 分，新 Unit 3 从 2028 年开始给分。能力目标权重上，现行版本是 AO1 40%、AO2 40%、AO3 20%，新版把 Unit 3 内部改成 AO1 2.5%、AO2 5%、AO3 2.5%，整科因此变成 38.5%、41%、20.5%。Higher Tier 覆盖 A* 到 D，Foundation Tier 覆盖 C 到 G。WJEC 的 2026 年关键日期表把实践考核窗口写在 1 月 5 日到 2 月 13 日，成绩发布在 8 月 20 日。
Unit 1 and Unit 2 are each 1 hour 45 minutes, 80 marks and 45% of the qualification; both mix short answers, structured questions, extended writing and data response items. Unit 3 holds the remaining 10%, and in the current specification it is a Practical Assessment of 30 marks taken in centre between January and February and marked externally; version 3 of the specification, published in February 2026, renames it Scientific Enquiry with a one-hour practical task and a one-hour written task worth 28 marks, first awarded in 2028. The current specification weights the objectives at 40% AO1, 40% AO2 and 20% AO3, while the updated unit moves the overall balance to 38.5%, 41% and 20.5%; Higher Tier runs A* to D and Foundation Tier C to G.
电荷与电流 Q = It
电流的定义是单位时间内通过某点的电荷量，公式写作 Q = It，Q 用库仑、I 用安培、t 用秒。WJEC 的题常把时间给成分钟，学生最常把 4 分钟直接代进公式，算出的电荷整整小了 60 倍。举例：一个灯泡的电流是 0.25 A，通电 4 分钟，电荷量 Q = 0.25 × 240 = 60 C，这里的 240 来自 4 乘 60。反过来，如果题目给出 90 C 的电荷量与 0.5 A 的电流，时间就是 180 秒，也就是 3 分钟。电荷量与电流是两回事，前者描述搬了多少电荷、后者描述搬得多快，答案里单位写错通常拿不到分。电子电荷是 1.6 × 10^-19 C，用总电荷除以它就能算电子个数，60 C 对应 3.75 × 10^20 个电子。
Current is the charge passing a point each second, so the working equation is Q = It with Q in coulombs, I in amperes and t in seconds; students most often lose the mark by substituting a time given in minutes. A lamp carrying 0.25 A for four minutes passes 60 C once the time is converted to 240 s, and working backwards from 90 C and 0.5 A gives 180 s, which is three minutes. Charge and current are different quantities, so the unit matters: dividing 60 C by the electron charge of 1.6 x 10^-19 C gives 3.75 x 10^20 electrons.
欧姆定律 V = IR 与 I-V 特性曲线
欧姆定律写成 V = I R，适用于温度不变的金属导体；12 Ω 的定值电阻接在 6 V 电源上，电流是 0.5 A，反过来 0.5 A 通过 12 Ω 时两端电压是 6 V。灯丝灯泡不守这条规律，通电后温度升高、电阻变大，I-V 图从原点出发先陡后平，同样的电压增量换来的电流增量越来越小。二极管的曲线更极端，正向电压要超过大约 0.6 V 才出现明显电流，反接时几乎没有电流，所以负半轴贴着横轴。常见错误是给灯泡的 I-V 图画成直线，阅卷时这类图形题的第一分往往就落在曲线的形状上。做这类题的顺序也固定：先读图取点，再用 V = I R 算电阻，最后用温度升高解释曲线的走向。
Ohm's law is written V = IR and holds for a metal conductor at constant temperature; a 12 ohm resistor across 6 V carries 0.5 A, and 0.5 A through 12 ohms gives 6 V across it. A filament lamp breaks the pattern because its temperature rises and its resistance rises with it, so the I-V curve begins steeply and flattens; a diode allows almost no current until the forward voltage passes about 0.6 V and blocks the reverse direction. Reading a point from the graph, calculating resistance with V = IR and explaining the shape through temperature is the order that collects the marks.
串联与并联的分压分流
串联电路里电流处处相同，电压按电阻比例分配，电阻直接相加：2 Ω 与 3 Ω 串联得 5 Ω，接 10 V 电源时电流是 2 A。并联电路里电压处处相同，电流按电阻反比分配，等值电阻比任何一条支路都小：6 Ω 与 3 Ω 并联得 2 Ω，接 12 V 时两条支路各走 2 A 与 4 A，干路是 6 A。这两条结论在 WJEC 的电路图题里几乎每份卷子都出现，学生最常做错的是把并联电阻直接相加。并联等值电阻用积除以和最快，6 × 3 ÷ (6 + 3) = 2 Ω，比倒数法省一步。电压表并联在被测元件两端、电流表串联在支路里，把电流表并联进主路会让它自己的内阻改变电流，读数就不再是原电路的读数。分压结论还能直接用于电位器题：10 V 电源上两个等值电阻各分得 5 V。
Series circuits carry the same current everywhere, so voltages split in proportion to resistance and the resistances simply add; 2 ohms and 3 ohms in series give 5 ohms, and a 10 V supply drives 2 A. Parallel branches share the same voltage, so currents split in inverse proportion to resistance and the combined resistance is smaller than either branch; 6 ohms and 3 ohms in parallel give 2 ohms, with 2 A and 4 A in the branches and 6 A in total. Ammeters belong in series with the component and voltmeters in parallel across it, because an ammeter placed in parallel changes the very current being measured.
电功率与电能 P = VI
电功率有三种写法：P = V I、P = I^2 R、P = V^2 ÷ R，题目给哪两个量就用哪一个。一个 230 V、4 A 的电水壶功率是 920 W，工作 3 小时消耗的电能 E = P t = 0.92 kW × 3 h = 2.76 kWh。把 kWh 换成焦耳要乘 3.6 × 10^6，2.76 kWh 等于 9.936 × 10^6 J。电费按每千瓦时单价乘度数算，以每千瓦时 0.25 英镑为例，2.76 kWh 大约是 0.69 英镑。很多学生会把功率与电能混着写，题目问消耗多少度电、答案里却出现 W，这种情况即使数字对了也拿不到最后一分。效率题往往和这组数字绑在一起：1 kg 水从 20 摄氏度升到 100 摄氏度需要 336000 J，把这个数与输入的电能相比就是热效率。
Electrical power can be written as P = VI, P = I^2 R or P = V^2 / R, and the question normally supplies the two quantities needed for one of them; a 230 V, 4 A kettle transfers 920 W. Running a 0.92 kW appliance for three hours uses 2.76 kWh, which becomes 9.936 x 10^6 J when multiplied by 3.6 x 10^6, and an assumed price of 0.25 pounds per kWh puts the cost near 0.69 pounds. Efficiency questions hang off the same figures, since heating 1 kg of water by 80 degrees needs 336000 J and that value can be compared with the electrical energy supplied.
比热容与能量守恒 E = mcΔT
比热容写成 E = m c ΔT，其中 E 是吸收或放出的能量、m 是质量、ΔT 是温度变化量，水的比热容是 4200 J/(kg °C)。2 kg 水从 20 摄氏度加热到 80 摄氏度要吸收 2 × 4200 × 60 = 504000 J 的能量，也就是 504 kJ。WJEC 的实验题常让学生用电热丝加热水并每隔一段时间读温度，误差来源要写具体：热量散失到空气里、容器本身也吸热、温度计读数有延迟。很多学生只写一句有误差却不交代方向，反而丢掉解释分；热量散失时测得水吸收的能量偏小，算出的比热容就偏小。1 kg 铝的比热容是 900 J/(kg °C)，同样升温 60 摄氏度只需要 54000 J，这就是金属锅比锅里的水升温快的原因。做这类题时 ΔT 用温度差而不是温度本身，反复出现的失分就是把 80 摄氏度直接当成 ΔT。
Specific heat capacity appears as E = mcΔT, where E is the energy transferred, m the mass and ΔT the temperature change, with water at 4200 J per kg per degree; heating 2 kg of water from 20 to 80 degrees therefore needs 504 kJ. In practical questions the heat lost to the room, the container and the thermometer all matter, and the direction of the error has to be stated, because energy escaping to the surroundings makes the calculated value too small. Aluminium needs only 900 J per kg per degree, so the same 60 degree rise in 1 kg takes 54000 J, which is why metal pans heat up faster than the water inside them.
波的性质 v = fλ
波速等于频率乘波长，v = f λ；用米做波长单位时，波速的单位就是 m/s。声速在空气中约 340 m/s，一个 250 Hz 的音叉产生的声波波长是 340 ÷ 250 = 1.36 m。反过来算也常见：1 MHz 的中波波长是 300 m。横波的振动方向与传播方向垂直，纵波的振动方向与传播方向平行，声波是纵波，光与水面波是横波。反射与折射要分清：反射时波速不变、角度对称；折射时波从空气进水，速度变小、波长变短，而频率保持不变，这一条最容易被写成频率变了。水波槽里数出 5 个波峰之间 4 段距离，把总距离除以 4 就是波长，这是最省时间的一次测量。
Wave speed is frequency multiplied by wavelength, v = f λ, so sound travelling at 340 m/s from a 250 Hz tuning fork has a wavelength of 1.36 m; a 1 MHz radio wave works out at 300 m. Transverse waves vibrate at right angles to the direction of travel and longitudinal waves vibrate along it, which is why sound is longitudinal while light and water ripples are transverse. Refraction changes speed and wavelength but never frequency, and reading four gaps between five crests in a ripple tank gives the wavelength by dividing the total distance by four.
电磁波谱的顺序、用途与危害
电磁波谱按波长从长到短是无线电波、微波、红外、可见光、紫外、X 射线与伽马射线，它们在真空中的速度都是 3 × 10^8 m/s。用途要与波段对上：微波用于卫星通信与加热，红外用于遥控器与热成像，紫外用于杀菌，X 射线用于成像，伽马射线用于检查和治疗。危害随能量升高而增加，紫外以上电离能力变强，过量照射会损伤细胞，所以 X 射线操作者要穿铅围裙并佩戴剂量计。可见光只占很窄的一段，波长大约 400 nm 到 700 nm，红光波长最长、紫光最短。常见错误是把微波炉加热解释成微波直接把食物分子加热，实际过程是微波使水分子转动、内能增加。答题时把用途、危害与能量高低三点一起写，分值通常就按这三点分开给。
The electromagnetic spectrum runs from radio waves through microwaves, infrared, visible light and ultraviolet to X-rays and gamma rays, and all of them travel at 3 x 10^8 m/s in a vacuum. Uses map onto the bands, with microwaves for satellite links and heating, infrared for remote controls and thermal imaging, ultraviolet for sterilising and X-rays for imaging; hazards grow with energy, so X-ray operators wear lead aprons and dosimeters. Visible light occupies only 400 nm to 700 nm, and a common mistake is to say a microwave oven heats food directly rather than by making water molecules rotate.
力与运动 F = ma 与终端速度
平均速度是路程除以时间，加速度是速度变化量除以时间，匀加速题常用 v = u + a t 与 s = u t + ½ a t^2。质量 1200 kg 的汽车受到 3000 N 的合力，加速度 a = 3000 ÷ 1200 = 2.5 m/s^2。重量与质量不同，W = m g，g 取 10 N/kg 时 5 kg 的物体重 50 N，同样 5 kg 在月球上只有 8 N，质量不变而重量变小。v-t 图的斜率是加速度、图下面积是位移，WJEC 2024 年的阅卷报告写明，只有很强的考生能用梯度、正比与比值完成这类计算，所以这两种读图方法要练到不用想。终端速度是最常考的定性题：下落时空气阻力随速度增大，阻力等于重力后合力为零，速度不再增大。这类题的分值通常要求先说阻力与重力的关系，再说合力与速度的关系。
Average speed is distance divided by time and acceleration is the change in velocity divided by time, so constant acceleration work uses v = u + at and s = ut + ½at^2; a 1200 kg car with a 3000 N resultant force accelerates at 2.5 m/s^2. Weight comes from W = mg, giving 50 N for 5 kg at 10 N/kg and about 8 N for the same mass on the Moon, where the mass itself is unchanged. The gradient of a velocity-time graph is the acceleration and the area beneath it is the displacement, and the WJEC 2024 report noted that only the strongest candidates handled gradients, proportionality and ratios confidently.
做功、动能与重力势能
做功写成 W = F d，力的方向与位移方向垂直时做功为零，提着箱子水平走的题每年都出现。动能公式是 ½ m v^2，2 kg 的物体速度 3 m/s 时动能是 9 J；重力势能是 m g h，2 kg 升高 3 m 时是 60 J。能量守恒题的写法是把起点与终点的能量列出来：小球从 1.8 m 高处落下，忽略空气阻力时 m g h = ½ m v^2 给出 v = 6 m/s，质量在两边约掉。汽车安全设计用同一套逻辑：crumple zone 把相同的动能分散到更长的距离与时间上，乘客受到的力因此变小。很多学生会写气囊减小了动量变化量，实际上动量变化量基本不变，变的是作用时间。做题时把这两个量分开写，力与时间的乘积才是解释的落点。
Work done is force multiplied by distance, and it is zero when the force acts at right angles to the motion, as with a case carried horizontally; kinetic energy is ½mv^2, giving 9 J for 2 kg at 3 m/s. Gravitational potential energy is mgh, so 2 kg lifted 3 m stores 60 J, and energy conservation links the two: a ball falling 1.8 m arrives at 6 m/s because the mass cancels on both sides. Crumple zones extend the collision distance and time for the same change in momentum, so the force on the occupants is smaller.
放射性、半衰期与核方程
alpha 粒子是氦核，电离能力最强、穿透能力最弱，一张纸就能挡住；beta 粒子是高速电子，几毫米铝板可以挡住；gamma 是电磁波，需要厚铅板才能明显减弱。半衰期题的套路固定：初始 80 Bq 的样品经过 3 个半衰期后剩 80 ÷ 2^3 = 10 Bq，若半衰期是 6 小时，3 个半衰期就是 18 小时。核方程要同时配平质量数与电荷数，碳 14 变成氮 14 时质量数不变、电荷数加 1，所以放出的是 beta 粒子。铀 238 放出 alpha 后质量数减 4、电荷数减 2，变成钍 234。本底辐射来自岩石、氡气、宇宙射线与食物，测样品计数率时要把本底值减掉，学生最常把本底当成样品本身的辐射。照射与污染要分清：照射是外部射线穿过身体，污染是放射性物质落在身上或被吸入体内。
Alpha particles are helium nuclei with the strongest ionising power and the weakest penetration, stopped by paper, while beta particles are fast electrons stopped by a few millimetres of aluminium and gamma rays need thick lead to be reduced. Half-life questions follow a fixed pattern: 80 Bq falls to 10 Bq after three half-lives, so a six-hour half-life means 18 hours have passed. Nuclear equations must balance mass and charge, which is why carbon-14 becomes nitrogen-14 by emitting a beta particle while uranium-238 loses four mass units to become thorium-234.
红移、大爆炸与恒星的生命周期
远处星系光谱里的特征谱线整体向红端移动，位移量用 Δλ ÷ λ 描述，红移说明这些星系在远离我们。星系越远退行速度越大，Planck 2018 的测量把哈勃常数定在 67.4 km/s/Mpc，对应的宇宙年龄约 138 亿年。大爆炸理论的三条证据是红移、微波背景辐射与轻元素丰度，微波背景辐射的温度约 2.7 K，那是早期高温宇宙冷却下来的余温。恒星的生命周期按质量分岔：像太阳这样的恒星最后成为白矮星，质量更大的恒星经超新星爆发后留下中子星或黑洞。很多学生会把红移解释成星系的光变红了，实际是波长被拉长，红光只是可见光里波长较长的一段。答题时把红移、距离与退行速度三者的关系写成一个链条，比单独描述现象更容易拿满分。
Spectra from distant galaxies show their lines shifted towards the red end, and the size of that shift indicates how fast those galaxies recede; the further away a galaxy is, the faster it moves. The Big Bang model rests on the red shift, the microwave background at about 2.7 K and the observed abundances of light elements, and Planck 2018 measurements put the Hubble constant at 67.4 km/s/Mpc with an age of about 13.8 billion years. Stellar life cycles split by mass, with Sun-like stars ending as white dwarfs and heavier stars leaving neutron stars or black holes after a supernova.
延展写作 QER 的落笔口径
Unit 1 与 Unit 2 都含延展写作题，官方叫 QER，阅卷按知识点、应用与写作质量分层给分。WJEC 2024 年夏季阅卷报告指出，回忆型小题仍然是多数考生丢分的地方，而给数据的应用型小题得分更高，定量能力整体好于定性能力。同一份报告列出的具体失分动作有：替换数字时没换算单位、结果进位方式不对、只给答案不写过程。报告也写明，只要过程可见就有机会拿到方法分，所以计算题把代入式写出来比直接给答案更稳。定性题必须带上数据，报告注意到考生被要求给出定性回答并包含数据时经常不写数据。有效数字要与题目给出的数据保持一致，不要随手截断。
Units 1 and 2 both include extended writing, known as QER questions, marked for knowledge, application and the quality of the writing; the WJEC Summer 2024 report found recall questions still costing candidates marks while data-based application questions scored better. The same report records substitutions with units that were never converted, answers rounded incorrectly and final answers written without working, which matters because method marks are available when the working is visible. Qualitative answers also need data included, since candidates who wrote only a description often left those marks behind.
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