📚 WJEC Year 11 Physics Unit 1 Mock Paper Analysis | WJEC 11年级物理第一单元模拟卷解析
This mock paper analysis is designed to help Year 11 students master the key concepts in WJEC GCSE Physics Unit 1 (Electricity, Energy and Waves). Each section breaks down a typical exam-style question, showing step-by-step working and examiner insights. By working through these model answers, you will strengthen your problem-solving skills and avoid common pitfalls in your real unit test.
本模拟卷解析旨在帮助11年级学生掌握 WJEC GCSE 物理第一单元（电学、能量与波）的核心概念。每个小节拆解一道典型考题，展示逐步解题过程与考官评析。通过梳理这些范例答案，你将提升解题能力，避开单元测试中的常见失分点。
1. Series Circuit Calculations | 串联电路计算
A 12 V battery is connected in series with a 4 Ω resistor and a 8 Ω resistor. Calculate (a) the total resistance, (b) the current flowing in the circuit, and (c) the potential difference across the 8 Ω resistor.
一个 12 V 电池与一个 4 Ω 电阻和一个 8 Ω 电阻串联。计算 (a) 总电阻，(b) 电路中的电流，以及 (c) 8 Ω 电阻两端的电压。
For resistors in series, total resistance Rₜ = R₁ + R₂ = 4 Ω + 8 Ω = 12 Ω.
串联电路总电阻 Rₜ = R₁ + R₂ = 4 Ω + 8 Ω = 12 Ω。
Using Ohm’s Law I = V / Rₜ = 12 V / 12 Ω = 1.0 A.
根据欧姆定律 I = V / Rₜ = 12 V / 12 Ω = 1.0 A。
resistor V₈ = I × R₈ = 1.0 A × 8 Ω = 8 V. The remaining 4 V drops across the 4 Ω resistor.
8 Ω 电阻两端的电压 V₈ = I × R₈ = 1.0 A × 8 Ω = 8 V。剩下的 4 V 降落在 4 Ω 电阻上。
2. Parallel Circuit and Current Splitting | 并联电路与电流分流
A 6 V battery is connected to two resistors in parallel: 3 Ω and 6 Ω. Find (a) the total resistance, (b) the current drawn from the battery, and (c) the current through the 3 Ω branch.
一个 6 V 电池与两个并联电阻连接：3 Ω 和 6 Ω。求 (a) 总电阻，(b) 电池输出的总电流，以及 (c) 流过 3 Ω 支路的电流。
For parallel resistors: 1/Rₜ = 1/R₁ + 1/R₂ = 1/3 + 1/6 = 2/6 + 1/6 = 3/6, so Rₜ = 2 Ω.
并联电阻：1/Rₜ = 1/R₁ + 1/R₂ = 1/3 + 1/6 = 2/6 + 1/6 = 3/6，所以 Rₜ = 2 Ω。
Total current Iₜ = V / Rₜ = 6 V / 2 Ω = 3 A.
总电流 Iₜ = V / Rₜ = 6 V / 2 Ω = 3 A。
Ω branch I₃ = V / R₃ = 6 V / 3 Ω = 2 A. The remaining 1 A flows through the 6 Ω resistor.
因为每个并联支路两端电压均为 6 V，流过 3 Ω 支路的电流 I₃ = V / R₃ = 6 V / 3 Ω = 2 A。剩下的 1 A 流过 6 Ω 电阻。
3. Investigating Ohm’s Law and Filament Lamp Behaviour | 探究欧姆定律与灯丝灯泡特性
A student investigates how the current through a filament lamp changes with voltage. The data are shown in the table below. Plot an I–V graph and explain why the line is not straight.
一名学生探究通过灯丝灯泡的电流随电压如何变化。数据如下表所示。绘制 I–V 图并解释图线为何不是直线。
| Voltage / V | Current / A |
|---|---|
| 0.0 | 0.00 |
| 2.0 | 0.50 |
| 4.0 | 0.85 |
| 6.0 | 1.10 |
| 8.0 | 1.30 |
The plotted graph shows a curve that flattens at higher voltages. A metal filament gets hotter as current increases; the metal atoms vibrate more, making it harder for electrons to flow, so resistance increases. This non-ohmic behaviour means the gradient (1/R) is not constant.
绘制的图形是一条在较高电压下变平缓的曲线。金属灯丝随电流增大而升温；金属原子振动加剧，使电子更难通过，因此电阻增大。这种非欧姆特性意味着斜率 (1/R) 不是常数。
For a perfect ohmic conductor at constant temperature, the I–V graph would be a straight line through the origin. Filament lamps are not ohmic because temperature changes significantly.
对于恒温下的理想欧姆导体，I–V 图是一条过原点的直线。灯丝灯泡不是欧姆导体，因为温度变化显著。
4. Electrical Safety – Fuses and Earthing | 用电安全 – 保险丝与接地
An electric heater uses a current of 10 A from the 230 V mains. Explain why a 13 A fuse is suitable and why the metal case must be earthed.
一台电暖器使用 230 V 市电，电流为 10 A。解释为何选用 13 A 保险丝合适，以及为何金属外壳必须接地。
A fuse rating should be slightly higher than the normal operating current to prevent nuisance blowing. 10 A is below 13 A, so the fuse will not melt during normal use but will blow if a fault causes an excessive current, preventing overheating and fire.
保险丝的额定电流应略高于正常工作电流，以避免误熔断。10 A 小于 13 A，因此正常使用时保险丝不会熔断，但若故障导致过大电流时便会熔断，防止过热和火灾。
If a live wire touches the metal case, a large fault current flows through the earth wire to the ground. This surges the current, melting the fuse and disconnecting the appliance, protecting the user from electric shock.
如果火线触碰金属外壳，巨大的故障电流会通过地线流向大地。这会使电流骤增，熔断保险丝并断开用电器，保护使用者免受电击。
5. Energy Transfers and Efficiency Calculations | 能量传递与效率计算
A kettle transfers 200 kJ of electrical energy. 180 kJ raises the water temperature; the rest is wasted as heat and sound. Calculate the efficiency of the kettle.
一个电水壶输入 200 kJ 电能，其中 180 kJ 用于提升水温，其余以热量和声音形式浪费。计算水壶的效率。
Efficiency = (useful energy output / total energy input) × 100% = (180 kJ / 200 kJ) × 100% = 90%.
效率 = (有用输出能量 / 总输入能量) × 100% = (180 kJ / 200 kJ) × 100% = 90%。
An efficient appliance converts most of the input energy into the desired form. Reducing wasted energy – for example, by better insulation – can improve efficiency and lower running costs.
高效率的电器将大部分输入能量转化为所需形式。减少浪费的能量——例如通过更好的隔热——可以提高效率并降低使用成本。
6. Renewable vs Non‑renewable Energy Resources | 可再生能源与不可再生能源
Compare fossil fuel power stations with wind farms in terms of reliability, environmental impact and start-up time.
从可靠性、环境影响和启动时间方面比较化石燃料发电站与风力发电场。
Fossil fuel plants are reliable and can generate continuously, but they release CO₂ and SO₂, contributing to climate change and acid rain. They have short start‑up times for gas, longer for coal.
化石燃料发电厂运行可靠，能够持续发电，但会排放 CO₂ 和 SO₂，加剧气候变化和酸雨。燃气电厂启动时间较短，煤电厂较长。
Wind farms produce no greenhouse gases during operation and are renewable. However, they depend on wind speed, so output is intermittent and unpredictable. They have very short start‑up times but cannot meet base‑load demand alone.
风力发电场运行中不产生温室气体，且可再生。但它们依赖风速，发电量间歇且难以预测。启动时间非常短，但无法单独满足基荷需求。
WJEC questions often ask for an evaluation linking power demand patterns to resource choice. Wind energy is clean but requires backup from reliable sources or storage systems.
WJEC 考试常要求学生将电力需求模式与资源选择相联系进行评估。风能清洁，但需要可靠电源或储能系统作为备用。
7. Wave Equation and Transverse Waves | 波动方程与横波
A water wave has a wavelength of 0.5 m and a frequency of 4 Hz. Calculate the wave speed and state the direction of particle oscillation relative to energy transfer.
一水波的波长为 0.5 m，频率为 4 Hz。计算波速，并指出质点振动方向与能量传递方向的关系。
lso be calculated from the wave equation v = distance / time if given a wavefront diagram.
波速 v = f × λ = 4 Hz × 0.5 m = 2 m/s。若给出波前图，也可根据 v = 距离 / 时间计算波速。
Water waves are transverse: particles oscillate perpendicular (up and down) to the direction of energy transfer. Longitudinal waves, such as sound, vibrate parallel to energy transfer.
水波是横波：质点的振动方向（上下）与能量传递方向垂直。纵波（如声波）的振动方向平行于能量传递方向。
8. Refraction and Snell’s Law Practice | 折射与斯涅尔定律练习
Light travels from air into glass. The angle of incidence is 30° and the refractive index of glass is 1.5. Calculate the angle of refraction.
光从空气射入玻璃。入射角为 30°，玻璃的折射率为 1.5。计算折射角。
30°. So sin θ₂ = (1 × sin 30°) / 1.5 = 0.5 / 1.5 = 1/3 ≈ 0.333. θ₂ = sin⁻¹(0.333) ≈ 19.5°.
斯涅尔定律：n₁ sin θ₁ = n₂ sin θ₂。此处 n₁ = 1（空气），n₂ = 1.5，θ₁ = 30°。因此 sin θ₂ = (1 × sin 30°) / 1.5 = 0.5 / 1.5 = 1/3 ≈ 0.333。θ₂ = sin⁻¹(0.333) ≈ 19.5°。
Always check that the angle in the denser medium (glass) is smaller than the angle in air. Refraction occurs because light changes speed at the boundary.
务必确认光密介质（玻璃）中的角度小于空气中的角度。折射的发生是因为光在界面处速度发生改变。
9. Electromagnetic Spectrum – Order and Uses | 电磁波谱 – 顺序与用途
List the electromagnetic spectrum from longest to shortest wavelength and give one typical use for each of radio waves, X‑rays and visible light.
按波长从长到短列出电磁波谱，并分别给出无线电波、X 射线和可见光的一种典型用途。
The order (longest to shortest): radio waves, microwaves, infrared, visible light, ultraviolet, X‑rays, gamma rays. A common mnemonic is ‘Raging Martians Invaded Venus Using X‑ray Guns’.
顺序（由长至短）：无线电波、微波、红外线、可见光、紫外线、X 射线、伽马射线。常用的助记口诀是‘Raging Martians Invaded Venus Using X‑ray Guns’。
Radio waves: broadcasting TV and radio signals (long-range communication).
无线电波：广播电视与无线电信号（长距离通信）。
Visible light: human vision and photography.
可见光：人类视觉与摄影。
X‑rays: medical imaging to view broken bones.
X 射线：医疗成像，观察骨折情况。
WJEC papers may also ask you to link properties (frequency, energy) to applications – higher frequency waves carry more energy and can be ionising.
WJEC 试卷也可能要求你将波的特性（频率、能量）与用途联系起来——频率越高的波携带的能量越多，可能具有电离作用。
10. Graph Skills – Interpreting Displacement–Time Graphs for Waves | 图表技能 – 波的位移–时间图解读
A displacement–time graph for a point on a rope wave shows a peak every 0.2 s. The wavelength is 0.6 m. Determine the frequency, period, and speed of the wave.
绳波上某点的位移–时间图显示每 0.2 秒出现一个峰值。波长为 0.6 m。求波的频率、周期和波速。
Period T = time between peaks = 0.2 s. Frequency f = 1 / T = 1 / 0.2 s = 5 Hz.
周期 T = 波峰之间的时间 = 0.2 s。频率 f = 1 / T = 1 / 0.2 s = 5 Hz。
d v = f × λ = 5 Hz × 0.6 m = 3 m/s. Always use the standard units: seconds, hertz, metres.
波速 v = f × λ = 5 Hz × 0.6 m = 3 m/s。始终使用标准单位：秒、赫兹、米。
In a displacement–distance graph the distance between two adjacent crests gives the wavelength. In a displacement–time graph, it gives the period – a common confusion to avoid.
在位移–距离图中，相邻两个波峰间的距离给出波长；而在位移–时间图中，这给出周期——这是一个需要避免的常见混淆点。
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