📚 Year 11 WJEC Physics Unit 1 Mock Paper Analysis | WJEC物理单元一模拟卷精讲
Mock exams are a crucial part of Year 11 preparation for the WJEC GCSE Physics Unit 1: Electricity, Energy and Waves. This article analyses a typical mock paper, guiding you through common question types, essential formulas, and examiner tips to boost your confidence and grades.
模拟考试是十一年级学生备考WJEC GCSE物理单元一（电学、能量与波）的关键环节。本文通过解析一份典型模拟卷，带领你理清常见题型、核心公式和评分要点，有效提升应试信心与成绩。
1. Series Circuits: Resistance and Potential Difference | 串联电路：电阻与电势差
Question: A 12 V battery is connected in series with a 4 Ω resistor and an 8 Ω resistor. Calculate (a) the total resistance, (b) the current in the circuit, and (c) the potential difference across each resistor.
题目： 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，(c) 每个电阻两端的电势差。
total resistance Rₜ is the sum of individual resistances: Rₜ = R₁ + R₂ = 4 Ω + 8 Ω = 12 Ω.
在串联电路中，总电阻 Rₜ 等于各个电阻之和：Rₜ = R₁ + R₂ = 4 Ω + 8 Ω = 12 Ω。
Using Ohm' = V / Rₜ = 12 V / 12 Ω = 1.0 A. The current is the same at all points in a series circuit.
应用欧姆定律，电流 I = V / Rₜ = 12 V / 12 Ω = 1.0 A。串联电路中各处电流相等。
= 1.0 A × 8 Ω = 8.0 V. Notice that their sum equals the supply voltage (4 V + 8 V = 12 V).
每个电阻的电压用 V = I × R 计算：V₁ = 1.0 A × 4 Ω = 4.0 V，V₂ = 1.0 A × 8 Ω = 8.0 V。注意它们之和等于电源电压（4 V + 8 V = 12 V）。
Always show the full calculation and unit conversions; marks are awarded for correct substitution and final unit.
务必展示完整计算和单位换算；正确的代入和最终单位都能得分。
2. Parallel Circuits: Total Resistance and Branch Currents | 并联电路：总电阻与支路电流
Question: A 6.0 V battery is connected to a 3 Ω resistor and a 6 Ω resistor in parallel. Determine (a) the total resistance, (b) the current drawn from the battery, and (c) the current in each branch.
题目： 一个6.0 V电池与一个3 Ω和一个6 Ω电阻并联。求 (a) 总电阻，(b) 电池提供的总电流，(c) 各支路电流。
, the total resistance Rₚ is given by 1/Rₚ = 1/R₁ + 1/R₂ = 1/3 + 1/6 = 3/6, so Rₚ = 2.0 Ω.
对于并联电阻，总电阻满足 1/Rₚ = 1/R₁ + 1/R₂ = 1/3 + 1/6 = 3/6，因此 Rₚ = 2.0 Ω。
The total current from the battery is Itotal = V / Rₚ = 6.0 V / 2.0 Ω = 3.0 A.
电池提供的总电流 Itotal = V / Rₚ = 6.0 V / 2.0 Ω = 3.0 A。
6.0 / 3 = 2.0 A, I₂ = 6.0 / 6 = 1.0 A. Check: 2.0 A + 1.0 A = 3.0 A, confirming Kirchhoff's first law.
支路电流通过 I = V / R 计算，因为每个并联支路两端电压均为6.0 V：I₁ = 6.0 / 3 = 2.0 A，I₂ = 6.0 / 6 = 1.0 A。检验：2.0 A + 1.0 A = 3.0 A，符合基尔霍夫第一定律。
3. Domestic Energy Consumption and Cost | 家庭电能消耗与费用计算
Question: A 2.5 kW oven is used for 1.5 hours. Electricity costs 12 p per kilowatt-hour. Calculate the energy transferred in kWh and the total cost.
题目： 一个2.5 kW的烤箱工作了1.5小时。电费为每千瓦时12便士。计算以kWh为单位的能量转移和总费用。
kW × 1.5 h = 3.75 kWh. This formula is key when dealing with domestic electricity meters.
电能 (kWh) = 功率 (kW) × 时间 (h)。E = 2.5 kW × 1.5 h = 3.75 kWh。处理家庭电表时该公式至关重要。
r kWh = 3.75 × 12 p = 45 p. Always express the final cost in the appropriate currency unit — here pence.
费用 = 电能 (kWh) × 单价 = 3.75 × 12 p = 45 p。最终费用需以正确货币单位表示——此处为便士。
ersion: 1 kW = 1000 W. If the power had been given in watts (2500 W), convert to kW first.
养成写出单位换算的习惯：1 kW = 1000 W。若功率以瓦特给出（2500 W），要先换算为千瓦。
4. Selecting the Correct Fuse for an Appliance | 为电器选择合适的保险丝
Question: A microwave oven is rated at 800 W, 230 V. Calculate the normal operating current and state which standard fuse rating (3 A, 5 A or 13 A) should be fitted in its plug.
题目： 一台微波炉额定功率800 W，额定电压230 V。计算其正常工作电流，并指出应该在插头中安装哪种标准额定值的保险丝（3 A、5 A或13 A）。
se 3.48 A > 3 A, while a 5 A fuse provides a safe margin and is the smallest standard rating above 3.48 A.
工作电流 I = P / V = 800 W / 230 V ≈ 3.48 A。3 A保险丝会因3.48 A > 3 A而熔断，而5 A保险丝提供了安全余量，且是大于3.48 A的最小标准额定值。
A 13 A fuse would be unsuitable because it allows a much larger current to flow before melting, reducing protection. The correct choice is 5 A.
13 A保险丝不合适，因为其允许大得多的电流通过才熔断，降低了保护作用。正确选择为5 A。
Always link the selection to the concept of overcurrent protection: the fuse melts and breaks the circuit if the current exceeds the rating, preventing overheating and fire.
务必把选择原因与过流保护概念联系起来：当电流超过额定值，保险丝熔断并切断电路，防止过热和火灾。
5. Specific Heat Capacity and Heating Time | 比热容与加热时间计算
Question: An electric heater raises the temperature of 2.0 kg of water from 15 °C to 95 °C. The specific heat capacity of water is 4200 J/kg°C. The heater is rated at 2500 W and operates at 90% efficiency. Calculate the time taken.
题目： 一个电加热器将2.0 kg水从15 °C加热到95 °C。水的比热容为4200 J/kg°C。加热器额定功率2500 W，效率为90％。计算所需时间。
mal energy needed: ΔE = m × c × Δθ = 2.0 × 4200 × (95 - 15) = 2.0 × 4200 × 80 = 672 000 J.
所需有用热能：ΔE = m × c × Δθ = 2.0 × 4200 × (95 - 15) = 2.0 × 4200 × 80 = 672 000 J。
With 90% energy input = useful energy / efficiency = 672 000 J / 0.90 = 746 667 J (approximately).
效率为90%，因此总输入电能 = 有用能量 / 效率 = 672 000 J / 0.90 ≈ 746 667 J。
t 5 minutes. Always check that the power is in watts and energy in joules for consistency.
最后，时间 t = 能量 / 功率 = 746 667 J / 2500 W ≈ 299 s，约5分钟。务必检查功率单位是瓦特、能量单位是焦耳，以保证单位一致。
6. Efficiency of an Energy Transfer | 能量转换的效率
Question: A crane motor has an input power of 4000 W and lifts a load, increasing its gravitational potential energy at a rate of 3000 J/s. Calculate the efficiency of the crane.
题目： 一台起重机电动机的输入功率为4000 W，提升重物时使重物的重力势能以3000 J/s的速率增加。计算起重机的效率。
Efficiency = (useful power output / total power input) × 100% = (3000 W / 4000 W) × 100% = 75%. The remaining 1000 W is dissipated as heat and sound energy.
效率 = (有用输出功率 / 总输入功率) × 100% = (3000 W / 4000 W) × 100% = 75%。剩余的1000 W以热量和声能形式耗散。
It is important to use the same units for both powers. Here, the rate of increasing potential energy (3000 J/s) is exactly 3000 W, so no conversion is needed.
两处的功率须使用相同单位。此处势能增加率（3000 J/s）正好就是3000 W，因此无需换算。
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