📚 SQA National 5 Physics Unit Test: Electricity & Energy – Mock Paper Walkthrough | SQA 物理单元测试：电与能模拟卷解析
Welcome to this detailed walkthrough of a mock unit test for the SQA National 5 Physics Electricity and Energy topic. This article will guide you through each question, explaining the solutions and the key physics concepts you need to master. Whether you are preparing for your final assessment or just consolidating learning, this breakdown will help boost your confidence.
欢迎阅读这份针对 SQA National 5 物理“电与能量”单元的模拟测试详细解析。本文将逐题引导你理解解答过程和必须掌握的核心物理概念。无论你是在准备期末评估还是巩固学习，这篇解析都能帮助你提升信心。
1. Current in a Simple Circuit | 简单电路中的电流
Question: A 12 V battery is connected across a 4 Ω resistor. What is the current in the circuit? A. 3 A B. 48 A C. 0.33 A D. 8 A
题目：一个 12 V 的电池连接在一个 4 Ω 的电阻两端。电路中的电流是多少？A. 3 A B. 48 A C. 0.33 A D. 8 A
We use Ohm’s Law, which states V = IR. Rearranging gives I = V/R. Substituting the given values: I = 12 V / 4 Ω = 3 A. The correct answer is therefore choice A. This is the most basic application of Ohm’s Law, where current is directly proportional to voltage and inversely proportional to resistance. A common mistake is to multiply V by R, yielding 48 A; always check the formula carefully.
我们使用欧姆定律 V = IR。变形后得到 I = V/R。代入已知数值：I = 12 V / 4 Ω = 3 A。因此正确答案是 A。这是欧姆定律最基础的应用，电流与电压成正比，与电阻成反比。一个常见错误是将 V 与 R 相乘得到 48 A，一定要仔细核对公式。
2. Power and Current in a Kettle | 电热水壶的功率与电流
Question: An electric kettle operates at 230 V and has a power rating of 2000 W. Calculate the current drawn from the mains and the resistance of the heating element.
题目：一个电热水壶在 230 V 电压下工作，额定功率为 2000 W。计算从市电获取的电流和发热元件的电阻。
The relationship between power, voltage, and current is P = IV. So the current I = P / V = 2000 W / 230 V ≈ 8.70 A. To find the resistance, we can use R = V / I = 230 V / 8.70 A ≈ 26.4 Ω. Alternatively, combine the formulas to use R = V² / P = 230² / 2000 ≈ 26.45 Ω. Both approaches give essentially the same resistance. Note that the current is quite high, which is why kettles often require a dedicated circuit or a higher-rated fuse.
功率、电压和电流的关系为 P = IV。因此电流 I = P / V = 2000 W / 230 V ≈ 8.70 A。再求电阻，可用 R = V / I = 230 V / 8.70 A ≈ 26.4 Ω。或者综合公式使用 R = V² / P = 230² / 2000 ≈ 26.45 Ω。两种方法给出的电阻基本一致。注意这里的电流相当大，这就是为什么电热水壶通常需要独立电路或更高额定值的保险丝。
3. Energy Transferred by a Lightbulb | 灯泡转换的能量
Question: A 60 W filament bulb is switched on for 5 minutes. Calculate the total energy transferred and state the main energy transformations that occur.
题目：一个 60 W 的白炽灯泡点亮了 5 分钟。计算总共转移的能量，并说明发生的主要能量转换。
First convert the time to seconds: 5 min × 60 s/min = 300 s. Then use the energy formula E = P t = 60 W × 300 s = 18 000 J (or 18 kJ). The main energy transformation is from electrical energy to light energy and thermal (heat) energy. In a filament bulb, most of the input energy is actually transferred as heat, with only a small fraction as visible light, making them very inefficient compared to LEDs.
首先将时间换算为秒：5 min × 60 s/min = 300 s。然后用能量公式 E = P t = 60 W × 300 s = 18 000 J（或 18 kJ）。主要的能量转换是从电能到光能和热（内）能。在白炽灯中，大部分输入能量实际上转化为了热，只有一小部分以可见光形式输出，因此与 LED 灯相比其效率很低。
4. Series Circuit Analysis | 串联电路分析
Question: Two resistors, R₁ = 5 Ω and R₂ = 10 Ω, are connected in series to a 9 V battery. Calculate the total resistance, the current through the circuit, and the voltage across each resistor.
题目：两个电阻，R₁ = 5 Ω 和 R₂ = 10 Ω，与一个 9 V 的电池串联。计算总电阻、电路中的电流以及每个电阻两端的电压。
For a series circuit, the total resistance is the sum: Rtotal = R₁ + R₂ = 5 Ω + 10 Ω = 15 Ω. The current is the same everywhere: I = Vsupply / Rtotal = 9 V / 15 Ω = 0.6 A. Then the voltage across R₁: V₁ = I × R₁ = 0.6 A × 5 Ω = 3 V. Across R₂: V₂ = I × R₂ = 0.6 A × 10 Ω = 6 V. As a check, 3 V + 6 V = 9 V, which matches the supply voltage. This shows the important principle that in a series circuit the supply voltage is shared between the components in proportion to their resistances.
对于串联电路，总电阻等于各电阻之和：R总 = R₁ + R₂ = 5 Ω + 10 Ω = 15 Ω。电流处处相等：I = V电源 / R总 = 9 V / 15 Ω = 0.6 A。然后 R₁ 两端电压：V₁ = I × R₁ = 0.6 A × 5 Ω = 3 V。R₂ 两端电压：V₂ = I × R₂ = 0.6 A × 10 Ω = 6 V。作为检验，3 V + 6 V = 9 V，与电源电压相符。这表明串联电路的一个重要原则：电源电压在各个元件之间按电阻比例分配。
5. Parallel Circuit Analysis | 并联电路分析
Question: Two resistors, R₁ = 6 Ω and R₂ = 3 Ω, are connected in parallel to a 12 V battery. Determine the total resistance of the circuit, the current through each branch, and the total current drawn from the battery.
题目：两个电阻，R₁ = 6 Ω 和 R₂ = 3 Ω，与一个 12 V 的电池并联。求电路的总电阻、每条支路的电流以及从电池流出的总电流。
In parallel, the total resistance is found from 1/Rtotal = 1/R₁ + 1/R₂. So 1/Rtotal = 1/6 + 1/3 = 1/6 + 2/6 = 3/6, therefore Rtotal = 2 Ω. Notice that the total resistance is smaller than either individual resistance. The voltage across each branch is the supply voltage, 12 V. Current through R₁: I₁ = V / R₁ = 12 V / 6 Ω = 2 A. Through R₂: I₂ = 12 V / 3 Ω = 4 A. Total current is the sum of branch currents: Itotal = I₁ + I₂ = 2 A + 4 A = 6 A. You can verify this using Rtotal: Itotal = V / Rtotal = 12 V / 2 Ω = 6 A. This demonstrates that current splits in parallel but voltage stays the same.
在并联电路中，总电阻由 1/R总 = 1/R₁ + 1/R₂ 得出。所以 1/R总 = 1/6 + 1/3 = 1/6 + 2/6 = 3/6，因此 R总 = 2 Ω。注意总电阻小于任意单个电阻。各支路两端的电压均为电源电压 12 V。通过 R₁ 的电流：I₁ = V / R₁ = 12 V / 6 Ω = 2 A。通过 R₂ 的电流：I₂ = 12 V / 3 Ω = 4 A。总电流等于支路电流之和：I总 = I₁ + I₂ = 2 A + 4 A = 6 A。你也可以用 R总 来验证：I总 = V / R总 = 12 V / 2 Ω = 6 A。这说明电流在并联电路中分流，而各支路电压保持不变。
6. Fuse Placement and Safety | 保险丝的安装与安全
Question: Explain why a fuse must always be placed in the live wire of a domestic circuit, rather than in the neutral wire.
题目：解释为什么家用电路中的保险丝必须接在火线（live wire）上，而不能接在零线上。
The live wire carries the high voltage into the appliance. If a fault causes an excessive current, the fuse in the live wire melts and disconnects the appliance completely from the high potential. If the fuse were placed in the neutral wire and it blew, the appliance would stop working but its internal metal parts could still be at 230 V relative to earth, creating a serious shock risk to anyone touching it. Placing the fuse on the live side ensures that the appliance is fully isolated from the supply when a fault occurs, maximising safety.
火线将高电压送入电器。如果故障导致过大的电流，接在火线中的保险丝会熔断，使电器完全脱离高电势。如果把保险丝装在零线上并熔断，电器会停止工作，但其内部金属部件仍可能相对于大地带有 230 V 的电压，任何人触碰到都有严重触电危险。将保险丝接在火线侧可以确保故障发生时电器完全与电源隔离，最大程度地保障安全。
7. Energy Consumption and Cost | 电能消耗与计费
Question: A 1500 W electric heater is used for 2 hours each day. Electricity costs 15 p per kilowatt-hour (kWh). Calculate the total energy consumed in kWh over a 30-day month and the total cost.
题目：一个 1500 W 的电暖器每天使用 2 小时。电费为每千瓦时（kWh）15 便士。计算 30 天一个月内消耗的总电能（以 kWh 表示）及总费用。
First, convert the power to kilowatts: 1500 W = 1.5 kW. Daily energy consumption: Eday = P × t = 1.5 kW × 2 h = 3 kWh. Over 30 days: Emonth = 3 kWh/day × 30 days = 90 kWh. The total cost is 90 kWh × £0.15/kWh = £13.50. Always check your unit conversions: power in kW, time in hours, and energy in kWh. This makes electricity billing calculations straightforward.
首先将功率换算为千瓦：1500 W = 1.5 kW。每天的能耗：E天 = P × t = 1.5 kW × 2 h = 3 kWh。30 天总计：E月 = 3 kWh/天 × 30 天 = 90 kWh。总费用为 90 kWh × £0.15/kWh = £13.50。务必检查单位换算：功率用千瓦，时间用小时，能量用千瓦时。这样就使得电费计算非常直观。
8. Determining Resistance Experimentally | 实验测定电阻
Question: Describe a suitable experiment to determine the resistance of an unknown resistor. Include a circuit diagram (in words), the measurements taken, and how you would ensure a reliable result.
题目：描述一个测定未知电阻阻值的合适实验。请用文字描述电路图、需要测量哪些数据，以及如何确保结果的可靠性。
Set up a circuit with the unknown resistor connected in series with an ammeter and a battery or power supply. A voltmeter must be connected in parallel across the resistor. Include a variable resistor (rheostat) in series to allow you to vary the current. For each setting, record the ammeter reading (I) and voltmeter reading (V). Calculate the resistance using R = V / I for each pair of readings. Repeat for at least five different current values, then calculate the mean resistance. A graph of V against I should produce a straight line through the origin, and the gradient gives the resistance. This method reduces random errors and allows you to spot any anomalous results.
搭建电路：将未知电阻与一个电流表和一个电池或电源串联。电压表必须并联在电阻两端。在电路中串联一个可变电阻（变阻器）以便改变电流。对于每次设置，记录电流表读数（I）和电压表读数（V）。对每一组数据使用 R = V / I 计算电阻。至少改变五次电流值进行重复，然后计算平均电阻。画出 V-I 图像应得到一条过原点的直线，其斜率即为电阻。这种方法可以减少随机误差并有助于发现异常数据。
9. Wire Materials: Copper vs Fuse Wire | 导线材料：铜与保险丝
Question: Explain why copper is used for domestic wiring while a lead-tin alloy is used for fuses. Refer to the properties of resistivity and melting point.
题目：解释为什么家用布线使用铜，而保险丝使用铅锡合金。请涉及电阻率和熔点的特性。
Copper has a very low resistivity, meaning it offers little opposition to current flow, so it wastes minimal energy as heat during normal operation. It also has a high melting point, so it can safely carry large currents without melting. Fuse wire, on the other hand, is made of an alloy with a higher resistivity and a relatively low melting point. When the current exceeds the rated value, the fuse wire heats up (P = I²R) and reaches its melting point quickly, breaking the circuit. This deliberate trade-off ensures that the fuse is the weakest link and protects the rest of the wiring from overheating.
铜的电阻率极低，意味着对电流的阻碍很小，正常工作期间几乎不消耗电能转换成热。同时它还有很高的熔点，因此能安全承载大电流而不熔化。而保险丝则由具有较高电阻率和相对较低熔点的合金制成。当电流超过额定值时，保险丝发热（P = I²R）并迅速达到熔点，从而断开电路。这种有意的特性取舍确保保险丝是最薄弱的环节，保护其余线路免于过热。
10. Thermistors and Temperature Sensing | 热敏电阻与温度传感
Question: A thermistor is used in a temperature-sensing circuit. State how its resistance changes as temperature rises (assuming an NTC thermistor) and explain how it can be used to trigger a warning light.
题目：一个热敏电阻被用于温度感应电路中。说明其电阻如何随温度升高而变化（假设为 NTC 热敏电阻），并解释如何利用它触发警示灯。
An NTC (Negative Temperature Coefficient) thermistor has a resistance that decreases as its temperature increases. In a potential divider circuit, the thermistor is connected in series with a fixed resistor. As temperature rises, the thermistor’s resistance falls, causing the voltage across it to drop and the voltage across the fixed resistor to rise. This change in voltage can be used to switch on a transistor or a comparator circuit, which then lights a warning LED or activates a buzzer when a preset temperature threshold is reached. This is a common application in fire alarms, oven monitors and greenhouse temperature control.
NTC（负温度系数）热敏电阻的阻值随温度升高而减小。在分压电路中，热敏电阻与一个固定电阻串联。随着温度升高，热敏电阻阻值下降，导致其两端电压减小，而固定电阻两端电压升高。这种电压变化可用来导通晶体管或比较器电路，在达到预设温度阈值时点亮警示 LED 或启动蜂鸣器。这是火灾报警器、烤箱监控和温室温控中的常见应用。
Published by TutorHao | Physics Revision Series | aleveler.com
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