📚 AS AQA Physics Unit 5 June 2019 Insert: Circuit Analysis and Exam Strategy | AQA AS物理第5单元2019年6月考题详解：电路分析与应考策略
The June 2019 insert for AQA AS Physics Unit 5 (Electricity) provides a rich context for understanding core circuit concepts. This article breaks down the key topics you need to master — from current and resistance to internal resistance and potential dividers — and connects them directly to the exam style of the AQA paper. You will also find practical tips for interpreting the insert and answering questions with confidence.
AQA AS物理第5单元（电学）2019年6月的试题插入材料为理解核心电路概念提供了丰富情境。本文将拆解你需要掌握的关键主题——从电流、电阻到内阻和电位分压器——并直接联系AQA试卷的出题风格。你还将获得解读插入材料并自信作答的实用技巧。
1. Electric Current and Charge Conservation | 电流与电荷守恒
Electric current is defined as the rate of flow of charge through a cross-sectional area of a conductor. In equation form, I = ΔQ / Δt, where I is current in amperes (A), ΔQ is charge in coulombs (C), and Δt is time in seconds (s). The direction of conventional current is from positive to negative terminal, while electron flow is opposite.
电流定义为电荷通过导体横截面的速率。公式为 I = ΔQ / Δt，其中 I 的单位为安培（A），ΔQ 的单位为库仑（C），Δt 的单位为秒（s）。习惯电流方向从正极到负极，而电子流动方向相反。
Charge conservation is a fundamental principle that leads directly to Kirchhoff’s first law: the total current entering a junction equals the total current leaving the junction. This is because charge cannot accumulate at a junction in steady state. In the June 2019 insert, you may be asked to apply this law to a circuit with multiple branches.
电荷守恒是基本定律，直接引出基尔霍夫第一定律：流入结点的总电流等于流出结点的总电流。这是因为在稳恒状态下电荷不能在结点累积。在2019年6月插入材料中，你可能会被要求将该定律应用于多支路电路。
I₁ = I₂ + I₃ (for a junction with three wires)
Example: If a junction receives 2.0 A and sends 1.2 A along one branch, the other branch must carry 0.8 A. Always check your answer by ensuring the total incoming flux equals total outgoing flux.
示例：若一个结点接受2.0 A电流，其中一条支路分流1.2 A，则另一条支路必定承载0.8 A。始终通过检查总流入电流等于总流出电流来验证答案。
2. Electromotive Force (EMF) and Internal Resistance | 电动势与内阻
The electromotive force (e.m.f.) of a source is the energy transferred per unit charge by the source to the charges passing through it. It is measured in volts (V). A real battery has internal resistance, r, which causes a drop in potential difference across the battery when current flows.
电源的电动势是指电源向通过它的电荷传递单位电荷的能量。单位为伏特（V）。真实电池具有内阻 r，当有电流通过时会在电池两端产生电势降。
When a battery of e.m.f. ε and internal resistance r is connected to an external resistor R, the terminal potential difference V is given by:
当电动势为 ε、内阻为 r 的电池连接外部电阻 R 时，端电压 V 为：
V = ε − I r
Rearranging gives ε = I(R + r). This equation is central to many AQA questions; you may be given a graph of V against I and asked to calculate ε and r. The intercept on the V-axis equals ε, and the gradient equals −r.
整理得 ε = I(R + r)。该方程是许多AQA题目的核心；你可能获得 V-I 图并要求计算 ε 和 r。V 轴截距等于 ε，斜率等于 −r。
In the June 2019 insert, a typical question might ask you to find the internal resistance from a circuit diagram or from experimental data. Ensure you understand how the terminal voltage drops as the external resistance decreases.
在2019年6月的插入材料中，一个典型问题可能要求你从电路图或实验数据中找出内阻。确保你理解端电压如何随外部电阻减小而下降。
3. Kirchhoff’s Second Law and Energy Conservation | 基尔霍夫第二定律与能量守恒
Kirchhoff’s second law states that the sum of the e.m.f.s around any closed loop in a circuit equals the sum of the potential differences across every component in that loop. This is a restatement of energy conservation: the energy gained per unit charge from sources must equal the energy lost per unit charge in resistors.
基尔霍夫第二定律指出，电路中任何闭合回路中的电动势之和等于该回路中每个元件两端的电势差之和。这是能量守恒的表述：单位电荷从电源获得的能量必须等于电阻器中单位电荷损失的能量。
Σε = ΣIR (for a closed loop)
For example, consider a loop containing a 6.0 V battery (internal resistance 0.5 Ω) and two resistors of 2.0 Ω and 3.0 Ω in series. The total resistance is 5.5 Ω, so the current is I = 6.0 / 5.5 = 1.09 A. The p.d. across each resistor can then be found.
例如，考虑一个包含6.0 V电池（内阻0.5 Ω）以及两个分别2.0 Ω和3.0 Ω串联电阻的回路。总电阻为5.5 Ω，因此电流为 I = 6.0 / 5.5 = 1.09 A。然后可求出每个电阻两端的电势差。
When applying Kirchhoff’s second law in the exam, it is important to assign direction to the loop polarity of each battery and current direction clearly on your circuit diagram. AQA marks often reward clear sign convention.
在考试中应用基尔霍夫第二定律时，重要的是在电路图上清晰标出每个电池的极性和电流方向。AQA评分常常对清晰的符号约定给予奖励。
4. Series and Parallel Circuits | 串联与并联电路
For resistors in series, the total resistance is simply the sum: R_total = R₁ + R₂ + R₃ + … . The current is the same through every component, and the potential differences divide according to resistance values.
对于串联电阻，总电阻为各电阻之和：R_total = R₁ + R₂ + R₃ + …。每个元件中的电流都相同，而电势差按电阻值分配。
For resistors in parallel, the total resistance is found from:
对于并联电阻，总电阻为：
1/R_total = 1/R₁ + 1/R₂ + 1/R₃ + …
The total resistance of a parallel combination is always less than the smallest individual resistance. The p.d. across each branch is identical, while currents divide inversely with resistance.
并联组合的总电阻总是小于最小的单个电阻。每个支路两端的电势差相同，而电流按电阻的反比分配。
In the June 2019 insert, you might be presented with a combination (e.g., two resistors in parallel, both in series with another). Break the problem down step by step: simplify parallel groups first, then treat the remaining as a series circuit. Always keep your working clear so partial marks can be awarded.
在2019年6月插入材料中，你可能会看到组合电路（例如两个并联电阻再与另一个串联）。逐步分解问题：先简化并联组，然后将剩余部分视为串联电路。始终保持计算过程清晰，以便获得部分分数。
5. Resistance, Resistivity and Conductivity | 电阻、电阻率与电导率
The resistance R of a conductor depends on its length L, cross-sectional area A, and the material’s resistivity ρ, according to:
导体的电阻 R 取决于其长度 L、横截面积 A 和材料的电阻率 ρ，公式为：
R = ρL / A
Resistivity is a material property measured in Ω·m. Metals have low resistivity; insulators have very high resistivity. Conductivity σ is the reciprocal of resistivity, σ = 1/ρ, with units S/m (siemens per metre).
电阻率是材料的属性，单位为 Ω·m。金属电阻率低，绝缘体电阻率非常高。电导率 σ 是电阻率的倒数，σ = 1/ρ，单位为西门子每米（S/m）。
A common exam question involves a wire of known length and diameter; you may need to calculate cross-sectional area from the diameter (A = πd²/4) and then determine resistivity from measured resistance. In the June 2019 insert, data may be given in a table format, requiring careful unit conversion.
常见考题涉及已知长度和直径的导线；你可能需要根据直径计算横截面积（A = π²/4），然后从测量的电阻中确定电阻率。在2019年6月插入材料中，数据可能以表格形式给出，需要仔细进行单位换算。
- Convert mm to m before calculating area (1 mm = 1 × 10⁻³ m).
- Use consistent units: length in metres, area in m², resistivity in Ω·m.
- Check if the question asks for percentage uncertainty — handle it systematically.
- 计算面积前要将毫米转换为米（1 mm = 1 × 10⁻³ m）。
- 使用一致的单位：长度用米，面积用m²，电阻率用Ω·m。
- 注意问题是否要求百分比不确定度——系统化处理。
6. Potential Dividers and Variable Resistors | 电位分压器与可变电阻
A potential divider allows you to obtain a desired fraction of a given voltage. It consists of a resistor chain connected across a power supply; the output voltage is taken across one of the resistors or a portion of a variable resistor.
电位分压器允许你获取给定电压的所需比例部分。它由连接在电源两端的电阻链组成；输出电压取自其中一个电阻或可变电阻的一部分。
V_out = V_in × [R₂ / (R₁ + R₂)]
This equation is worth committing to memory for AQA. If R₁ = R₂, the output is half the input. If R₂ increases, V_out increases (assuming V_in remains fixed).
这个公式值得牢记以应对AQA。若 R₁ = R₂，则输出为输入的一半。若 R₂ 增大，V_out 增大（假设 V_in 保持不变）。
In the June 2019 insert, you might see a potential divider used as a sensing circuit with an LDR or thermistor. As light intensity or temperature changes, the resistance of the sensor changes, altering V_out. You must explain the direction of the voltage change clearly.
在2019年6月插入材料中，你可能会看到使用光敏电阻或热敏电阻作为传感器的分压电路。当光照强度或温度变化时，传感器的电阻变化，从而改变 V_out。你必须清楚解释电压变化的方向。
For example, in a potential divider with a thermistor and a fixed resistor, if the thermistor is in the top position and its resistance decreases when heated, then the p.d. across the bottom fixed resistor increases (since total current increases and the current flows through both).
例如，在包含热敏电阻和固定电阻的分压器中，如果热敏电阻位于上方，温度升高时其阻值减小，则下方固定电阻两端的电势差增大（因为总电流增大且电流通过两者）。
7. Current–Voltage (I–V) Characteristics | 电流-电压（I-V）特性
For a metallic conductor obeying Ohm’s law at constant temperature, the I–V graph is a straight line through the origin: V = IR. The gradient gives the reciprocal of resistance (1/R).
对于在恒温下遵循欧姆定律的金属导体，I-V图是通过原点的直线：V = IR。斜率给出电阻的倒数（1/R）。
For a filament lamp, the resistance increases with temperature, so the I–V graph curves away from the voltage axis at higher currents. For a semiconductor diode, current flows easily in one direction but almost not in the reverse direction; beyond a small threshold voltage, the forward current rises steeply.
对于白炽灯，电阻随温度升高而增加，因此I-V图在高电流时偏离电压轴向。对于半导体二极管，电流在一个方向容易通过，而反向几乎不通；超过阈值电压后，正向电流急剧上升。
The June 2019 insert likely includes an I–V graph for a component; you may be asked to identify the type of component, find resistance at a given voltage, or explain why the graph is not linear. Remember: for non-linear components, V/I gives the static resistance at a point, while ΔV/ΔI gives the dynamic resistance.
2019年6月插入材料可能包含某个元件的I-V图；可能要求你识别元件类型、在给定电压下求电阻，或解释为什么图形不是线性的。记住：对于非线性元件，V/I 给出该点的静态电阻，而 ΔV/ΔI 给出动态电阻。
Static resistance: R = V/I; Dynamic resistance: r_dynamic = ΔV/ΔI
Be sure to read the axes carefully — sometimes the vertical axis is current in mA, so convert to A before calculating resistance.
务必仔细阅读坐标轴——有时纵轴电流单位为毫安，计算电阻前需转换为安培。
8. Power, Energy and Heating Effects | 功率、能量与热效应
Electrical power delivered to a component is the product of current and voltage:
传递给元件的电功率是电流与电压的乘积：
P = VI = I²R = V²/R
The energy converted in a circuit is given by E = Pt = VIt. For a resistor, the energy is dissipated as heat (Joule heating). In a filament lamp, only a small fraction of energy is radiated as visible light; the rest is thermal energy.
电路中转换的能量为 E = Pt = VIt。对于电阻器，能量以热（焦耳热）形式耗散。在钨丝灯中，只有一小部分能量以可见光辐射，其余为热能。
In the June 2019 insert, you may need to compare power dissipation in different resistors. Since P = I²R, for resistors in series (same current), the larger resistor dissipates more power. For resistors in parallel (same voltage), the smaller resistor dissipates more power because P = V²/R.
在2019年6月插入材料中，你可能需要比较不同电阻器的功率耗散。由于 P = I²R，对于串联电阻（电流相同），电阻更大的耗散更多功率。对于并联电阻（电压相同），电阻更小的耗散更多功率，因为 P = V²/R。
Another common task is calculating the cost of electrical energy. Given the power rating of an appliance and usage time, the kilowatt-hour (kWh) consumption and cost can be determined. While AQA AS focuses on physics principles, a brief familiarity with such applications helps understand the context.
另一常见任务是计算电能成本。给定用电器的功率额定值和使用时间，可确定千瓦时（kWh）消耗量和成本。虽然AQA AS重点在物理原理上，但简单了解此类应用有助于理解情境。
9. Interpreting the Insert: Data Tables and Graphs | 解读插入材料：数据表与图表
The insert for June 2019 contains diagrams of circuits and possibly data tables or graphs. When you encounter a table of current and voltage values, always look for relationships: does the resistance stay constant? Does the temperature affect results?
2019年6月的插入材料包含电路图以及可能的数表或图表。当你遇到电流与电压值表时，始终寻找关系：电阻是否保持不变？温度是否影响结果？
If you see a graph of V against I for a battery, remember the intercept is ε and the gradient is −r. If the graph is a curve, check whether it is due to a changing internal resistance or external resistance — this is unusual but possible in some experimental situations.
如果你看到电池的 V-I 图，记住截距为 ε，斜率为 −r。如果图形是曲线，检查是否由于内阻或外阻变化——这在某些实验情境中可能但不常见。
For circuit diagrams in the insert, annotate directly: label currents, potentials, resistances, and battery polarity. This prevents sign errors and makes it easier to apply Kirchhoff’s laws correctly.
对于插入材料中的电路图，直接在上面做标记：标出电流、电势、电阻和电池极性。这能防止符号错误，也更易正确应用基尔霍夫定律。
When reading fine print on the insert, pay attention to component values like “internal resistance 0.25 Ω” or “battery of e.m.f. 9.0 V”. Missing such details is the most common cause of incorrect calculations.
阅读插入材料上的小字时，注意元件值，如“内阻0.25 Ω”或“电动势9.0 V的电池”。漏看这些细节是计算错误的最常见原因。
10. Common Pitfalls and Exam Techniques | 常见陷阱与应考技巧
One frequent error is forgetting to account for internal resistance when calculating current in a complete circuit. Always use total resistance (R + r), not just the external resistance R, unless the internal resistance is explicitly ignored.
一个常见错误是在计算整个电路电流时忘记计及内阻。始终使用总电阻（R + r），而不仅仅是外阻 R，除非明确忽略内阻。
Another pitfall involves unit conversions. A current of 40 mA is 0.040 A, not 0.4 A. A resistance of 2.0 kΩ is 2000 Ω. Make a habit of converting every quantity to SI units before substituting into equations.
另一个陷阱涉及单位换算。40 mA 是0.040 A，不是0.4 A。2.0 kΩ 为2000 Ω。养成在代入公式前将所有量转换为国际单位制的习惯。
In circuit analysis, the “voltage” reading on a voltmeter is the potential difference across the points it is connected to, not necessarily the battery e.m.f. If the battery has internal resistance, the terminal voltage is less than the e.m.f. when current flows.
在电路分析中，电压表读数是指其连接两点之间的电势差，不一定是电池电动势。如果电池有内阻，当有电流通过时端电压小于电动势。
AQA rewards clear working and logical structure. Write down every step: define variables, write the formula, substitute values, calculate the result, include units. Even if the final numerical answer is wrong, you may gain method marks.
AQA注重清晰的计算过程和逻辑结构。写出每个步骤：定义变量、写出公式、代入数值、计算结果、包含单位。即使最终数值算错，你也可能获得方法分数。
11. Worked Example: Finding Internal Resistance (Jun 19 style) | 实例演练：求内阻（2019年6月风格）
A battery has an e.m.f. of 12.0 V. When a 5.0 Ω resistor is connected across its terminals, the current in the circuit is 2.0 A. Calculate the internal resistance.
一块电池的电动势为12.0 V。当在其两端连接一个5.0 Ω电阻时，电路中的电流为2.0 A。计算其内阻。
ε = I(R + r)
Substitute the known values: 12.0 = 2.0 × (5.0 + r). Dividing both sides by 2.0 gives 6.0 = 5.0 + r, hence r = 1.0 Ω.
代入已知值：12.0 = 2.0 × (5.0 + r)。两边除以2.0得6.0 = 5.0 + r，因此 r = 1.0 Ω。
Now suppose the same battery is connected to a 2.0 Ω resistor. What is the terminal potential difference?
现在假设同一电池连接到2.0 Ω电阻。那么端电压是多少？
Total resistance = 2.0 + 1.0 = 3.0 Ω. Current I = 12.0 / 3.0 = 4.0 A. Terminal p.d. V = ε − Ir = 12.0 − (4.0 × 1.0) = 8.0 V. Alternatively, V = IR = 4.0 × 2.0 = 8.0 V, which agrees.
总电阻 = 2.0 + 1.0 = 3.0 Ω。电流 I = 12.0 / 3.0 = 4.0 A。端电压 V = ε − Ir = 12.0 − (4.0 × 1.0) = 8.0 V。也可以通过 V = IR = 4.0 × 2.0 = 8.0 V 得到相同结果。
This example illustrates the exam technique: identify what is asked, choose the correct formula, and use both methods to check consistency when possible.
此例说明应考技巧：确定所求，选择正确公式，并在可能时用两种方法验证一致性。
12. Summary and Final Revision Pointers | 总结与最后复习要点
The June 2019 insert covers fundamental electricity topics that are frequently tested in AQA AS Paper 2. To score well, you must be comfortable with: charge conservation, internal resistance, Kirchhoff’s laws, series/parallel combinations, resistivity, potential dividers, I–V graphs, and power calculations.
2019年6月插入材料覆盖了AQA AS试卷2中频繁考查的基础电学主题。要获得高分，你必须熟练掌握：电荷守恒、内阻、基尔霍夫定律、串并联组合、电阻率、电位分压器、I-V图和功率计算。
Practice past paper questions with a timer. After each session, review the mark scheme to understand exactly where marks are given — especially for written explanations of voltage changes or component behaviour.
计时练习历年真题。每次练习后，对照评分方案仔细审查得分点——特别是关于电压变化或元件行为的文字解释。
Finally, build a single-page formula sheet that you can annotate and memorise over time. Include: I = Q/t, V = IR, R = ρL/A, P = VI = I²R = V²/R, ε = I(R+r), and V_out = V_in R₂/(R₁+R₂). Stick it on your wall and test yourself weekly.
最后，制作一页公式汇总表，随时注解和记忆。包括：I = Q/t、V = IR、R = ρL/A、P = VI = I²R = V²/R、ε = I(R+r) 以及 V_out = V_in R₂/(R₁+R₂)。把它贴在墙上，每周自测一次。
Published by TutorHao | Physics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导