# 电路中的能量与电压

> 爱德思国际GCSE物理 · 4PH1（2017版）
> 来源: https://www.owlsprep.com/zh/study/edexcel-igcse-physics-s2-energy-and-voltage-in-circuits/

本指南覆盖爱德思IGCSE物理（4PH1）中电路能量与电压相关的全部核心考点，包含电路类型、I-V特性、核心公式与元件特性，助力你备考。

**先修:** [基础电路符号与元件识别](https://www.owlsprep.com/zh/study/edexcel-igcse-physics-s2-circuit-fundamentals/)

## 学习目标

- 解释串联与并联电路在包括家用照明在内的各类应用中的适用性
- 描述导线、电阻器、灯丝灯与二极管的电流随电压的变化规律，包含I-V特性的实验方法
- 识记并运用核心公式：V=IR、Q=It、E=QV进行电路计算
- 描述光敏电阻（LDR）的阻值随光照强度的变化规律，以及NTC热敏电阻的阻值随温度的变化规律
- 计算串联连接的两个电阻元件的电流、电压与电阻值
- 解释节点处的电流守恒规律与并联电路的电压规则

## 串并联电路的特性与应用

电路可分为两类：串联（所有元件连接在单一流路中）与并联（元件分别连接在独立支路上）。两类电路拥有不同的特性，因此适用于不同场景。

**串联电路与并联电路** — 串联：单一流路，所有元件通过的电流相等，电源电压在各元件间分配。并联：多支路，所有支路两端的电压相等，电源总电流在各支路间分配。

- Series circuits are used for low-power fairy lights, where multiple bulbs share the supply voltage to reduce power use
- Parallel circuits are used for domestic lighting, so each bulb can be switched independently and receives full mains voltage

**例题:** 将一个2Ω电阻与一个3Ω电阻串联接入10V电源，计算总电阻、总电流与每个电阻两端的电压。

1. 串联电路的总电阻为各元件电阻之和：

   $$R_{total} = R_1 + R_2 = 2 + 3 = 5\Omega$$
2. 使用欧姆定律计算总电流：

   $$I = \frac{V}{R} = \frac{10}{5} = 2A$$
3. 所有串联元件通过的电流相等，因此可计算每个电阻两端的电压：

   $$V_1 = I \times R_1 = 2 \times 2 = 4V$$
4. $$V_2 = I \times R_2 = 2 \times 3 = 6V$$
5. 验证：4V + 6V = 10V，与电源电压一致。

> **考试提示:** 当考题要求你解释家用照明电路的设计原理时，务必提及独立开关控制、每个灯泡都能获得额定全电压这两个并联电路布局的核心优势。

*计算器:* allowed

## 电路核心公式与物理量

本主题有三个核心公式不会印在考试公式表上，必须熟记。计算前请务必将所有单位转换为国际标准单位。

**电压** — 电路中两点之间每单位转移电荷对应的能量。1伏特等于1焦耳每库仑（1V = 1J/C）。

$$V = I \times R \text{ (Ohm's Law: Voltage = Current } \times \text{ Resistance)}$$

$$Q = I \times t \text{ (Charge = Current } \times \text{ Time)}$$

$$E = Q \times V \text{ (Energy Transferred = Charge } \times \text{ Voltage)}$$

> **tip**
>
> 电流是电荷的流动速率。在金属导体中，电流是带负电的电子从电池负极向正极移动形成的。

**例题:** 一个12V电池提供2A电流持续30秒，计算电池的总电荷流量和转移的总能量。

1. 使用公式Q = I × t计算电荷流量：

   $$Q = 2 \times 30 = 60C$$
2. 使用公式E = Q × V计算转移的能量：

   $$E = 60 \times 12 = 720J$$

*计算器:* allowed

## 常见元件的伏安（I-V）特性

I-V图将元件的电流随电压的变化关系绘制成图，展示电阻随电压的变化规律。你需要掌握四种常见元件的I-V图形状，以及测量这些特性的实验方法。

**欧姆导体** — 一类元件（例如恒定温度下的定值电阻、导线），其电流与电压成正比，I-V图为过原点的直线。

| Component | I-V Graph Shape | Reason for Shape |
| --- | --- | --- |
| Fixed Resistor (Ohmic) | Straight line through origin | Resistance constant at constant temperature |
| Filament Lamp | Curved line, gradient decreases at high voltage | Resistance increases as filament temperature rises with current |
| Diode | Zero current for negative voltage, sharp rise in current for positive voltage | Only conducts in forward bias (positive direction) |

**例题:** 描述一个测量灯丝灯I-V特性的实验。

1. 搭建串联电路，包含电池、滑动变阻器、电流表、灯丝灯，将电压表并联在灯丝灯两端。
2. 调节滑动变阻器改变灯丝灯两端的电压，记录至少5组电流表的电流读数和电压表的电压读数。
3. 反转电池的接线，获取负电压和负电流的读数。
4. 绘制以电流为y轴、电压为x轴的图像，得到灯丝灯的I-V特性曲线。

> **考试提示:** 如果题目要求解释灯丝灯的曲线，一定要将曲线形状与温度升高导致电阻增大联系起来，这会降低高电压下电流的增长速率。

*计算器:* allowed

## 可变电阻元件：光敏电阻（LDR）、热敏电阻与二极管

特殊电阻的阻值会随外部条件变化而改变，因此适用于传感器电路。灯泡和发光二极管（LED）也可用于指示电流的存在，当有电流流过时它们就会发光。

**光敏电阻与NTC热敏电阻** — LDR（光敏电阻）：阻值随光照强度升高而降低。NTC热敏电阻：阻值随温度升高而降低。

> **mnemonic**
>
> 光敏电阻：光照越强，阻值越低；NTC热敏电阻：温度越高，阻值越低。这两条规律经常在1分的应用题中考查。

**例题:** 一个光敏电阻被用于自动门廊灯电路。解释为什么天黑时灯会亮起。

1. 当光照强度很低（天黑）时，光敏电阻的阻值会增大。
2. 更高的阻值会降低控制电路中的电流。
3. 这会触发继电器开关闭合，点亮门廊灯。

*计算器:* allowed

## 电流守恒与节点定则

所有电路中电流都是守恒的，这意味着并联电路的节点处不会有电荷损失。并联元件两端的电压与电源电压完全相等。

**节点定则** — 流入并联电路某一节点的总电流等于流出该节点的总电流，因为电荷不能被创造也不能被消灭。

**例题:** 在并联电路的某个节点处，有3A的电流流入该节点，其中1A通过一条支路流出。计算通过第二条支路流出的电流。

1. 应用节点定则：总流入电流 = 总流出电流

   $$I_{in} = I_{1} + I_{2}$$
2. 变形公式求解第二条支路的电流：

   $$I_{2} = 3A - 1A = 2A$$

> **考试提示:** 对于并联电路，你只需要定性描述变化趋势：不要尝试计算并联总电阻，这不属于爱德思IGCSE考纲的要求范围。

*计算器:* allowed

## 常见错误

- **错误做法:** Assuming current is identical across all parallel branches
  - 原因: Current splits at junctions, only voltage is identical across parallel branches
  - 正确做法: Use the junction rule: total current into a junction equals total current out
- **错误做法:** Using the parallel combined resistance formula for calculations
  - 原因: Edexcel IGCSE only requires series resistance calculations for two components; parallel resistance calculations are out of scope
  - 正确做法: Only add resistances directly for series circuits, describe parallel behavior qualitatively
- **错误做法:** Drawing the filament lamp I-V graph as a straight line
  - 原因: Filament temperature rises with current, increasing resistance, so the graph curves
  - 正确做法: Draw an S-shaped curve for the filament lamp, with decreasing gradient at high voltages
- **错误做法:** Defining voltage as a 'push of current' in exam questions
  - 原因: Mark schemes require the precise definition: energy transferred per unit charge, 1V = 1J/C
  - 正确做法: Use the exact syllabus definition for all 2+ mark voltage explanation questions
- **错误做法:** Mixing up LDR and thermistor resistance trends
  - 原因: Both components have resistance that falls with an increase in their trigger stimulus, but students often mix up the stimulus for each
  - 正确做法: Use the mnemonic: Lighter = Lower LDR resistance, Hotter = Lower thermistor resistance

## 速查表

| Quantity/Rule | Formula/Statement | Key Detail | Exam Tip |
| --- | --- | --- | --- |
| Ohm's Law | V = I × R | Units: V (V), I (A), R (Ω) | Memorize, not provided on formula sheet |
| Charge Flow | Q = I × t | Units: Q (C), t (s) | Convert time to seconds before calculation |
| Energy Transferred | E = Q × V | 1V = 1 J/C | Use exact definition for voltage explain questions |
| Series Circuits | I same, V_total = V1+V2, R_total = R1+R2 | Only for 2 resistive components | Add resistances directly only for series |
| Parallel Circuits | V same across branches, I_in = I_out at junctions | No parallel resistance calculations needed | Domestic lighting uses parallel for independent control |
| I-V Graphs | Ohmic: straight line, Filament: curved, Diode: forward only | Diode only conducts with positive voltage | Draw correct shapes for graph questions |
| Variable Resistors | LDR: higher light = lower R, Thermistor: higher temp = lower R | Only NTC thermistors are assessed | Use the mnemonic to avoid mixing trends |

## 下一步

Now you have mastered core energy and voltage in circuits content for Edexcel IGCSE Physics, you are ready to progress to mains electricity and circuit safety, which builds on your current and voltage knowledge to explain fuse operation and the P=IV power formula. You can also move to the next electricity sub-topic, electrostatic charge. This content is assessed in both Paper 1 and Paper 2, as well as the Double Award (4SD0) specification. Make sure you practice past paper questions focusing on series circuit calculations and I-V characteristic descriptions, and memorize the three core formulae as they are not provided on the exam formula sheet.

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