📚 Cambridge IGCSE Science: Electricity and Circuits | 剑桥 IGCSE 科学：电与电路
Electricity is a core topic in Cambridge IGCSE Science. Understanding electric charge, current, voltage, resistance, and circuit behaviour is essential for both written papers and practical assessments. This article reviews the key ideas, equations, and safety points you need to master for the Cambridge Science syllabus.
电是剑桥 IGCSE 科学的核心主题。理解电荷、电流、电压、电阻以及电路行为，对笔试和实验评估都至关重要。本文回顾剑桥科学大纲中必须掌握的关键概念、公式和安全要点。
1. Electric Charge and Current | 电荷与电流
Electric current is the rate of flow of electric charge. In a metallic conductor, current is carried by free electrons moving from a region of negative potential to positive potential. The unit of current is the ampere (A). One ampere is one coulomb of charge passing a point per second. The equation is I = Q / t, where I is current in amperes, Q is charge in coulombs, and t is time in seconds. Conventional current flows from positive to negative, opposite to electron flow.
电流是电荷流动的速率。在金属导体中，电流由自由电子从低电位区域向高电位区域移动而形成。电流的单位是安培（A）。1 安培表示每秒钟通过导体某一点的电荷量为 1 库仑。公式为 I = Q ÷ t，其中 I 为电流（安培），Q 为电荷（库仑），t 为时间（秒）。规定电流方向为正电荷移动方向，与电子实际运动方向相反。
I = Q ÷ t
2. Voltage and Potential Difference | 电压与电势差
Voltage, or potential difference (p.d.), measures the energy transferred per unit charge between two points. It is measured in volts (V). One volt is one joule of energy per coulomb. A battery or power supply provides the potential difference that pushes charges around a circuit. The equation is V = E / Q, where V is potential difference in volts, E is energy in joules, and Q is charge in coulombs.
电压又称电势差（p.d.），衡量单位电荷在两点之间转移的能量。单位为伏特（V）。1 伏特表示每库仑电荷转移 1 焦耳能量。电池或电源提供电势差，推动电荷在电路中流动。公式为 V = E ÷ Q，其中 V 为电势差（伏特），E 为能量（焦耳），Q 为电荷（库仑）。
V = E ÷ Q
3. Resistance and Ohm’s Law | 电阻与欧姆定律
Resistance opposes the flow of electric current. It is measured in ohms (Ω). Ohm’s law states that the current through a conductor is directly proportional to the potential difference across it, provided temperature remains constant. The formula is V = IR, where V is voltage in volts, I is current in amperes, and R is resistance in ohms. Components with a constant resistance are called ohmic conductors.
电阻阻碍电流的流动，单位为欧姆（Ω）。欧姆定律指出，在温度不变的条件下，通过导体的电流与导体两端的电势差成正比。公式为 V = IR，其中 V 为电压（伏特），I 为电流（安培），R 为电阻（欧姆）。电阻恒定的元件称为欧姆导体。
V = IR
4. Series Circuits | 串联电路
In a series circuit, components are connected end-to-end, so the same current flows through each component. The total resistance is the sum of individual resistances: Rtotal = R₁ + R₂ + R₃ + … The total potential difference from the supply is shared across the components, so Vtotal = V₁ + V₂ + V₃. If one component breaks, the whole circuit stops working.
在串联电路中，元件首尾相连，因此每个元件中流过相同的电流。总电阻等于各电阻之和：Rtotal = R₁ + R₂ + R₃ + …。电源的总电势差在各元件之间分配，即 Vtotal = V₁ + V₂ + V₃。如果其中一个元件发生断路，整个电路都会停止工作。
Rtotal = R₁ + R₂ + R₃
5. Parallel Circuits | 并联电路
In a parallel circuit, components are connected in separate branches across the same two points. The potential difference across each branch is the same as the supply voltage. The total current splits among the branches: Itotal = I₁ + I₂ + I₃. The total resistance is less than the smallest individual resistance, and is calculated using 1/Rtotal = 1/R₁ + 1/R₂ + 1/R₃.
在并联电路中，各元件分别连接在相同的两个节点之间，形成独立支路。每条支路两端的电势差与电源电压相同。总电流分配于各支路：Itotal = I₁ + I₂ + I₃。总电阻小于最小的单个电阻，并用公式 1/Rtotal = 1/R₁ + 1/R₂ + 1/R₃ 计算。
1/Rtotal = 1/R₁ + 1/R₂ + 1/R₃
6. Electrical Power and Energy | 电功率与电能
Electrical power is the rate at which electrical energy is transferred. It is measured in watts (W). The formula is P = VI, where P is power in watts, V is voltage in volts, and I is current in amperes. Combining with Ohm’s law gives P = I²R and P = V² / R. Energy transferred is E = P × t, measured in joules (J) or kilowatt-hours (kWh) for mains supply.
电功率是电能转移的速率，单位为瓦特（W）。公式为 P = VI，其中 P 为功率（瓦特），V 为电压（伏特），I 为电流（安培）。结合欧姆定律可得 P = I²R 和 P = V² / R。电能转移量 E = P × t，单位为焦耳（J）；市电中常用千瓦时（kWh）。
P = VI | P = I²R | E = P × t
7. Circuit Components and Symbols | 电路元件与符号
Common circuit components include cells, batteries, switches, lamps, resistors, variable resistors, ammeters, voltmeters, diodes, and light-dependent resistors (LDRs). Standard symbols must be used in circuit diagrams. A diode allows current to flow in one direction only. An LDR has a resistance that decreases as light intensity increases.
常见电路元件包括电池、电池组、开关、灯泡、定值电阻、可变电阻、电流表、电压表、二极管和光敏电阻（LDR）。电路图中必须使用标准符号。二极管只允许电流沿一个方向流动。光敏电阻的阻值随光照强度增大而减小。
- Cell 电池
- Battery 电池组
- Switch 开关
- Lamp 灯泡
- Resistor 电阻
- Variable resistor 可变电阻
- Ammeter 电流表
- Voltmeter 电压表
- Diode 二极管
- LDR 光敏电阻
8. Measuring Current and Voltage | 测量电流与电压
An ammeter measures current and must be connected in series with the component so the same current flows through it. A voltmeter measures potential difference and must be connected in parallel across the component being tested. Digital meters have very high resistance, so they draw negligible current and do not significantly affect the circuit.
电流表测量电流，必须与被测元件串联，以使相同电流通过。电压表测量电势差，必须并联在被测元件两端。数字电表具有很高的内阻，因此吸取的电流可以忽略，不会对电路产生明显影响。
9. Mains Electricity and Safety | 市电与安全
Mains electricity in many countries is supplied as alternating current (a.c.) at 230 V with a frequency of 50 Hz, or 120 V at 60 Hz in others. Safety features include fuses, circuit breakers, earthing, and double insulation. A fuse melts when too much current flows, breaking the circuit and preventing overheating. Circuit breakers are reusable electromagnetic switches.
许多国家的市电为 230 V、50 Hz 的交流电（a.c.），有些国家为 120 V、60 Hz。安全装置包括保险丝、断路器、接地和双重绝缘。保险丝在电流过大时熔断，切断电路，防止过热。断路器是可以重复使用的电磁开关。
10. Calculating Resistance in Circuits | 电路中的电阻计算
For series circuits, total resistance is the simple sum of resistances. For parallel circuits, the reciprocal formula applies. Worked example: two resistors of 6 Ω and 3 Ω are connected in parallel. 1/Rtotal = 1/6 + 1/3 = 1/6 + 2/6 = 3/6, so Rtotal = 2 Ω. Then if connected to a 12 V supply, the total current is I = V / Rtotal = 12 V / 2 Ω = 6 A.
对于串联电路，总电阻为各电阻之和。对于并联电路，应用倒数公式。例题：两个电阻分别为 6 Ω 和 3 Ω 并联，1/Rtotal = 1/6 + 1/3 = 1/6 + 2/6 = 3/6，故 Rtotal = 2 Ω。若接到 12 V 电源上，总电流为 I = V ÷ Rtotal = 12 V ÷ 2 Ω = 6 A。
1/Rtotal = 1/6 Ω + 1/3 Ω ⇒ Rtotal = 2 Ω
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