📚 IB Physics B.5 Current and circuits SL: Deriving Formulas and Answers | IB 物理 B.5 电流与电路 SL：公式推导与解答
The IB Physics B.5 Current and circuits topic for Standard Level focuses on the fundamental principles that govern the flow of electric charge. Understanding how key equations are derived—from the definition of current to Kirchhoff’s laws—is essential for solving circuit problems confidently. In this article, we walk through the derivations of all essential SL formulas, explain their physical meaning, and show how to apply them to typical examination-style questions.
IB 物理标准水平 B.5 电流与电路专题聚焦于支配电荷流动的基本原理。掌握从电流的定义到基尔霍夫定律等关键公式的推导过程，对自信地解决电路问题至关重要。本文将逐步推导所有核心 SL 公式，解释其物理意义，并展示如何将它们应用于典型的考试类题目。
1. Electric Current – Definition and Derivation | 电流 – 定义与推导
Electric current is defined as the rate of flow of electric charge past a given cross-section of a conductor. If an amount of charge Δq passes through a point in a time interval Δt, the average current I is given by:
I = Δq / Δt
电流定义为电荷流过导体某截面的速率。若在时间间隔 Δt 内有电荷量 Δq 通过某一点，则平均电流 I 为上述公式。
The SI unit of current is the ampere (A), where 1 A = 1 C s⁻¹. Conventional current flows from positive to negative potential, opposite to the direction of electron flow. In metallic conductors, the current can also be expressed microscopically as I = nAvq, where n is the number density of charge carriers, A is the cross-sectional area, v is the drift speed, and q is the charge on each carrier.
电流的国际单位是安培（A），1 A = 1 C s⁻¹。传统电流从正电位流向负电位，与电子流动方向相反。在金属导体中，电流也可用微观公式 I = nAvq 表示，其中 n 是载流子数密度，A 为横截面积，v 为漂移速度，q 为每个载流子的电荷。
2. Potential Difference and Energy Transfer | 电位差与能量转移
Potential difference (p.d.) between two points is the work done per unit charge to move a charge between those points. If work W is required to move a charge q, the potential difference V is:
V = W / q
两点之间的电位差是将单位电荷从一点移动到另一点所做的功。若移动电荷 q 需要做功 W，则电位差 V 如上式所示。
This relationship shows that 1 volt (V) = 1 joule per coulomb (J C⁻¹). When a charge q passes through a potential drop V, the electrical energy transferred to the component is ΔE = qV, and the rate of energy transfer is power. This fundamental link between energy, charge, and voltage underpins many circuit derivations.
这一关系表明 1 伏特（V）= 1 焦耳每库仑（J C⁻¹）。当电荷 q 经过电位降 V 时，传递给元件的电能为 ΔE = qV，能量传递的速率即功率。能量、电荷与电压之间的这一基本联系为许多电路推导奠定了基础。
3. Ohm’s Law – V = IR and Resistance | 欧姆定律 – V = IR 与电阻
For many conductors at constant temperature, the current flowing through the conductor is directly proportional to the potential difference across it. This is Ohm’s law, and materials that obey it are called ohmic conductors. The constant of proportionality is the resistance R:
V = IR
对许多导体而言，在温度恒定条件下，流过导体的电流与其两端电位差成正比，这就是欧姆定律，遵从该定律的材料称为欧姆导体。比例常数即电阻 R。
Resistance R is defined by R = V / I and has units of ohms (Ω), where 1 Ω = 1 V A⁻¹. The derivation of this relationship from the definition of potential difference and current shows that resistance quantifies how much potential difference is needed to drive a unit of current. For ohmic devices, a graph of V against I is a straight line through the origin, with slope equal to R.
电阻 R 由 R = V / I 定义，单位为欧姆（Ω），1 Ω = 1 V A⁻¹。从电位差和电流的定义可推导出这一关系，表明电阻衡量驱动单位电流所需的电位差大小。对于欧姆器件，V-I 图是一条通过原点的直线，斜率等于 R。
4. Resistivity – R = ρL / A | 电阻率 – R = ρL / A
The resistance of a uniform wire depends on its length L, cross-sectional area A, and the material’s resistivity ρ. Through experiments, it is found that R is proportional to L and inversely proportional to A, leading to:
R = ρL / A
均匀导线的电阻取决于其长度 L、横截面积 A 以及材料的电阻率 ρ。实验表明 R 与 L 成正比，与 A 成反比，从而得出上式。
Resistivity ρ is a material property with units of ohm metres (Ω m). It can be derived from the microscopic picture: a longer path increases collisions, raising resistance, while a larger area reduces resistance by offering more parallel paths for charge flow. This formula is essential for understanding how dimensions and material choice affect circuit components.
电阻率 ρ 是材料的固有属性，单位为欧姆米（Ω m）。从微观角度看，更长的路径增加碰撞，使电阻增大；更大的面积提供更多并联电荷通道，使电阻减小。该公式对于理解尺寸和材料选择如何影响电路元件至关重要。
5. Electrical Power – P = IV, I²R, V²/R | 电功率 – P = IV、I²R、V²/R
Power is the rate at which energy is transferred. When a charge Δq moves through a potential difference V, the work done is ΔW = VΔq. Dividing by time Δt gives power P = V (Δq/Δt) = IV. Thus:
P = IV
功率是能量传递的速率。当电荷 Δq 通过电位差 V 时，所做的功为 ΔW = VΔq。除以时间 Δt 得功率 P = V (Δq/Δt) = IV。因此得到上式。
Substituting Ohm’s law (V = IR) into P = IV gives P = I²R. Alternatively, substituting I = V/R yields P = V²/R. These three equivalent forms are applied depending on which quantities are known. They describe the rate at which a resistor dissipates energy as heat (Joule heating).
将欧姆定律（V = IR）代入 P = IV 得到 P = I²R；或将 I = V/R 代入得到 P = V²/R。这三种等价形式可根据已知量选择使用，它们描述了电阻器以焦耳热的形式耗散能量的速率。
6. Electromotive Force (EMF) and Internal Resistance | 电动势 (EMF) 与内阻
A cell or battery provides an electromotive force ε (not a force, but a voltage) that does work on charges. The EMF is the total energy supplied per unit charge. Inside a real cell, internal resistance r causes some energy to be dissipated, so the terminal potential difference V is less than ε when current I flows:
V = ε − Ir
电池或电源提供电动势 ε（并非力，而是电压）对电荷做功。电动势是每单位电荷提供的总能量。在实际电池内部，内阻 r 会耗散部分能量，因此有电流 I 时，端电压 V 小于 ε。
Considering the whole circuit with external resistance R, the total resistance is R + r, so the current is I = ε / (R + r). Combining these gives ε = I(R + r). This derivation is crucial for understanding why the terminal voltage drops under load and for calculating maximum power transfer.
考虑包含外电阻 R 的整个电路，总电阻为 R + r，因此电流 I = ε / (R + r)。联立可得 ε = I(R + r)。这一推导对于理解为何端电压在有负载时下降以及计算最大功率传输至关重要。
7. Resistors in Series – Derivation of R_total | 串联电阻 – 总电阻推导
For resistors connected in series, the current I is the same through each. The total potential difference V across the combination equals the sum of individual voltages: V = V₁ + V₂ + V₃ + …. Using V = IR for each, we have:
IR_total = IR₁ + IR₂ + IR₃ + … ⇒ R_total = R₁ + R₂ + R₃ + …
对于串联电阻，通过每个电阻的电流 I 相等。组合两端的总电位差 V 等于各电阻电压之和：V = V₁ + V₂ + V₃ + …。对每个电阻应用 V = IR，可得上述公式。
This shows the equivalent resistance of series resistors is simply the sum of their resistances. Adding resistors in series increases total resistance, which decreases current for a fixed applied voltage.
这表明串联电阻的等效电阻就是各电阻值之和。串联增加电阻会使总电阻增大，在电压固定的情况下电流减小。
8. Resistors in Parallel – Derivation of 1/R_total | 并联电阻 – 总电阻推导 (1/R)
For resistors in parallel, the potential difference V across each resistor is the same. The total current I from the source equals the sum of branch currents: I = I₁ + I₂ + I₃ + …. Using I = V/R for each branch:
V/R_total = V/R₁ + V/R₂ + V/R₃ + … ⇒ 1/R_total = 1/R₁ + 1/R₂ + 1/R₃ + …
对于并联电阻，每个电阻两端的电位差 V 相同。从电源流出的总电流 I 等于各支路电流之和：I = I₁ + I₂ + I₃ + …。对每个支路使用 I = V/R，可得上述公式。
The reciprocal formula means that the equivalent resistance is always smaller than the smallest individual resistance. This derivation highlights charge conservation and the fact that adding parallel paths provides more routes for current, reducing overall resistance.
倒数公式表明等效电阻总是小于最小的单个电阻。这一推导突显了电荷守恒，以及增加并联支路为电流提供更多通路而降低总电阻的事实。
9. Potential Divider – Derivation of Output Voltage | 电位分压器 – 输出电压推导
A potential divider consists of two resistors, R₁ and R₂, connected in series across a supply voltage V_in. The current through both resistors is I = V_in / (R₁ + R₂). The voltage across R₂, which serves as the output V_out, is then:
V_out = I × R₂ = V_in × (R₂ / (R₁ + R₂))
电位分压器由串联在电源电压 V_in 上的两个电阻 R₁ 和 R₂ 组成。流过两个电阻的电流 I = V_in / (R₁ + R₂)。作为输出的 R₂ 两端的电压 V_out 由上式给出。
This formula is derived directly from Ohm’s law and the series current. It allows a continuously variable output voltage if one resistor is replaced by a variable resistor or a sensor such as an LDR or thermistor. The derivation illustrates a simple yet powerful application of series circuit principles.
该公式直接由欧姆定律和串联电流推导得出。若将其中一个电阻换为可变电阻或传感器（如光敏电阻或热敏电阻），即可得到连续可调的输出电压。这一推导展示了串联电路原理简单而有力的应用。
10. Kirchhoff’s Laws – Current and Voltage Rules | 基尔霍夫定律 – 电流定律与电压定律
Kirchhoff’s current law (KCL) states that the total current entering a junction equals the total current leaving it: ∑I_in = ∑I_out. This is a consequence of charge conservation. For circuit analysis we write ∑I = 0, treating currents entering as positive and currents leaving as negative, or vice versa.
基尔霍夫电流定律（KCL）表明，流入某一节点的电流之和等于流出该节点的电流之和：∑I_in = ∑I_out。这是电荷守恒的结果。在电路分析中，我们将流入电流取正、流出电流取负，写作 ∑I = 0，反之亦可。
Kirchhoff’s voltage law (KVL) states that the algebraic sum of potential differences around any closed loop is zero: ∑V = 0. This arises from energy conservation: a test charge moving around a complete loop must return to its starting potential. When applying KVL, we add EMFs and subtract IR drops according to a consistent sign convention.
基尔霍夫电压定律（KVL）表明，沿任一闭合回路的电位差代数和为零：∑V = 0。这源自能量守恒：试探电荷绕闭合回路一周必然回到起始电位。应用 KVL 时，按一致的符号约定将电动势相加、电阻压降减去。
These two laws are the foundation for analysing complex circuits where simple series/parallel simplifications are not possible. They allow us to set up simultaneous equations to solve for unknown currents and voltages.
这两条定律是分析无法用简单串并联简化的复杂电路的基础。它们使我们能够列出联立方程，求解未知电流与电压。
11. Worked Example: Applying Derivations – Cell with Internal Resistance and Parallel Load | 应用推导示例：带内阻的电池与并联负载
A cell of EMF ε = 12.0 V and internal resistance r = 1.0 Ω is connected to an external circuit consisting of a 6.0 Ω resistor and a 3.0 Ω resistor in parallel. Derive the total current, the terminal voltage of the cell, and the current in each branch.
一个电动势 ε = 12.0 V、内阻 r = 1.0 Ω 的电池连接到一个由 6.0 Ω 和 3.0 Ω 电阻并联组成的外部电路。试推导总电流、电池端电压以及各支路电流。
First, find the equivalent resistance of the parallel pair: 1/R_parallel = 1/6.0 + 1/3.0 = 1/6.0 + 2/6.0 = 3/6.0, so R_parallel = 2.0 Ω. Then total circuit resistance R_total = R_parallel + r = 2.0 + 1.0 = 3.0 Ω. The current supplied by the cell is I = ε / R_total = 12.0 / 3.0 = 4.0 A.
首先求并联部分的等效电阻：1/R_parallel = 1/6.0 + 1/3.0 = 1/6.0 + 2/6.0 = 3/6.0，因此 R_parallel = 2.0 Ω。接着，电路总电阻 R_total = R_parallel + r = 2.0 + 1.0 = 3.0 Ω。电池输出的电流 I = ε / R_total = 12.0 / 3.0 = 4.0 A。
The terminal voltage V across the cell terminals (and across the parallel pair) is V = ε – Ir = 12.0 – (4.0 × 1.0) = 8.0 V. Using the potential divider idea, or simply V = I × R_parallel, we also get 4.0 A × 2.0 Ω = 8.0 V.
电池端电压（也即并联部分两端的电压）V = ε – Ir = 12.0 – (4.0 × 1.0) = 8.0 V。利用分压概念，或直接用 V = I × R_parallel，也可得 4.0 A × 2.0 Ω = 8.0 V。
Finally, branch currents: I₁ (through 6.0 Ω) = V / R₁ = 8.0 / 6.0 ≈
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