📚 AS Physics Formula Derivation Guide (PH02 January 2023 Paper) | AS物理公式推导指南（PH02 2023年1月试卷）
This article delivers a systematic breakdown of essential formula derivations tested in the Edexcel International AS Physics Unit 2 (PH02) paper, with particular reference to the January 2023 sitting. Each section pairs theory with step-by-step reasoning, ensuring you can reproduce every derivation confidently and apply it to novel problem contexts. The derivations span electricity, waves, and the nature of light, which form the core of the PH02 specification.
本文系统梳理了 Edexcel 国际版 AS 物理 Unit 2 (PH02) 考试中必考的重要公式推导，并结合 2023 年 1 月试卷的命题特点进行讲解。每一节都将理论与逐步推导配对呈现，确保你能够独立复现推导过程，并灵活迁移到新的题目情境中。推导内容覆盖电学、波动和光的本质，构成 PH02 考试的核心主干。
1. Resistors in Series | 串联电阻
When resistors are connected end-to-end, the total resistance is the sum of individual resistances because the same current passes through each and the total potential difference divides across them.
当多个电阻首尾相接时，由于经过每个电阻的电流相同而总电压在各电阻上分配，总电阻等于各电阻之和。
By conservation of energy, the total p.d. V = V₁ + V₂ + V₃. Using Ohm’s law, V = IR_total, V₁ = IR₁, V₂ = IR₂, V₃ = IR₃. Substituting gives IR_total = IR₁ + IR₂ + IR₃. Since current I is common, cancelling I yields the series rule.
根据能量守恒，总电压 V = V₁ + V₂ + V₃。应用欧姆定律 V = IR_total，V₁ = IR₁，V₂ = IR₂，V₃ = IR₃。代入得 IR_total = IR₁ + IR₂ + IR₃。因为电流 I 处处相等，约去 I 即得到串联公式。
R_total = R₁ + R₂ + R₃
This derivation is frequently examined in PH02, and the January 2023 paper featured a question requiring you to combine series circuits with internal resistance of a cell.
该推导在 PH02 中常考，2023 年 1 月试卷出现了需要将串联电路与电池内阻结合的题目。
2. Resistors in Parallel | 并联电阻
For a parallel arrangement, the p.d. across each branch is identical, but the total current splits among the branches. Applying charge conservation at a junction yields the reciprocal formula.
在并联连接中，各支路两端电压相同，但总电流在各支路中分配。将电荷守恒应用于节点即可得出倒数关系。
Total current I = I₁ + I₂ + I₃. With the same p.d. V, Ohm’s law gives I = V/R_total, I₁ = V/R₁, I₂ = V/R₂, I₃ = V/R₃. Substituting, V/R_total = V/R₁ + V/R₂ + V/R₃. Cancelling V (V ≠ 0) produces the familiar parallel-resistance formula.
总电流 I = I₁ + I₂ + I₃。由于各支路电压 V 相同，由欧姆定律 I = V/R_total，I₁ = V/R₁，I₂ = V/R₂，I₃ = V/R₃。代入得 V/R_total = V/R₁ + V/R₂ + V/R₃。约去非零电压 V，即得熟悉的并联电阻公式。
1/R_total = 1/R₁ + 1/R₂ + 1/R₃
The January 2023 paper asked candidates to find the combined resistance of a parallel network and then compute the current drawn from a battery, directly testing this derivation logic.
2023 年 1 月试卷要求考生计算并联网络的总电阻，进而求出电池输出的电流，直接考查了此推导逻辑。
3. Resistivity Formula | 电阻率公式
Resistance of a uniform conductor depends on its length L, cross-sectional area A, and the intrinsic property of the material called resistivity ρ. The formula is derived by assuming the conductor acts as a series of identical resistive segments.
均匀导体的电阻取决于其长度 L、横截面积 A 以及材料的本征特性电阻率 ρ。该公式的推导基于把导体视为一系列相同电阻段串联的模型。
Increasing length L adds more segments in series, so resistance is proportional to L. Increasing cross-sectional area A effectively adds more paths in parallel, so resistance is inversely proportional to A. Thus R ∝ L/A. Introducing a constant of proportionality, resistivity ρ, yields R = ρL/A.
增加长度 L 相当于串联更多电阻段，因此电阻与 L 成正比。增大横截面积 A 等效于增加并联支路，因此电阻与 A 成反比。于是 R ∝ L/A。引入比例常数电阻率 ρ，即得 R = ρL/A。
R = ρL/A
In the PH02 paper, you may be asked to determine ρ from a wire’s dimensions and measured resistance, or to explain how temperature changes affect ρ.
在 PH02 试卷中，你可能需要根据导线的尺寸和测得的电阻求 ρ，或解释温度变化如何影响 ρ。
4. Temperature Dependence of Resistance | 电阻的温度依赖性
For a metallic conductor, resistance increases with temperature because the metal ions vibrate more vigorously, causing more frequent collisions with flowing electrons. A linear approximation gives R ≈ R₀(1 + αθ), where θ is the temperature rise and α is the temperature coefficient of resistance.
对于金属导体，电阻随温度升高而增大，原因是金属离子振动加剧，导致与自由电子的碰撞频率增大。线性近似给出 R ≈ R₀(1 + αθ)，其中 θ 为温升，α 是电阻温度系数。
Starting from the resistivity form R = ρL/A, and noting that ρ changes almost linearly with temperature for many metals over a moderate range: ρ ≈ ρ₀(1 + αθ). Substituting into R gives R = (ρ₀L/A)(1 + αθ) = R₀(1 + αθ), assuming L and A are constant (thermal expansion neglected).
从电阻率形式 R = ρL/A 出发，并注意到在不太宽的温区内多数金属的 ρ 近似随温度线性变化：ρ ≈ ρ₀(1 + αθ)。代入 R 得 R = (ρ₀L/A)(1 + αθ) = R₀(1 + αθ)，假定 L 和 A 不变（忽略热膨胀）。
R ≈ R₀(1 + αθ)
January 2023 candidates needed to interpret a graph of resistance against temperature and extract α, making this derivation crucial for both theoretical and practical questions.
2023 年 1 月的考生需要解读电阻–温度图像并提取 α，因此该推导对理论和实验题都至关重要。
5. Balanced Wheatstone Bridge | 平衡惠斯通电桥
The Wheatstone bridge is balanced when the ratio of resistances in one branch equals the ratio in the other, resulting in zero current through the galvanometer. The condition is derived from equal potential at the mid-points.
当惠斯通电桥的一个支路电阻比等于另一支路的电阻比时电桥平衡，此时通过检流计的电流为零。这一条件可从中间节点等电位推出。
Consider four resistors R₁, R₂, R₃, R₄ arranged in a quadrilateral with a galvanometer between the junction of R₁-R₂ and R₃-R₄. When balanced, no current flows through the galvanometer, so the potential at the two junctions is equal. Hence the p.d. across R₁ equals that across R₃, and p.d. across R₂ equals that across R₄. Writing I₁R₁ = I₃R₃ and I₂R₂ = I₄R₄. Since the same current flows through R₁ and R₂ (I₁ = I₂) and through R₃ and R₄ (I₃ = I₄) in the absence of galvanometer current, dividing the two equations yields R₁/R₂ = R₃/R₄.
考虑四个电阻 R₁、R₂、R₃、R₄ 组成四边形，检流计接在 R₁-R₂ 和 R₃-R₄ 的结点之间。平衡时检流计无电流，故两结点电位相等。因此 R₁ 两端电压等于 R₃ 两端电压，R₂ 两端电压等于 R₄ 两端电压。写出 I₁R₁ = I₃R₃ 和 I₂R₂ = I₄R₄。由于检流计无电流，流过 R₁ 和 R₂ 的电流相同 (I₁ = I₂)，流过 R₃ 和 R₄ 的电流也相同 (I₃ = I₄)。两式相除即得 R₁/R₂ = R₃/R₄。
R₁/R₂ = R₃/R₄
The January 2023 paper included a practical setup where an unknown resistance was determined using a metre-bridge variant of this principle, making the derivation highly relevant.
2023 年 1 月试卷有一道实验题用到了该原理的滑线电桥变体来测定未知电阻，使得该推导非常切题。
6. The Wave Equation | 波动方程
The relationship between wave speed v, frequency f, and wavelength λ is fundamental to all wave phenomena. It is derived from the definition of speed in terms of distance travelled per unit time.
波速 v、频率 f 和波长 λ 之间的关系是所有波动现象的基础。该关系可从速度的定义（单位时间传播的距离）推出。
In one period T, a wave advances by exactly one wavelength λ. Therefore wave speed v = distance/time = λ/T. Since frequency f = 1/T, substituting gives v = fλ. This applies to both transverse and longitudinal waves, and to mechanical as well as electromagnetic waves.
在一个周期 T 内，波恰好前进一个波长 λ 的距离。因此波速 v = 距离/时间 = λ/T。因为频率 f = 1/T，代入即得 v = fλ。该式既适用于横波与纵波，也适用于机械波与电磁波。
v = fλ
In PH02 January 2023, this equation was used to interpret oscilloscope traces of sound waves and to calculate the speed of light in a double-slit experiment, so clear derivation understanding is expected.
在 PH02 2023 年 1 月考试中，该方程被用于解读声波的示波器波形以及计算双缝实验中的光速，因此清晰的推导理解是必须的。
7. Young’s Double-Slit Fringe Spacing | 杨氏双缝干涉条纹间距
The spacing Δy between adjacent bright (or dark) fringes on a screen, produced by two coherent point sources separated by a distance d and observed at a distance D, is derived using geometry and path-difference arguments.
由间距为 d 的两个相干点光源在相距 D 的屏上产生的相邻亮（或暗）条纹间距 Δy，可通过几何和光程差论证推出。
For a bright fringe at position yₙ from the centre, the path difference between waves from the two slits is nλ. From the geometry of the double-slit arrangement, this path difference ≈ d sinθ, and for small angles sinθ ≈ tanθ = yₙ/D. Hence d (yₙ/D) = nλ, so yₙ = nλD/d. The fringe spacing Δy = yₙ₊₁ − yₙ = (n+1)λD/d − nλD/d = λD/d.
设距中心 yₙ 处为第 n 级亮纹，则两缝光波的光程差为 nλ。由双缝几何关系，光程差 ≈ d sinθ，且小角度下 sinθ ≈ tanθ = yₙ/D。因此 d (yₙ/D) = nλ，解得 yₙ = nλD/d。条纹间距 Δy = yₙ₊₁ − yₙ = (n+1)λD/d − nλD/d = λD/d。
Δy = λD/d
The January 2023 paper required candidates to rearrange this expression to find λ given measured Δy, D, and d, reinforcing the importance of mastering the derivation to handle variations.
2023 年 1 月试卷要求考生重组此式，由测得的 Δy、D 和 d 求 λ，强化了掌握推导以应对变式的重要性。
8. Snell’s Law from Huygens’ Principle | 由惠更斯原理推导斯涅尔定律
Snell’s law n₁ sinθ₁ = n₂ sinθ₂ describes how light refracts at a boundary. It can be derived using Huygens’ principle, which treats each point on a wavefront as a source of secondary wavelets.
斯涅尔定律 n₁ sinθ₁ = n₂ sinθ₂ 描述了光在界面处的折射规律。它可以利用惠更斯原理推导，该原理将波前上的每一点视为次波源。
Consider a plane wavefront incident at angle θ₁ on a boundary where speed changes from v₁ to v₂. In time Δt, the wavefront at the faster side travels a distance v₁Δt while the lagging side travels v₂Δt. From the geometry of the right triangles formed, the shared hypotenuse of length L gives sinθ₁ = v₁Δt / L and sinθ₂ = v₂Δt / L. Dividing: sinθ₁/sinθ₂ = v₁/v₂. Since refractive index n = c/v, v₁/v₂ = n₂/n₁. Hence sinθ₁/sinθ₂ = n₂/n₁, or n₁ sinθ₁ = n₂ sinθ₂.
考虑一平面波前以入射角 θ₁ 射到界面，光速由 v₁ 变为 v₂。在时间 Δt 内，较快一侧的波前沿界面方向行进 v₁Δt，而滞后侧行进 v₂Δt。由形成的直角三角形几何关系，共享的斜边长 L 满足 sinθ₁ = v₁Δt / L，sinθ₂ = v₂Δt / L。两式相除得 sinθ₁/sinθ₂ = v₁/v₂。由于折射率 n = c/v，有 v₁/v₂ = n₂/n₁，因此 sinθ₁/sinθ₂ = n₂/n₁，即 n₁ sinθ₁ = n₂ sinθ₂。
n₁ sinθ₁ = n₂ sinθ₂
Candidates sitting the January 2023 exam needed to apply Snell’s law to fibre optics and to find critical angles, so a solid grasp of the derivation helped with multi-step problems.
参加 2023 年 1 月考试的学生需要将斯涅尔定律应用到光纤和求临界角中，因此牢固掌握推导有助于解决多步问题。
9. Critical Angle and Total Internal Reflection | 临界角与全内反射
The critical angle C is the angle of incidence in an optically denser medium for which the angle of refraction is 90°. This is derived directly from Snell’s law and underpins fibre optics.
临界角 C 是光密介质中使折射角为 90° 的入射角。它由斯涅尔定律直接导出，是光纤通信的基础。
Using Snell’s law at the boundary from medium of index n₁ (denser) to n₂ (rarer, often air with n₂ = 1): n₁ sinθ₁ = n₂ sinθ₂. When θ₂ = 90°, sinθ₂ = 1, and the incident angle θ₁ becomes the critical angle C. Thus n₁ sin C = n₂ × 1, so sin C = n₂/n₁. For an interface with air, sin C = 1/n.
在光密介质（折射率 n₁）到光疏介质（折射率 n₂，通常为空气 n₂=1）的边界应用斯涅尔定律：n₁ sinθ₁ = n₂ sinθ₂。当 θ₂ = 90° 时，sinθ₂ = 1，入射角 θ₁ 即为临界角 C。因此 n₁ sin C = n₂ × 1，即 sin C = n₂/n₁。对于与空气的界面，sin C = 1/n。
sin C = 1/n
In the PH02 January 2023 paper, a diagram showed a light ray travelling through a step-index fibre, requiring use of the critical angle condition to ensure total internal reflection.
在 PH02 2023 年 1 月试卷中，有一幅光线在阶跃折射率光纤中传播的示意图，要求用临界角条件确保全内反射。
10. Einstein’s Photoelectric Equation | 爱因斯坦光电方程
The photoelectric effect is explained by Einstein’s photon theory: each photon of frequency f carries energy hf, and a single photon interacts with one electron to eject it from a metal surface. The maximum kinetic energy of the ejected electron is derived from energy conservation.
光电效应由爱因斯坦光子理论解释：每个频率为 f 的光子携带能量 hf，一个光子与一个电子相互作用将其从金属表面击出。逸出电子的最大动能从能量守恒中导出。
When a photon hits the surface, its energy hf is used in two ways: part of it overcomes the work function φ (minimum energy to free the electron), and the remainder appears as the electron’s kinetic energy. Hence hf = φ + E_k_max. Since E_k_max = ½ m v_max² (where v_max is maximum speed), we can write hf = φ + ½ m v_max². If the photon energy is exactly equal to φ, the threshold frequency f₀ satisfies hf₀ = φ.
当光子撞击表面时，其能量 hf 被用于两方面：一部分克服逸出功 φ（释放电子的最小能量），剩余部分转化为电子动能。因此 hf = φ + E_k_max。由于 E_k_max = ½ m v_max²（v_max 为最大速率），可写为 hf = φ + ½ m v_max²。若光子能量恰好等于 φ，则极限频率 f₀ 满足 hf₀ = φ。
hf = φ + ½ m v_max²
The January 2023 paper presented a graph of maximum kinetic energy against frequency and required linking the gradient to Planck’s constant, directly testing the significance of this derivation.
2023 年 1 月试卷给出了一张最大动能 – 频率图，要求将斜率与普朗克常数关联起来，直接考查了此推导的物理意义。
11. Standing Waves on a String: Node–Antinode Spacing | 弦上的驻波：波节与波腹间距
A standing wave formed by superposition of two identical progressive waves travelling in opposite directions exhibits nodes (zero displacement) and antinodes (maximum displacement). The distance between adjacent nodes or adjacent antinodes is λ/2, derived from the waveform.
由两列相同行波反向传播叠加形成的驻波表现出波节（位移始终为零）和波腹（位移最大）。相邻波节或相邻波腹之间的距离为 λ/2，这由波形导出。
Consider a standing wave described by y = 2A sin(kx) cos(ωt). Nodes occur where sin(kx) = 0, i.e., kx = nπ, n = 0, 1, 2… With k = 2π/λ, we have (2π/λ)x = nπ ⇒ x = nλ/2. Thus the positions of successive nodes differ by Δx = (n+1)λ/2 − nλ/2 = λ/2. Antinodes lie halfway between nodes, so the same spacing applies.
考虑驻波方程 y = 2A sin(kx) cos(ωt)。波节出现在 sin(kx) = 0 处，即 kx = nπ，n = 0,1,2…。由 k = 2π/λ 得 (2π/λ)x = nπ ⇒ x = nλ/2。因此相邻波节的位置差 Δx = (n+1)λ/2 − nλ/2 = λ/2。波腹位于两波节中间，间距相同。
Node spacing = Antinode spacing = λ/2
In the January 2023 exam, a question showed a stationary wave on a string fixed at both ends and asked to deduce the wavelength from the number of loops, directly using this spacing rule.
在 2023 年 1 月考试中，一题展示了两端固定的弦上的驻波，要求由段数推断波长，直接应用了此间距规则。
12. Power in Electrical Circuits – P = IV, P = I²R, P = V²/R | 电路中的功率 – P = IV, P = I²R, P = V²/R
Electrical power is the rate at which energy is transferred by a circuit element. The three equivalent forms arise from the definitions of voltage and Ohm’s law.
电功率是电路元件传递能量的速率。三种等价形式源于电压的定义和欧姆定律。
By definition, power P = energy transferred / time = QV / t = (Q/t)V = IV, because current I = Q/t. This yields P = IV. Substituting Ohm’s law V = IR gives P = I × (IR) = I²R. Alternatively, substituting I = V/R gives P = V × (V/R) = V²/R. All three are used in PH02 for analysing circuits and heating effects.
由定义，功率 P = 转移能量/时间 = QV/t = (Q/t)V = IV，因为电流 I = Q/t。由此得 P = IV。代入欧姆定律 V = IR 得 P = I × (IR) = I²R。或者代入 I = V/R 得 P = V × (V/R) = V²/R。这三种形式在 PH02 中都用于分析电路和热效应。
P = IV = I²R = V²/R
The January 2023 paper had a circuit where a heater was connected to a source with internal resistance; candidates needed to calculate power delivered to the load and power wasted internally, making these derivations essential for efficiency questions.
2023 年 1 月试卷有一道题描述了接有内阻的电源向加热器供电的电路；考生需计算负载功率和内阻损耗功率，这些推导对于效率问题必不可少。
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