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Combined Resistance: Series and Parallel Connections | 组合电阻：串联与并联连接

2026-08-02 16:21 · 本文部分内容由AI辅助创作

📚 Combined Resistance: Series and Parallel Connections | 组合电阻：串联与并联连接

In A-Level Physics, understanding how to analyse and calculate the combined resistance of circuits containing both series and parallel connections is essential. Whether you are simplifying a complex network or predicting current and power, a clear grasp of resistance combination rules allows you to break down any circuit into a single equivalent resistor. This article explores the key principles, step-by-step reduction strategies, and worked examples – including a network that simplifies to a combined resistance of 0.54 Ω – all aligned with the Edexcel specification.

在A-Level物理中，掌握如何分析和计算包含串联与并联连接的电路组合电阻至关重要。无论是简化一个复杂网络，还是预测电流与功率，清楚理解电阻组合规则都能让你将任何电路化简为单一等效电阻。本文探讨关键原理、逐步化简策略以及详细例题——包括一个最终化简为组合电阻0.54 Ω的网络——内容完全贴合Edexcel课程要求。

1. Understanding Electrical Resistance | 理解电阻

emperature, the potential difference V across it is directly proportional to the current I flowing through it, and the constant of proportionality is the resistance R: V = I × R. The unit of resistance is the ohm (Ω).

电阻是衡量电流在电气元件中流动时所受阻力的物理量。根据欧姆定律，对于恒定温度下的金属导体，其两端的电势差V与流过它的电流I成正比，比例常数即为电阻R：V = I × R。电阻的单位是欧姆（Ω）。

V = I R

V = I R

When multiple resistors are connected together, the overall opposition offered by the network is called the combined resistance or equivalent resistance. The method for combining them depends on whether they are arranged in series, in parallel, or in a mixture of both.

当多个电阻器连接在一起时，整个网络所呈现的总阻碍作用被称为组合电阻或等效电阻。组合它们的方法取决于电阻是串联、并联还是混联。

2. Resistors in Series | 串联电阻

Resistors are in series when they are connected end-to-end so that the same current passes through each one. The total or combined resistance Rₛ is simply the sum of the individual resistances:

当电阻器首尾相连、流过每个电阻器的电流相同时，它们就是串联连接。总电阻或组合电阻Rₛ就是各个电阻值的代数和：

Rₛ = R₁ + R₂ + R₃ + …

Rₛ = R₁ + R₂ + R₃ + …

Key characteristics of series circuits include:

串联电路的关键特征包括：

This additive property makes series connections useful when you need to increase resistance or create voltage divider circuits.

这种相加特性使得串联连接在需要增大电阻或构建分压电路时非常有用。

3. Resistors in Parallel | 并联电阻

Resistors are in parallel when they are connected across the same two points, providing separate paths for current. In a parallel arrangement, the reciprocal of the combined resistance Rₚ equals the sum of the reciprocals of the individual resistances:

当电阻器连接在两个公共点之间并提供独立电流路径时，它们就是并联连接。在并联组合中，组合电阻Rₚ的倒数等于各电阻倒数之和：

1/Rₚ = 1/R₁ + 1/R₂ + 1/R₃ + …

1/Rₚ = 1/R₁ + 1/R₂ + 1/R₃ + …

For two resistors in parallel, this simplifies to:

对于两个并联的电阻，可以简化为：

Rₚ = (R₁ × R₂) / (R₁ + R₂)

Rₚ = (R₁ × R₂) / (R₁ + R₂)

Important properties of parallel circuits:

并联电路的重要性质：

Parallel connections are widely used in household wiring to ensure that appliances operate independently at the same voltage.

并联连接广泛应用于家庭配线，以确保电器在相同电压下独立工作。

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4. Combined Series-Parallel Networks | 串并联组合网络

Many practical circuits cannot be classified as purely series or purely parallel; they contain combinations of both. Calculating the combined resistance for such networks requires a systematic reduction approach: you repeatedly replace series or parallel clusters with their equivalent resistance until a single resistor remains.

许多实际电路并不能简单地归类为纯串联或纯并联；它们包含了两种连接方式的组合。计算这类网络的组合电阻需要采用系统化简方法：反复用等效电阻替换串联或并联的子模块，直到最终只剩下一个电阻。

A typical Edexcel exam question might present a network with three or four resistors arranged in a mixed configuration and ask you to find the total combined resistance, the current drawn from the supply, or the p.d. across a particular resistor. The ability to fluently combine resistors is therefore essential.

Edexcel考试中的典型题目可能会给出一个由三四个电阻器混联而成的网络，要求你求出总的组合电阻、电源输出的电流或某一电阻两端的电压。因此，熟练掌握电阻组合方法至关重要。

5. Step-by-Step Reduction Strategy | 逐步化简策略

Follow these steps to simplify a complex resistor network:

遵循以下步骤化简复杂的电阻网络：

In some networks, you may need to deal with nested parallel sections inside a series loop. Always work from the innermost group outward.

在一些网络中，你可能需要处理串联回路内部的嵌套并联部分。始终从最内层的组合开始向外化简。

6. Worked Example 1: Simple Series-Parallel Circuit | 例题1：简单的串并联电路

Consider a circuit where a 4.0 Ω resistor and a 12.0 Ω resistor are connected in parallel, and this parallel combination is then connected in series with a 6.0 Ω resistor. Calculate the total combined resistance.

假设一个电路中，一个4.0 Ω和一个12.0 Ω的电阻器并联，随后该并联组合与一个6.0 Ω的电阻器串联。计算总的组合电阻。

Step 1: Combine the parallel 4.0 Ω and 12.0 Ω. Using the product/sum formula for two resistors:
Rₚ = (4.0 × 12.0) / (4.0 + 12.0) = 48.0 / 16.0 = 3.0 Ω.

步骤1： 合并并联的4.0 Ω和12.0 Ω。使用两电阻乘积/和公式：
Rₚ = (4.0 × 12.0) / (4.0 + 12.0) = 48.0 / 16.0 = 3.0 Ω。

Step 2: with the 6.0 Ω resistor. Therefore, the total combined resistance Rₜ = 3.0 + 6.0 = 9.0 Ω.

步骤2： 这个3.0 Ω等效电阻与6.0 Ω电阻串联。因此，总的组合电阻Rₜ = 3.0 + 6.0 = 9.0 Ω。

, and the voltage across the 6.0 Ω resistor would be 6.0 V, while the parallel block would share 3.0 V.

如果电源电压为9.0 V，电路中的电流将是I = V / Rₜ = 1.0 A，6.0 Ω电阻两端的电压将是6.0 V，而并联部分将分得3.0 V。

7. Worked Example 2: Complex Network with 0.54 Ω Result | 例题2：复杂网络得出0.54 Ω

This more elaborate network is designed to illustrate how multi-step reduction leads to a combined resistance of 0.54 Ω. The circuit consists of four resistors: a 1.0 Ω and a 2.0 Ω resistor connected in parallel, that parallel block connected in parallel with another 1.0 Ω resistor, and finally a 0.14 Ω resistor connected in series with the resulting group.

这个更复杂的网络旨在展示多步化简如何得到0.54 Ω的组合电阻。电路由四个电阻组成：一个1.0 Ω和一个2.0 Ω电阻并联，该并联组再与另一个1.0 Ω电阻并联，最后再串联一个0.14 Ω电阻。

Reduction process:

化简过程：