📚 Combined Resistance Masterclass: Problem 170 | 复合电阻大师课：问题170
In A-Level Physics, the ability to simplify combined resistor networks is a foundational skill that bridges simple circuit theory and more advanced concepts like potential dividers and internal resistance. Problem 170 represents a typical Edexcel exam-style challenge where resistors are arranged in both series and parallel, demanding a methodical, step-by-step approach. This article will guide you through the principles, worked examples, and common pitfalls, ensuring you can confidently tackle any combined resistance question on your examination.
在A-Level物理中，化简复合电阻网络是一项基本技能，它连接了简单的电路理论与分压器、内阻等更高级的概念。问题170代表了一种典型的爱德思考试风格的难题，其中电阻器以串联和并联混合排列，要求采用有条理、逐步的方法。本文将引导你掌握原理、例题分析及常见陷阱，确保你能够自信地应对考试中任何复合电阻的问题。
1. Series Combination | 串联组合
When resistors are connected end-to-end, the total or equivalent resistance is simply the sum of individual resistances. This is because the same current must flow through each resistor, but the voltage drops add up.
当电阻器首尾相连时，总电阻或等效电阻就是各个电阻值的简单加和。这是因为相同的电流必须流过每个电阻，但电压降会相加。
The series resistance formula is:
串联电阻公式为：
Rtotal = R1 + R2 + R3 + ...
For example, three resistors of 2 Ω, 3 Ω and 5 Ω in series give a total of 10 Ω. This rule is intuitive and forms the basis of more complex network simplification.
例如，三个电阻分别为2 Ω、3 Ω和5 Ω串联，总电阻为10 Ω。这条规则直观易懂，是化简更复杂网络的基础。
2. Parallel Combination | 并联组合
In a parallel arrangement, resistors are connected across the same two points, so they all share the same potential difference. The total current splits among the branches, and the reciprocal of the total resistance equals the sum of the reciprocals of individual resistances.
在并联排列中，电阻器连接在相同的两个点之间，因此它们共享相同的电势差。总电流在各支路中分流，总电阻的倒数等于各电阻倒数之和。
The parallel resistance formula is:
并联电阻公式为：
1/Rtotal = 1/R1 + 1/R2 + 1/R3 + ...
For two resistors in parallel, a convenient shortcut is Rtotal = (R1 × R2) / (R1 + R2). Remember that the total resistance in parallel is always less than the smallest individual resistor.
对于两个并联电阻，一个方便的快速公式是 R总 = (R1 × R2) / (R1 + R2)。请记住，并联总电阻总是小于最小的单个电阻。
3. Identifying Series and Parallel Sections | 识别串联与并联部分
Before applying formulas, you must analyse the circuit diagram carefully. Trace the path of current from one terminal to the other. Resistors that share exactly the same current without junctions are in series. If they share the same two nodes, they are in parallel.
在应用公式之前，你必须仔细分析电路图。从一端到另一端追踪电流路径。若电阻之间没有节点而流过相同的电流，它们就是串联的。如果它们共享相同的两个节点，它们就是并联的。
In many Edexcel problems, a network appears tangled at first, but by redrawing the circuit using standard symbols and straight lines, the series and parallel blocks become obvious. Practise redrawing—it is a crucial exam skill.
在许多爱德思考题中，网络起初看起来错综复杂，但通过用标准符号和直线重新绘制电路，串并联模块就会变得明显。练习重新作图——这是一项关键的考试技巧。
4. Step-by-Step Simplification | 逐步化简方法
The strategy for combined networks is to identify a small group of resistors that are purely in series or purely in parallel, replace them with their equivalent resistance, and then redraw the circuit. Repeat this process until you obtain a single equivalent resistance.
处理复合网络的策略是，找出一小组纯粹串联或纯粹并联的电阻，用它们的等效电阻替换，然后重新绘制电路。重复这一过程，直到获得单个等效电阻。
For example, if a 6 Ω and a 3 Ω are in parallel, combine them to get 2 Ω. Then that 2 Ω may be in series with another 4 Ω, giving 6 Ω total. This gradual reduction is often called "circuit reduction."
例如，如果一个6 Ω和一个3 Ω并联，合并后得到2 Ω。然后这个2 Ω可能与另一个4 Ω串联，总电阻为6 Ω。这种逐步归并常被称为“电路简缩”。
5. Worked Example: Problem 170 Circuit | 例题分析：问题170电路
Let us apply this method to Problem 170. Imagine a network where a 10 Ω resistor is in series with a parallel combination of two 15 Ω resistors, and that whole block is in parallel with a 20 Ω resistor. The task is to find the total resistance across the supply.
让我们将这个方法应用到问题170。设想一个网络：一个10 Ω电阻与两个15 Ω电阻的并联组合串联，而整个这一块又与一个20 Ω电阻并联。要求求出电源两端的总电阻。
Step 1: The two 15 Ω in parallel give Rp ep 2: This 7.5 Ω is in series with 10 Ω, yielding 17.5 Ω. Step 3: Now this 17.5 Ω is in parallel with 20 Ω. Using the parallel formula: Rtotal = (17.5 × 20)/(17.5 + 20) = 350/37.5 ≈ 9.33 Ω.
步骤1：两个15 Ω并联，得到 Rp = (15×15)/(15+15) = 225/30 = 7.5 Ω。步骤2：这个7.5 Ω与10 Ω串联，得到17.5 Ω。步骤3：现在这个17.5 Ω与20 Ω并联。使用并联公式：R总 = (17.5×20)/(17.5+20) = 350/37.5 ≈ 9.33 Ω。
Notice how each step reduces the circuit complexity. Always double-check whether resistors are truly in parallel or series after each reduction.
注意每一步如何降低电路复杂度。每次化简后，都要仔细检查电阻是否真正构成纯并联或纯串联。
6. Kirchhoff's Laws for More Complex Networks | 更复杂网络中的基尔霍夫定律
Some networks, such as unbalanced Wheatstone bridges, cannot be simplified using simple series-parallel reduction. In those cases, you must use Kirchhoff's current and voltage laws to set up simultaneous equations.
有些网络，比如不平衡的惠斯通电桥，无法用简单的串并联化简来处理。在这些情况下，必须使用基尔霍夫电流定律和电压定律来建立联立方程组。
Kirchhoff's current law (KCL) states that the sum of currents entering a junction equals the sum leaving it. Kirchhoff's voltage law (KVL) states that the sum of emfs around any closed loop equals the sum of potential drops.
基尔霍夫电流定律指出，进入一个节点的电流之和等于离开该节点的电流之和。基尔霍夫电压定律指出，沿任一闭合回路的电动势之和等于电势降之和。
For Edexcel A-Level, you are expected to apply KCL and KVL to circuits with more than one power source or complex branching. These laws underpin the behaviour of combined networks that cannot be reduced easily.
对于爱德思A-Level考试，你需要将基尔霍夫定律应用于包含多个电源或复杂分支的电路。这些定律支撑着那些无法轻易化简的复合网络的行为。
7. Potential Divider with Combined Load | 带组合负载的分压器
A common exam question involves a potential divider where the output voltage is connected to a load resistance that is a combined network. The load affects the output voltage because it appears in parallel with part of the divider.
常见的考题涉及分压器，其输出电压连接到一个由组合网络构成的负载电阻。负载会影响输出电压，因为它并联在分压器的一部分上。
To solve such problems, first simplify the load network to a single equivalent resistance RL. Then treat this RL as being in parallel with the lower resistor of the potential divider. Compute the parallel combination, and then apply the potential divider formula with the new effective resistance.
解决此类问题时，首先将负载网络简化为单个等效电阻 RL。然后将这个 RL 视为与分压器的下电阻并联。计算并联组合，然后用新的有效电阻应用分压器公式。
For instance, if the lower resistor is 100 Ω and the load network simplifies to 200 Ω, their parallel equivalent is 66.7 Ω. This changes the division ratio significantly compared to a no-load scenario.
例如，如果下电阻为100 Ω，负载网络简化为200 Ω，它们的并联等效值为66.7 Ω。与空载情况相比，这会显著改变分压比。
8. Internal Resistance and Combined External Load | 内阻与组合外负载
A source of emf, such as a battery, has internal resistance r. When connected to an external combined network, the terminal pd is less than the emf. The total circuit resistance is r plus the equivalent external resistance Rext.
电动势源（如电池）具有内阻 r。当连接到外部复合网络时，端电压低于电动势。电路的总电阻为 r 加上外部等效电阻 Rext。
If the external network is a mix of series and parallel resistors, you must first find Rext by reduction. The current drawn from the battery is then I = emf / (r + Rext), and power calculations follow naturally.
如果外部网络是串并联电阻的混合，首先必须通过化简求出 Rext。从电池获取的电流为 I = 电动势 / (r + Rext)，随后自然可以进行功率计算。
This type of problem often appears in practical-style questions, where you may be asked to explain why the terminal voltage drops as more parallel branches are added—each new branch reduces the total external resistance.
这类问题常出现在实验风格题目中，可能要求你解释为何随着增加更多并联支路，端电压会下降——每增加一个支路都会降低总外电阻。
9. Power Dissipation in Combined Networks | 复合网络中的功率耗散
R or P = V²/R, but you must use the correct current or voltage for that specific resistor.
一旦求出等效电阻和总电流，你就可以分析各个元件的功率。记住 P = I²R 或 P = V²/R，但必须使用该特定电阻的正确电流或电压。
In a combined network, current splits in parallel branches, so resistors in different branches dissipate different powers even if they have the same resistance. This is a common pitfall—always calculate branch currents first.
在复合网络中，电流在并联支路中分流，因此即使电阻值相同，不同支路中的电阻耗散的功率也不同。这是一个常见陷阱——务必首先计算支路电流。
om the supply, determine the branch current through the parallel 15 Ω pair and apply P = Ibranch² × 15.
对于问题170，你可能被要求求出其中一个15 Ω电阻的耗散功率。在求出电源的总电流后，确定通过15 Ω并联对的支路电流，然后应用 P = I_支路² × 15。
10. Common Mistakes to Avoid | 应避免的常见错误
One frequent error is applying the series formula to resistors that are not in series because of a junction in between. Always trace the current path. Another mistake is using the parallel shortcut formula for more than two resistors incorrectly.
一个常见错误是将串联公式应用于因中间有节点而非串联的电阻。务必追踪电流路径。另一个错误是错误地对两个以上电阻使用并联简化公式。
Students often forget that the total resistance of a parallel combination is always smaller than the smallest resistor in that group—use this as a quick sanity check. Also, misreading the circuit diagram by not redrawing it neatly can lead to incorrect identification.
学生们常常忘记并联组合的总电阻总是小于该组中最小的电阻——可将此作为快速合理性检查。此外，未整洁地重新绘制电路图而导致误判也是一个问题。
Finally, failing to convert correctly when resistances are given in kΩ or mΩ leads to order-of-magnitude errors. Keep everything in the same unit, preferably ohms, during your calculation steps.
最后，当电阻以 kΩ 或 mΩ 给出时未能正确转换会导致数量级错误。在计算步骤中，将所有量保持为同一单位，最好是欧姆。
11. Exam Technique and Practice | 考试技巧与练习
Edexcel papers often present a "Show that the total resistance is ..." question, which provides a target value—use this to check your reduction steps. If you arrive at the given value, you are likely correct. If not, retrace your steps from the innermost combination outward.
爱德思考卷常出现“证明总电阻为……”的问题，它提供了一个目标值——利用这个目标值检查你的化简步骤。如果得出给定值，你很可能正确。如果不对，从最内部的组合开始向外重新检查各步骤。
Practise with networks that have three or four levels of nesting. Build up your confidence by starting from simple series and parallel exercises, then gradually introduce mixed networks. Timed practice is essential—combined resistance questions can appear in both multiple-choice and structured sections.
练习具有三到四层嵌套的网络。从简单的串并联练习入手，逐步引入混合网络以建立信心。限时练习必不可少——复合电阻问题既可能出现在选择题部分，也可能出现在结构化问题中。
It is also useful to familiarise yourself with the concept of equivalent resistance as a black-box replacement; this abstraction helps in solving problems involving maximum power transfer or matching impedances, which are hinted at in extension topics.
同样有用的是将等效电阻概念熟悉为黑箱替代；这种抽象有助于解决涉及最大功率传输或阻抗匹配的问题，这些在拓展主题中有所暗示。
12. Summary and Key Formulae | 总结与核心公式
To master combined resistance questions like Problem 170, remember these core ideas: simplify in stages, redraw the circuit frequently, and verify each step with physical intuition. Keep the following formula table in mind.
要掌握问题170这类复合电阻题目，请记住这些核心思想：分阶段化简，经常重新绘制电路，并用物理直觉验证每一步。记住以下公式表格。
| Configuration | Formula |
| Series | Rtotal = R1 + R2 + ... |
| Parallel (two) | Rtotal = (R1R2)/(R1+R2) |
| Parallel (general) | 1/Rtotal = 1/R1 + 1/R2 + 1/R3 + ... |
| Power dissipation | P = I²R = V²/R = IV |
With methodical practice, you will find these networks less intimidating and more like a logical puzzle. Keep going, and soon you'll be solving Problem 170 and its variations with ease.
通过有条理的练习，你会发现这些网络不那么令人生畏，而更像是一个逻辑谜题。继续努力，你很快就能轻松解决Problem 170及其变体。
Published by TutorHao | Edexcel Physics Revision Series | aleveler.com
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