📚 Electromagnets 1.2.2 – Parallel Circuits Derivation | 电磁铁 1.2.2 – 并联电路公式推导
Parallel circuits are everywhere in modern electronics, from the wiring in your home to the control boards inside electromagnetic devices. Understanding how resistors combine in parallel is not just a textbook exercise – it is essential for designing safe and efficient circuits for electromagnets, relays, and sensors. This article derives the equivalent resistance formula step by step, explores the current division rule, and connects the theory directly to real-world electromagnetic applications.
并联电路在现代电子设备中无处不在，从家庭布线到电磁装置内部的控制电路。理解电阻在并联中的组合方式不仅是一个教科书上的练习，对于设计安全高效的电磁铁、继电器和传感器电路也至关重要。本文将一步步推导等效电阻公式，深入分析分流原理，并将理论直接与实际电磁应用联系起来。
1. Understanding Parallel Circuits | 理解并联电路
A parallel circuit is one where two or more components are connected across the same two points, creating multiple paths for current to flow. Each component forms a separate branch, and the total current is shared among these branches.
并联电路是指两个或更多元件连接在相同的两个点之间，为电流提供了多条路径。每个元件构成一个独立的支路，总电流在这些支路间进行分配。
In the context of electromagnets, parallel connections are often used when multiple coils need to operate from a single power supply without affecting each other’s voltage. For example, an array of small electromagnets in a sorting machine may be wired in parallel so that each receives the full supply voltage and can be controlled individually.
在电磁铁的应用中，当多个线圈需要从同一电源供电且互不影响电压时，常采用并联连接。例如，分拣机中的小型电磁铁阵列可以并联布线，使每个电磁铁都获得完整的电源电压，并可以独立控制。
2. Key Rules for Parallel Circuits | 并联电路的关键规则
There are two fundamental laws governing parallel circuits: the voltage across every branch is identical, and the total current is the sum of the branch currents. These rules are direct consequences of Kirchhoff’s Voltage Law (KVL) and Kirchhoff’s Current Law (KCL).
并联电路有两条基本定律：每个支路两端的电压完全相同；总电流等于各支路电流之和。这些规则是基尔霍夫电压定律和基尔霍夫电流定律的直接推论。
Because each component is connected directly between the same two nodes, the potential difference across any resistor in parallel must equal the source voltage V. Hence, V₁ = V₂ = V₃ = V.
由于每个元件都直接连接在两个相同的节点之间，任何并联电阻两端的电势差必然等于电源电压 V。因此，V₁ = V₂ = V₃ = V。
Meanwhile, the charge flowing into a junction must equal the charge flowing out, so the total current I_total is the sum of the individual branch currents: I_total = I₁ + I₂ + I₃ + …
同时，流入节点的电荷必须等于流出的电荷，因此总电流 I_total 是各支路电流之和：I_total = I₁ + I₂ + I₃ + …
3. Deriving the Equivalent Resistance Formula | 推导等效电阻公式
To find a single resistor R_eq that can replace all parallel resistors without changing the total current drawn from the source, we start with the current sum rule and apply Ohm’s law to each branch.
为了找到一个可以替代所有并联电阻而不改变电源输出总电流的单个等效电阻 R_eq，我们从电流求和规则开始，并对每个支路应用欧姆定律。
For resistor 1, I₁ = V / R₁; for resistor 2, I₂ = V / R₂; for resistor 3, I₃ = V / R₃, and so on. Since all branch voltages are V, the total current is:
对于电阻 1，I₁ = V / R₁；对于电阻 2，I₂ = V / R₂；对于电阻 3，I₃ = V / R₃，依此类推。由于所有支路电压都是 V，总电流为：
I_total = V/R₁ + V/R₂ + V/R₃ + …
Factor out V from the right-hand side: I_total = V × (1/R₁ + 1/R₂ + 1/R₃ + …)
从右边提取公因子 V：I_total = V × (1/R₁ + 1/R₂ + 1/R₃ + …)
Now, the equivalent resistance R_eq is defined by the relationship V = I_total × R_eq. Rearranging gives I_total = V / R_eq. Substituting into the equation above yields:
现在，等效电阻 R_eq 由关系式 V = I_total × R_eq 定义。整理得 I_total = V / R_eq。代入上面的方程可得：
V / R_eq = V × (1/R₁ + 1/R₂ + 1/R₃ + …)
Cancel V from both sides (since V is non-zero) and we obtain the familiar reciprocal formula for parallel resistance:
两边消去 V（因为 V 非零），我们便得到了熟悉的并联电阻倒数公式：
1 / R_eq = 1 / R₁ + 1 / R₂ + 1 / R₃ + …
This result shows that the equivalent conductance (1/R_eq) is the sum of individual conductances. This derivation is elegant and highlights why parallel resistors always produce a smaller total resistance than the smallest individual resistor.
这一结果表明，等效电导（1/R_eq）是各个电导之和。这个推导十分简洁，并强调了为何并联电阻的总阻值总是小于其中最小的单个电阻。
4. Two-Resistor Special Case | 两个电阻的特例
For the common situation with only two resistors in parallel, the general formula can be rearranged into a product-over-sum form that is often quicker to use in calculations.
对于只有两个电阻并联的常见情况，通用公式可以变形为乘积除以和的形式，这在计算中往往更加快捷。
Starting from 1/R_eq = 1/R₁ + 1/R₂, find a common denominator: 1/R_eq = (R₂ + R₁) / (R₁R₂). Taking the reciprocal on both sides gives:
从 1/R_eq = 1/R₁ + 1/R₂ 出发，求公分母：1/R_eq = (R₂ + R₁) / (R₁R₂)。两边取倒数可得：
R_eq = (R₁ × R₂) / (R₁ + R₂)
This formula is valid only for two resistors. It is a useful shortcut, but must not be applied directly to three or more resistors without iterative steps.
这个公式仅适用于两个电阻。它是一个有用的捷径，但不可直接用于三个或更多电阻的情况，除非采用分步计算。
5. Conductance Approach | 电导方法
When dealing with multiple parallel branches, especially in electromagnetic coil circuits where resistances may be very low, it can be more intuitive to work with conductance, denoted by G. Conductance is the reciprocal of resistance: G = 1 / R, measured in siemens (S).
当处理多个并联支路时，特别是在电磁线圈电路中电阻可能极低的情况下，使用电导 G 会更直观。电导是电阻的倒数：G = 1 / R，单位为西门子（S）。
For parallel resistors, the total conductance G_eq is simply the sum of all branch conductances:
G_eq = G₁ + G₂ + G₃ + …
Then the equivalent resistance is R_eq = 1 / G_eq. This method eliminates fractions and is particularly helpful when analysing heating effects in electromagnet windings that share a common voltage rail.
那么等效电阻为 R_eq = 1 / G_eq。这种方法消除了分数运算，在分析共用一个电压母线的电磁铁绕组的热效应时特别有用。
6. Current Division Principle | 分流原理
In a parallel circuit, knowing the total current is not enough; we often need to find the current in each branch. The current divider rule allows us to calculate branch currents without first finding the voltage.
在并联电路中，只知道总电流是不够的；我们常常需要计算每个支路中的电流。分流器法则可以在不必先求电压的情况下计算出支路电流。
For a two-branch parallel circuit, the current in resistor R₁ is given by:
I₁ = I_total × (R₂ / (R₁ + R₂))
Similarly, the current in R₂ is:
I₂ = I_total × (R₁ / (R₁ + R₂))
Notice that the larger current flows through the smaller resistor. This is exactly the opposite of the voltage divider rule and a common source of confusion.
需要注意的是，较大的电流流过较小的电阻。这与分压器法则正好相反，是常见的混淆点。
For more than two resistors, the general form is I_x = I_total × (R_eq / R_x), where R_eq is the equivalent parallel resistance. This formula is derived directly from the fact that the voltage V = I_total × R_eq is shared, so I_x = V / R_x.
对于多于两个电阻的情况，一般形式为 I_x = I_total × (R_eq / R_x)，其中 R_eq 是并联等效电阻。该公式直接源自电压 V = I_total × R_eq 被共享，因此 I_x = V / R_x 这一事实。
7. Power Dissipation in Parallel | 并联电路中的功率消耗
Each resistor in a parallel network dissipates power according to P = V² / R, since the voltage across it is constant. The total power dissipated is the sum of the powers in each branch.
并联网络中的每个电阻根据 P = V² / R 消耗功率，因为其两端电压是恒定的。总消耗功率等于各支路功率之和。
Thus, P_total = V² / R₁ + V² / R₂ + V² / R₃ + … = V² × (1/R₁ + 1/R₂ + 1/R₃ + …) = V² / R_eq.
因此，P_total = V² / R₁ + V² / R₂ + V² / R₃ + … = V² × (1/R₁ + 1/R₂ + 1/R₃ + …) = V² / R_eq。
This makes sense: equivalent resistance correctly models both current and power characteristics of the original parallel network. When designing an electromagnet array, this ensures the power supply can handle the total load.
这是有意义的：等效电阻正确地描述了原始并联网络的电流和功率特性。在设计电磁铁阵列时，这可以确保电源能够承受总负载。
8. Real-World Application: Electromagnets in Parallel | 实际应用：并联电磁铁
Electromagnets are essentially coils of wire with a certain resistance. In automated machinery, multiple electromagnets are often wired in parallel to the same DC supply. A classic example is a relay bank controlling several pneumatic valves.
电磁铁本质上就是具有一定电阻的线圈。在自动化机械中，多个电磁铁通常并联在同一直流电源上。一个经典例子是控制多个气动阀的继电器组。
If each coil has a resistance of 120 Ω and the supply is 24 V, a single coil draws 0.2 A. If four are placed in parallel, the equivalent resistance becomes 120/4 = 30 Ω, and the total current rises to 0.8 A. The power supply must be rated to deliver this current without voltage drop, otherwise the electromagnets may fail to actuate reliably.
如果每个线圈的电阻为 120 Ω，电源为 24 V，单个线圈将吸取 0.2 A 电流。若将四个线圈并联，等效电阻变为 120/4 = 30 Ω，总电流升至 0.8 A。电源必须能够提供如此大的电流而不会造成电压下降，否则电磁铁可能无法可靠动作。
Understanding parallel resistance also helps when coils are connected in series-parallel combinations to achieve specific magnetic fields while balancing current draw.
理解并联电阻也有助于在需要特定磁场并平衡电流消耗的情况下，将线圈进行串并联组合。
9. Step-by-Step Worked Example | 逐步求解实例
Problem: Three resistors, R₁ = 6 Ω, R₂ = 3 Ω, and R₃ = 2 Ω, are connected in parallel across a 12 V battery. Calculate (a) the equivalent resistance, (b) the total current, (c) the current through each resistor, and (d) the total power dissipation.
问题：三个电阻 R₁ = 6 Ω、R₂ = 3 Ω 和 R₃ = 2 Ω 并联在 12 V 电池上。计算 (a) 等效电阻，(b) 总电流，(c) 每个电阻中的电流，(d) 总功率消耗。
Step (a): 1/R_eq = 1/6 + 1/3 + 1/2 = 1/6 + 2/6 + 3/6 = 6/6 = 1 ⇒ R_eq = 1 Ω.
步骤 (a)：1/R_eq = 1/6 + 1/3 + 1/2 = 1/6 + 2/6 + 3/6 = 6/6 = 1 ⇒ R_eq = 1 Ω。
Step (b): I_total = V / R_eq = 12 V / 1 Ω = 12 A.
步骤 (b)：I_total = V / R_eq = 12 V / 1 Ω = 12 A。
Step (c): Using I = V / R for each branch: I₁ = 12/6 = 2 A; I₂ = 12/3 = 4 A; I₃ = 12/2 = 6 A. Check sum: 2+4+6 = 12 A, which matches I_total.
步骤 (c)：对每个支路使用 I = V / R：I₁ = 12/6 = 2 A；I₂ = 12/3 = 4 A；I₃ = 12/2 = 6 A。检验求和：2+4+6 = 12 A，与 I_total 吻合。
Step (d): P_total = V × I_total = 12 V × 12 A = 144 W. Alternatively, P_total = V² / R_eq = 144/1 = 144 W.
步骤 (d)：P_total = V × I_total = 12 V × 12 A = 144 W。或者，P_total = V² / R_eq = 144/1 = 144 W。
This example underscores how quickly total current escalates when low-resistance branches are added in parallel, an important safety consideration in electromagnetic circuit design.
这个例子凸显了当低电阻支路并联加入时，总电流会迅速增加，这是电磁电路设计中一个重要的安全考量。
10. Common Mistakes and Misconceptions | 常见错误和误解
Mistake 1: Adding resistances directly in parallel. Many beginners erroneously think R_eq = R₁ + R₂ for parallel connections, confusing it with series rules. Always remember that parallel equivalent resistance is always less than the smallest individual resistance.
错误 1：直接相加并联电阻。许多初学者错误地认为并联时 R_eq = R₁ + R₂，混淆了串并联规则。务必记住，并联等效电阻总是小于最小的单个电阻。
Mistake 2: Forgetting to take the reciprocal at the end. After summing 1/R values, you must invert the result to get R_eq. Leaving the answer as a conductance value is a common slip.
错误 2：忘记最后取倒数。在求和了 1/R 值之后，必须将结果取倒数才能得到 R_eq。直接把答案留作电导值是一个常见的疏漏。
Mistake 3: Applying product-over-sum to more than two resistors directly. The formula R_eq = (R₁R₂R₃…)/(R₁+R₂+R₃+…) is mathematically incorrect. Always use the reciprocal sum.
错误 3：将乘积除以和直接用于两个以上电阻的情况。公式 R_eq = (R₁R₂R₃…)/(R₁+R₂+R₃+…) 在数学上是不正确的。务必使用倒数求和的方法。
Mistake 4: Assuming current is the same in all branches. In parallel circuits, currents divide inversely with resistance. A branch with higher resistance carries less current, which is critical for ensuring uniform magnetic pull in parallel electromagnets with slightly different coil resistances.
错误 4：假设所有支路的电流相等。在并联电路中，电流与电阻成反比分配。电阻较高的支路流过的电流较小，这对于确保具有略微不同线圈电阻的并联电磁铁产生均匀的磁吸力至关重要。
11. Summary | 总结
The parallel circuit is a cornerstone of circuit analysis. Its governing equations – 1/R_eq = Σ(1/R_i), V same across all branches, I_total = ΣI_i – are derived directly from charge and energy conservation. Mastering these principles enables you to analyse complex networks and design reliable electromagnetic systems where multiple loads share a common voltage source.
并联电路是电路分析的基石。其支配方程——1/R_eq = Σ(1/R_i)、所有支路电压相等、I_total = ΣI_i——直接来自电荷和能量守恒。掌握这些原理使你能够分析复杂网络，并设计多个负载共用同一电压源的可靠电磁系统。
Remember to practise switching between resistance and conductance views, and always check that your calculated equivalent resistance is smaller than the smallest branch resistance. With this solid foundation, you will confidently tackle any parallel circuit problem, from exam questions to real-life electromagnetic applications.
记得在电阻观点和电导观点之间灵活切换，并始终检查计算出的等效电阻是否小于最小的支路电阻。有了这个坚实的基础，你将自信地应对任何并联电路问题，无论是试题还是现实生活中的电磁应用。
Published by TutorHao | Physics Revision Series | aleveler.com
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