Capacitors in Series | 电容器的串联

📚 Capacitors in Series | 电容器的串联

When capacitors are connected end-to-end so that the same path carries charge from one device to the next, they are said to be in series. This arrangement is common in A-Level circuit analysis because it produces a smaller equivalent capacitance and divides the supply voltage across the individual capacitors.

当电容器首尾相连，使同一条路径从一个电容器流向下一个电容器时，就称为串联。这种连接在 A-Level 电路分析中很常见，因为它会产生更小的等效电容，并把电源电压分配到各个电容器上。

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1. What It Means to Connect Capacitors in Series | 电容器串联的含义

In a series connection, the negative plate of the first capacitor is connected to the positive plate of the next capacitor, forming a single conducting chain. There is only one path for charge to flow, so the same current passes through each capacitor during charging or discharging.

在串联连接中，第一个电容器的负极板连接到下一个电容器的正极板，形成一条单一的导电链。电荷只有一条路径可走，因此在充电或放电过程中，通过每个电容器的电流相同。

Because charge cannot leave the internal wire between two series capacitors, any charge displaced from one plate must appear on the opposite plate of the next capacitor. This leads to the important rule that series capacitors store equal charge.

由于电荷不能离开两个串联电容器之间的内部导线，从一个极板移出的电荷必然出现在下一个电容器的相对极板上。这就得出串联电容器储存相等电荷的重要规则。

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2. Equal Charge on Series Capacitors | 串联电容器上的电荷相等

For capacitors in series, the magnitude of charge on every capacitor is the same: Q = Q₁ = Q₂ = Q₃. This is because the inner plates are electrically isolated and can only gain charge by induction from the adjacent plates.

对于串联电容器，每个电容器上的电荷量都相同：Q = Q₁ = Q₂ = Q₃。这是因为内部极板是电隔离的，只能通过相邻极板的感应获得电荷。

If one capacitor stored a different amount of charge, current would have to flow through the open circuit between the plates, which cannot happen. Therefore charge equality is a direct consequence of conservation of charge.

如果某个电容器储存了不同的电荷量，电流就必须在极板之间的开路中流动，而这是不可能发生的。因此电荷相等是电荷守恒的直接结果。

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3. Voltage Sharing in a Series Circuit | 串联电路中的电压分配

The total potential difference across a series combination is the sum of the p.d.s across each capacitor: V = V₁ + V₂ + V₃. Since V = Q/C for each capacitor, the supply voltage is shared in inverse proportion to capacitance.

串联组合两端的总电势差等于每个电容器两端电势差之和：V = V₁ + V₂ + V₃。由于每个电容器都有 V = Q/C，因此电源电压按电容的反比进行分配。

A smaller capacitor has a larger voltage across it for the same charge, while a larger capacitor has a smaller voltage. This is often tested in CIE questions that ask you to compare voltmeter readings across series capacitors.

对于相同的电荷，较小的电容器两端电压较大，而较大的电容器两端电压较小。CIE 试题经常要求比较串联电容器上电压表的读数，考查这一点。

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4. Deriving the Equivalent Capacitance Formula | 等效电容公式推导

For the series combination, the total voltage is V = V₁ + V₂ + V₃. Substituting V = Q/C for each capacitor gives Q/Cₑ = Q/C₁ + Q/C₂ + Q/C₃, where Cₑ is the equivalent capacitance.

对于串联组合，总电压为 V = V₁ + V₂ + V₃。将每个电容器的 V = Q/C 代入，得到 Q/Cₑ = Q/C₁ + Q/C₂ + Q/C₃，其中 Cₑ 是等效电容。

Dividing through by Q gives the series formula. Notice that the equivalent capacitance is always smaller than the smallest individual capacitance in the chain.

两边同时除以 Q 得到串联公式。注意，等效电容总是小于串联链中最小的单个电容。

1/Cₑ = 1/C₁ + 1/C₂ + 1/C₃ + …

This reciprocal relationship is the opposite of the parallel rule. It also resembles the series resistor rule, which may help you remember it.

这种倒数关系与并联规则相反。它与串联电阻规则相似，这可能有助于记忆。

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5. Special Case: Two Capacitors in Series | 特例：两个电容器串联

For two capacitors, the reciprocal formula can be rearranged into a product-over-sum form: Cₑ = C₁C₂ / (C₁ + C₂). This is often faster to use in numerical problems.

对于两个电容器，倒数公式可以整理为“乘积除以和”的形式：Cₑ = C₁C₂ / (C₁ + C₂)。在数值计算题中，这个公式通常更快。

Cₑ = C₁C₂ / (C₁ + C₂)

For example, 6 μF and 3 μF in series give Cₑ = (6 × 3)/(6 + 3) = 18/9 = 2 μF, which is less than both 6 μF and 3 μF.

例如，6 μF 和 3 μF 串联，得到 Cₑ = (6 × 3)/(6 + 3) = 18/9 = 2 μF，既小于 6 μF 也小于 3 μF。

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6. Energy Stored in Series Capacitors | 串联电容器储存的能量

The energy stored by a capacitor can be written as E = ½QV = ½CV² = ½Q²/C. For a series combination, the total energy is the sum of the energies stored in the individual capacitors.

电容器储存的能量可以写为 E = ½QV = ½CV² = ½Q²/C。对于串联组合，总能量是各个电容器储存的能量之和。

Because each capacitor in series has the same charge Q, the energy stored in capacitor i is Eᵢ = ½Q²/Cᵢ. Smaller capacitors store more energy for the same charge.

由于串联中每个电容器的电荷 Q 相同，第 i 个电容器储存的能量为 Eᵢ = ½Q²/Cᵢ。对于相同的电荷，较小的电容器储存更多能量。

You can also find the total energy by using the equivalent capacitance

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