📚 Potentiometer Circuits: Principle and Worked Examples | 电位器电路：原理与实例分析
A potentiometer is a three-terminal resistor with a sliding or rotating contact that forms an adjustable voltage divider. In A-Level Physics, it is one of the most examinable applications of Ohm’s law and Kirchhoff’s rules, appearing in questions on variable potential dividers, sensor circuits, and the comparison of unknown e.m.f. sources.
电位器是一种带滑动或旋转触头的三端电阻元件，可构成可调电压分配器。在 A-Level 物理中，它是欧姆定律和基尔霍夫定律最常考的应用之一，经常出现在可变压分压电路、传感器电路以及比较未知电动势的题目中。
1. Structure and Symbol | 结构与符号
A potentiometer consists of a resistive track with a fixed total resistance Rtot, and a wiper that can slide along the track. The wiper divides the track into two portions: R1 between the top terminal and the wiper, and R2 between the wiper and the bottom terminal, such that R1 + R2 = Rtot.
电位器由一条总电阻为 Rtot 的电阻轨道和一个可沿轨道滑动的触头组成。触头把轨道分成两部分：顶端与触头之间的 R1，以及触头与底端之间的 R2，满足 R1 + R2 = Rtot。
When used as a potential divider, the total supply voltage Vs is applied across the whole track. The output voltage is taken between the wiper and one fixed end, often the bottom terminal.
当作为分压器使用时，总电源电压 Vs 加在整条轨道两端。输出电压取自触头与某一固定端（通常为底端）之间。
The circuit symbol shows a resistor with an arrow indicating the movable wiper. In CIE exam papers, the potentiometer may be drawn as a rectangle with an arrow, or as a zigzag resistor with a sliding arrow.
电路符号用一个带箭头的电阻表示活动触头。在 CIE 试卷中，电位器可能画成带箭头的矩形，或者带滑动箭头的锯齿形电阻。
2. Principle of Operation | 工作原理
Assuming the track is uniform, the resistance per unit length is constant. Therefore, the resistance of each portion is proportional to its length. If the total length is L and the wiper is at distance x from the bottom end, then:
假设轨道均匀，则单位长度的电阻恒定。因此，每一部分的电阻与其长度成正比。若总长度为 L，触头距底端的距离为 x，则：
R2 / Rtot = x / L
Since the same current flows through both parts (when no current is drawn from the output), the voltage across R2 is proportional to R2. From the potential divider equation:
由于在输出端不取电流时，流过两部分的电流相同，因此 R2 两端的电压正比于 R2。根据分压公式：
Vout = Vs × R2 / (R1 + R2) = Vs × x / L
Thus, the output voltage can be varied continuously from 0 (when the wiper is at the bottom, x = 0) to Vs (when the wiper is at the top, x = L).
因此，输出电压可以从 0（触头在底端，x = 0）连续变化到 Vs（触头在顶端，x = L）。
3. The Potentiometer as a Variable Potential Divider | 电位器作为可变压分压器
In this mode, the potentiometer is connected across a power supply, and the output feeds another circuit. The output voltage is set simply by moving the wiper.
在这种模式下，电位器跨接在电源两端，输出电压供给其他电路。只需移动触头即可设定输出电压。
When a load resistor RL is connected across the output, the situation changes. The load is in parallel with R2, so the effective resistance between the wiper and the bottom terminal becomes:
当输出端并联一个负载电阻 RL 时，情况会发生变化。负载与 R2 并联，因此触头与底端之间的等效电阻变为：
Rp = (R2 × RL) / (R2 + RL)
Since Rp is always less than R2, the output voltage decreases. This is called the loading effect. In the ideal case, the load current should be negligibly small compared with the current through the potentiometer track.
由于 Rp 总是小于 R2，输出电压会下降。这称为负载效应。在理想情况下，负载电流应远小于流过电位器轨道的电流。
4. Worked Example: No Load | 实例一：空载情况
A potentiometer of total resistance 100 Ω is connected across a 6.0 V battery. The wiper is set so that R2 = 30 Ω. Calculate the output voltage when no load is connected.
一个总电阻为 100 Ω 的电位器接在 6.0 V 电池两端。触头位置使 R2 = 30 Ω。计算空载时的输出电压。
Vout = 6.0 × 30 / (70 + 30) = 6.0 × 0.30 = 1.8 V
The output is exactly 1.8 V, as determined by the ratio of resistances.
输出电压恰好为 1.8 V，由电阻比值决定。
5. Worked Example: With Load | 实例二：带负载情况
Now connect a 50 Ω resistor across the output. The wiper position is unchanged, so R1 = 70 Ω and R2 = 30 Ω. Calculate the new output voltage.
现在在输出端并联一个 50 Ω 的电阻。触头位置不变，因此 R1 = 70 Ω，R2 = 30 Ω。计算新的输出电压。
First calculate the parallel combination:
首先计算并联等效电阻：
Rp = (30 × 50) / (30 + 50) = 1500 / 80 = 18.75 Ω
Now apply the potential divider formula:
然后应用分压公式：
Vout = 6.0 × 18.75 / (70 + 18.75) = 6.0 × 18.75 / 88.75 ≈ 1.27 V
The output voltage has fallen from 1.8 V to about 1.27 V. This demonstrates the loading effect clearly: the lower the load resistance, the greater the reduction in output voltage.
输出电压从 1.8 V 下降到约 1.27 V。这清楚地展示了负载效应：负载电阻越小，输出电压下降越多。
6. The Potentiometer as a Measuring Instrument | 电位器作为测量仪器
Historically, a potentiometer was used to measure an unknown e.m.f. by comparing it with a known standard cell. In this method, the unknown cell is connected in series with a sensitive galvanometer, and the wiper is adjusted until the galvanometer reads zero.
传统上，电位器通过将未知电动势与已知标准电池进行比较来测量未知电动势。此方法中，未知电池与灵敏电流计串联，调节触头直到电流计读数为零。
At the balance point, no current flows through the unknown cell. This means the potential difference across R2 exactly equals the unknown e.m.f. Ex. Therefore:
在平衡点，通过未知电池的电流为零。这意味着 R2 两端的电势差恰好等于未知电动势 Ex。因此：
Ex = Vs × x / L
Because no current is drawn from the unknown cell at balance, the measurement is free from the error caused by the cell’s internal resistance — a key advantage over using a voltmeter directly.
由于平衡时未知电池中无电流通过，测量不受电池内阻引起的误差影响——这是相对于直接使用电压表测量的关键优势。
7. Comparison of Two Unknown E.M.F.s | 比较两个未知电动势
The same potentiometer can be used to determine the ratio of two unknown e.m.f.s by finding the balance length for each. If the balance lengths are x1 and x2, then:
同一个电位器可用以确定两个未知电动势之比：分别找出它们各自的平衡长度。若平衡长度为 x1 和 x2，则：
E1 / E2 = x1 / x2
This is because the supply voltage and the total length cancel out in the ratio. If one of the cells is a standard cell of known e.m.f., the other can be determined precisely.
这是因为电源电压和总长度在比值中相互抵消。如果其中一个电池是已知电动势的标准电池，另一个就可以被精确测定。
8. Worked Example: Finding an Unknown E.M.F. | 实例三：求未知电动势
A potentiometer has a total length of 80.0 cm. A standard cell of e.m.f. 1.02 V balances at a length of 51.0 cm. An unknown cell balances at a length of 63.0 cm. Calculate the unknown e.m.f.
一个电位器总长度为 80.0 cm。电动势为 1.02 V 的标准电池在 51.0 cm 处达到平衡。未知电池在 63.0 cm 处平衡。求未知电动势。
Ex / 1.02 = 63.0 / 51.0 ⇒ Ex = 1.02 × 63.0 / 51.0 = 1.26 V
The unknown e.m.f. is 1.26 V. Note that the total length of 80.0 cm is not needed once the standard cell balance is known, because the ratio of lengths directly gives the ratio of e.m.f.s.
未知电动势为 1.26 V。注意：一旦已知标准电池的平衡位置，总长度 80.0 cm 就不再需要，因为长度之比直接给出电动势之比。
9. Internal Resistance and the Potentiometer | 内阻与电位器
If a cell in the potentiometer circuit has appreciable internal resistance r, the terminal potential difference falls when current is drawn. However, in the balance method, no current flows through the cell being measured at the balance point, so its internal resistance does not affect the result.
如果电位器电路中的电池具有不可忽略的内阻 r，则在输出电流时其端电压会下降。然而，在平衡法中，平衡时被测电池中没有电流通过，因此其内阻不影响测量结果。
For the driving cell (supplying the potentiometer track), its internal resistance does not affect the ratio of balance lengths, provided it remains steady during the experiment. The ratio R2/Rtot is independent of the actual value of the driving current.
对于驱动电池（为电位器轨道供电），只要在实验过程中保持稳定，其内阻不会影响平衡长度之比。比值 R2/Rtot 与驱动电流的实际大小无关。
10. Potentiometer in Sensor Circuits | 电位器在传感器电路中的应用
In CIE A-Level questions, the potentiometer often appears as part of a sensing device. For example, a light-dependent resistor (LDR) or a thermistor may replace one section of a potential divider. When the physical quantity changes, the resistance changes, altering the output voltage.
在 CIE A-Level 题目中，电位器常作为传感装置的一部分出现。例如，光敏电阻或热敏电阻可以替代分压器的某一段。当物理量变化时，电阻改变，从而改变输出电压。
Consider a circuit where a thermistor RT is connected in series with a fixed resistor R, across a supply Vs. The output voltage across the thermistor is:
考虑一个电路：热敏电阻 RT 与固定电阻 R 串联，接在电源 Vs 两端。热敏电阻两端的输出电压为：
Vout = Vs × RT / (R + RT)
As temperature rises, RT decreases, so Vout decreases. This voltage change can be fed into an op-amp comparator or an analogue-to-digital converter for further processing.
当温度升高时，RT 减小，因此 Vout 减小。此电压变化可送入运算放大器比较器或模数转换器作进一步处理。
11. Operational Amplifier with Potentiometer Feedback | 电位器反馈的运算放大器
In some higher-level questions, a potentiometer is used to set the gain of an inverting amplifier. The feedback resistance Rf is taken from the wiper of a potentiometer, allowing the gain to be adjusted continuously.
在一些较高层次的问题中，电位器用于设定反相放大器的增益。反馈电阻 Rf 取自电位器的触头，从而可以连续调节增益。
For an inverting amplifier, the voltage gain is:
对于反相放大器，电压增益为：
Gain = −Rf / Rin
By moving the wiper, Rf varies from 0 to the full track resistance, giving a gain range from 0 to −Rtot/Rin. The negative sign indicates a 180° phase inversion.
通过移动触头，Rf 从 0 变化到整个轨道电阻，增益范围从 0 到 −Rtot/Rin。负号表示 180° 的相位反转。
12. Summary of Key Equations | 关键公式总结
The table below summarizes the most important relationships for the potentiometer in A-Level examinations.
下表汇总了 A-Level 考试中电位器最重要的关系式。
| Quantity | 物理量 | Equation | 公式 | Condition | 条件 |
| Output voltage (no load) | 输出电压（空载） | Vout = Vs × R2/(R1+R2) | Iload = 0 |
| Output by length | 按长度求输出 | Vout = Vs × x/L | Uniform track | 均匀轨道 |
| With load RL | 带负载 RL | Rp = R2RL/(R2+RL) | Parallel combination | 并联 |
| Balance condition | 平衡条件 | Ex = Vs × x/L | Ig = 0 |
| Ratio of e.m.f.s | 电动势之比 | E1/E2 = x1/x2 | Same driving cell | 同一驱动电池 |
Mastering these equations and understanding when each condition applies is essential for scoring full marks in potentiometer questions.
掌握这些公式并理解每种条件的适用情境，是在电位器题目中获得满分的关键。
Published by TutorHao | Physics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导