📚 AS AQA Physics Unit 4 Insert June 2019: Electrical Circuit Analysis | AS AQA 物理 Unit4 2019年6月插入材料：电路分析
The insert provided in the AQA AS Physics Unit 4 paper from June 2019 contained a table of experimental results from an investigation into a fixed resistor. A student measured the current through the resistor for different values of applied voltage. The data table is reproduced below.
2019年6月AQA AS物理Unit4试卷中的插入材料提供了一组实验数据表，内容是关于一个固定电阻的研究。学生测量了在不同外加电压下通过电阻的电流。下面的数据表转载了该内容。
| Voltage / V | 0.0 | 1.0 | 2.0 | 3.0 | 4.0 | 5.0 |
| Current / mA | 0.0 | 4.2 | 8.0 | 12.5 | 16.1 | 20.0 |
1. Background and Data | 背景与数据
The insert provides a set of voltage and current readings for a single fixed resistor. The resistor is connected to a variable power supply, and the current is measured with an ammeter. The voltage is measured with a voltmeter connected in parallel.
插入材料提供了一组单个固定电阻的电压和电流读数。电阻连接到可调电源，电流用电流表测量，电压用并联的电压表测量。
Notice that the current values are given in milliamperes (mA). To use them in calculations with the ohm and the volt, you must convert them to amperes (A) by dividing by 1000.
注意电流值以毫安（mA）为单位。要利用欧姆和伏特进行计算，必须将它们除以1000转换为安培（A）。
The data can be plotted on a graph of current against voltage. For an ohmic conductor, the graph is a straight line through the origin. The gradient of this line gives 1/R, where R is the resistance.
数据可以绘制成电流-电压图像。对于欧姆导体，该图像是经过原点的一条直线。直线的斜率给出1/R，其中R是电阻。
2. Ohm’s Law and Resistance | 欧姆定律与电阻
Ohm’s law states that the current through a conductor is directly proportional to the potential difference across it, provided the physical conditions (such as temperature) remain constant.
欧姆定律指出：在物理条件（如温度）保持恒定的情况下，通过导体的电流与导体两端的电势差成正比。
Mathematically, this is written as:
V = I × R
where V is the potential difference in volts, I is the current in amperes, and R is the resistance in ohms (Ω).
其中V是电势差（单位：伏特），I是电流（单位：安培），R是电阻（单位：欧姆，Ω）。
Using the first non-zero data point (V = 1.0 V, I = 4.2 mA = 0.0042 A), we can calculate the resistance:
利用第一个非零数据点（V = 1.0 V，I = 4.2 mA = 0.0042 A），我们可以计算电阻：
R = V / I = 1.0 / 0.0042 ≈ 238 Ω
This value is close to the resistance calculated from other points. Slight variations arise from experimental uncertainties and the resolution of the meters.
该值接近从其他点计算的电阻。轻微差异源于实验不确定性和仪表的精度。
3. Series and Parallel Circuits | 串联与并联电路
In the insert, the resistor is part of a simple circuit. However, questions often extend to circuits with two or more resistors. In a series circuit, the total resistance is the sum of the individual resistances:
在插入材料中，电阻是一个简单电路的一部分。然而，问题常常扩展到具有两个或更多电阻的电路。在串联电路中，总电阻等于各电阻之和：
R_total = R₁ + R₂ + …
In a parallel circuit, the total resistance is found from:
在并联电路中，总电阻由下式计算：
1 / R_total = 1 / R₁ + 1 / R₂ + …
For two resistors in parallel with equal values, the total resistance is half the value of one resistor. For unequal values, use the product-over-sum rule for two resistors:
对于两个阻值相等的电阻并联，总电阻是单个电阻的一半。对于阻值不同的两个电阻，可使用两电阻之积除以之和的公式：
R_total = R₁ × R₂ / (R₁ + R₂)
4. Potential Difference and Voltage Division | 电势差与分压
When two resistors are connected in series with a battery, the total potential difference of the battery is shared between the resistors. The voltage across each resistor is proportional to its resistance.
当两个电阻与电池串联时，电池的总电势差在电阻之间分配。每个电阻两端的电压与其电阻成正比。
This is known as the potential divider rule. For resistors R₁ and R₂ in series, the voltage across R₁ is:
这被称为分压法则。对于串联的R₁和R₂，R₁两端的电压为：
V₁ = V_total × R₁ / (R₁ + R₂)
This relation is extremely useful in sensor circuits and in the analysis of complex circuits.
这个关系在传感器电路和复杂电路分析中非常有用。
5. Resistivity | 电阻率
The resistance of a wire depends on its length, cross-sectional area and the material from which it is made. This relationship is expressed by the resistivity equation:
导线的电阻取决于其长度、横截面积和制造材料。这种关系由电阻率方程表示：
R = ρ × L / A
where ρ (rho) is the resistivity in ohm-metres (Ω m), L is the length in metres, and A is the cross-sectional area in square metres.
其中ρ（rho）是电阻率（单位：欧姆·米，Ω·m），L是以米为单位的长度，A是以平方米为单位的横截面积。
For a wire of radius r, the area is given by A = π r². If the radius is doubled, the area increases by a factor of four, and the resistance decreases by a factor of four.
对于半径为r的导线，面积由A = π r²给出。如果半径加倍，面积增大四倍，电阻减小到原来的四分之一。
6. Power and Energy | 功率与能量
When a current flows through a resistor, electrical energy is converted into thermal energy. The power dissipated by a resistor is given by:
当电流流过电阻时，电能转化为热能。电阻耗散的功率由下式给出：
P = V × I
Using Ohm’s law, this can also be written as:
利用欧姆定律，这也可以写成：
P = I² × R or P = V² / R
In the insert data, if the resistor carries a current of 20.0 mA (0.020 A) at 5.0 V, the power is:
在插入数据中，如果电阻在5.0 V电压下通过20.0 mA（0.020 A）电流，则功率为：
P = 5.0 × 0.020 = 0.10 W
This calculation is often asked as a follow-up question in the exam.
这个计算经常作为考试中的后续问题出现。
7. Experimental Error and Uncertainty | 实验误差与不确定性
The readings in the insert show small deviations from a perfect straight line. These deviations are due to random errors in reading the ammeter and voltmeter, and possibly due to temperature changes in the resistor.
插入材料中的读数与完美直线存在微小偏差。这些偏差是由于读取电流表和电压表时的随机误差，以及电阻温度可能发生变化所致。
To reduce random errors, the student could repeat readings and calculate the mean. Systematic errors can be reduced by checking the zero calibration of the meters.
为减少随机误差，学生可以重复读数并计算平均值。系统误差可通过检查仪表零点校准来减少。
The uncertainty in the resistance can be estimated using the percentage uncertainties in voltage and current. For example, if the voltmeter has an uncertainty of ±0.1 V and the ammeter ±0.2 mA, the percentage uncertainties are:
电阻的不确定性可以通过电压和电流的百分比不确定性来估计。例如，如果电压表的不确定性为±0.1 V，电流表为±0.2 mA，则百分比不确定性为：
- Voltage: 0.1 / 5.0 × 100% = 2%
- 电压：0.1 / 5.0 × 100% = 2%
- Current: 0.2 / 20.0 × 100% = 1%
- 电流：0.2 / 20.0 × 100% = 1%
Adding these gives a total percentage uncertainty of about 3% in the resistance.
将它们相加得到电阻的总百分比不确定性约为3%。
8. Graphical Analysis and Gradient | 图像分析与斜率
Plotting the data from the insert on a current-voltage graph allows the resistance to be determined from the gradient. The graph should be a straight line through the origin if the resistor is ohmic.
将插入材料中的数据绘制成电流-电压图像，可以通过斜率确定电阻。如果电阻是欧姆性的，图像应是通过原点的一条直线。
The gradient is calculated as:
斜率计算如下：
gradient = ΔI / ΔV = 1 / R
Using the first and last points:
使用第一个和最后一个点：
ΔI = 0.020 A – 0 A = 0.020 A
ΔV = 5.0 V – 0 V = 5.0 V
gradient = 0.020 / 5.0 = 0.004 Ω⁻¹
Therefore R = 1 / 0.004 = 250 Ω. This is slightly different from the point-by-point calculation because of rounding of the measured currents.
因此R = 1 / 0.004 = 250 Ω。这比逐点计算稍有不同，原因是实测电流的舍入。
9. Common Pitfalls | 常见陷阱
Students often make avoidable errors in circuit questions. Here are the most common ones:
学生在电路问题中经常犯可避免的错误。以下是最常见的错误：
- Forgetting to convert milliamperes to amperes before calculation.
- 忘记在计算前将毫安转换为安培。
- Using the resistance formula incorrectly: R = V/I, not R = I/V.
- 错误使用电阻公式：R = V/I，而不是R = I/V。
- Confusing series and parallel formulas for total resistance.
- 混淆串联和并联总电阻公式。
- Drawing the line of best-fit through the origin when the data clearly suggest it should not.
- 当数据明显表明不经过原点时，仍将最佳拟合线画过原点。
Always check units and estimate expected answers to avoid these errors.
始终检查单位并估计预期答案，以避免这些错误。
10. Practice Questions | 练习题
To master the topic, try answering these questions based on the insert data:
为掌握该主题，请根据插入数据尝试回答以下问题：
- Calculate the resistance at V = 3.0 V.
- 计算V = 3.0 V时的电阻。
- A second identical resistor is placed in parallel with the original resistor. Calculate the total resistance of the combination.
- 将一个相同的电阻与原来的电阻并联，并计算组合的总电阻。
- If the current through the original resistor is 16.1 mA, how much charge passes through it in 2 minutes?
- 如果通过原电阻的电流为16.1 mA，2分钟内有多少电荷通过它？
Answers: 1. 240 Ω (using 3.0 / 0.0125); 2. 120 Ω; 3. Q = I × t = 0.0161 × 120 = 1.93 C.
答案：1. 240 Ω（使用3.0 / 0.0125）；2. 120 Ω；3. Q = I × t = 0.0161 × 120 = 1.93 C。
11. Summary | 总结
The insert data from the June 2019 AQA AS Physics Unit 4 paper provide a classic example of analysing an ohmic conductor. By converting units, applying Ohm’s law, and understanding graphical and power formulas, students can confidently answer circuit-based questions.
2019年6月AQA AS物理Unit4试卷中的插入数据提供了分析欧姆导体的经典示例。通过转换单位、应用欧姆定律并理解图像和功率公式，学生可以自信地回答电路相关问题。
Remember to check units, use the correct formulas, and interpret the graph gradient correctly. These skills will help you achieve full marks in the exam.
记住检查单位、使用正确的公式并正确解释图像斜率。这些技能将帮助你在考试中获得满分。
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