📚 IB Physics: Circuit Analysis and Ohm’s Law | IB物理：电路分析与欧姆定律
Circuit analysis forms the backbone of IB Physics Topic 5 (Electricity and Magnetism). Mastering Ohm’s law, Kirchhoff’s rules, and their applications to series and parallel networks is essential for both Paper 1 and Paper 2 success. This guide walks through the key concepts, common pitfalls, and exam-style strategies to help you secure full marks.
电路分析是IB物理Topic 5（电与磁）的核心内容。熟练掌握欧姆定律、基尔霍夫定律及其在串联和并联电路中的应用，对Paper 1和Paper 2都至关重要。本指南将梳理关键概念、常见误区及应试策略，助你冲击满分。
1. Ohm’s Law and Resistance | 欧姆定律与电阻
Ohm’s law states that the current through a metallic conductor is directly proportional to the potential difference across it, provided the physical conditions (temperature, strain, etc.) remain constant. Mathematically, \( V = IR \), where \( V \) is the potential difference in volts (V), \( I \) is the current in amperes (A), and \( R \) is the resistance in ohms (Ω). This linear relationship only holds for ohmic conductors.
欧姆定律指出：在物理条件（温度、形变等）保持恒定时，通过金属导体的电流与导体两端的电势差成正比。数学表达式为 \( V = IR \)，其中 \( V \) 为电势差（单位：伏特 V），\( I \) 为电流（单位：安培 A），\( R \) 为电阻（单位：欧姆 Ω）。这一线性关系仅适用于欧姆导体。
Resistance is defined as the ratio of potential difference to current: \( R = \frac{V}{I} \). It quantifies how strongly a component opposes the flow of charge. For a uniform wire, resistance depends on length \( L \), cross-sectional area \( A \), and resistivity \( \rho \) of the material: \( R = \rho\frac{L}{A} \).
电阻定义为电势差与电流之比：\( R = \frac{V}{I} \)。它衡量元件对电荷流动的阻碍程度。对于均匀导线，电阻取决于长度 \( L \)、横截面积 \( A \) 和材料的电阻率 \( \rho \)：\( R = \rho\frac{L}{A} \)。
V = IR
R = ρL/A
2. Resistivity and Temperature | 电阻率与温度
Resistivity \( \rho \) is an intrinsic property of a material, independent of its shape or size. For metals, resistivity increases with temperature because lattice vibrations scatter conduction electrons more frequently. This results in a positive temperature coefficient of resistance.
电阻率 \( \rho \) 是材料的固有属性，与形状和尺寸无关。对于金属，电阻率随温度升高而增大，因为晶格振动更频繁地散射传导电子。这导致金属具有正的温度电阻系数。
For semiconductors (e.g., silicon, germanium) and carbon, resistivity decreases as temperature rises. Higher temperatures liberate more charge carriers, reducing resistance. This negative temperature coefficient is exploited in thermistors, which are often used in temperature-sensing circuits.
对于半导体（如硅、锗）和碳，电阻率随温度升高而减小。温度升高释放更多载流子，从而降低电阻。负温度系数被应用于热敏电阻，常用于温度传感电路。
In IB exams, you may be asked to sketch or interpret \( I-V \) graphs for different components. A straight line through the origin indicates an ohmic conductor. A curve that flattens (for a filament lamp) or steepens (for a thermistor) indicates non-ohmic behaviour.
IB考试中，你可能会被要求绘制或解读不同元件的 \( I-V \) 图像。过原点的直线表示欧姆导体；曲线趋于平缓（白炽灯）或变陡（热敏电阻）则表示非欧姆行为。
3. Series Circuits | 串联电路
In a series circuit, components are connected end-to-end, forming a single path for current. The current is identical at every point in the circuit: \( I = I_1 = I_2 = \cdots \). The total potential difference across the battery is equal to the sum of potential differences across each component: \( V_{\text{total}} = V_1 + V_2 + \cdots \).
在串联电路中，元件首尾相连，形成单一电流路径。电路中各点电流相同：\( I = I_1 = I_2 = \cdots \)。电池两端的总电势差等于各元件两端电势差之和：\( V_{\text{total}} = V_1 + V_2 + \cdots \)。
The equivalent resistance of resistors in series is simply the sum: \( R_{\text{eq}} = R_1 + R_2 + \cdots \). This is because each resistor adds its own opposition to the flow of current. For \( n \) identical resistors of value \( R \), the equivalent resistance is \( nR \).
串联电阻的等效电阻等于各电阻之和：\( R_{\text{eq}} = R_1 + R_2 + \cdots \)。因为每个电阻都增加了对电流的阻碍。对于 \( n \) 个阻值均为 \( R \) 的相同电阻，等效电阻为 \( nR \)。
V_total = V₁ + V₂ + …
R_eq = R₁ + R₂ + …
A common exam question involves a voltage divider in series: a potential divider consisting of two resistors connected across a battery. The output voltage \( V_{\text{out}} \) across one resistor \( R_2 \) is given by \( V_{\text{out}} = \frac{R_2}{R_1 + R_2}V_{\text{in}} \).
常见考题涉及串联分压：由两个电阻构成的电位分压器跨接在电池两端。电阻 \( R_2 \) 上的输出电压 \( V_{\text{out}} \) 为 \( V_{\text{out}} = \frac{R_2}{R_1 + R_2}V_{\text{in}} \)。
4. Parallel Circuits | 并联电路
In a parallel circuit, components are connected across the same two nodes, providing multiple paths for current. The potential difference across each branch is equal to the applied voltage: \( V = V_1 = V_2 = \cdots \). The total current from the source splits among branches: \( I_{\text{total}} = I_1 + I_2 + \cdots \).
在并联电路中，元件连接在同一对节点之间，为电流提供多条路径。每条支路两端的电势差都等于外加电压：\( V = V_1 = V_2 = \cdots \)。电源提供的总电流在各支路中分流：\( I_{\text{total}} = I_1 + I_2 + \cdots \)。
The equivalent resistance of parallel resistors is found by:
并联电阻的等效电阻由下式计算：
1/R_eq = 1/R₁ + 1/R₂ + …
For two resistors in parallel, this simplifies to \( R_{\text{eq}} = \frac{R_1R_2}{R_1 + R_2} \). The equivalent resistance is always less than the smallest individual resistance. Adding more parallel branches always reduces the total resistance.
对于两个并联电阻，简化为 \( R_{\text{eq}} = \frac{R_1R_2}{R_1 + R_2} \)。等效电阻总是小于其中最小的电阻值。增加并联支路会降低总电阻。
A common question: two resistors 6 Ω and 3 Ω are connected in parallel. The equivalent resistance is \( \frac{6 \times 3}{6 + 3} = 2\ \Omega \). If a 12 V battery drives this combination, the total current is \( I = \frac{12}{2} = 6\ \text{A} \), with 4 A through the 3 Ω resistor and 2 A through the 6 Ω resistor.
典型问题：两个电阻6 Ω和3 Ω并联，等效电阻为 \( \frac{6 \times 3}{6 + 3} = 2\ \Omega \)。若由12 V电池供电，总电流为 \( I = \frac{12}{2} = 6\ \text{A} \)，其中通过3 Ω电阻的电流为4 A，通过6 Ω电阻的电流为2 A。
5. Kirchhoff’s Laws | 基尔霍夫定律
Kirchhoff’s Current Law (KCL) states that the total current entering a junction equals the total current leaving it. This reflects conservation of charge. For example, if 5 A enters a junction and splits into 3 A and 2 A, KCL is satisfied.
基尔霍夫电流定律（KCL）指出：流入节点的总电流等于流出节点的总电流。这体现了电荷守恒。例如，5 A流入一个节点并分成3 A和2 A，KCL成立。
Kirchhoff’s Voltage Law (KVL) states that the sum of the electromotive forces (emf) and potential differences around any closed loop is zero. When traversing a loop, a rise in potential is positive, and a drop is negative. This reflects conservation of energy.
基尔霍夫电压定律（KVL）指出：沿任意闭合回路，电动势与电势差之和为零。沿回路行进时，电势升高为正，电势降低为负。这体现了能量守恒。
ΣI_enter = ΣI_leave
ΣV_rise = ΣV_drop
When applying KVL, first assign a direction to each loop (clockwise or counterclockwise). For each resistor, if the loop direction matches the assumed current direction, the voltage drop is \( IR \); otherwise it is \( -IR \). For a battery, if the loop goes from negative to positive terminal, the emf is positive; if from positive to negative, it is negative.
应用KVL时，先为每个回路指定方向（顺时针或逆时针）。对于每个电阻，若回路方向与假设电流方向一致，则电压降为 \( IR \)；否则为 \( -IR \)。对于电池，若回路从负极端到正极端，电动势为正；若从正极端到负极端，则为负。
A typical IB problem involves two loops with one or two batteries. Set up simultaneous equations and solve for unknown currents. Always check that the final currents are consistent with KCL at junctions.
典型IB题目包含两个回路、一个或两个电池。列出联立方程并求解未知电流，最后务必用KCL检查节点处电流是否一致。
6. Potential Dividers | 电位分压器
A potential divider is a circuit that produces a fraction of the input voltage. It consists of two or more resistors in series. The output voltage across one resistor is proportional to its resistance relative to the total resistance.
电位分压器是一种产生输入电压一部分的电路，由两个或多个串联电阻组成。某个电阻上的输出电压与其电阻占总电阻的比例成正比。
For a divider with resistors \( R_1 \) and \( R_2 \), the output voltage \( V_{\text{out}} \) across \( R_2 \) is:
对于由 \( R_1 \) 和 \( R_2 \) 构成的分压器，\( R_2 \) 上的输出电压 \( V_{\text{out}} \) 为：
V_out = V_in × R₂/(R₁+R₂)
Potential dividers are widely used in sensors. A light-dependent resistor (LDR) in a divider changes its resistance with light intensity, producing a variable output voltage that can trigger a transistor or comparator. Similarly, a thermistor in a divider is used in temperature control systems.
电位分压器广泛应用于传感器。光敏电阻（LDR）在分压器中的阻值随光照强度变化，产生可变的输出电压，从而触发晶体管或比较器。类似地，热敏电阻在分压器中用于温度控制系统。
Exam tip: When a load is connected across \( R_2 \), the effective resistance of that branch drops, and the output voltage changes. You must recalculate using the parallel combination of \( R_2 \) and the load.
考试提示：当在 \( R_2 \) 两端接入负载时，该支路的等效电阻会下降，输出电压随之改变。你必须用 \( R_2 \) 与负载的并联组合重新计算。
7. Internal Resistance and EMF | 内电阻与电动势
A real battery is modelled as an ideal emf source \( \mathcal{E} \) in series with an internal resistance \( r \). When current \( I \) flows through the external circuit (load resistance \( R \)), the terminal voltage \( V_{\text{terminal}} \) is the emf minus the voltage drop across the internal resistance:
真实电池可建模为理想电动势源 \( \mathcal{E} \) 与内电阻 \( r \) 的串联。当电流 \( I \) 流过外部电路（负载电阻 \( R \)）时，端电压 \( V_{\text{terminal}} \) 等于电动势减去内电阻上的电压降：
V_terminal = ℰ − Ir
When no current flows (open circuit), the terminal voltage equals the emf. As current increases, the internal resistance causes the terminal voltage to decrease. If a battery is short-circuited (\( R = 0 \)), the maximum current is \( I_{\text{short}} = \mathcal{E}/r \).
无电流时（开路），端电压等于电动势。随着电流增大，内电阻导致端电压下降。若电池短路（\( R = 0 \)），最大电流为 \( I_{\text{short}} = \mathcal{E}/r \)。
In the lab, plotting \( V_{\text{terminal}} \) against \( I \) yields a straight line with intercept \( \mathcal{E} \) on the voltage axis and slope \( -r \). This is a standard IB Data-based question: students read the emf from the y-intercept and the internal resistance from the (negative) gradient.
在实验中，绘制 \( V_{\text{terminal}} \) 与 \( I \) 的图像，得到一条直线，电压轴截距为 \( \mathcal{E} \)，斜率为 \( -r \)。这是IB数据型题目的标准考点：从y截距读电动势，从（负）斜率读内电阻。
Energy consideration: the total power supplied by the emf is \( P_{\text{total}} = \mathcal{E}I \). The power dissipated in the internal resistance is \( P_{\text{internal}} = I^2r \), and the useful power delivered to the load is \( P_{\text{load}} = I^2R \). Maximum power transfer occurs when \( R = r \).
能量角度：电动势提供的总功率为 \( P_{\text{total}} = \mathcal{E}I \)。内电阻消耗的功率为 \( P_{\text{internal}} = I^2r \)，传递给负载的有用功率为 \( P_{\text{load}} = I^2R \)。最大功率传输发生在 \( R = r \) 时。
8. Power and Energy in Circuits | 电路中的功率与能量
The power dissipated by a component is the rate at which electrical energy is converted to other forms (heat, light, sound, etc.). Three equivalent expressions are:
元件消耗的功率是电能转化为其他形式能量（热、光、声等）的速率。三个等价表达式为：
P = VI = I²R = V²/R
Use the form that is most convenient for the given known quantities. For example, if current and resistance are given, use \( P = I^2R \). If voltage and resistance are given, use \( P = V^2/R \).
根据已知量选择最方便的表达式。例如，已知电流和电阻时用 \( P = I^2R \)；已知电压和电阻时用 \( P = V^2/R \)。
Electrical energy \( E \) is given by \( E = Pt = VIt = I^2Rt \). The unit is the joule (J), but the kilowatt-hour (kWh) is commonly used for household energy consumption: 1 kWh = 3.6 × 10⁶ J.
电能 \( E \) 由 \( E = Pt = VIt = I^2Rt \) 给出，单位是焦耳（J）。但家庭用电中常用千瓦时（kWh）：1 kWh = 3.6 × 10⁶ J。
A classic IB question: a 12 V battery drives a current of 2 A through a resistor for 5 minutes. The energy dissipated is \( E = VIt = 12 \times 2 \times 300 = 7200\ \text{J} \). If the resistor is 4 Ω, the power is \( P = I^2R = 2^2 \times 4 = 16\ \text{W} \), which also equals \( VI = 12 \times 2 = 24\ \text{W} \) only if the battery is ideal; otherwise the difference is the internal heat loss.
经典IB题：12 V电池在5分钟内驱动2 A电流通过电阻。消耗的能量为 \( E = VIt = 12 \times 2 \times 300 = 7200\ \text{J} \)。若电阻为4 Ω，功率为 \( P = I^2R = 2^2 \times 4 = 16\ \text{W} \)，这与 \( VI = 12 \times 2 = 24\ \text{W} \) 相等仅在理想电池时成立；否则差值即为内阻热损耗。
9. Ammeters and Voltmeters | 安培计与伏特计
An ideal ammeter has zero resistance and is connected in series with the component whose current is being measured. In practice, ammeters have a small but non-zero resistance, which slightly reduces the current in the circuit. IB questions often ask you to explain why ammeters are connected in series and why their resistance should be very low.
理想安培计内阻为零，应串联在被测电流的支路中。实际安培计具有很小但非零的电阻，会略微减小电路电流。IB题目常要求解释安培计为何串联连接以及内阻为何要非常小。
An ideal voltmeter has infinite resistance and is connected in parallel with the component whose voltage is being measured. A real voltmeter draws a small current, which can affect the circuit, especially if the circuit resistance is high. Therefore, voltmeters should have as high a resistance as possible.
理想伏特计内阻为无穷大，应并联在被测电压的元件两端。真实伏特计会分走微小电流，尤其在电路电阻较大时会影响测量结果。因此，伏特计内阻应尽可能大。
Common exam trap: Placing a voltmeter in series or an ammeter in parallel will produce misleading readings. A voltmeter in series breaks the circuit, and an ammeter in parallel creates a short circuit path for current.
常见考试陷阱：伏特计串联或安培计并联会产生误导读数。伏特计串联会断开电路，安培计并联会造成电流短路。
10. Non-Ohmic Components | 非线性元件
Non-ohmic components do not obey Ohm’s law; their resistance changes with the applied voltage or current. The \( I-V \) graph is not a straight line. Key examples include filament lamps, diodes, thermistors, and LDRs.
非线性元件不满足欧姆定律；其电阻随外加电压或电流变化。\( I-V \) 图像不是直线。典型例子包括白炽灯、二极管、热敏电阻和光敏电阻。
A filament lamp’s resistance increases as the filament heats up. At low voltages, the graph is nearly linear; at higher voltages, the curve flattens because resistance grows. The dynamic resistance is found from the gradient of the tangent at a given point.
白炽灯的电阻随灯丝温度升高而增大。低电压时图像近似线性；高电压时曲线变得平缓，因为电阻增大。动态电阻由某点切线的斜率给出。
A diode conducts current in one direction only (forward bias) and blocks current in the reverse direction (reverse bias) until breakdown. Its \( I-V \) graph shows a sharp rise in current after the forward threshold voltage (about 0.7 V for silicon). In IB questions, you may be asked to interpret such graphs and explain the shape.
二极管只允许单向导通（正向偏置），反向截止（反向偏置）直至击穿。其 \( I-V \) 图像在正向阈值电压（硅管约0.7 V）之后电流急剧上升。IB题目中可能要求解读此类图像并解释形状。
For a thermistor or LDR, the \( I-V \) curve is symmetrical (no polarity) but non-linear. The resistance at a given voltage can be read from \( V/I \), while the dynamic resistance is the local slope \( dV/dI \).
对于热敏电阻或光敏电阻，\( I-V \) 曲线是对称的（无极性）但非线性。某电压下的电阻可由 \( V/I \) 读出，而动态电阻是局部斜率 \( dV/dI \)。
11. Analysing Complex Circuits | 复杂电路分析
When faced with a complex circuit, follow a systematic approach: (1) Simplify any obvious series or parallel combinations; (2) Redraw the circuit to make the structure clearer; (3) Apply Ohm’s law to find branch currents; (4) Use KVL and KCL for remaining unknowns.
面对复杂电路时，采用系统化步骤：(1) 简化明显的串联或并联组合；(2) 重绘电路使结构更清晰；(3) 用欧姆定律求支路电流；(4) 对剩余未知量用KVL和KCL求解。
Wheatstone bridge circuits are common in IB extended-response questions. The bridge balances when the potential difference across the central galvanometer (or voltmeter) is zero, which occurs when \( \frac{R_1}{R_2} = \frac{R_3}{R_4} \). At balance, no current flows through the bridge, so the branches can be treated as independent series pairs.
惠斯通电桥常见于IB扩展答题中。当中央检流计（或伏特计）两端电势差为零时，电桥平衡，此时 \( \frac{R_1}{R_2} = \frac{R_3}{R_4} \)。平衡时桥路无电流，各支路可视为独立的串联对。
Another recurring task: determining the current through a specific resistor when multiple batteries are present. Draw labelled currents for each branch, then write KVL equations for two independent loops. Solve the linear system algebraically. If a computed current is negative, it simply means the actual direction is opposite to the assumed one.
另一类常见任务：当存在多个电池时求某个电阻中的电流。为各支路标出电流方向，然后对两个独立回路写出KVL方程，联立求解线性方程组。若计算的电流为负值，仅表示实际方向与假设方向相反。
12. Exam Strategy and Common Errors | 应试策略与常见错误
Error 1: Forgetting unit conversions. Always convert mA to A, kJ to J, and Ω to kΩ as needed. A current of 200 mA must be written as 0.2 A in \( V = IR \).
错误1：忘记单位换算。 始终将mA换算为A，kJ换算为J，Ω与kΩ按需转换。200 mA必须写成0.2 A才能代入 \( V = IR \)。
Error 2: Confusing series and parallel formulas. Resistances add in series but reciprocals add in parallel. A common mistake is using \( R_{\text{eq}} = \frac{1}{R_1} + \frac{1}{R_2} \) directly without taking the reciprocal of the sum.
错误2：混淆串联与并联公式。 串联电阻相加，并联电阻倒数相加。常见错误是直接写 \( R_{\text{eq}} = \frac{1}{R_1} + \frac{1}{R_2} \) 而忘记对总和取倒数。
Error 3: Wrong sign convention in KVL. When moving through a battery, always check the direction from negative to positive (gain) or positive to negative (loss). Inconsistent sign conventions are the leading cause of Kirchhoff errors.
错误3：KVL符号约定错误。 通过电池时，务必判断是从负极到正极（电势升）还是从正极到负极（电势降）。符号约定不一致是基尔霍夫解题错误的主要原因。
Error 4: Ignoring internal resistance. In any real battery question, remember the terminal voltage is \( \mathcal{E} – Ir \), not simply \( \mathcal{E} \). The total resistance in the circuit includes the internal resistance \( r \).
错误4：忽略内电阻。 在真实电池问题中，端电压是 \( \mathcal{E} – Ir \)，而不只是 \( \mathcal{E} \)。电路总电阻包含内阻 \( r \)。
For Paper 2 calculations, show every step with units, and state the direction of currents explicitly. For data-based questions, always read the intercept and gradient from the graph carefully, including the units on both axes.
Paper 2计算题中，每一步都要写单位，并明确指出电流方向。数据型题目中，仔细从图像读取截距和斜率，注意两个坐标轴的单位。
Strategy: In multiple-choice questions, estimate before calculating. If the answer is about 5 Ω, you can quickly eliminate options that are orders of magnitude different. This saves time and catches arithmetic slips.
策略： 选择题中，先估算再计算。如果答案约为5 Ω，可迅速排除数量级偏差过大的选项。这既节省时间又能发现运算失误。
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