📚 IB Physics: B.5 Current and Circuits SL — Techniques for Solving Application Questions | IB 物理：B.5 电流与电路 SL 应用题解题技巧
Application questions in IB Physics B.5 require more than just recalling formulas — they demand a systematic approach to analysing circuits, interpreting data, and applying concepts to novel situations. This guide introduces key problem-solving techniques for Current and Circuits at SL, helping you tackle exam questions with confidence.
IB 物理 B.5 的应用题不仅需要记住公式，更需要系统分析电路、解读数据并将概念应用于新情境。本指南介绍 SL 电流与电路的关键解题技巧，助你自信应对考试题目。
1. Understanding Circuit Diagrams and Identifying Series/Parallel Connections | 理解电路图并识别串联与并联
Always begin by redrawing the given circuit in a simplified, conventional layout. Use coloured pens to trace current paths from the positive terminal of the battery back to the negative terminal. Two components are in series if they share exactly one common node and the current has no alternative path between them. Two components are in parallel if their two ends are connected to the same two nodes, providing separate branches for the current.
始终先以简化的常规布局重绘给定电路。用彩色笔从电池正极开始勾勒电流路径回到负极。如果两个元件恰好共享一个节点且电流在它们之间没有其他路径，则它们串联。如果两个元件的两端分别连接在相同的两个节点上，形成独立的电流支路，则它们是并联。
When a circuit appears tangled, label every junction with a letter and redraw, placing the battery at the top or left. This technique reveals series and parallel groups immediately and prevents mistakes when calculating equivalent resistance.
当电路看起来混乱时，为每个节点标上字母并重绘，将电池放在顶部或左侧。这种方法能立即揭示串并联组合，避免计算等效电阻时出错。
2. Step-by-Step Calculation of Equivalent Resistance | 逐步计算等效电阻
For series resistors, the equivalent resistance is the sum:
对于串联电阻，等效电阻等于各电阻之和：
Rseries = R₁ + R₂ + R₃ + …
For parallel resistors, use the reciprocal formula. For two resistors the product-over-sum shortcut applies:
对于并联电阻，使用倒数公式。两个电阻时可用积和比公式：
1/Rparallel = 1/R₁ + 1/R₂ or Rparallel = (R₁ × R₂) / (R₁ + R₂)
In mixed circuits, replace a clearly parallel or series sub-group with its equivalent resistance, redrawing the circuit step by step until only a single equivalent resistance remains. Always work from the innermost group outwards. Double-check that the new simplified diagram still respects the original connections at the terminals.
在混联电路中，用一个明确的并联或串联子组的等效电阻替换该部分，逐步重绘电路，直到只剩一个等效电阻。始终从最内层的组合向外计算。务必再次确认简化后的电路图仍保留了原接线端子的连接关系。
3. Using the Voltage Divider Rule for Series Resistors | 应用串联电阻分压法则
For a series combination connected across a voltage source Vtotal, the voltage across one resistor R₁ is:
对于跨接在电压源Vtotal上的串联组合，电阻R₁两端的电压为：
V₁ = Vtotal × (R₁ / (R₁ + R₂ + …))
This rule is a powerful shortcut: you do not need to calculate the current first. Just form the fraction of the target resistance over the total series resistance and multiply by the source voltage. It is especially useful in potential divider problems and when a sensor such as an LDR or thermistor forms part of a series chain.
这一法则是非常有用的捷径：无需先计算电流。只需用目标电阻占串联总电阻的比值乘以电源电压。它在分压器问题以及光敏电阻或热敏电阻作为串联链一部分时尤其有用。
4. Using the Current Divider Rule for Parallel Resistors | 应用并联电阻分流法则
When a total current Itotal enters a parallel branch of two resistors R₁ and R₂, the current through R₁ is:
当总电流Itotal流入两个电阻R₁和R₂的并联支路时，流过R₁的电流为：
I₁ = Itotal × (R₂ / (R₁ + R₂))
Notice that the opposite resistance appears in the numerator — current prefers the path of lower resistance. For more than two parallel resistors, first compute the equivalent parallel resistance Rp, then use I₁ = Itotal × (Rp / R₁). This avoids manipulating multiple reciprocal terms during timed exams.
注意分子中出现的是另一个电阻——电流偏爱低电阻路径。对于超过两个并联电阻的情况，先计算等效并联电阻Rp，然后使用I₁ = Itotal × (Rp / R₁)。这样可以在限时考试中避免处理多个倒数项。
5. Applying Ohm’s Law in Mixed Circuits: V = IR | 在混联电路中应用欧姆定律：V = IR
In a circuit combining series and parallel sections, the safest strategy is: (1) Reduce the entire network to a single equivalent resistance Req. (2) Use the battery emf (or terminal voltage) to find the total current Itot = V / Req. (3) Expand the circuit stepwise, distributing voltage and current using series and parallel rules at each stage.
对于包含串联和并联部分的电路，最安全的策略是：(1) 将整个网络化简为单个等效电阻Req。(2) 用电池电动势（或端电压）求出总电流Itot = V / Req。(3) 逐步恢复电路，每一步都用串并联法则分配电压和电流。
A frequent exam mistake is applying V = IR to a whole circuit while using the resistance of only one component — always ensure the V, I and R in the formula refer to the same branch or component. Write the subscript consistently: V₁ = I₁ R₁.
考试中常见的错误是将 V = IR 应用于整个电路却只代入一个元件的电阻值——务必确保公式中的 V、I 和 R 对应同一支路或元件。要一致地书写下标：V₁ = I₁ R₁。
6. Power Calculations and Efficiency in Circuits | 电路中的功率计算与效率
Three equivalent power expressions are available — choose the one that matches the given quantities:
有三种等价的功率表达式——选择与已知量匹配的那个：
P = I V | P = I² R | P = V² / R
If a question asks for the energy dissipated over time, simply multiply power by time: E = P t, with energy in joules when power is in watts and time in seconds. In circuits with internal resistance, some power is wasted inside the battery as heat; the useful output power is Pout = I Vterminal. The efficiency is η = (Pout / Ptotal) × 100%. Maximum power transfer to an external load occurs when the load resistance equals the internal resistance, a relation sometimes tested qualitatively.
如果问题要求计算一段时间内耗散的能量，只需将功率乘以时间：E = P t，功率用瓦特、时间用秒时能量单位是焦耳。在有内阻的电路中，部分功率在电池内部以热的形式浪费掉；有用输出功率为Pout = I Vterminal。效率为η = (Pout / Ptotal) × 100%。当负载电阻等于内阻时，传输到外部负载的功率最大，这一关系有时会做定性考查。
7. Dealing with Internal Resistance and Terminal Voltage | 处理内阻与端电压
A real battery is modelled as an ideal emf ε in series with a small internal resistance r. When current I flows, the terminal voltage drops:
真实电池可建模为一个理想电动势ε与一个小内阻r串联。当电流I流过时，端电压下降：
Vterminal = ε – I r
To determine ε and r experimentally, IB problems often provide a graph of terminal voltage versus current. The y-intercept gives ε, and the gradient’s magnitude gives r. Always convert the graph’s equation to the form V = -r I + ε to read these values directly. If a problem asks for the power delivered to a load, use the terminal voltage — not the emf — when the current is known.
为通过实验确定 ε 和 r，IB 题目常给出端电压随电流变化的图像。图像在 y 轴上的截距为 ε，斜率绝对值为 r。始终将图像方程化为V = -r I + ε的形式以便直接读取这些值。如果问题要求计算传递给负载的功率，在已知电流时应使用端电压而非电动势。
8. Constructing and Analyzing Potential Divider Circuits | 构建和分析分压器电路
A potential divider (or voltage divider) typically consists of two resistors in series across a supply. The output voltage Vout is taken across one resistor:
分压器通常由两个电阻串联跨接在电源上构成。输出电压Vout取自其中一个电阻两端：
Vout = Vin × (R₂ / (R₁ + R₂))
This circuit appears frequently with a sensor (LDR, thermistor) as R₁ or R₂. As the sensor’s resistance changes with light or temperature, Vout varies, which can be used to switch a transistor or trigger an alarm. When drawing or interpreting such circuits, always check which resistor the output is measured across and whether the sensor forms the upper or lower part of the divider — this determines whether Vout increases or decreases when the physical quantity changes.
这种电路常以传感器（LDR、热敏电阻）作为 R₁ 或 R₂ 出现。随着光照或温度变化，传感器电阻改变，Vout 随之变化，可用于开关晶体管或触发报警器。绘制或分析此类电路时，务必检查输出电压取自哪个电阻两端，以及传感器是分压器的上半部分还是下半部分——这决定了当物理量变化时 Vout 是增加还是减少。
9. Solving Problems with Ammeters and Voltmeters | 解决涉及电流表和电压表的问题
An ideal ammeter has zero resistance and is placed in series so that the current to be measured passes through it completely. An ideal voltmeter has infinite resistance and is connected in parallel with the component across which the potential difference is to be measured. In practice, meters have finite resistances that can affect the circuit. If a problem gives an ammeter’s internal resistance, treat it as a small series resistor and re-calculate the total current. If a voltmeter’s resistance is given and is not vastly greater than the component’s resistance, the voltmeter draws a non-negligible current, reducing the voltage reading slightly. To minimise this error, always use a voltmeter with a resistance much larger than the component being measured, and place the ammeter in a position where its low resistance does not significantly alter the branch current.
理想电流表内阻为零，串联在电路中，使待测电流完全通过它。理想电压表内阻无限大，并联在待测电位差的两端。实际上，电表有有限的电阻，会影响电路。如果题目给出电流表的内阻，将其视为一个小串联电阻，重新计算总电流。如果给出电压表电阻且它并非远大于元件电阻，电压表会流过不可忽略的电流，导致读数略微偏低。为减少这种误差，应始终使用电阻远大于被测元件的电压表，并将电流表放置在低电阻不会明显改变支路电流的位置。
10. Battery Combinations and EMF Calculations | 电池组合与电动势计算
When identical cells are connected in series (positive to negative), the total emf is the sum of the individual emfs, and the total internal resistance is the sum of their internal resistances. This arrangement increases the terminal voltage and is useful when a higher voltage is needed. When identical cells are connected in parallel (all positive terminals together, all negatives together), the total emf remains the same as that of a single cell, but the total internal resistance decreases — the internal resistances combine in parallel. This provides a larger maximum current capacity and reduces the voltage drop under load. IB problems often ask for the combined emf and internal resistance of mixed arrays: treat each parallel group first, then add the series contributions.
当相同的电池串联（正极接负极）时，总电动势为各电池电动势之和，总内阻为各内阻之和。这种接法能提高端电压，适用于需要更高电压的场合。当相同的电池并联（正极相连、负极相连）时，总电动势与单个电池相同，但总内阻减小——各内阻并联。这能提供更大的最大电流能力并减小负载下的电压跌落。IB 题目常要求计算混合阵列的组合电动势和内阻：先处理每一并联组，再叠加串联部分。
11. Analyzing I‑V Characteristics and Non‑Ohmic Behaviour | 分析 I‑V 特性曲线与非欧姆行为
An ohmic conductor gives a straight-line I‑V graph through the origin; its resistance is constant and equals the inverse slope of the graph. For a non‑ohmic device like a filament lamp, the graph curves because the resistance increases with temperature. When answering questions about such curves, read the potential difference and current at the specific point of interest and apply R = V / I. Do not simply use the gradient of a tangent — for DC circuit operation the ratio V/I gives the resistance at that operating point. For a diode, show that current rises sharply above a threshold voltage in forward bias while remaining near zero in reverse bias. Always label axes clearly and use a ruler for straight sections.
欧姆导体的 I‑V 图是一条通过原点的直线；其电阻恒定并等于图像斜率倒数。对于白炽灯等非欧姆器件，曲线弯曲是因为电阻随温度升高而增加。在回答有关此类曲线的问题时，读取关心点的电压和电流并使用R = V / I。不要直接使用切线斜率——在直流电路运行中，V/I 的比值给出该工作点的电阻。对于二极管，需展示正向偏置下电压超过阈值后电流急剧上升，而反向偏置下电流接近零。务必清晰地标记坐标轴，直线部分使用直尺绘制。
12. Common Pitfalls and Practical Tips for Exam Success | 常见陷阱与考试实用建议
Always convert units to the standard SI base: milliamperes to amperes, kilo-ohms to ohms, minutes to seconds. Forgetting this step leads to numerical answers that are off by orders of magnitude. When writing down the final answer, give it to the appropriate number of significant figures consistent with the data. In multi‑step problems, store intermediate results in your calculator rather than rounding prematurely. Read the question carefully to see whether it asks for power dissipated in a component or power delivered by the battery — they are not the same when internal resistance is present. Finally, practise drawing clear, well‑labelled circuits: IB examiners expect components to be represented by standard symbols and connections to be unambiguous. Drawing a messy circuit is often the first step towards a wrong solution.
始终将单位换算为标准国际单位：毫安培换算为安培，千欧姆换算为欧姆，分钟换算为秒。忘记这一步会导致数值答案出现数量级错误。写出最终答案时，根据数据保留适当的有效数字。在多步问题中，将中间结果储存在计算器中，避免过早四舍五入。仔细读题，看清是求元件耗散的功率还是电池输出的功率——有内阻存在时两者并不相同。最后，练习绘制清晰、标注完整的电路图：IB 考官期望使用标准符号表示元件，连接关系明确。画出杂乱电路往往是导致解题错误的第一步。
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