Unit B5 · The Particulate Nature of MatterUnit B5 · 物质的粒子本质
Current and Circuits电流与电路
IB-Style Practice QuestionsIB 风格练习题
MEDIUMHARDPaper 1Paper 1BPaper 2HL ONLY
Syllabus B5.1 to B5.6考纲 B5.1 至 B5.6PHYSICS HL
Name:姓名：Date:日期：
PART I · PAPER 1 STYLE第一部分 · 第一卷风格Short structured · calculator · 30 marks短结构题 · 可用计算器 · 30 分
Short Structured Items短结构题
Show all working in the space below each question. Marks are awarded for correct method as well as final answers. Convert times to seconds before using $I = \Delta q/\Delta t$. Give numerical answers to an appropriate number of significant figures.在每题下方空白处写出全部解题过程。方法分（method marks）与最终答案同等重要。使用 $I = \Delta q/\Delta t$ 前先把时间换算成秒。数值答案保留适当的有效数字。
A copper wire of cross-sectional area $1.0 \times 10^{-6}\ \mathrm{m^{2}}$ carries a steady current. Copper has $n = 8.5 \times 10^{28}$ free electrons per $\mathrm{m^{3}}$.横截面积 $1.0 \times 10^{-6}\ \mathrm{m^{2}}$ 的铜导线中通过稳恒电流。铜的自由电子数密度 $n = 8.5 \times 10^{28}\ \mathrm{m^{-3}}$。
(a)The current is $0.50\ \mathrm{A}$ for $3.0\ \mathrm{min}$. Calculate the charge that passes a cross-section and the number of electrons this represents.电流为 $0.50\ \mathrm{A}$，持续 $3.0\ \mathrm{min}$。计算通过某横截面的电荷量及对应的电子数目。[2]
(b)When the current is $8.5\ \mathrm{A}$, calculate the electron drift speed in the wire.当电流为 $8.5\ \mathrm{A}$ 时，计算导线中的电子漂移速率。[2]
Q2MEDIUMPaper 1power, energy and emf vs pd功率、能量与电动势-电势差[6 marks]
A resistor of resistance $8.0\ \Omega$ carries a current of $1.5\ \mathrm{A}$.阻值 $8.0\ \Omega$ 的电阻中通过 $1.5\ \mathrm{A}$ 电流。
(a)Calculate the power dissipated in the resistor using a form of the power equation, and state the energy converted in $1.0\ \mathrm{minute}$.用功率方程的某一形式计算电阻的耗散功率，并写出 $1.0\ \mathrm{min}$ 内转化的能量。[2]
(b)A $2.0\ \mathrm{kW}$ heater runs for $4.0\ \mathrm{hours}$. Calculate the energy it transfers, in kilowatt-hours and in joules.一台 $2.0\ \mathrm{kW}$ 的电暖器运行 $4.0\ \mathrm{h}$。计算它转移的能量，分别以千瓦时和焦耳计。[2]
(c)Distinguish between the electromotive force (emf) of a source and the potential difference across a resistor, referring to the direction of energy transfer.结合能量传递方向，区分电源的电动势（emf）与电阻两端的电势差。[2]
A nichrome wire has length $1.5\ \mathrm{m}$, diameter $0.40\ \mathrm{mm}$, and resistivity $\rho = 1.1 \times 10^{-6}\ \mathrm{\Omega\,m}$.一根镍铬合金导线长 $1.5\ \mathrm{m}$，直径 $0.40\ \mathrm{mm}$，电阻率 $\rho = 1.1 \times 10^{-6}\ \mathrm{\Omega\,m}$。
(a)Calculate the resistance of the wire.计算该导线的电阻。[3]
(b)The wire is now stretched uniformly to twice its original length, its volume remaining constant. Determine the new resistance, justifying how both the length and the cross-sectional area change.现把导线均匀拉伸到原长的两倍，体积保持不变。求新的电阻，并说明长度与横截面积各自如何变化。[3]
A $2.0\ \Omega$ resistor is connected in series with a parallel combination of a $6.0\ \Omega$ resistor and a $3.0\ \Omega$ resistor. The network is driven by an ideal $12\ \mathrm{V}$ supply.一个 $2.0\ \Omega$ 电阻与 $6.0\ \Omega$、$3.0\ \Omega$ 两电阻的并联组合串联。网络由理想 $12\ \mathrm{V}$ 电源驱动。
(a)Determine the equivalent resistance of the whole network.求整个网络的等效电阻。[3]
(b)Calculate the total current drawn from the supply.计算电源输出的总电流。[1]
(c)Calculate the current in the $6.0\ \Omega$ resistor and in the $3.0\ \Omega$ resistor.计算 $6.0\ \Omega$ 与 $3.0\ \Omega$ 电阻中的电流。[2]
Q5HARDPaper 1HL ONLYKirchhoff's laws, two cells基尔霍夫定律，双电源[8 marks]
Two ideal cells drive current through a shared resistor. The left branch contains a cell of emf $\varepsilon_{1} = 6.0\ \mathrm{V}$ in series with $R_{1} = 2.0\ \Omega$; the middle branch contains a cell of emf $\varepsilon_{2} = 4.0\ \mathrm{V}$ in series with $R_{2} = 1.0\ \Omega$. Both branches join at the same pair of nodes and drive current down through a shared resistor $R_{3} = 2.0\ \Omega$. Let $I_{1}$, $I_{2}$ be the currents the two cells push toward the top node, so the current through $R_{3}$ is $I_{3} = I_{1} + I_{2}$.两个理想电源通过一个公共电阻驱动电流。左支路含电动势 $\varepsilon_{1} = 6.0\ \mathrm{V}$ 的电源与 $R_{1} = 2.0\ \Omega$ 串联；中支路含电动势 $\varepsilon_{2} = 4.0\ \mathrm{V}$ 的电源与 $R_{2} = 1.0\ \Omega$ 串联。两支路接在同一对节点上，共同向下驱动电流流过公共电阻 $R_{3} = 2.0\ \Omega$。设 $I_{1}$、$I_{2}$ 为两电源推向上节点的电流，则流过 $R_{3}$ 的电流为 $I_{3} = I_{1} + I_{2}$。
(a)State Kirchhoff's junction law and loop law, and name the conservation principle behind each.写出基尔霍夫节点定律与回路定律，并说出各自背后的守恒原理。[2]
(b)Apply the loop law to each of the two source loops to obtain two simultaneous equations, and solve them for $I_{1}$ and $I_{2}$.对两个含电源的回路分别应用回路定律，得到两个联立方程，并解出 $I_{1}$ 与 $I_{2}$。[4]
(c)Determine the current $I_{3}$ through $R_{3}$ and the potential difference across it, then verify that the loop law is satisfied for the left loop.求流过 $R_{3}$ 的电流 $I_{3}$ 及其两端电势差，再验证左回路满足回路定律。[2]
PART II · PAPER 1B / DATA ANALYSIS第二部分 · 第一卷 B / 数据分析Graphs · data · uncertainties · 22 marks图像 · 数据 · 不确定度 · 22 分
Graph and Data Questions图像与数据题
These items reward correct reading of $I$-$V$ characteristics, gradients and intercepts, and careful handling of uncertainties. Quote percentage uncertainties to an appropriate precision and use $R = V/I$ at each operating point, never the slope of a curve.这些题考查对 $I$-$V$ 特性、斜率与截距的正确读取，以及对不确定度的细致处理。百分比不确定度保留适当精度；对每个工作点用 $R = V/I$，切勿用曲线的斜率。
Q6HARDPaper 1BI-V characteristics: ohmic vs non-ohmicI-V 特性：欧姆与非欧姆[8 marks]
A student measures the current $I$ through two components at several potential differences $V$. Component X is a fixed resistor; component Y is a filament lamp.学生测量两个元件在若干电势差 $V$ 下通过的电流 $I$。元件 X 为定值电阻；元件 Y 为灯丝灯泡。
$V\ /\ \mathrm{V}$
$2.0$
$4.0$
$6.0$
$I_{\mathrm{X}}\ /\ \mathrm{A}$
$0.50$
$1.00$
$1.50$
$I_{\mathrm{Y}}\ /\ \mathrm{A}$
$0.40$
$0.52$
$0.60$
(a)For component X, calculate the resistance at $V = 2.0\ \mathrm{V}$ and at $V = 6.0\ \mathrm{V}$, and state whether X is ohmic.对元件 X，计算 $V = 2.0\ \mathrm{V}$ 与 $V = 6.0\ \mathrm{V}$ 处的电阻，并判断 X 是否为欧姆元件。[2]
(b)For component Y, calculate the resistance at $V = 2.0\ \mathrm{V}$ and at $V = 6.0\ \mathrm{V}$, and explain the trend in terms of the filament's temperature.对元件 Y，计算 $V = 2.0\ \mathrm{V}$ 与 $V = 6.0\ \mathrm{V}$ 处的电阻，并结合灯丝温度解释其变化趋势。[3]
(c)The current reading $I_{\mathrm{X}} = 1.50\ \mathrm{A}$ has an absolute uncertainty of $\pm 0.05\ \mathrm{A}$. Calculate its percentage uncertainty.电流读数 $I_{\mathrm{X}} = 1.50\ \mathrm{A}$ 的绝对不确定度为 $\pm 0.05\ \mathrm{A}$。计算其百分比不确定度。[1]
(d)Describe the shape of the $I$-$V$ characteristic of a diode, and state how it differs from that of component X.描述二极管的 $I$-$V$ 特性曲线的形状，并说明它与元件 X 的特性有何不同。[2]
Q7HARDPaper 1BHL ONLYemf and internal resistance from a graph由图线测电动势与内阻[14 marks]
To find the emf $\varepsilon$ and internal resistance $r$ of a cell, a student varies the external load and records the terminal potential difference $V$ at several values of the current $I$:为测定某电池的电动势 $\varepsilon$ 与内阻 $r$，学生改变外负载，记录若干电流 $I$ 下的端电压 $V$：
$I\ /\ \mathrm{A}$
$0.50$
$1.00$
$1.50$
$2.00$
$V\ /\ \mathrm{V}$
$5.60$
$5.20$
$4.80$
$4.40$
(a)Starting from $\varepsilon = I(R + r)$, show that the terminal pd obeys $V = \varepsilon - rI$, and hence explain why a graph of $V$ against $I$ is a straight line.从 $\varepsilon = I(R + r)$ 出发，证明端电压满足 $V = \varepsilon - rI$，并由此说明为何 $V$ 对 $I$ 的图为直线。[3]
(b)Calculate the gradient of the line and hence determine the internal resistance $r$.计算直线斜率，由此求内阻 $r$。[3]
(c)By extrapolating to $I = 0$, determine the emf $\varepsilon$ of the cell, and explain why this intercept gives the emf.通过外推到 $I = 0$，求电池的电动势 $\varepsilon$，并解释为何该截距即为电动势。[3]
(d)Calculate the power dissipated inside the cell when the current is $2.00\ \mathrm{A}$, and the maximum (short-circuit) current the cell could deliver.计算电流为 $2.00\ \mathrm{A}$ 时电池内部耗散的功率，以及电池所能输出的最大（短路）电流。[3]
(e)The reading $V = 4.40\ \mathrm{V}$ has an absolute uncertainty of $\pm 0.10\ \mathrm{V}$. Calculate its percentage uncertainty.读数 $V = 4.40\ \mathrm{V}$ 的绝对不确定度为 $\pm 0.10\ \mathrm{V}$。计算其百分比不确定度。[2]
PART III · PAPER 2 STYLE第三部分 · 第二卷风格Extended structured · calculator · 30 marks长结构题 · 可用计算器 · 30 分
Extended Structured Problems长结构问题
Set up each problem with a clear circuit diagram. Method marks dominate the longer items; carry intermediate values to extra figures and round only the final answer. Check current and energy conservation where you can.每题先画清晰的电路图。长题中方法分占比最大；中间值多保留几位，仅在最终答案处取舍有效数字。尽可能核查电流与能量守恒。
Q8HARDPaper 2full network with power含功率的完整网络[12 marks]
An ideal $24\ \mathrm{V}$ supply is connected to the following network. A $4.0\ \Omega$ resistor ($R_{1}$) is in series with a parallel combination of a $6.0\ \Omega$ resistor ($R_{2}$) and a $3.0\ \Omega$ resistor ($R_{3}$); this is then in series with a $2.0\ \Omega$ resistor ($R_{4}$).理想 $24\ \mathrm{V}$ 电源接到如下网络。一个 $4.0\ \Omega$ 电阻（$R_{1}$）与 $6.0\ \Omega$ 电阻（$R_{2}$）、$3.0\ \Omega$ 电阻（$R_{3}$）的并联组合串联；该组合再与一个 $2.0\ \Omega$ 电阻（$R_{4}$）串联。
(a)Calculate the equivalent resistance of the network and the current drawn from the supply.计算网络的等效电阻以及电源输出的电流。[3]
(b)Calculate the potential difference across the parallel combination.计算并联组合两端的电势差。[2]
(c)Calculate the current in the $6.0\ \Omega$ resistor and in the $3.0\ \Omega$ resistor, and verify that they sum to the supply current.计算 $6.0\ \Omega$ 与 $3.0\ \Omega$ 电阻中的电流，并验证二者之和等于电源电流。[3]
(d)Calculate the power dissipated in the $4.0\ \Omega$ resistor and the total power delivered by the supply.计算 $4.0\ \Omega$ 电阻耗散的功率以及电源输出的总功率。[2]
(e)State, with a reason, which single resistor dissipates the most power.写出哪一个电阻耗散功率最大，并说明理由。[2]
A battery of emf $12\ \mathrm{V}$ and internal resistance $0.50\ \Omega$ is connected to an external load resistor $R$.一个电动势 $12\ \mathrm{V}$、内阻 $0.50\ \Omega$ 的电池接到外负载电阻 $R$。
(a)The load is $R = 5.5\ \Omega$. Calculate the current in the circuit and the terminal potential difference of the battery.负载为 $R = 5.5\ \Omega$。计算电路中的电流以及电池的端电压。[3]
(b)Calculate the power dissipated in the load $R$ and the power dissipated inside the battery, and hence find the efficiency of energy transfer to the load.计算负载 $R$ 上的耗散功率与电池内部的耗散功率，由此求向负载传递能量的效率。[3]
(c)The load is now reduced to $R = 2.5\ \Omega$. Calculate the new current and the new terminal pd, and comment on how the terminal pd has changed.现把负载减小为 $R = 2.5\ \Omega$。计算新的电流与新的端电压，并说明端电压如何变化。[2]
(d)Determine the maximum current the battery could supply, and explain in one sentence the condition under which it occurs.求电池所能供给的最大电流，并用一句话说明其发生条件。[2]
Q10HARDPaper 2HL ONLYpotential divider with thermistor含热敏电阻的分压器[8 marks]
A $9.0\ \mathrm{V}$ supply is connected across a potential divider consisting of a fixed resistor $R_{1} = 3.0\ \mathrm{k\Omega}$ in series with a thermistor (a resistor whose resistance falls as its temperature rises). The output voltage $V_{\text{out}}$ is taken across the thermistor.一个 $9.0\ \mathrm{V}$ 电源接在分压器上，分压器由固定电阻 $R_{1} = 3.0\ \mathrm{k\Omega}$ 与热敏电阻（其电阻随温度升高而减小）串联组成。输出电压 $V_{\text{out}}$ 取自热敏电阻两端。
(a)At a cool temperature the thermistor has resistance $6.0\ \mathrm{k\Omega}$. Calculate the output voltage $V_{\text{out}}$ across the thermistor.在较低温度下热敏电阻阻值为 $6.0\ \mathrm{k\Omega}$。计算热敏电阻两端的输出电压 $V_{\text{out}}$。[3]
(b)The temperature rises and the thermistor resistance falls to $1.0\ \mathrm{k\Omega}$. Calculate the new output voltage.温度升高，热敏电阻阻值降到 $1.0\ \mathrm{k\Omega}$。计算新的输出电压。[2]
(c)State how $V_{\text{out}}$ responds to a rise in temperature, and explain how this circuit could be used as part of a temperature-sensing system.说明 $V_{\text{out}}$ 如何随温度升高而响应，并解释该电路如何用作温度感应系统的一部分。[3]