Topic 2: Electricity
AQA · GCSE · Physics Topic 2 12:01 English narration · English + 中文 subtitles burned in
Chapters
Transcript
Start with a familiar switch and lamp.
从熟悉的开关和灯开始。
Trace whether there is a complete conducting loop and a source of potential difference. Then distinguish charge flow from energy transfer.
判断是否有完整导电回路和电势差来源，再区分电荷流动与能量转移。
The stated current is steady.
题目给定的是稳定电流。
The lamp does not consume the charge that reaches it.
灯不会消耗到达它的电荷。
In this steady series loop the current before and after it is the same.
这个稳定串联回路中，灯前后的电流相同。
Energy is transferred while charge continues around the circuit.
电荷继续沿回路流动，同时发生能量转移。
Identify the long positive plate of the cell and the repeated plate pairs of the battery.
认出电池单元的长正极线及电池组重复的长短线对。
Compare the open and closed switch.
比较开关断开和闭合的符号。
Draw an ammeter in the loop and a voltmeter across the lamp, using these standard symbols.
用标准符号把电流表串联接入回路，把电压表并联在灯两端。
The variable resistor has an arrow through it; the thermistor has a hooked diagonal line.
可变电阻有穿过符号的箭头，热敏电阻有带折角的斜线。
Light arrows point into an LDR and out of an LED.
光箭头指向光敏电阻内部，并从发光二极管向外。
A diode has one cathode bar.
二极管有一条阴极线。
Cover the names and redraw the six symbols before checking.
遮住名称，画出六种符号后再核对。
Current measures a rate, not the amount of charge alone.
电流表示速率，不是单独的电荷量。
Calculate coulombs per second and use the steady single-loop condition at the second point.
计算每秒通过的库仑数，并用稳定单回路的条件判断第二点。
Then compare equal charge passing in a longer time.
再比较相同电荷通过更长时间的情况。
Four coulombs in two seconds is two amperes.
两秒内通过四库仑，对应两安培。
The second point in the steady series loop has the same charge flow in the same time.
同一稳定串联回路的第二点，在相同时间内通过相同电荷。
Four coulombs in four seconds is one ampere: a longer time means a lower rate.
四秒内通过四库仑，对应一安培：时间更长，流动速率更小。
Calculate before checking.
先计算再核对。
This is teacher-written practice, with no claimed mark allocation.
这是教师编写的练习，不声称对应某个分值。
Convert minutes to seconds, calculate charge, and explain the effect of doubling time at the same current.
把分钟换成秒，计算电荷量，再解释电流相同时把时间加倍的影响。
Twenty-five minutes is fifteen hundred seconds.
二十五分钟是一千五百秒。
Charge equals zero point nine zero amperes times fifteen hundred seconds, giving thirteen hundred and fifty coulombs.
电荷量等于零点九零安培乘一千五百秒，得到一千三百五十库仑。
At the same current, doubling time doubles charge.
电流相同时，时间加倍，电荷量也加倍。
The first task is the actual exam calculation; the second is a teacher-written extension.
第一项是真实考试计算题，第二项是教师编写的延伸问题。
Convert milliseconds to seconds, rearrange charge equals current times time, then substitute with units.
先把毫秒换算成秒，再对电荷量等于电流乘时间的公式变形，最后带单位代入。
The current is five point zero amperes, matching the official scheme.
电流为五点零安培，与官方评分标准一致。
Four hundred milliseconds is zero point four zero zero seconds, not four hundred seconds.
四百毫秒是零点四零零秒，不是四百秒。
The unit conversion changes the denominator by a factor of one thousand.
单位换算使分母相差一千倍。
Use the handout example.
使用讲义例题。
Select the rearrangement and check that a current in amperes is used. Then predict current when resistance changes at the same potential difference.
先移项，确认电流单位为安培，再预测电势差不变时电阻改变会怎样影响电流。
Divide potential difference by current to obtain sixty ohms.
电势差除以电流得到六十欧。
Twice the resistance gives half the current only when potential difference is unchanged.
只有电势差不变，电阻加倍才使电流减半。
These readings describe the operating point; they do not prove that resistance is constant at all temperatures.
这些读数描述当前工作点，并不能证明所有温度下电阻都不变。
Trace the loop through the cell, switch, ammeter and selected wire.
沿电池、开关、电流表和选定电阻丝追踪回路。
The voltmeter branch connects to the same two clips that define length.
电压表两端接到定义长度的同一对夹子。
There is no conducting bypass across the wire.
没有导线短接这段电阻丝。
Open the switch between readings.
读数之间断开开关。
Measure length between contact points, not from an assumed zero on the ruler.
长度从实际接触点量起，不是假定从尺的零点开始。
Keep material, cross-sectional area and temperature constant.
材料、横截面积和温度保持不变。
Read both meters for each length, repeat and calculate resistance.
每个长度都读两个电表，重复测量并计算电阻。
Low potential difference and switching off limit heating.
低电压和间歇断电可减少发热。
These ideal teacher-written data are an analysis exercise, not a record of a real experiment.
这些是教师编写的理想分析数据，不是真实实验记录。
Calculate resistance in each row and predict the resistance at eighty centimetres.
计算每行电阻，并预测八十厘米时的电阻。
The potential differences differ, so comparing currents alone would be misleading.
各行电势差不同，只比较电流会产生误导。
The resistances are two, four and six ohms.
电阻分别为二、四、六欧。
Resistance divided by length is the same in all three rows.
每行电阻与长度的比值相同。
A graph of resistance against length is a straight line through the origin for these ideal data.
这些理想数据的电阻与长度图像为过原点的直线。
The eighty-centimetre prediction is eight ohms, assuming the same wire and temperature.
同一导线且温度相同时，八十厘米预测为八欧。
Distinguish an observed intercept from a diagnosed zero error.
区分观察到的截距和已确认的零误差。
A contact contribution and a length offset are different possible causes.
接触电阻和长度起点偏移是不同的可能原因。
Decide what to check before changing data.
先决定检查什么，再考虑修正数据。
A non-zero intercept does not identify its own cause.
非零截距本身不能说明原因。
Recheck the contact-to-contact length and the connections.
重新检查接触点间长度和接线。
Correct a demonstrated length offset using the measured offset, rather than subtracting a resistance intercept blindly.
只有确认长度起点偏移，才按实测偏移修正，不能盲目减去电阻截距。
Heating raises the resistance of this metal wire and changes the experiment.
金属丝发热会增大电阻，改变实验条件。
The official practical includes both changing wire length and comparing series and parallel resistors.
官方实验同时包括改变电阻丝长度和比较串并联电阻。
Here use two equal resistors and predict what total-current measurements would show at the same potential difference.
这里用两个相同电阻，预测相同总电压下总电流的测量结果。
In series resistances add to twenty ohms and total current is point one ampere.
串联总电阻二十欧，总电流零点一安。
In parallel each ten-ohm branch has two volts and point two amperes; total current is point four amperes, giving five ohms.
并联时每个十欧支路电压二伏，电流零点二安；总电流零点四安，总电阻五欧。
Measure total current before the branch split and potential difference across the entire combination.
电流表在分流前测总电流，电压表接在整个组合两端。
Trace the series path through supply, ammeter and lamp.
沿电源、电流表和灯泡追踪串联回路。
The voltmeter measures across the lamp alone.
电压表只测灯泡两端的电势差。
A variable dc supply can change the potential difference without a variable resistor.
可调直流电源可改变电势差，不必另接可变电阻。
This is a low-voltage laboratory circuit.
这是低压实验电路。
The actual June twenty twenty-three method question requires a working circuit and a method for negative readings.
二零二三年六月的方法题要求可工作的电路和取得负读数的方法。
Start at zero, change the potential difference, record both meters and repeat.
从零开始改变电势差，每次读取两表并重复。
Switch off before reversing the supply.
先断电再反接电源。
Resistor temperature should remain approximately constant; lamp heating is what produces its curved characteristic.
电阻器温度应近似不变；灯丝发热正是曲线产生的原因。
These schematic curves share current on the vertical axis and potential difference on the horizontal axis.
这些示意曲线纵轴均为电流，横轴为电势差。
The ohmic resistor has a constant ratio of voltage to current.
欧姆电阻的电压电流比恒定。
The lamp becomes less steep away from zero as its filament warms.
灯丝变热后，远离原点的曲线逐渐变平。
Diode reverse current is negligible on this scale.
这个尺度下二极管反向电流可忽略。
This is the official AQA June twenty twenty-three Figure six.
这是AQA二零二三年六月试卷图六。
Read the current at plus three volts before using voltage divided by current.
先读出正三伏时的电流，再用电压除以电流。
Compare it with the reading at six volts.
与六伏的读数比较。
A resistance calculated from a point is not the tangent gradient of this curved graph.
从曲线上一点计算的电阻，不是该点切线斜率。
At three volts the graph reads about point one six amperes, so resistance is eighteen point seven five ohms, about nineteen ohms.
三伏时图上电流约零点一六安，因此电阻十八点七五欧，约十九欧。
At six volts, point two one amperes gives about twenty-nine ohms.
六伏时电流零点二一安，电阻约二十九欧。
Voltage doubles but current does not; heating raises the filament resistance.
电压加倍而电流没有加倍，因为发热使灯丝电阻增大。
The six-volt resistance comparison is a teacher extension.
六伏电阻比较是教师拓展。
The diode method has an extra component for protection.
二极管实验另需保护元件。
Explain what the protective resistor does and why the voltmeter must not include it.
解释保护电阻的作用，以及电压表为何不能连同它一起测量。
A sensitive current meter measures small current; it does not replace current limiting.
灵敏电流表测小电流，不能代替限流。
Use the protective resistor to limit current.
保护电阻限制电流。
Connect the voltmeter directly across the diode so the voltage and current pair belongs to that component.
电压表直接接二极管两端，使电压和电流数据对应同一元件。
Reverse current is negligible in the GCSE model, corresponding to very high reverse resistance.
GCSE模型下反向电流可忽略，对应很高的反向电阻。
Use a suitable low-voltage supply within component ratings.
使用合适低压电源并遵守元件额定值。
These are not current-voltage graphs.
这不是电流电压图。
The thermistor plot changes temperature and the LDR plot changes light intensity.
热敏电阻图改变温度，光敏电阻图改变光强。
Both show falling resistance as the horizontal quantity rises.
两图都显示横轴量增加时电阻减小。
An application also needs a circuit and a stated switching rule.
应用还需要电路和明确的开关控制规则。
Use the stated high-resistance switching rule to decide when the lamp turns on. Then use the same-potential-difference condition to predict sensor current.
根据给出的高电阻触发规则判断灯何时亮，再用传感器电势差相同的条件预测电流。
Do not transfer that condition automatically to an unspecified series circuit.
不要把这一条件自动套用于未说明的串联电路。
Darkness raises LDR resistance, so this high-resistance controller switches the lamp on.
黑暗使光敏电阻增大，因此这个高电阻触发控制器开灯。
Bright light lowers resistance and it switches off.
强光使电阻减小，灯关闭。
A warming thermistor has lower resistance; at fixed sensor potential difference its current increases.
热敏电阻变热时电阻减小；只有传感器电势差不变时，电流才增大。
The circuit determines output behaviour, so the component trend alone does not determine whether an appliance turns on.
输出由电路决定，元件趋势本身不能决定电器开关。
The left circuit has one main path through both lamps; its voltmeter branch measures lamp one.
左图的主回路依次经过两个灯，电压表支路测灯一。
On the right the ammeter is before the split, so it measures total current.
右图电流表在分流前，测总电流。
The two lamps connect between the same pair of junctions and have the same potential difference.
两个灯连接于同一对节点之间，因此电势差相同。
The voltmeter is a measuring branch, not a conducting bypass.
电压表支路用于测量，不是导线短接。
Use fixed resistors rather than pretending filament lamps have one resistance at every voltage.
这里使用定值电阻，不假定白炽灯在所有电压下电阻相同。
Trace the one loop, add the resistances and find current before finding each voltage share.
先识别单回路，求总电阻与电流，再求每个电阻的电压。
Series resistance is sixteen ohms and the current is point seven five amperes through both resistors.
串联总电阻十六欧，两个电阻中电流都是零点七五安。
Multiplying this current by eight ohms gives six volts across each.
电流乘八欧得到各六伏。
They share equally because their resistances are equal; unequal resistors need not have equal voltage shares.
均分电压是因为电阻相同，不同电阻不一定均分。
The same two resistors are now in independent parallel branches.
把相同两个电阻改接为独立并联支路。
The ideal supply keeps twelve volts between the branch junctions.
理想电源保持两节点间十二伏。
Work out each branch current, then add them. Explain the resistance trend without needing a reciprocal formula.
先求支路电流，再相加；不需要倒数公式即可解释总电阻趋势。
Each branch has twelve volts, so each carries one point five amperes. The supply current is three amperes.
每条支路都是十二伏，电流各一点五安，总电流三安。
Compared with one resistor, current doubles at the same potential difference; total resistance is smaller.
与单个电阻相比，同一电势差下总电流加倍，所以总电阻较小。
The specification does not require a reciprocal-resistance formula for this comparison.
这个比较不需要倒数电阻公式。
Use the current rule and voltage rule separately.
分别应用电流与电压规则。
Series components share the same current, while parallel branches share the same voltage.
串联元件电流相同，并联支路电压相同。
Equal voltage shares in series are a special case for equal resistances, not the general rule.
串联均分电压只是相同电阻的特殊情况，不是一般规则。
Distinguish breaking the only main loop from breaking one parallel branch.
区分唯一主回路断开与一条并联支路断开。
The question specifies an ideal supply with unchanged potential difference so the remaining branch prediction is justified.
题目明确电源电势差不变，因此可以判断另一支路是否仍工作。
In series the only conducting path is broken, so current through both lamps stops.
串联唯一导电路径断开，两个灯都无电流。
In parallel the unaffected branch still connects across the same fixed supply potential difference and continues working.
并联时另一支路仍接于同一固定电势差两端，继续工作。
Total supply current falls because one branch current is now zero.
总电流减小，因为断开支路电流为零。
This prediction assumes the supply voltage remains fixed.
结论以电源电压不变为前提。
This adapts the June twenty twenty-four question by supplying its graph reading.
这里改编二零二四年六月题目，并给出图上读数。
The thermistor is eighty ohms at twenty degrees Celsius, in series with four hundred ohms across twelve volts.
二十摄氏度时热敏电阻八十欧，与四百欧定值电阻串联接十二伏。
Find common current before the thermistor voltage.
先求共同电流，再求热敏电阻电压。
The original question also asks students to read the resistance graph.
原题还要求读电阻图。
Total resistance is four hundred eighty ohms; current is point zero two five amperes.
总电阻四百八十欧，电流零点零二五安。
This gives two volts across the thermistor and ten volts across the fixed resistor.
热敏电阻两端二伏，定值电阻十伏。
The official scheme confirms two volts.
官方评分方案确认二伏。
Current is the same through both, but their voltages differ because their resistances differ.
两者电流相同，电阻不同所以电压不同。
Start from the actual June twenty twenty-four question: give frequency and potential difference in the correct order with units.
先回答二零二四年六月原题，按顺序写频率、电势差及单位。
Then explain alternating potential difference and distinguish a cycle from a single direction change.
再解释交变电势差，并区分一个完整周期与一次方向改变。
These graphs show polarity against time with no numerical voltage amplitude.
这两图显示极性随时间变化，没有给出电压幅值。
The mains curve has fifty complete cycles per second, one cycle in twenty milliseconds.
市电每秒五十个完整周期，一个周期二十毫秒。
The battery example stays one polarity.
电池例子保持同一极性。
The quoted two hundred thirty volts is not drawn as the curve peak.
图中没有把额定二百三十伏当作峰值。
The official recall answer is fifty hertz and two hundred thirty volts.
官方记忆题答案为五十赫兹和二百三十伏。
Alternating means potential difference repeatedly reverses polarity.
交流表示电势差反复改变极性。
Fifty hertz counts complete cycles.
五十赫兹计的是完整周期。
A battery gives direct potential difference, which does not reverse polarity; direct does not require every possible supply to have a perfectly constant magnitude.
电池产生直流电势差，不反转极性；直流并不要求所有电源的幅值完全恒定。
Retrieve colour and function together.
把颜色和作用一起回忆。
Neutral and earth are both normally near zero volts relative to earth, but they are not interchangeable: one completes the load circuit and one is a protective connection.
零线和地线通常对地都接近零伏，但不能混为一谈：一个完成负载回路，另一个作保护连接。
Use the printed colour names to identify the insulation. Brown is live, blue is neutral, and green and yellow is protective earth.
按图上的颜色名称识别绝缘层：棕色为火线，蓝色为零线，黄绿色为保护地线。
Follow each leader to its own core.
沿引线找到各自的芯线。
Live supplies alternating potential difference.
火线提供交变电势差。
Neutral completes the normal load circuit and is close to earth potential.
零线完成正常负载回路，电势接近地。
The protective earth connects exposed metal to the protective system and carries no normal load current in this classroom model.
保护地线将外露金属与保护系统连接，在课堂模型的正常工作中不承载负载电流。
The open switch breaks the main loop, so the lamp is off.
开关断开使主回路中断，灯不亮。
Point A still connects directly to the live supply. Point B is after the open switch on the load side.
A点仍直接接火线电源，B点位于断开的开关后方负载侧。
The diagram separates those points so we do not claim that every lamp terminal remains live after a correctly wired switch opens.
图中区分两点，不能声称正确火线开关断开后所有灯端子仍带电。
Use point A in the previous diagram. Explain its potential difference relative to earth and the current path if a person makes a conducting connection.
使用前图A点，解释它对地的电势差及人体形成导电连接时的电流路径。
Do not assume that the same path resistance applies to a body and to a wire.
不能假定人体和金属导线的电阻相同。
Point A remains at mains potential relative to earth.
A点仍对地具有市电电势差。
A conducting path through a person can carry a dangerous current.
经人体形成的导电路径可能产生危险电流。
A human path is not automatically a low-resistance wire, and a body current below an appliance fuse rating can still cause severe injury.
人体路径不能直接当成低阻导线，小于电器保险丝额定电流的电流仍可严重伤人。
A fuse does not guarantee protection against touching live parts.
保险丝不能保证接触带电部位时不触电。
This is a specified metal-case fault with a sound earth path and a correctly selected fuse.
这里明确给出金属外壳故障、完好的接地路径及适当保险丝。
Explain the physical sequence, then decide what can change if the protective earth is missing.
解释物理过程，再判断保护地线缺失会怎样。
The protective earth provides a low-resistance path for the case fault.
保护地线为外壳故障提供低阻路径。
Large fault current heats and melts the suitable fuse, breaking the live supply.
较大故障电流使适当保险丝发热熔断，切断火线。
Without the earth path the case may become live without enough current to blow the fuse.
没有接地路径，外壳可能带电而电流不足以熔断保险丝。
This is different from saying every possible shock current operates a fuse.
这与所有触电电流都能使保险丝动作的说法不同。
This is the ideal single-phase model from sheet two point five.
这是练习二点五中理想单相模型。
The neutral return is broken between the lamp and supply.
灯与电源之间的零线回路断开。
The load-side neutral terminal remains connected through the lamp to live, with no other return path.
负载侧零线端经灯仍连到火线，没有其他回路。
Compare voltage across the lamp with voltage to earth.
比较灯两端电压与各端对地电压。
The open neutral stops normal current.
零线断开使正常电流停止。
Without a return path the load-side terminal is at approximately the same potential as live through the lamp.
没有回路时，负载侧端经灯与火线处于近似相同电势。
Both can be about two hundred thirty volts relative to earth while the voltage between them is approximately zero.
两端对地都可约二百三十伏，彼此之间电压却约零伏。
The lamp is unlit, but a person providing an earth path can still receive a shock.
灯不亮，但人体若提供对地路径仍可触电。
Read a rating as joules per second at the working potential difference.
额定功率表示工作电势差下每秒转移的焦耳数。
The energy goes into different pathways in a kettle, motor and lamp.
水壶、电动机和灯的能量去向不同。
A rating does not by itself tell us the total energy used or the useful output.
仅凭额定功率不能确定总耗能或有用输出。
Choose by what you know and what you need.
按已知量和目标选公式。
With watts use seconds to calculate joules.
瓦配合秒才能得到焦耳。
Current and resistance give power for the resistive model.
电流与电阻可计算电阻模型的功率。
If a first step gives the missing quantity, chain equations rather than choosing by a single familiar number.
若先求出中间量，就连接两个公式，不要只凭熟悉数字选公式。
The actual question asks for energy.
原题要求能量。
Try the power route first, then compare it with charge times potential difference.
先试功率路线，再与电荷乘电势差比较。
Keep intermediate values and label the units.
保留中间值，并写单位。
First convert thirty minutes into eighteen hundred seconds.
先把三十分钟换为一千八百秒。
The two electrical readings give one point two six watts.
电压与电流给出一点二六瓦。
Multiplying that rate by time gives two thousand two hundred sixty-eight joules.
速率乘时间得到二千二百六十八焦耳。
Round the final result, not the intermediate power.
最后才舍入，不先舍入功率。
Current times time is three hundred seventy-eight coulombs.
电流乘时间为三百七十八库仑。
Each coulomb transfers six joules across the lamp, so the same energy follows.
每库仑在灯两端转移六焦耳，因此得到相同能量。
The official scheme accepts both equation chains; this is a useful independent check.
官方评分方案接受两条公式链，可相互核对。
The paper prescribes a resistance-based calculation.
这道题使用给定电阻模型。
Power divided by resistance gives current squared, so remember the square root.
功率除以电阻得到电流平方，必须开平方根。
Then charge comes from current and time.
再由电流和时间求电荷。
Do not generalise the exam model into a claim that all real motor input is heating.
不能把题目模型推广为真实电机所有输入都变成发热。
The official scheme gives zero point nine amperes, eighteen hundred seconds, and sixteen hundred twenty coulombs.
官方方案得到零点九安、一千八百秒及一千六百二十库仑。
The intermediate current is essential: four point eight six divided by six is zero point eight one amperes squared, not the current.
中间电流很关键：四点八六除以六得到零点八一安培平方，不是电流。
Explain what went wrong before correcting it.
先解释错误再改正。
One student mixed minutes and watts; another missed a square root; the third used the unit of power for energy.
第一位把分钟与瓦混用；第二位漏开平方根；第三位把功率单位用于能量。
The repair depends on the type of error.
改正要针对错误类型。
Converting time changes the energy value; taking a square root changes the current; checking the requested quantity fixes the unit.
换算时间改变能量值；开平方根改变电流；核对所求物理量纠正单位。
These are checks to apply to your own working.
把这些检查用于自己的步骤。
These teacher-written lines show input energy against time at constant power.
这些教师编写的直线表示恒定功率下输入能量随时间变化。
The steeper two kilowatt line reaches one hundred twenty kilojoules in sixty seconds; the one kilowatt line takes one hundred twenty seconds.
较陡的二千瓦线六十秒达到一百二十千焦；一千瓦线要一百二十秒。
Compare the endpoints before making a claim about total energy.
比较端点，再判断总能量。
There are two comparisons.
有两种比较。
First keep input energy fixed; then keep duration fixed.
先固定输入能量，再固定时间。
State what stays the same each time.
每次说明不变量。
Total cost at a fixed tariff follows total electrical energy, so the rating alone is insufficient.
相同电价下总费用取决于总电能，所以仅看额定功率不够。
The input energy is fixed at one hundred twenty kilojoules.
输入能量固定为一百二十千焦。
Divide energy by power: one kilowatt takes one hundred twenty seconds; two kilowatts takes sixty.
能量除以功率：一千瓦需要一百二十秒，二千瓦需要六十秒。
Compare useful output only with efficiency and losses specified.
比较有用输出还需明确效率及损耗。
Now duration is fixed at sixty seconds.
现在时间固定为六十秒。
The input energies are sixty and one hundred twenty kilojoules.
输入能量分别为六十及一百二十千焦。
Greater power gives more energy at equal time, but total use for a job also depends on duration and useful output.
同样时间内功率大则能量多，但一项任务的总耗能还取决于时间和有用输出。
Follow the route from left to right.
从左向右追踪。
The step-up transformer raises the potential difference before transmission. The step-down transformer lowers it for consumers.
升压变压器在输电前提高电势差，降压变压器为用户降低电势差。
The arrows show the system-level energy route, not individual circuit wires.
箭头表示系统层面的能量传输路线，不是具体电路导线。
Use the actual exam figure.
使用原题图。
Explain the change in potential difference, then current, then cable heating, then efficiency.
依次解释电势差、电流、电缆发热及效率的变化。
For Y, explain the lower consumer potential difference.
对Y说明用户端电势差降低。
The official scheme requires the physical chain.
官方方案要求说明物理链条。
Higher potential difference reduces the current for the same transmitted power.
同样传输功率下，电势差高则电流小。
Lower current reduces heating of the unchanged cables, so transmission efficiency increases.
电缆不变时，电流小则发热损耗少，所以效率提高。
Do not say that all loss disappears.
不要说所有损耗消失。
Transformer Y lowers the potential difference to a more suitable and safer value for consumers.
Y将电势差降至更适合用户且相对较安全的值。
In the UK the domestic value is about two hundred thirty volts.
英国家庭约二百三十伏。
Safer than transmission voltage does not mean harmless to touch.
比输电电压较安全，不等于可以触碰。
These are teacher-written numbers in a simplified resistive-line model.
这些是教师编写的简化电阻线路数据。
Hold power entering the line fixed, not the output after losses.
固定进入线路的功率，不是损耗后的输出。
Doubling potential difference should halve current and quarter the loss if cable resistance is unchanged.
电缆电阻不变时，电势差加倍应使电流减半、损耗变为四分之一。
Five hundred kilowatts is five hundred thousand watts.
五百千瓦为五十万瓦。
At ten thousand volts, current is fifty amperes.
一万伏时，电流为五十安。
Squaring fifty and multiplying by two gives five thousand watts of cable heating.
五十平方乘二，得到五千瓦电缆发热损耗。
At twenty kilovolts the current is twenty-five amperes. The loss is one thousand two hundred fifty watts, a quarter of the previous value.
二万伏时电流为二十五安，损耗一千二百五十瓦，是原来的四分之一。
Input remains fixed while output grows because cable heating is smaller.
输入固定而输出增加，因为电缆发热减少。
This is the existing sheet comparison.
这是已有练习的比较。
Calculate the two losses, but separate the valid comparison from an unsupported claim about absolute efficiency.
求两种损耗，但要区分有效比较与没有根据的绝对效率判断。
We need the total input to calculate a loss fraction.
损耗比例还需要总输入。
The two losses are one hundred sixty megawatts and ten megawatts. The loss ratio is one sixteenth.
两种损耗为一亿六千万瓦和一千万瓦，比例为十六分之一。
These deliberately large numerical losses are not by themselves evidence of a small percentage loss: input power has not been supplied.
仅凭这些很大的损耗数字不能证明损耗百分比很小，因为没有给输入功率。
Here the total input and efficiency are supplied, so we can calculate useful output.
这里给出了总输入和效率，可以求有用输出。
Both sides use gigajoules; the energy unit cancels in the ratio.
两边都用吉焦，能量单位在比值中约去。
Multiply zero point nine nine two by thirty-four point two to get thirty-three point nine two six four gigajoules, approximately thirty-three point nine.
零点九九二乘三十四点二，得到三十三点九二六四吉焦，约三十三点九。
The official scheme gives this value.
与官方方案一致。
The useful output is smaller than the input, as expected.
有用输出小于输入，符合预期。
Attempt the actual explanation before checking it.
先回答这道真题，再核对。
Track electrons from the cloth to the rod; do not move positive charge.
追踪电子从布到棒的转移；不要说正电荷转移。
The official scheme requires electron transfer, the negative sign of electrons, and the resulting excess positive charge on the cloth.
官方评分要求电子转移、电子带负电，以及布上剩余正电荷较多。
Any transfer of positive charge is rejected in this context.
这里不能说正电荷转移。
The left arrow tracks electron transfer; the right arrows describe the direction of force on a positive test charge.
左侧箭头表示电子转移，右侧箭头表示正试探电荷的受力方向。
They represent different ideas. Nearer field lines indicate a stronger field.
两者含义不同；电场线越密，场越强。
Draw before looking back at the diagram.
先画图，再看示意图。
Field arrows show the force on a positive charge, not necessarily the motion of electrons.
电场箭头表示正电荷受力方向，不一定表示电子运动方向。
The June2024 paper separately tests the field definition and decreasing strength with distance.
2024年6月试卷分别考查电场定义和场强随距离增大而减小。
Link each feature of the drawing to those meanings.
把图中各特征与这些含义联系起来。
Connect transfer, the resulting charge, and repulsion.
把电子转移、头发所带电荷和相斥联系起来。
This is an exam explanation, not a classroom demonstration instruction.
这是试题解释，不是演示实验操作指引。
The three linked steps follow the official June2024 scheme.
三个连续步骤依据2024年6月官方评分。
Keep the neutral-object distinction in mind when interpreting attraction.
解释吸引时，要记住中性物体也可能被吸引。
A gap is not an absolute barrier.
空气间隙并非绝对屏障。
Use the potential difference and strong field to explain the brief conducting path through air.
用电势差和强电场解释空气中短暂形成的导电通路。
The June2022 scheme accepts strong-field breakdown of air and then electron flow from rod to conductor.
2022年6月评分接受强电场使空气击穿，再说明电子从棒流向导体。
The earthed conductor is not simply called positive.
不能简单说接地导体带正电。
This joins static electricity to energy transfer.
这题把静电与能量转移联系起来。
Choose E equals QV and convert microcoulombs before calculation.
选用E等于QV，计算前把微库仑换算成库仑。
Keep the equation, rearrangement, conversion and substitution visible.
保留公式、变形、单位换算和代入步骤。
The official answer is three hundred thousand volts.
官方答案是三十万伏。
Evaluate the claim without treating this as practical advice.
判断这句话，不把它当作实际操作建议。
The overhead supply remains at high potential even if the nearby cable is earthed.
附近电缆即使接地，高压线仍处于高电势。
The old worksheet conclusion was false.
原练习结论错误。
An earthed object can be the destination of a spark, exactly as in the actual exam question.
接地物体仍可以成为火花放电的终点，正如真题所示。
Use this as retrieval rather than a list of things to recognise.
把它当作提取练习，不是看着清单觉得熟悉。
Explain each decision before checking the next frame.
核对下一页前，解释每一步。
If you used different charges for the two lamps, revisit the distinction between charge flow and energy transfer. Check both meter connections.
如果两个灯的电荷量不同，要复习电荷流动与能量转移的区别，同时核对仪表连接。
Retrieve the causal chains.
提取因果链。
A formula or a key word alone is not a complete explanation.
只写一个公式或关键词，并不能构成完整解释。
Use any error to choose the sheet to revisit: components, mains, Grid or static electricity.
根据错误选择复习元件、市电、电网或静电练习。
For RP4, also redraw its circuit and all three I–V shapes from memory.
对于必做实验4，还应凭记忆重画电路和三个I–V图像。