4.2 Electricity — revision question pack

12 specification points · notes, questions, answers and worked methods

Checked against AQA 8463 section 4.2. Review basis: the qualification registry sourced from the AQA GCSE Physics (8463) specification; registry verification recorded 17 July 2026.

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4.2.1.1 · Standard circuit diagram symbols

Explanation

Circuit diagrams use standard symbols so that the components and their connections are unambiguous. Follow the conducting lines to decide which components are in the same loop and where branches begin.
An ammeter measures the current through a component, so connect it in series with that component.
A voltmeter measures the potential difference between two points, so connect it in parallel across the component.
A working measuring circuit also needs a source of potential difference and a closed conducting path.
The examiner expects recognisable standard symbols joined by lines, not pictures of apparatus.

A measuring circuit with the ammeter in series and voltmeter in parallel across the resistor.

Worked example

Describe how to connect meters to measure the current through and potential difference across a lamp.

1.Place the ammeter in the same series path as the lamp.
2.Connect the voltmeter between the two terminals of the lamp, forming a parallel branch.

Answer: The ammeter is in series with the lamp and the voltmeter is in parallel across it.

Common mistakes

Don't fall into the trap of connecting the ammeter in parallel across the resistor.
Don't fall into the trap of putting the voltmeter in the main series loop instead of across the component.
Don't fall into the trap of drawing pictures of bulbs and cells rather than their standard circuit symbols.

Exam tip

For a ‘draw the circuit’ question, trace the main loop first, then add the voltmeter branch across the measured component.

Tier 1 · Easy

Draw a circuit diagram containing one cell, one open switch and one lamp, all connected in a single loop.
[3 marks]
Total for this question: 3
A circuit diagram shows component X in the single main loop beside a lamp. Component Y is on a separate branch connected across the lamp. X measures the lamp current and Y measures the potential difference across it. Identify X and Y.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Draw a circuit that can measure both the current through a fixed resistor and the potential difference across it. Include a battery and a switch.
[4 marks]
Total for this question: 4
A circuit for testing a motor already contains a cell, closed switch, motor and ammeter in one loop. Choose one modification that allows the potential difference across the motor to be measured. State how the new component must be connected and explain why.
[3 marks]
Total for this question: 3
Circuit A has a cell, closed switch, lamp and motor connected in one loop. Circuit B has the same four standard symbols in one loop, but the lamp and motor appear in the opposite order on the page. State whether the circuits are electrically equivalent. Justify your answer.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A student draws a resistance-measuring circuit with the ammeter connected across the resistor and the voltmeter inserted in the main loop. State both corrections and describe the corrected circuit.
[4 marks]
Total for this question: 4
A student wants one switch to disconnect two lamps at once. The lamps are on separate parallel branches, but the switch is placed in only one lamp branch. Describe how the circuit diagram should be modified and explain how the new switch position controls both branches.
[4 marks]
Total for this question: 4
Two identical lamps are connected on parallel branches across a cell that maintains a constant potential difference. Ammeter A is in the main line, ammeter B is in the first lamp branch, and a voltmeter is across the second lamp. State what happens to each meter reading and to the brightness of the second lamp when the first lamp is removed. Explain your answer.
[4 marks]
Total for this question: 4
A circuit diagram contains four labelled standard symbols. W is a small rectangle with a line through it, X is a resistor rectangle crossed by a diagonal arrow, Y is a diode symbol with two arrows pointing away from it, and Z is a resistor rectangle crossed by a diagonal line with a short bar. The cell and W are on the main line. After W, the circuit splits: X and Y are together on the upper branch, while Z is alone on the lower branch. State what W, X, Y and Z represent. State which components are in the same complete loop as X and which are in the same complete loop as Z.
[5 marks]
Total for this question: 5

4.2.1.2 · Electrical charge and current

Explanation

Electric current is the rate of flow of electrical charge.
Charge flow, current and time are linked by $Q=It$ , where $Q$ is in coulombs, $I$ is in amperes and $t$ is in seconds.
A source of potential difference and a closed circuit are required for charge to flow.
At every point in a single closed loop, the current has the same value because charge cannot accumulate at one point.
Current is not the amount of charge present; it tells you how many coulombs pass a point each second.

Worked example

$240\,\text{C}$ passes a point in $3.0$ minutes. Calculate the current.

1.Convert the time: $3.0\,\text{min}=180\,\text{s}$ .
2.Rearrange $Q=It$ to $I=\dfrac{Q}{t}$ .
3. $I=\dfrac{240}{180}=1.33\,\text{A}$ .

Answer: $1.3\,\text{A}$ to two significant figures

Common mistakes

Don't fall into the trap of using $3.0$ directly as the time when the equation requires seconds.
Don't fall into the trap of treating current as a stored quantity of charge rather than charge flow per second.
Don't fall into the trap of claiming current is used up as charge travels around a single loop.

Exam tip

For $Q=It$ , convert time to seconds before substituting and give charge in coulombs.

Tier 1 · Easy

A current of $0.35\,\text{A}$ flows for $40\,\text{s}$ . Calculate the charge that flows.
[2 marks]
Total for this question: 2
Sensor A records $18\,\text{C}$ passing in $6.0\,\text{s}$ . Sensor B records $20\,\text{C}$ passing in $10\,\text{s}$ . Identify the sensor with the larger current and calculate that current.
[2 marks]
Total for this question: 2

Tier 2 · Standard

$240\,\text{C}$ of charge passes a point in a circuit in $3.0$ minutes. Calculate the current.
[3 marks]
Total for this question: 3
A charge sensor gives cumulative readings of $14\,\text{C}$ at $20\,\text{s}$ and $38\,\text{C}$ at $50\,\text{s}$ . Calculate the mean current during this interval.
[3 marks]
Total for this question: 3
A logger table contains three records: P: $0.40\,\text{A}$ for $30\,\text{s}$ transfers $12\,\text{C}$ ; Q: $0.25\,\text{A}$ for $80\,\text{s}$ transfers $24\,\text{C}$ ; R: $1.2\,\text{A}$ for $15\,\text{s}$ transfers $18\,\text{C}$ . Identify the inconsistent record and correct its charge value.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A device carries $0.75\,\text{A}$ for $80\,\text{s}$ and then $0.30\,\text{A}$ for a further $150\,\text{s}$ . Determine the total charge transferred.
[4 marks]
Total for this question: 4
A pump transfers a total charge of $390\,\text{C}$ . It draws $1.5\,\text{A}$ for the first $2.0$ minutes, then a constant unknown current for the next $3.5$ minutes. Calculate the unknown current.
[4 marks]
Total for this question: 4
A controller repeats the same one-minute cycle: it draws $0.60\,\text{A}$ for $20\,\text{s}$ and $0.15\,\text{A}$ for the remaining $40\,\text{s}$ . The total charge transferred is $540\,\text{C}$ . Calculate the number of complete cycles and the operating time in minutes.
[5 marks]
Total for this question: 5
A portable radio draws $0.18\,\text{A}$ for $25$ minutes and then $0.040\,\text{A}$ in standby for $3.0$ hours. Calculate the total charge transferred and the mean current over the whole operating time.
[5 marks]
Total for this question: 5
A cordless vacuum operates for $6.0$ minutes. Its indicator draws $0.20\,\text{A}$ throughout, while its motor draws an additional $4.0\,\text{A}$ only when cleaning. The total charge transferred is $1056\,\text{C}$ . Calculate how long the motor runs, in seconds and minutes.
[5 marks]
Total for this question: 5

4.2.1.3 · Current, resistance and potential difference

Explanation

The current through a component depends on its resistance and the potential difference across it. Use $V=IR$ , where potential difference is in volts, current in amperes and resistance in ohms.
For a fixed potential difference, increasing resistance decreases current.
Measure current with an ammeter in series and potential difference with a voltmeter in parallel; corresponding readings give $R=V/I$ .
Potential difference describes energy transferred per unit charge, while resistance describes how strongly the component opposes current.
Use the values for the same component and the same operating condition.

Worked example

A $15\,\Omega$ resistor has a potential difference of $6.0\,\text{V}$ across it. Calculate the current.

1.Rearrange $V=IR$ to $I=\dfrac{V}{R}$ .
2. $I=\dfrac{6.0}{15}=0.40\,\text{A}$ .

Answer: $0.40\,\text{A}$

Common mistakes

Don't fall into the trap of calculating resistance as $I/V$ instead of $V/I$ .
Don't fall into the trap of using a voltmeter reading from a different component from the ammeter reading.
Don't fall into the trap of saying greater resistance gives greater current at fixed potential difference.

Exam tip

Write $V=IR$ first and rearrange symbolically before substituting the values.

Tier 1 · Easy

A current of $0.50\,\text{A}$ passes through an $8.0\,\Omega$ resistor. Calculate the potential difference across the resistor.
[2 marks]
Total for this question: 2
A voltmeter reads $4.8\,\text{V}$ across a component while an ammeter in series reads $0.30\,\text{A}$ . A results table has a blank resistance entry. Calculate the missing value.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A component carries $0.24\,\text{A}$ when the potential difference across it is $3.6\,\text{V}$ . Its resistance remains constant. Calculate its resistance and the current when the potential difference is $9.0\,\text{V}$ .
[4 marks]
Total for this question: 4
A student investigates how the resistance of a wire depends on its length while keeping its temperature constant. For lengths $20$ , $40$ , $60$ and $80\,\text{cm}$ , the measured resistances are $0.82$ , $1.61$ , $3.70$ and $3.22\,\Omega$ . Identify the anomalous result, state what the student should do about it, and describe the relationship supported by the other readings.
[4 marks]
Total for this question: 4
A student puts $2.0\,\text{V}$ across wire A and $4.0\,\text{V}$ across wire B, then claims B has the greater resistance because its potential difference is larger. Explain why the evidence is insufficient and state the measurements and calculation needed for a valid comparison.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A $0.80\,\text{m}$ uniform wire at constant temperature has $3.0\,\text{V}$ across it and carries $0.25\,\text{A}$ . The wire is replaced by $0.50\,\text{m}$ of the same wire. Assume resistance is proportional to length. Calculate the new resistance and current at $3.0\,\text{V}$ .
[5 marks]
Total for this question: 5
A graph of measured resistance against wire length has gradient $0.040\,\Omega/\text{cm}$ . Its $0.30\,\Omega$ vertical intercept is caused by the connecting leads. Calculate the wire length needed for the wire itself to have resistance $2.4\,\Omega$ , and explain why the intercept is not included.
[4 marks]
Total for this question: 4
Before a circuit is connected, its ammeter reads $+0.05\,\text{A}$ when the current should be zero. At a fixed potential difference of $3.0\,\text{V}$ , repeated ammeter readings are $0.66\,\text{A}$ , $0.64\,\text{A}$ and $0.68\,\text{A}$ . Explain why the zero error is systematic but the spread is random. Correct the readings, then calculate the mean corrected current and the resistance.
[5 marks]
Total for this question: 5
A $12\,\text{V}$ alarm buzzer has a resistance of $28\,\Omega$ . Its current must not exceed $0.30\,\text{A}$ . Calculate the current without an extra resistor, determine the minimum total circuit resistance, and calculate the resistance that must be added in series.
[5 marks]
Total for this question: 5
A $12\,\text{V}$ tyre inflator contains a $3.0\,\Omega$ motor and is connected by two leads, each with resistance $0.50\,\Omega$ . Calculate the circuit current, the potential difference across the motor, and the potential difference across each lead. Give the sum of these three potential differences.
[6 marks]
Total for this question: 6

4.2.1.4 · Resistors

Explanation

At constant temperature, an ohmic conductor has current directly proportional to potential difference, so its resistance is constant and its current–potential difference graph is a straight line through the origin.
A filament lamp heats as current increases, so its resistance rises and the graph becomes less steep.
A diode conducts readily in one direction but has very high resistance in reverse.
Thermistor resistance decreases as temperature increases, while LDR resistance decreases as light intensity increases.
Investigate a component using an ammeter in series, a voltmeter in parallel and paired readings as the supply is varied.

Characteristic current–potential difference graphs for a resistor, filament lamp and diode.

Worked example

Explain why a filament lamp's current–potential difference graph becomes less steep at larger currents.

1.A larger current heats the filament to a higher temperature.
2.The hotter filament has greater resistance.
3.Current therefore rises by less for each further increase in potential difference.

Answer: Heating increases the filament's resistance, so the graph's gradient decreases.

Common mistakes

Don't fall into the trap of calling a curved lamp graph an experimental error instead of linking it to heating and rising resistance.
Don't fall into the trap of saying thermistor resistance increases when temperature increases.
Don't fall into the trap of drawing substantial reverse current for a diode.

Exam tip

When explaining a characteristic graph, describe how its gradient changes and link that to resistance.

Tier 1 · Easy

An automatic garden light must switch on when it becomes dark. State how the resistance of its LDR changes as darkness increases.
[1 mark]
Total for this question: 1
A cold-store alarm needs a sensor whose resistance rises when the temperature falls. Choose a thermistor or an LDR for the sensor and justify the choice.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A filament lamp carries $0.40\,\text{A}$ at $2.0\,\text{V}$ and $0.80\,\text{A}$ at $6.0\,\text{V}$ . Calculate its resistance at each potential difference and explain the change.
[4 marks]
Total for this question: 4
A student is investigating the current–potential difference characteristic of a diode. In the circuit, the diode is connected in the reverse direction and the variable resistor is never adjusted. State the change needed to the diode connection and to the variable-resistor setting, and state the two readings taken at each setting.
[4 marks]
Total for this question: 4
Component P gives these current readings: $+0.20\,\text{A}$ at $+1.0\,\text{V}$ , $+0.40\,\text{A}$ at $+2.0\,\text{V}$ , $-0.20\,\text{A}$ at $-1.0\,\text{V}$ and $-0.40\,\text{A}$ at $-2.0\,\text{V}$ . Choose P from: ohmic conductor, filament lamp or diode. Justify your choice by calculating its resistance at both potential-difference magnitudes.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Describe an investigation of how the resistance of a thermistor changes with temperature. Include the circuit, measurements, one control and the processing of results.
[6 marks]
Total for this question: 6
A thermistor is connected across a constant $6.0\,\text{V}$ supply. Its resistance is $3.0\,\text{k}\Omega$ at $10\,^{\circ}\text{C}$ and $0.75\,\text{k}\Omega$ at $50\,^{\circ}\text{C}$ . Calculate the current at each temperature in milliamperes and explain the change.
[5 marks]
Total for this question: 5
An LDR and a $1.0\,\text{k}\Omega$ fixed resistor are connected in series across a $9.0\,\text{V}$ supply. The measured current is $3.0\,\text{mA}$ . Calibration data give LDR resistances of $0.40\,\text{k}\Omega$ in bright light, $2.0\,\text{k}\Omega$ in ordinary indoor light and $8.0\,\text{k}\Omega$ in darkness. Calculate the LDR resistance and identify the light condition.
[4 marks]
Total for this question: 4
An LDR calibration table gives these pairs of light-meter reading and resistance: $100\,\text{lux}$ , $8.0\,\text{k}\Omega$ ; $300\,\text{lux}$ , $4.0\,\text{k}\Omega$ ; $500\,\text{lux}$ , $2.0\,\text{k}\Omega$ . State the trend. At $400\,\text{lux}$ , choose the only prediction consistent with the table: $1.0$ , $3.0$ , $5.0$ or $9.0\,\text{k}\Omega$ . Explain why this LDR calibration cannot be used to identify a temperature change and name the component that should be used instead.
[5 marks]
Total for this question: 5

4.2.2 · Series and parallel circuits

Explanation

In a series circuit, current is the same through every component, the supply potential difference is shared, and resistances add: $R_{\text{total}}=R_1+R_2+\ldots$ .
In a parallel circuit, the potential difference is the same across every branch and the total current equals the sum of the branch currents.
Adding a parallel branch gives charge another path, so total resistance decreases and is less than the smallest branch resistance.
Use junctions to decide whether components share one path or sit on separate branches; do not apply series rules just because symbols are drawn next to each other.

Two resistors in series and two resistors on parallel branches.

Worked example

Two series resistors are $8\,\Omega$ and $12\,\Omega$ . The current is $0.30\,\text{A}$ . Calculate the total resistance and supply potential difference.

1. $R_{\text{total}}=8+12=20\,\Omega$ .
2. $V=IR=0.30\times20=6.0\,\text{V}$ .

Answer: $20\,\Omega$ and $6.0\,\text{V}$

Common mistakes

Don't fall into the trap of adding branch currents and then claiming that same total flows through each parallel branch.
Don't fall into the trap of adding parallel resistances as though the components were in series.
Don't fall into the trap of sharing supply potential difference between parallel branches.

Exam tip

Mark junctions first: current splits at a junction, while potential difference is common across branches.

Tier 1 · Easy

Two resistors of $4\,\Omega$ and $7\,\Omega$ are connected in series. Calculate their total resistance.
[1 mark]
Total for this question: 1
Two branches are connected in parallel. The current entering the junction is $0.86\,\text{A}$ and one branch carries $0.32\,\text{A}$ . Determine the ammeter reading in the other branch.
[1 mark]
Total for this question: 1

Tier 2 · Standard

A $2.0\,\Omega$ resistor and a $4.0\,\Omega$ resistor are connected in series to a $12\,\text{V}$ supply. Calculate the circuit current and the potential difference across each resistor.
[5 marks]
Total for this question: 5
Two lamps are connected in series, so removing either lamp turns both off. Choose a circuit modification that lets either lamp remain lit when the other is removed, and justify the new arrangement using potential difference.
[3 marks]
Total for this question: 3
A circuit contains one resistor. Predict and explain how the total resistance changes when an identical resistor is added first in series and then, in a separate circuit, in parallel with the original resistor.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Two branches are connected in parallel across a $12\,\text{V}$ supply. One branch contains a $6.0\,\Omega$ resistor. The other branch carries $3.0\,\text{A}$ . Calculate the resistance in the second branch and the current from the supply.
[5 marks]
Total for this question: 5
A $9.0\,\text{V}$ supply drives a $0.75\,\text{A}$ current through resistor R₀ before the circuit splits into two parallel branches. The potential difference across R₀ is $3.0\,\text{V}$ and one branch carries $0.25\,\text{A}$ . Calculate the resistance of R₀, the current in the other branch and the resistance of that branch.
[6 marks]
Total for this question: 6
Two identical resistors are tested in two configurations using a $6.0\,\text{V}$ supply. Configuration P draws $0.50\,\text{A}$ and configuration Q draws $2.0\,\text{A}$ . Calculate each total resistance, determine the resistance of one resistor, and identify which configuration is series and which is parallel.
[5 marks]
Total for this question: 5
A $12\,\text{V}$ supply is connected across two parallel branches. Branch A contains $4.0\,\Omega$ and $8.0\,\Omega$ resistors in series. Branch B contains one $6.0\,\Omega$ resistor. Calculate the current in each branch, the total supply current, and the potential difference across the $8.0\,\Omega$ resistor.
[6 marks]
Total for this question: 6
Describe how to construct and check circuits containing two unequal resistors first in series and then in parallel. Include where to place ammeters and voltmeters, the readings that should match in each arrangement, and how the measurements show the difference between the arrangements.
[6 marks]
Total for this question: 6

4.2.3.1 · Direct and alternating potential difference

Explanation

A direct potential difference has one polarity, so it drives current in one direction; cells and batteries provide dc.
An alternating potential difference repeatedly reverses polarity, so the current also repeatedly reverses direction.
The UK mains supply is approximately $230\,\text{V}$ ac at $50\,\text{Hz}$ .
A frequency of $50\,\text{Hz}$ means $50$ complete cycles each second, with a period of $1/50=0.020\,\text{s}$ .
On a potential difference–time graph, dc stays on one side of zero, whereas ac crosses zero and alternates between positive and negative values.

Direct and alternating potential difference plotted against time.

Worked example

The mains frequency is $50\,\text{Hz}$ . Calculate the duration of one complete cycle.

1.Period $T=\dfrac{1}{f}$ .
2. $T=\dfrac{1}{50}=0.020\,\text{s}$ .

Answer: $0.020\,\text{s}$

Common mistakes

Don't fall into the trap of saying alternating potential difference only changes size and never reverses polarity.
Don't fall into the trap of interpreting $50\,\text{Hz}$ as one cycle lasting $50$ seconds.
Don't fall into the trap of drawing a dc trace that alternates above and below zero.

Exam tip

A comparison must mention direction: dc keeps one polarity, while ac repeatedly reverses.

Tier 1 · Easy

State whether a battery supplies direct or alternating potential difference, and give the defining feature of that supply.
[2 marks]
Total for this question: 2
A data logger records the potential difference from a source as $+6$ , $+3$ , $0$ , $-3$ , $-6$ , $-3$ , $0$ , $+3$ and $+6\,\text{V}$ at equal time intervals. Identify the supply type and give the evidence.
[2 marks]
Total for this question: 2

Tier 2 · Standard

The UK mains supply has a frequency of $50\,\text{Hz}$ . Calculate the number of complete cycles in $0.30\,\text{s}$ and describe what happens to the polarity.
[3 marks]
Total for this question: 3
An oscilloscope trace shows one complete cycle lasting $0.025\,\text{s}$ . Determine the frequency and state whether the supply could be UK mains.
[3 marks]
Total for this question: 3
Two indicator branches contain diodes facing in opposite directions. With source X, only indicator A stays lit. With source Y, A and B light alternately. Identify each source as direct or alternating and explain the observations using polarity.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Source A maintains a potential difference of $+6.0\,\text{V}$ . Source B varies between positive and negative values and completes $25$ cycles in $0.50\,\text{s}$ . Identify each supply type, calculate the frequency of B and compare the current directions they produce.
[5 marks]
Total for this question: 5
A technician counts 30 polarity reversals in $0.24\,\text{s}$ . Each complete cycle contains two reversals. Calculate the supply frequency and state whether the source is a $50\,\text{Hz}$ UK mains supply.
[4 marks]
Total for this question: 4
Three sources are monitored with the same terminal labelled positive on the logger. P remains at $+6.0\,\text{V}$ . Q changes repeatedly between $+2.0$ and $+5.0\,\text{V}$ . R changes through $+4.0$ , $0$ and $-4.0\,\text{V}$ . State whether each source is direct or alternating and correct the claim that every changing potential difference is alternating.
[5 marks]
Total for this question: 5
An alternating supply has a frequency of $25\,\text{Hz}$ . Calculate its period and the number of complete cycles produced in $0.20\,\text{s}$ . State what the period represents.
[4 marks]
Total for this question: 4
Four supplies are described. A is rated at $230\,\text{V}$ and $50\,\text{Hz}$ but keeps the same polarity. B is rated at $230\,\text{V}$ , completes $50$ cycles each second and repeatedly reverses polarity. C is rated at $110\,\text{V}$ , completes $50$ cycles each second and repeatedly reverses polarity. D is rated at $230\,\text{V}$ , completes $60$ cycles each second and repeatedly reverses polarity. Identify the supply that could be UK mains. Justify your choice using all three relevant properties.
[4 marks]
Total for this question: 4

4.2.3.2 · Mains electricity

Explanation

A three-core mains cable contains a brown live wire, blue neutral wire and green-and-yellow earth wire.
The live wire carries the alternating potential difference from the supply; the neutral completes the circuit and is close to $0\,\text{V}$ .
The earth is a safety wire connected to a metal case and normally carries no current.
If a fault makes the case live, a large current flows through the low-resistance earth path so the fuse melts or circuit breaker opens.
The live wire is dangerous even when a switch is open, so switches and fuses must be placed in the live wire.

The three insulated conductors inside a mains cable.

Worked example

Explain how the earth wire and fuse protect a person if a fault connects the live wire to a metal appliance case.

1.The earth wire provides a low-resistance path from the case to Earth.
2.A large fault current flows through the live wire, case, earth wire and fuse.
3.The fuse melts and disconnects the live supply.

Answer: The case is earthed and the large fault current causes the fuse to break the circuit.

Common mistakes

Don't fall into the trap of swapping the colours of the live and neutral wires.
Don't fall into the trap of saying the earth wire normally carries the appliance current.
Don't fall into the trap of putting a switch or fuse in the neutral wire and leaving the appliance connected to live.

Exam tip

For an electrical-safety explanation, trace the fault current through earth to the fuse or circuit breaker.

Tier 1 · Easy

State the insulation colour of the live, neutral and earth wires in a UK three-core mains cable.
[3 marks]
Total for this question: 3
Fault A exposes the copper of a blue mains wire. Fault B exposes the copper of a brown mains wire. Identify which fault normally presents the greater shock danger and explain using potential difference relative to earth.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A fault makes the metal case of a mains appliance touch the live wire. Explain how the earth wire reduces the danger to a user.
[4 marks]
Total for this question: 4
Three conductors are hidden behind labels X, Y and Z. X is about $230\,\text{V}$ relative to earth and carries current normally. Y is close to $0\,\text{V}$ and completes the normal circuit. Z is at $0\,\text{V}$ and carries current only during a fault. Identify X, Y and Z.
[3 marks]
Total for this question: 3
A mains appliance's switch is open, but damaged insulation exposes part of the brown wire between the plug and the switch. Explain why this wire is dangerous to touch even though the appliance is switched off, and explain the danger of a connection between this wire and earth.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A mains lamp is wired so that its switch opens the neutral wire rather than the live wire. The lamp goes out when the switch is opened. Explain why the lamp holder may still be dangerous to touch.
[4 marks]
Total for this question: 4
Appliance P has exposed live copper. In appliance Q, the live wire touches its metal case, but the earth wire and fuse both work. Compare the two faults by tracing the possible current paths and explaining the likely protective outcome.
[5 marks]
Total for this question: 5
A metal-cased appliance works normally even though its earth connection is broken. Later, a live wire touches the case and the protective device does not disconnect the supply. Explain why the broken earth did not affect normal operation, why the case is now dangerous, and how restoring the earth connection changes the fault current path.
[5 marks]
Total for this question: 5
A $2.3\,\text{kW}$ appliance is connected to the $230\,\text{V}$ mains. Calculate its normal operating current. Choose a $3\,\text{A}$ or $13\,\text{A}$ fuse, explain why the other rating is unsuitable, and state which wire the fuse must be placed in.
[5 marks]
Total for this question: 5
Current measurements are taken from a metal-cased electric iron. In test A, the brown wire carries $4.2\,\text{A}$ , the blue wire carries $4.2\,\text{A}$ , and the green-and-yellow wire carries no current. In test B, the brown and green-and-yellow wires each carry $18\,\text{A}$ while the blue wire carries no current. Identify the normal test and the fault test. Explain both current paths and the expected protective response.
[5 marks]
Total for this question: 5

4.2.4.1 · Power

Explanation

Power is the rate at which energy is transferred, so one watt is one joule per second.
For an electrical component, use $P=VI$ when potential difference and current are known.
Combining this with $V=IR$ gives $P=I^2R$ , useful when current and resistance are known.
A larger power rating means the appliance transfers more energy each second, not that it necessarily transfers more energy overall; total energy also depends on operating time.
Use potential difference in volts, current in amperes and resistance in ohms to obtain power in watts.

Worked example

A heater has resistance $24\,\Omega$ and carries $5.0\,\text{A}$ . Calculate its power.

1.Use $P=I^2R$ because current and resistance are given.
2. $P=5.0^2\times24=25\times24$ .

Answer: $600\,\text{W}$

Common mistakes

Don't fall into the trap of using $P=IR$ and omitting the square on current.
Don't fall into the trap of treating watts as a unit of energy rather than energy transfer per second.
Don't fall into the trap of comparing power ratings without considering how long appliances operate.

Exam tip

Choose the power equation that uses only the given quantities, then show the squared current explicitly.

Tier 1 · Easy

A motor has a potential difference of $12\,\text{V}$ across it and a current of $3.0\,\text{A}$ . Calculate its power.
[2 marks]
Total for this question: 2
Two devices are each connected across $12\,\text{V}$ . Device A draws $0.50\,\text{A}$ and device B draws $1.5\,\text{A}$ . Identify which device transfers energy faster and calculate the factor by which its power is greater.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A heating resistor has resistance $8.0\,\Omega$ and carries $2.5\,\text{A}$ . Calculate the power transferred.
[2 marks]
Total for this question: 2
A test table gives device X a potential difference of $24\,\text{V}$ and current of $1.2\,\text{A}$ , while device Y has $12\,\text{V}$ and $2.0\,\text{A}$ . Calculate both powers and identify the higher-power device.
[3 marks]
Total for this question: 3
An appliance transfers $7200\,\text{J}$ of energy in $180\,\text{s}$ while drawing a current of $2.0\,\text{A}$ . Calculate its power and the potential difference across it. State what the power value means in joules transferred each second.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A mains heating element is rated at $1.8\,\text{kW}$ when connected to $230\,\text{V}$ . Calculate its current and resistance at this operating point.
[5 marks]
Total for this question: 5
A heater has two resistance settings, $52\,\Omega$ and $42\,\Omega$ , on a $230\,\text{V}$ supply protected by a $5.0\,\text{A}$ fuse. Calculate the current and power for each setting, then determine which setting can operate without exceeding the fuse rating.
[5 marks]
Total for this question: 5
A $20\,\Omega$ heating element has a mean power of $320\,\text{W}$ for $4.0$ minutes. Calculate the energy transferred, current and potential difference. Give the current and potential difference to two significant figures.
[5 marks]
Total for this question: 5
An $18\,\text{V}$ cordless drill draws $4.0\,\text{A}$ while turning freely and $12\,\text{A}$ when briefly stalled. Calculate its electrical input power in each condition. Calculate the electrical energy transferred from the battery in $5.0\,\text{s}$ in each condition, then compare the rates of electrical energy transfer.
[6 marks]
Total for this question: 6
A low-voltage heated blanket has two heating sections connected in series to a $24\,\text{V}$ supply. The current is $3.0\,\text{A}$ and section A transfers power at $18\,\text{W}$ . Calculate the potential difference and resistance of section A, then calculate the potential difference, resistance and power of section B.
[6 marks]
Total for this question: 6

4.2.4.2 · Energy transfers in everyday appliances

Explanation

Electrical appliances transfer energy from the mains or a battery to other energy stores. A motor increases a kinetic energy store, while a heater increases a thermal energy store.
Calculate transferred energy using $E=Pt$ , with power in watts and time in seconds for energy in joules.
The alternative equation $E=QV$ links energy transferred to charge flow and potential difference.
Appliance efficiency depends on how much of the input energy reaches the intended store; unwanted transfers usually heat the surroundings.
A high-power appliance transfers energy quickly, but energy used also depends on its operating time.

Worked example

A $1.8\,\text{kW}$ kettle operates for $150\,\text{s}$ . Calculate the energy transferred.

1.Convert power: $1.8\,\text{kW}=1800\,\text{W}$ .
2. $E=Pt=1800\times150$ .

Answer: $270000\,\text{J}$ or $270\,\text{kJ}$

Common mistakes

Don't fall into the trap of using $1.8$ as the power without converting kilowatts to watts.
Don't fall into the trap of treating a power rating as the total energy transferred.
Don't fall into the trap of naming thermal energy as useful for every appliance, including a motor.

Exam tip

For $E=Pt$ , convert kilowatts to watts and minutes to seconds before calculating joules.

Tier 1 · Easy

A $60\,\text{W}$ fan runs for $5.0$ minutes. Calculate the energy transferred.
[3 marks]
Total for this question: 3
Two appliances are both rated at $800\,\text{W}$ . Appliance A runs for three minutes and appliance B for five minutes. Without calculating, identify which transfers more energy and explain why.
[2 marks]
Total for this question: 2

Tier 2 · Standard

$9000\,\text{C}$ of charge flows through a $12\,\text{V}$ motor. Calculate the energy transferred and name the main useful energy transfer.
[3 marks]
Total for this question: 3
A mixer transfers $84\,\text{kJ}$ of energy in $7.0$ minutes. Calculate its average power in watts and name the main useful energy store increased by the motor.
[4 marks]
Total for this question: 4
Three appliances have these operating data: kettle, $1800\,\text{W}$ for $80\,\text{s}$ ; microwave, $900\,\text{W}$ for $240\,\text{s}$ ; television, $120\,\text{W}$ for $1500\,\text{s}$ . Calculate the energy transferred by each appliance and give the appliances in order of energy transferred, starting with the greatest.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A $1.5\,\text{kW}$ appliance operates for $18$ minutes from a $230\,\text{V}$ supply. Calculate the energy transferred and the charge that flows through the appliance.
[5 marks]
Total for this question: 5
A battery transfers $216\,\text{kJ}$ of energy through a potential difference of $12\,\text{V}$ while a motor runs for $30$ minutes. Calculate the charge flow and the average current.
[5 marks]
Total for this question: 5
A heater runs for $10$ minutes. One logger records a charge flow of $4200\,\text{C}$ through a potential difference of $24\,\text{V}$ . A second logger records a constant power of $168\,\text{W}$ . Calculate the energy transferred from each logger's data and determine whether the records are consistent.
[5 marks]
Total for this question: 5
Appliance A transfers energy at power $P$ for a time $t$ . Appliance B transfers energy at twice the power for half the time. State how the energy transferred by B compares with the energy transferred by A. Explain your answer using an equation.
[3 marks]
Total for this question: 3

4.2.4.3 · The National Grid

Explanation

The National Grid is the network of cables and transformers linking power stations to consumers. A step-up transformer increases potential difference before transmission.
For a given power, $P=VI$ shows that a higher potential difference allows a smaller current.
Cable heating depends on $P_{\text{loss}}=I^2R$ , so reducing current greatly reduces energy wasted heating the cables.
Near consumers, step-down transformers reduce the potential difference to safer, useful values.
Transformers do not create energy: they change the potential difference and current so electrical energy can be transferred efficiently over long distances.

Potential difference is stepped up for transmission and stepped down before consumers.

Worked example

A transmission line carries $2.0\,\text{MW}$ at $200\,\text{kV}$ . Calculate the current.

1.Convert prefixes: $P=2.0\times10^6\,\text{W}$ and $V=2.00\times10^5\,\text{V}$ .
2. $I=\dfrac{P}{V}=\dfrac{2.0\times10^6}{2.00\times10^5}$ .

Answer: $10\,\text{A}$

Common mistakes

Don't fall into the trap of saying a step-up transformer increases power or creates energy.
Don't fall into the trap of claiming a high transmission current reduces cable heating.
Don't fall into the trap of forgetting that the step-down transformer is used before electricity reaches consumers.

Exam tip

A full efficiency explanation links higher potential difference to lower current and then to smaller $I^2R$ losses.

Tier 1 · Easy

Name the transformer used before long-distance transmission and the transformer used before electricity enters homes.
[2 marks]
Total for this question: 2
Two transmission plans transfer the same power through identical cables. Plan A uses a higher potential difference and a smaller current than plan B. Choose the plan with the smaller cable-heating loss and justify the choice.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A transmission line transfers $6.0\,\text{MW}$ at $300\,\text{kV}$ . Calculate the current in the line and explain why this is preferable to transmitting the same power at a much lower potential difference.
[4 marks]
Total for this question: 4
A transmission table states that line A carries $2.4\,\text{MW}$ at $120\,\text{kV}$ with a current of $20\,\text{A}$ , while line B carries $3.0\,\text{MW}$ at $100\,\text{kV}$ with a current of $45\,\text{A}$ . Identify the inconsistent row and calculate its correct current.
[3 marks]
Total for this question: 3
A student claims, ‘A step-up transformer increases the potential difference, so it increases the power delivered.’ Evaluate this claim using $P=VI$ and conservation of energy.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A cable of resistance $0.80\,\Omega$ transfers $2.4\,\text{MW}$ . Compare the power lost in the cable when transmission is at $12\,\text{kV}$ and at $240\,\text{kV}$ .
[6 marks]
Total for this question: 6
A grid link transfers $18\,\text{MW}$ through cables with total resistance $3.2\,\Omega$ . The cable-heating loss must be no more than $1.0\%$ of the transferred power. Calculate the minimum transmission potential difference. Give the answer in kilovolts to two significant figures.
[6 marks]
Total for this question: 6
A grid link transfers $8.00\,\text{MW}$ at $200\,\text{kV}$ . The power delivered beyond the cables is $7.90\,\text{MW}$ . Calculate the transmission current, the power lost in the cables and the total cable resistance.
[5 marks]
Total for this question: 5
A transmission cable transfers $9.0\,\text{MW}$ at $150\,\text{kV}$ and has resistance $50\,\Omega$ . Calculate the cable current, the power wasted by heating, and the energy wasted during $2.0$ hours. Calculate the percentage of the transferred power that is wasted.
[6 marks]
Total for this question: 6
Grid route A receives $10.0\,\text{MW}$ and loses $0.20\,\text{MW}$ in its cables. Route B receives $4.0\,\text{MW}$ and also loses $0.20\,\text{MW}$ . Calculate the power delivered and the percentage power loss for each route. Determine which route is more efficient and explain why comparing only the loss values in megawatts gives the wrong conclusion.
[5 marks]
Total for this question: 5

4.2.5.1 · Static charge (physics only)

Explanation

When certain insulating materials are rubbed together, electrons transfer from one material to the other. The material gaining electrons becomes negatively charged; the material losing electrons is left with an equal positive charge.
Protons do not move between the materials.
Like charges repel and unlike charges attract through non-contact electrostatic forces.
Because charge cannot easily flow away from an insulator, it may build up until a spark transfers charge through the air.
Static discharge can be useful, but it can also ignite flammable vapours or damage electronic components.

Worked example

A cloth loses electrons when rubbed against a plastic rod. State the charge on each material.

1.The rod gains electrons, so it has an excess of negative charge.
2.The cloth loses electrons, so it has an electron deficit and is positive.

Answer: The plastic rod becomes negative and the cloth becomes equally positive.

Common mistakes

Don't fall into the trap of saying positive charge particles move from the rod to the cloth.
Don't fall into the trap of reversing the signs so the object that gains electrons is called positive.
Don't fall into the trap of describing attraction between like charges.

Exam tip

Always state which material gains electrons before assigning the two charge signs.

Tier 1 · Easy

During rubbing, material X gains electrons from material Y. State the charge left on each material.
[2 marks]
Total for this question: 2
A balloon is rubbed on a jumper and becomes negatively charged. Name the particles that moved between them, and explain why the jumper is left with an equal positive charge.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A negatively charged insulating strip repels a second charged strip without touching it. What does this show about the charge on the second strip and the type of force involved?
[3 marks]
Total for this question: 3
An unknown charged strip repels a negatively charged probe and attracts an uncharged suspended ball. Determine the strip charge and identify which observation gives conclusive evidence for the sign.
[3 marks]
Total for this question: 3
All four insulating strips A, B, C and D are known to be charged. D is positive. C repels D, B attracts C, and A repels B. Determine the sign of the charge on A, B and C.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A student rubs an insulating rod with a cloth and the rod becomes negatively charged. The rod is then brought close to a metal object and a spark occurs. Explain the charging and the spark in terms of electrons.
[5 marks]
Total for this question: 5
Insulating sheet A loses electrons to insulating sheet B when they are rubbed together. Predict both charge signs and the force between the sheets. A spark then removes some excess electrons from B, but B remains charged. Predict how the force changes and explain why.
[5 marks]
Total for this question: 5
A negatively charged metal rod has excess electrons and is supported by an insulating handle. A person first touches the charged rod with a bare hand, providing a path to earth. A second identical charged rod is moved only by its insulating handle. State what happens to the charge on each rod and explain why, including the direction in which electrons move.
[5 marks]
Total for this question: 5
Describe an investigation using suspended insulating rods to demonstrate both electrostatic repulsion and attraction without contact. State how to charge the rods, the observations required, and how electron transfer explains the charge signs and forces.
[6 marks]
Total for this question: 6
Three neutral insulating strips A, B and C are used in two rubbing steps. Six equal groups of electrons transfer from A to B. B then transfers two of these groups to C. Determine the final charge sign and relative charge amount on each strip, predict the force between A and B and between B and C, and explain how charge is conserved.
[6 marks]
Total for this question: 6

4.2.5.2 · Electric fields (physics only)

Explanation

A charged object creates an electric field around itself. Another charged object placed in that field experiences a force; the interaction is stronger where the field is stronger.
Field lines show the direction of force on a positive test charge.
Around an isolated positive sphere they point radially outwards, while around a negative sphere they point radially inwards.
Lines drawn closer together represent a stronger field, so the field is strongest near the charged object.
Field lines never cross because the force on a positive test charge at one point has only one direction.

Electric field directions around isolated positive and negative spheres.

Worked example

Describe the electric field around an isolated negatively charged sphere.

1.A positive test charge would be attracted towards the negative sphere.
2.Draw radial field lines with arrowheads pointing inwards.
3.Place the lines closer together near the sphere to show the stronger field.

Answer: Radial field lines point towards the sphere and are most closely spaced near it.

Common mistakes

Don't fall into the trap of drawing arrows towards a positive sphere or away from a negative sphere.
Don't fall into the trap of allowing electric field lines to cross.
Don't fall into the trap of spacing field lines farther apart near the sphere even though the field is stronger there.

Exam tip

Field arrows show the force direction on a positive test charge: away from positive and towards negative.

Tier 1 · Easy

Draw the electric field pattern around an isolated positively charged sphere.
[2 marks]
Total for this question: 2
A diagram of the electric field around an isolated negatively charged sphere shows arrows pointing away from the sphere and field lines most widely spaced next to the sphere. State the two corrections needed.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A small positive test charge is placed first near a positively charged sphere and then farther away. State the force direction in both positions and compare the force sizes.
[3 marks]
Total for this question: 3
Measurements near a charged sphere give forces of $1.8\,\text{N}$ at $2.0\,\text{cm}$ , $0.46\,\text{N}$ at $4.0\,\text{cm}$ and $0.20\,\text{N}$ at $6.0\,\text{cm}$ . Use the evidence to give a justified range for the force at $3.0\,\text{cm}$ and describe the field-strength trend.
[3 marks]
Total for this question: 3
A diagram of an isolated positively charged sphere marks point P $2\,\text{cm}$ to its left, point Q $2\,\text{cm}$ to its right and point R $5\,\text{cm}$ to its right. State the electric-field direction at each point, explain why the field strength at P equals that at Q, and state the force direction on a negative test charge placed at Q.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A charged metal dome is brought progressively closer to an earthed metal sphere until a spark crosses the gap. Use the electric-field model to explain why there is a force before contact and why a spark becomes more likely as the gap decreases.
[5 marks]
Total for this question: 5
A positively charged sphere A is left of point P, and sphere B is right of P. A positive test charge at P experiences a resultant force to the left, even though A alone would push it to the right. Identify the sign of B, compare the two field effects at P, and predict what happens to the resultant as P moves closer to B.
[5 marks]
Total for this question: 5
Explain why electric field lines never cross and why the field-line pattern around an isolated charged sphere is radial.
[4 marks]
Total for this question: 4

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

4.2.1.1 · Standard circuit diagram symbols

Tier 1 · Easy

Mark scheme for 4.2.1.1 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	A single complete loop using the standard cell, open-switch and lamp symbols, with no extra branches.	Draw one cell symbol, then an open-switch symbol and a lamp symbol in series. Join the components with straight conducting lines to make one loop; the switch contacts remain separated.	3
Total Question 1			3
02.1	X is an ammeter; Y is a voltmeter.	An ammeter measures the current through the lamp, so it is in the same series loop and must be X. A voltmeter measures the potential difference between the lamp terminals, so it is on the parallel branch and must be Y.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.2.1.1 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	Battery, switch, ammeter and resistor in series, with a voltmeter connected in parallel across the resistor; all components shown by standard symbols.	Make the main loop from the battery, switch, ammeter and resistor so the ammeter carries the resistor current. Add a separate branch containing the voltmeter between the two ends of the resistor, so it measures the resistor's potential difference.	4
Total Question 1			4
02.1	Add a voltmeter in parallel across the motor so it measures the potential difference between the motor's terminals.	Choose the standard voltmeter symbol and add a separate branch joining the two sides of the motor. These are the two points whose potential difference is required, so the voltmeter must be connected across them rather than inserted into the main loop.	3
Total Question 2			3
03.1	The circuits are electrically equivalent because every component is in the same single series loop; changing where a component is drawn, or the order of series components, does not change the connections.	Trace the conducting path rather than comparing the positions of symbols. Each diagram contains one closed path through the same cell, switch, lamp and motor, with no junctions. The connectivity is therefore unchanged, so the circuits are equivalent.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.2.1.1 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Move the ammeter into series with the resistor. Connect the voltmeter in parallel across the resistor; the source, switch, ammeter and resistor form the main loop.	Current through a component is measured by making that current pass through the ammeter, so the ammeter belongs in the main series path. Potential difference is measured between the component's two terminals, so the voltmeter must form a parallel branch across the resistor.	4
Total Question 1			4
02.1	Move the switch to the unbranched main line before the junction or after the branches rejoin; opening it then breaks the only supply path to both lamp branches.	A switch within one branch interrupts only that branch. Redraw the switch in series with the supply on a section of wire shared by the two branch currents. When this switch opens, neither branch has a complete path through the source, so both lamps turn off.	4
Total Question 2			4
03.1	A decreases to the current in the second branch, B falls to zero, and the voltmeter reading stays the same. The second lamp stays at the same brightness.	Removing the first lamp opens that branch, so no current passes through B. Ammeter A no longer carries the first-branch current and therefore decreases to the current in the remaining branch. The cell maintains the potential difference across the parallel branches, so the voltmeter reading and the current through, and hence brightness of, the second lamp are unchanged.	4
Total Question 3			4
04.1	W is a fuse, X is a variable resistor, Y is an LED and Z is a thermistor. The complete loop through X also contains the cell, fuse and LED. The complete loop through Z contains the cell, fuse and thermistor; it does not contain X or Y.	Match each description to the standard symbol: fuse, variable resistor, light-emitting diode and thermistor. Then trace one path at a time from one cell terminal back to the other. The upper path goes through the shared fuse, then X and Y; the lower path goes through the shared fuse, then Z. Components on the other parallel branch are not part of the selected loop.	5
Total Question 4			5

4.2.1.2 · Electrical charge and current

Tier 1 · Easy

Mark scheme for 4.2.1.2 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	$14\,\text{C}$	Use $Q=It$ : $Q=(0.35)(40)=14\,\text{C}$ .	2
Total Question 1			2
02.1	Sensor A; its current is $3.0\,\text{A}$ .	Current is charge flow per unit time. Sensor A gives $I=18/6.0=3.0\,\text{A}$ , whereas sensor B gives $20/10=2.0\,\text{A}$ , so A records the larger current.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.2.1.2 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	$1.3\,\text{A}$	Convert the time: $3.0\,\text{min}=180\,\text{s}$ . Rearrange $Q=It$ to $I=Q/t$ , so $I=240/180=1.33\,\text{A}$ , which is $1.3\,\text{A}$ to two significant figures.	3
Total Question 1			3
02.1	$0.80\,\text{A}$	Use the changes between the two readings: $\Delta Q=38-14=24\,\text{C}$ and $\Delta t=50-20=30\,\text{s}$ . Therefore $I=\Delta Q/\Delta t=24/30=0.80\,\text{A}$ .	3
Total Question 2			3
03.1	Record Q is inconsistent; its charge should be $20\,\text{C}$ .	Check each record using $Q=It$ . P gives $(0.40)(30)=12\,\text{C}$ and R gives $(1.2)(15)=18\,\text{C}$ , so both are consistent. Q gives $(0.25)(80)=20\,\text{C}$ , not $24\,\text{C}$ .	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.2.1.2 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	$105\,\text{C}$	Calculate each charge separately. During the first interval, $Q_1=(0.75)(80)=60\,\text{C}$ . During the second, $Q_2=(0.30)(150)=45\,\text{C}$ . Therefore $Q_{\text{total}}=60+45=105\,\text{C}$ .	4
Total Question 1			4
02.1	$1.0\,\text{A}$	Convert the times: $2.0\,\text{min}=120\,\text{s}$ and $3.5\,\text{min}=210\,\text{s}$ . The first interval transfers $Q_1=It=(1.5)(120)=180\,\text{C}$ , leaving $390-180=210\,\text{C}$ . The unknown current is $I=210/210=1.0\,\text{A}$ .	4
Total Question 2			4
03.1	$30$ complete cycles and an operating time of $30$ minutes.	The charge in the first part of a cycle is $(0.60)(20)=12\,\text{C}$ and in the second part it is $(0.15)(40)=6.0\,\text{C}$ . One complete cycle transfers $18\,\text{C}$ , so the number of cycles is $540/18=30$ . Each cycle lasts one minute, giving $30$ minutes in total.	5
Total Question 3			5
04.1	Total charge $=702\,\text{C}$ ; mean current $=0.057\,\text{A}$ to two significant figures.	Convert the times: $25\,\text{min}=1500\,\text{s}$ and $3.0\,\text{h}=10\,800\,\text{s}$ . The charges are $Q_1=(0.18)(1500)=270\,\text{C}$ and $Q_2=(0.040)(10\,800)=432\,\text{C}$ . Thus $Q_{\text{total}}=702\,\text{C}$ over $12\,300\,\text{s}$ , so $I_{\text{mean}}=702/12\,300=0.057073\ldots\,\text{A}=0.057\,\text{A}$ to two significant figures.	5
Total Question 4			5
05.1	The motor runs for $246\,\text{s}$ , or $4.1$ minutes.	The indicator runs for $6.0\,\text{min}=360\,\text{s}$ and transfers $Q=(0.20)(360)=72\,\text{C}$ . The motor therefore transfers $1056-72=984\,\text{C}$ . From $t=Q/I$ , its running time is $984/4.0=246\,\text{s}$ . Dividing by $60$ gives $4.1$ minutes.	5
Total Question 5			5

4.2.1.3 · Current, resistance and potential difference

Tier 1 · Easy

Mark scheme for 4.2.1.3 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	$4.0\,\text{V}$	Use $V=IR$ . Therefore $V=0.50\times8.0=4.0\,\text{V}$ .	2
Total Question 1			2
02.1	$16\,\Omega$	The two meters refer to the same component, so use $R=V/I$ . Hence $R=4.8/0.30=16\,\Omega$ .	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.2.1.3 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	Resistance $=15\,\Omega$ Current $=0.60\,\text{A}$	First use $R=V/I=3.6/0.24=15\,\Omega$ . With the resistance unchanged, $I=V/R=9.0/15=0.60\,\text{A}$ .	4
Total Question 1			4
02.1	The $3.70\,\Omega$ reading at $60\,\text{cm}$ is anomalous; repeat that measurement and use repeat readings to calculate a representative mean after checking anomalies. The other readings support resistance being proportional to wire length at constant temperature.	The $20$ , $40$ and $80\,\text{cm}$ readings are close to doubling when the length doubles, but $3.70\,\Omega$ at $60\,\text{cm}$ is too high for that pattern. Repeat the $60\,\text{cm}$ measurement with the same wire and temperature, take further repeats, and compare or average the consistent values before drawing the trend conclusion.	4
Total Question 2			4
03.1	Potential difference alone does not determine resistance. Measure the current through each wire as well as the potential difference across it under the same controlled conditions, then compare values calculated using $R=V/I$ .	Resistance depends on the ratio of potential difference to current, so a larger potential difference does not by itself show a larger resistance. Record paired voltmeter and ammeter readings for each wire, keep relevant conditions such as temperature constant, and calculate each resistance from $V/I$ .	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.2.1.3 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	New resistance $=7.5\,\Omega$ New current $=0.40\,\text{A}$	The original resistance is $R=V/I=3.0/0.25=12\,\Omega$ . Scale by the length ratio: $R_{\text{new}}=12(0.50/0.80)=7.5\,\Omega$ . Then $I_{\text{new}}=V/R=3.0/7.5=0.40\,\text{A}$ .	5
Total Question 1			5
02.1	$60\,\text{cm}$ ; the intercept is the resistance of the leads, not the test wire.	For the wire alone, $R_{\text{wire}}=(0.040\,\Omega/\text{cm})L$ . Therefore $L=2.4/0.040=60\,\text{cm}$ . The $0.30\,\Omega$ intercept is excluded because the required $2.4\,\Omega$ belongs only to the wire; it would be added only when finding the total measured resistance including the leads.	4
Total Question 2			4
03.1	The corrected readings are $0.61\,\text{A}$ , $0.59\,\text{A}$ and $0.63\,\text{A}$ . The mean corrected current is $0.61\,\text{A}$ and the resistance is $4.9\,\Omega$ , both to two significant figures.	The $+0.05\,\text{A}$ zero error shifts every measurement in the same direction, so it is systematic; the unpredictable differences among repeat readings are random. Subtract $0.05\,\text{A}$ from each reading to obtain $0.61$ , $0.59$ and $0.63\,\text{A}$ . Their mean is $1.83/3=0.61\,\text{A}$ exactly. $R=V/I=3.0/0.61=4.918\ldots\,\Omega=4.9\,\Omega$ to two significant figures.	5
Total Question 3			5
04.1	The current without the extra resistor is $0.43\,\text{A}$ ; the minimum total resistance is $40\,\Omega$ ; add at least $12\,\Omega$ in series.	Without an extra resistor, $I=V/R=12/28=0.4286\ldots\,\text{A}$ , which exceeds $0.30\,\text{A}$ . At the maximum permitted current, $R_{\text{total}}=V/I=12/0.30=40\,\Omega$ . Series resistances add, so the added resistance must be at least $40-28=12\,\Omega$ .	5
Total Question 4			5
05.1	Current $=3.0\,\text{A}$ ; motor potential difference $=9.0\,\text{V}$ ; each lead has $1.5\,\text{V}$ across it; $9.0+1.5+1.5=12\,\text{V}$ .	All three resistances are in series, so $R_{\text{total}}=3.0+0.50+0.50=4.0\,\Omega$ . The current is $I=V/R=12/4.0=3.0\,\text{A}$ . For the motor, $V=IR=(3.0)(3.0)=9.0\,\text{V}$ . For each lead, $V=(3.0)(0.50)=1.5\,\text{V}$ . Their sum is $9.0+1.5+1.5=12\,\text{V}$ , equal to the supply potential difference.	6
Total Question 5			6

4.2.1.4 · Resistors

Tier 1 · Easy

Mark scheme for 4.2.1.4 Tier 1 · Easy
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01.1	The LDR's resistance increases as the light intensity decreases.	An LDR has lower resistance in brighter light. Therefore reducing the light intensity makes its resistance increase.	1
Total Question 1			1
02.1	Choose a thermistor because its resistance increases when temperature decreases.	A thermistor responds to temperature, whereas an LDR responds to light intensity. Since thermistor resistance decreases as temperature rises, it rises when the cold store becomes colder.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.2.1.4 Tier 2 · Standard
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01.1	Resistance rises from $5.0\,\Omega$ to $7.5\,\Omega$ because the filament becomes hotter.	At $2.0\,\text{V}$ , $R=V/I=2.0/0.40=5.0\,\Omega$ . At $6.0\,\text{V}$ , $R=6.0/0.80=7.5\,\Omega$ . The larger current heats the filament more, so its resistance increases.	4
Total Question 1			4
02.1	Reverse the diode so it is connected in the opposite direction. Adjust the variable resistor through a range of resistance settings. Record the current through the diode at each setting. Record the potential difference across the diode at each setting.	Switch off before reversing the diode, then keep the ammeter in series and the voltmeter in parallel across the diode. Change the variable resistance through a range of settings. At each setting, record the paired ammeter and voltmeter readings.	4
Total Question 2			4
03.1	P is an ohmic conductor. Its resistance is $5.0\,\Omega$ at both $1.0\,\text{V}$ and $2.0\,\text{V}$ magnitudes, and current reverses with potential difference while remaining proportional to it.	At magnitude $1.0\,\text{V}$ , $R=V/I=1.0/0.20=5.0\,\Omega$ . At magnitude $2.0\,\text{V}$ , $R=2.0/0.40=5.0\,\Omega$ . The same ratio at both magnitudes, with equal reversed values for negative potential difference, shows that current is directly proportional to potential difference, so P is an ohmic conductor.	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.2.1.4 Tier 3 · Hard
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01.1	Use a thermistor with an ammeter in series and voltmeter in parallel, measure temperature and paired current-potential difference readings over a range, control the supply potential difference, calculate $R=V/I$ , and plot resistance against temperature.	Place the thermistor in a water bath with a thermometer. Connect it in a circuit with an ammeter in series and a voltmeter across it. Keep the supply potential difference constant. Change the bath temperature gradually, allow the reading to settle, and record temperature, current and potential difference at each point. Calculate each resistance using $R=V/I$ and plot resistance against temperature; the downward trend shows that thermistor resistance decreases as temperature increases.	6
Total Question 1			6
02.1	$2.0\,\text{mA}$ at $10\,^{\circ}\text{C}$ and $8.0\,\text{mA}$ at $50\,^{\circ}\text{C}$ ; heating lowers the thermistor resistance, so the current rises at constant potential difference.	Convert the resistances: $3.0\,\text{k}\Omega=3000\,\Omega$ and $0.75\,\text{k}\Omega=750\,\Omega$ . At $10\,^{\circ}\text{C}$ , $I=V/R=6.0/3000=0.0020\,\text{A}=2.0\,\text{mA}$ . At $50\,^{\circ}\text{C}$ , $I=6.0/750=0.0080\,\text{A}=8.0\,\text{mA}$ . The higher temperature reduces resistance, so the unchanged potential difference drives a larger current.	5
Total Question 2			5
03.1	The LDR resistance is $2.0\,\text{k}\Omega$ , so the sensor is in ordinary indoor light.	Convert the current: $3.0\,\text{mA}=0.0030\,\text{A}$ . The total series resistance is $R=V/I=9.0/0.0030=3000\,\Omega=3.0\,\text{k}\Omega$ . Subtract the fixed resistance to obtain $R_{\text{LDR}}=3.0-1.0=2.0\,\text{k}\Omega$ . This matches the ordinary-indoor-light calibration value.	4
Total Question 3			4
04.1	The LDR resistance decreases as the light-meter reading increases. The consistent prediction at $400\,\text{lux}$ is $3.0\,\text{k}\Omega$ . An LDR responds to light, so this calibration cannot distinguish a temperature change; a thermistor should be used to detect temperature.	Read down the table: increasing the light level from $100$ to $500\,\text{lux}$ reduces the resistance from $8.0$ to $2.0\,\text{k}\Omega$ . A value at $400\,\text{lux}$ must lie between the neighbouring $4.0$ and $2.0\,\text{k}\Omega$ readings, so only $3.0\,\text{k}\Omega$ is consistent. Light, not temperature, is the input quantity for an LDR; temperature sensing requires a thermistor.	5
Total Question 4			5

4.2.2 · Series and parallel circuits

Tier 1 · Easy

Mark scheme for 4.2.2 Tier 1 · Easy
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01.1	$11\,\Omega$	Series resistances add, so $R_{\text{total}}=4+7=11\,\Omega$ .	1
Total Question 1			1
02.1	$0.54\,\text{A}$	The current entering a junction equals the sum leaving through the branches. Therefore the hidden branch current is $0.86-0.32=0.54\,\text{A}$ .	1
Total Question 2			1

Tier 2 · Standard

Mark scheme for 4.2.2 Tier 2 · Standard
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01.1	Current $=2.0\,\text{A}$ Potential differences are $4.0\,\text{V}$ and $8.0\,\text{V}$ respectively.	The total resistance is $2.0+4.0=6.0\,\Omega$ , so $I=V/R=12/6.0=2.0\,\text{A}$ . The drops are $V_1=IR_1=(2.0)(2.0)=4.0\,\text{V}$ and $V_2=(2.0)(4.0)=8.0\,\text{V}$ ; they sum to the $12\,\text{V}$ supply.	5
Total Question 1			5
02.1	Reconnect the lamps on separate parallel branches; each lamp then has the full supply potential difference and one branch remains complete if the other lamp is removed.	Create a junction before the lamps and another after them so each lamp has its own path between the same two supply points. Parallel branches have the same potential difference as the supply. Breaking one branch therefore leaves a complete second branch for the other lamp.	3
Total Question 2			3
03.1	Adding the resistor in series increases total resistance because the resistances add. Adding it in parallel decreases total resistance because an extra path is available for current; the parallel total is less than the resistance of either resistor.	In series, the two resistances add, so the total becomes larger. In parallel, the added branch provides another route for charge flow, increasing the total current for a given supply potential difference. Since $R=V/I$ , this means the total resistance is smaller than that of either individual resistor.	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.2.2 Tier 3 · Hard
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01.1	Second-branch resistance $=4.0\,\Omega$ Supply current $=5.0\,\text{A}$	Each parallel branch has the full $12\,\text{V}$ . For the second branch, $R=V/I=12/3.0=4.0\,\Omega$ . The first-branch current is $I=12/6.0=2.0\,\text{A}$ . Branch currents add, so the supply current is $2.0+3.0=5.0\,\text{A}$ .	5
Total Question 1			5
02.1	R₀ is $4.0\,\Omega$ ; the other branch current is $0.50\,\text{A}$ ; the other branch resistance is $12\,\Omega$ .	For R₀, $R=V/I=3.0/0.75=4.0\,\Omega$ . At the junction, the other branch current is $0.75-0.25=0.50\,\text{A}$ . The parallel combination has $9.0-3.0=6.0\,\text{V}$ across it, so the unknown branch resistance is $R=6.0/0.50=12\,\Omega$ .	6
Total Question 2			6
03.1	P has total resistance $12\,\Omega$ and is the series configuration. Each resistor is $6.0\,\Omega$ . Q has total resistance $3.0\,\Omega$ and is the parallel configuration.	Use $R=V/I$ . For P, $R=6.0/0.50=12\,\Omega$ ; two identical series resistors therefore have $6.0\,\Omega$ each. For Q, $R=6.0/2.0=3.0\,\Omega$ . The larger total is the series arrangement, while the $3.0\,\Omega$ total is less than one $6.0\,\Omega$ resistor and therefore identifies the parallel arrangement.	5
Total Question 3			5
04.1	Branch A current $=1.0\,\text{A}$ ; branch B current $=2.0\,\text{A}$ ; supply current $=3.0\,\text{A}$ ; potential difference across the $8.0\,\Omega$ resistor $=8.0\,\text{V}$ .	Branch A has series resistance $4.0+8.0=12\,\Omega$ and the full $12\,\text{V}$ across it, so its current is $12/12=1.0\,\text{A}$ . Branch B also has $12\,\text{V}$ across it, giving $I=12/6.0=2.0\,\text{A}$ . The supply current is the sum, $1.0+2.0=3.0\,\text{A}$ . The $8.0\,\Omega$ resistor carries the branch-A current, so $V=IR=(1.0)(8.0)=8.0\,\text{V}$ .	6
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05.1	In series, place ammeters at different points in the single loop and voltmeters across each resistor and the supply: the currents should match and the two resistor potential differences should add to the supply value. In parallel, put an ammeter in each branch and in the main line and a voltmeter across each branch: the branch potential differences should match the supply, and the branch currents should add to the main current.	Build the series circuit from a dc supply, switch and the two resistors in one loop. Check that ammeters placed at different points give the same current, then measure both resistor potential differences and show that their sum equals the supply value. Reconnect the resistors on separate branches between the same two supply points. Measure each branch current and the main current, and check that the branch currents add to the main current. Measure across both branches to show that each has the supply potential difference.	6
Total Question 5			6

4.2.3.1 · Direct and alternating potential difference

Tier 1 · Easy

Mark scheme for 4.2.3.1 Tier 1 · Easy
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01.1	Direct potential difference; its polarity does not reverse.	A battery keeps the same positive and negative terminals, so it supplies a direct potential difference and drives current in one direction.	2
Total Question 1			2
02.1	It is an alternating supply because the potential difference changes sign, so the polarity reverses.	The readings include both positive and negative values and repeat. This means the two terminals repeatedly exchange polarity, which defines an alternating potential difference.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.2.3.1 Tier 2 · Standard
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01.1	$15$ complete cycles; the polarity repeatedly reverses.	Frequency is cycles per second, so the number of cycles is $50\times0.30=15$ . Because the supply is alternating, its polarity reverses during every cycle.	3
Total Question 1			3
02.1	$f=1/0.025=40\,\text{Hz}$ It could not be UK mains because $40\,\text{Hz}$ is $10\,\text{Hz}$ below the $50\,\text{Hz}$ mains frequency.	Use $f=1/T$ . Substituting $T=0.025\,\text{s}$ gives $f=1/0.025=40\,\text{Hz}$ . This differs from the $50\,\text{Hz}$ UK mains frequency by $10\,\text{Hz}$ , so the supply could not be UK mains.	3
Total Question 2			3
03.1	X is direct because its fixed polarity allows current through only one diode. Y is alternating because its reversing polarity makes the conducting direction alternate, so the indicators light in turn.	A diode conducts in only one direction. A direct supply retains one polarity, so only the correctly oriented branch conducts and A remains lit. An alternating supply repeatedly reverses polarity, so first one diode and then the oppositely oriented diode conducts.	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.2.3.1 Tier 3 · Hard
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01.1	A is dc; B is ac at $50\,\text{Hz}$ ; A drives current in one direction whereas B repeatedly reverses it.	A keeps a fixed polarity, so it is direct. B changes polarity, so it is alternating. Its frequency is $f=25/0.50=50\,\text{Hz}$ . A therefore produces current in one direction, while B produces a current that reverses direction.	5
Total Question 1			5
02.1	$63\,\text{Hz}$ to two significant figures; it is not a $50\,\text{Hz}$ UK mains supply.	The reversal count corresponds to $30/2=15$ cycles. Hence $f=15/0.24=62.5\,\text{Hz}$ , which rounds to $63\,\text{Hz}$ to two significant figures. This differs from the UK mains frequency of $50\,\text{Hz}$ .	4
Total Question 2			4
03.1	P and Q are direct; R is alternating. Q changes in size but never reverses polarity, whereas R changes sign and therefore reverses polarity.	State the supply type from direction, not merely whether the value changes. P has fixed magnitude and polarity, so it is direct. Q varies but remains positive at the labelled terminal, so it still drives current in one direction and is direct. R takes positive and negative values, showing repeated polarity reversal, so it is alternating.	5
Total Question 3			5
04.1	The period is $0.040\,\text{s}$ , and the supply produces $5.0$ complete cycles in $0.20\,\text{s}$ . The period is the time taken for one complete cycle.	Use $T=1/f=1/25=0.040\,\text{s}$ . Then use $N=ft=(25)(0.20)=5.0$ complete cycles. The period is the duration of one complete repeating cycle.	4
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05.1	B could be UK mains because it is alternating, its potential difference is about $230\,\text{V}$ and its frequency is $50\,\text{Hz}$ .	Check polarity, potential difference and frequency together. A is direct because its polarity does not reverse. C has the wrong potential difference, and D has the wrong frequency. Only B satisfies all three UK-mains criteria.	4
Total Question 5			4

4.2.3.2 · Mains electricity

Tier 1 · Easy

Mark scheme for 4.2.3.2 Tier 1 · Easy
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01.1	Live: brown Neutral: blue Earth: green and yellow stripes	Recall the standard identification: brown is live, blue is neutral, and green-and-yellow striped insulation identifies earth.	3
Total Question 1			3
02.1	Fault B is normally more dangerous because brown is live at about $230\,\text{V}$ relative to earth, whereas blue is neutral at or close to $0\,\text{V}$ .	Use the wire colours to identify brown as live and blue as neutral. A person connected to earth who touches the exposed live conductor can have about $230\,\text{V}$ across them, while the neutral conductor is normally close to earth potential.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.2.3.2 Tier 2 · Standard
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01.1	The earth wire provides a low-resistance path for a large fault current, keeps the case near $0\,\text{V}$ and causes the protective device to disconnect the supply.	The earth wire connects the case to earth potential. If live contacts the case, current flows through the earth wire rather than through a person. The large fault current makes a fuse or circuit breaker disconnect the live supply, so the case does not remain live.	4
Total Question 1			4
02.1	X is live; Y is neutral; Z is earth.	The live conductor provides the alternating potential difference and is about $230\,\text{V}$ relative to earth, so it is X. Neutral completes the normal circuit near earth potential, so it is Y. The earth conductor is a $0\,\text{V}$ safety path used only during a fault, so it is Z.	3
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03.1	The exposed wire is still live at about $230\,\text{V}$ relative to earth because it is before the open switch. Touching it can drive current through a person's body to earth. A live-to-earth connection can also produce a large fault current, causing heating until a protective device disconnects the supply.	Opening the switch breaks the circuit after this section of the live wire; it does not disconnect the wire between the plug and switch from the mains supply. A person touching the conductor while connected to earth may complete a current path through the body. Connecting live directly to earth creates a low-resistance fault path and hence a large current, so the fuse or circuit breaker should disconnect the supply.	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.2.3.2 Tier 3 · Hard
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01.1	The open switch stops the current but leaves the lamp holder connected to the live wire at about $230\,\text{V}$ relative to earth, so touching it while earthed could complete a dangerous path.	Opening the neutral wire breaks the normal circuit, so the lamp turns off. However, the live connection has not been isolated. Parts of the holder can therefore remain at the live potential relative to earth, and a person touching them could provide a current path to earth.	4
Total Question 1			4
02.1	Touching P while connected to earth can make the user the current path. In Q, the earth wire provides a low-resistance fault path, producing a large current that melts the fuse and disconnects the live supply.	P leaves a live conductor exposed at about $230\,\text{V}$ relative to earth. A person touching it may complete a path through their body to earth. In Q, fault current instead flows from live through the case and the earth wire. The low-resistance path makes the current large enough for the fuse to break the circuit, so the case should not remain live.	5
Total Question 2			5
03.1	The appliance normally uses the live and neutral wires, so a broken earth does not stop it working. The fault can leave the case at about $230\,\text{V}$ relative to earth with no low-resistance earth path. Restoring the earth gives a large fault current through the case and earth wire, allowing the protective device to disconnect the live supply.	Trace the normal circuit from live through the appliance to neutral; the earth wire is not part of this path. With the earth broken, live-to-case contact can leave the metal accessible at live potential, so a person may provide a path to earth. A sound earth conductor supplies a much lower-resistance fault path, producing the large current needed to operate the protective device.	5
Total Question 3			5
04.1	The normal current is $10\,\text{A}$ , so use the $13\,\text{A}$ fuse. A $3\,\text{A}$ fuse would melt during normal operation. The fuse must be in the live wire.	Convert $2.3\,\text{kW}$ to $2300\,\text{W}$ and rearrange $P=VI$ : $I=P/V=2300/230=10\,\text{A}$ . The fuse rating must exceed the normal current, so $13\,\text{A}$ is suitable and $3\,\text{A}$ is too low. Placing the fuse in the live wire means that when it melts, it disconnects the appliance from the live supply.	5
Total Question 4			5
05.1	A is normal operation: equal current travels through live, the iron, and neutral, with none in earth. B is a live-to-case fault: current travels from live through the case and earth wire, producing a large fault current that should make the protective device disconnect the supply.	In normal operation, charge flows into the iron through brown live and returns through blue neutral, so their currents match and the earth wire carries none. In test B, the zero neutral current and equal live and earth currents show that current has taken the fault path through the case and green-and-yellow earth conductor. The much larger current should operate the fuse or circuit breaker and break the circuit.	5
Total Question 5			5

4.2.4.1 · Power

Tier 1 · Easy

Mark scheme for 4.2.4.1 Tier 1 · Easy
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01.1	$36\,\text{W}$	Use $P=VI$ : $P=(12)(3.0)=36\,\text{W}$ .	2
Total Question 1			2
02.1	Device B; its power is three times device A's power.	At the same potential difference, $P=VI$ makes power proportional to current. The current ratio is $1.5/0.50=3$ , so B transfers energy three times as quickly.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.2.4.1 Tier 2 · Standard
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01.1	$50\,\text{W}$	Use $P=I^2R$ : $P=(2.5)^2(8.0)=6.25\times8.0=50\,\text{W}$ .	2
Total Question 1			2
02.1	X has power $29\,\text{W}$ and Y has power $24\,\text{W}$ , so X has the higher power.	Use $P=VI$ for each row. For X, $P=(24)(1.2)=28.8\,\text{W}$ , which is $29\,\text{W}$ to two significant figures. For Y, $P=(12)(2.0)=24\,\text{W}$ . Therefore X transfers energy faster.	3
Total Question 2			3
03.1	Power $=40\,\text{W}$ . Potential difference $=20\,\text{V}$ . $40\,\text{W}$ means the appliance transfers $40\,\text{J}$ of energy each second.	First calculate the energy-transfer rate: $P=E/t=7200/180=40\,\text{W}$ . Then rearrange $P=VI$ to $V=P/I$ , giving $V=40/2.0=20\,\text{V}$ . Since one watt is one joule per second, $40\,\text{W}$ means $40\,\text{J}$ transferred each second.	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.2.4.1 Tier 3 · Hard
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01.1	Current $=7.8\,\text{A}$ Resistance $=29\,\Omega$	Convert power: $1.8\,\text{kW}=1800\,\text{W}$ . From $P=VI$ , $I=P/V=1800/230=7.83\,\text{A}$ . Then $P=I^2R$ gives $R=P/I^2=1800/(7.83)^2=29.4\,\Omega$ . To two significant figures, the results are $7.8\,\text{A}$ and $29\,\Omega$ .	5
Total Question 1			5
02.1	$52\,\Omega$ : $4.4\,\text{A}$ and $1.0\,\text{kW}$ ; $42\,\Omega$ : $5.5\,\text{A}$ and $1.3\,\text{kW}$ . Only the $52\,\Omega$ setting stays below $5.0\,\text{A}$ .	For $52\,\Omega$ , $I=230/52=4.423\ldots\,\text{A}$ and $P=VI=(230)(4.423\ldots)=1017\ldots\,\text{W}$ , giving $4.4\,\text{A}$ and $1.0\,\text{kW}$ to two significant figures. For $42\,\Omega$ , $I=230/42=5.476\ldots\,\text{A}$ and $P=(230)(5.476\ldots)=1259\ldots\,\text{W}$ , giving $5.5\,\text{A}$ and $1.3\,\text{kW}$ . Compare the unrounded currents with $5.0\,\text{A}$ : only the $52\,\Omega$ setting is below the rating.	5
Total Question 2			5
03.1	$76\,800\,\text{J}$ , $4.0\,\text{A}$ and $80\,\text{V}$ .	Convert the time: $4.0\,\text{min}=240\,\text{s}$ . The energy transferred is $E=Pt=(320)(240)=76\,800\,\text{J}$ . From $P=I^2R$ , $I=\sqrt{320/20}=4.0\,\text{A}$ . Then $V=P/I=320/4.0=80\,\text{V}$ to two significant figures.	5
Total Question 3			5
04.1	Turning freely: electrical input power $=72\,\text{W}$ and energy transferred $=360\,\text{J}$ . Stalled: electrical input power $=216\,\text{W}$ and energy transferred $=1080\,\text{J}$ . The stalled drill receives electrical energy three times as quickly.	When turning freely, $P=VI=(18)(4.0)=72\,\text{W}$ , so $E=Pt=(72)(5.0)=360\,\text{J}$ . When stalled, $P=(18)(12)=216\,\text{W}$ , so $E=(216)(5.0)=1080\,\text{J}$ . The same operating time makes the energy ratio equal to the power ratio: $216/72=3$ .	6
Total Question 4			6
05.1	Section A: $6.0\,\text{V}$ and $2.0\,\Omega$ . Section B: $18\,\text{V}$ , $6.0\,\Omega$ and $54\,\text{W}$ .	For section A, $V=P/I=18/3.0=6.0\,\text{V}$ and $R=V/I=6.0/3.0=2.0\,\Omega$ . Series potential differences add, so section B has $24-6.0=18\,\text{V}$ across it. Its resistance is $R=18/3.0=6.0\,\Omega$ , and its power is $P=VI=(18)(3.0)=54\,\text{W}$ .	6
Total Question 5			6

4.2.4.2 · Energy transfers in everyday appliances

Tier 1 · Easy

Mark scheme for 4.2.4.2 Tier 1 · Easy
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01.1	$18\,000\,\text{J}$ or $18\,\text{kJ}$	Convert the time: $5.0\,\text{min}=300\,\text{s}$ . Then $E=Pt=(60)(300)=18\,000\,\text{J}=18\,\text{kJ}$ .	3
Total Question 1			3
02.1	Appliance B transfers more energy because the powers are equal and B operates for longer.	Power is energy transferred per second. Since both appliances transfer energy at the same rate, the one operating for the greater time transfers the greater total energy; therefore B does.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.2.4.2 Tier 2 · Standard
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01.1	$108\,000\,\text{J}$ Electrical energy is transferred to kinetic energy.	Use $E=QV$ : $E=(9000)(12)=108\,000\,\text{J}$ . A motor's intended output is movement, so the useful transfer is to kinetic energy.	3
Total Question 1			3
02.1	$200\,\text{W}$ ; the motor increases the kinetic energy store of the moving parts.	Convert both units: $84\,\text{kJ}=84\,000\,\text{J}$ and $7.0\,\text{min}=420\,\text{s}$ . Then $P=E/t=84\,000/420=200\,\text{W}$ . A motor's useful output is movement, so it increases a kinetic energy store.	4
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03.1	Kettle: $144\,\text{kJ}$ ; microwave: $216\,\text{kJ}$ ; television: $180\,\text{kJ}$ . From greatest to least energy transferred, the order is microwave, television, kettle.	Use $E=Pt$ for each row. The kettle transfers $(1800)(80)=144\,000\,\text{J}=144\,\text{kJ}$ . The microwave transfers $(900)(240)=216\,000\,\text{J}=216\,\text{kJ}$ . The television transfers $(120)(1500)=180\,000\,\text{J}=180\,\text{kJ}$ . Comparing these energies gives microwave first, then television, then kettle. A higher power does not always give more energy because operating time also affects the energy transferred.	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.2.4.2 Tier 3 · Hard
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01.1	Energy $=1.62\times10^6\,\text{J}$ Charge $=7.0\times10^3\,\text{C}$	Convert the quantities: $P=1500\,\text{W}$ and $t=18\times60=1080\,\text{s}$ . Then $E=Pt=(1500)(1080)=1.62\times10^6\,\text{J}$ . From $E=QV$ , $Q=E/V=(1.62\times10^6)/230=7.04\times10^3\,\text{C}$ , which is $7.0\times10^3\,\text{C}$ to two significant figures.	5
Total Question 1			5
02.1	Charge flow $=18\,000\,\text{C}$ ; current $=10\,\text{A}$	Convert energy: $216\,\text{kJ}=216\,000\,\text{J}$ . From $E=QV$ , $Q=E/V=216\,000/12=18\,000\,\text{C}$ . Convert time: $30\,\text{min}=1800\,\text{s}$ . Then $I=Q/t=18\,000/1800=10\,\text{A}$ .	5
Total Question 2			5
03.1	Both routes give $100\,800\,\text{J}$ , so the records are consistent.	From the first logger, $E=QV=(4200)(24)=100\,800\,\text{J}$ . Convert the operating time: $10\,\text{min}=600\,\text{s}$ . From the second logger, $E=Pt=(168)(600)=100\,800\,\text{J}$ . The independently calculated values agree exactly, so the records are consistent.	5
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04.1	The appliances transfer the same energy. For A, $E_A=Pt$ ; for B, $E_B=(2P)(t/2)=Pt$ .	Use $E=Pt$ . Doubling power multiplies the transfer rate by two, while halving time multiplies the operating duration by one half. These factors cancel, so both appliances transfer $Pt$ joules.	3
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4.2.4.3 · The National Grid

Tier 1 · Easy

Mark scheme for 4.2.4.3 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Step-up transformer before transmission Step-down transformer before homes	The transmission potential difference is raised by a step-up transformer. Near consumers it is reduced to a much lower domestic value by a step-down transformer.	2
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02.1	Plan A; the smaller current causes less heating in the same cables.	For the same transferred power, using a higher potential difference permits a lower current. Cable-heating power depends on current squared, so plan A's smaller current produces the smaller loss.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.2.4.3 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	Current $=20\,\text{A}$ The smaller current causes less heating loss in the cables.	Convert units and use $P=VI$ : $I=P/V=(6.0\times10^6)/(300\times10^3)=20\,\text{A}$ . For the same power, a lower potential difference would require a larger current. Since cable heating is proportional to $I^2R$ , the high-potential-difference transmission wastes less power.	4
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02.1	Line B is inconsistent; its correct current is $30\,\text{A}$ .	Convert the prefixes and apply $I=P/V$ . Line A gives $(2.4\times10^6)/(120\times10^3)=20\,\text{A}$ , so its row is consistent. Line B should give $(3.0\times10^6)/(100\times10^3)=30\,\text{A}$ , not $45\,\text{A}$ .	3
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03.1	A transformer does not create power. Increasing the potential difference is accompanied by a decrease in current, so $VI$ is approximately unchanged; conservation of energy means output power cannot exceed input power and is lower when there are losses. The higher transmission potential difference reduces current and cable heating, so it reduces wasted power rather than increasing power by transformation.	Apply $P=VI$ : for approximately the same transferred power, raising $V$ requires $I$ to fall. Conservation of energy rules out a transformer delivering more power than it receives; real losses make the output power smaller. The grid benefit is that the reduced transmission current produces less heating in the cables, so a greater fraction of the input power reaches consumers.	4
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Tier 3 · Hard

Mark scheme for 4.2.4.3 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	At $12\,\text{kV}$ the loss is $32\,\text{kW}$ ; at $240\,\text{kV}$ it is $80\,\text{W}$ , so the higher potential difference reduces the loss by a factor of $400$ .	At $12\,\text{kV}$ , $I=P/V=(2.4\times10^6)/(12\times10^3)=200\,\text{A}$ , so $P_{\text{loss}}=I^2R=(200)^2(0.80)=32\,000\,\text{W}=32\,\text{kW}$ . At $240\,\text{kV}$ , $I=10\,\text{A}$ and $P_{\text{loss}}=(10)^2(0.80)=80\,\text{W}$ . The loss ratio is $32\,000/80=400$ .	6
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02.1	$76\,\text{kV}$	Convert the transferred power: $18\,\text{MW}=18\times10^6\,\text{W}$ . The maximum loss is $0.010(18\times10^6)=1.8\times10^5\,\text{W}$ . From $P_{\text{loss}}=I^2R$ , the maximum current is $I=\sqrt{(1.8\times10^5)/3.2}=237.170\ldots\,\text{A}$ . The minimum potential difference is therefore $V=P/I=(18\times10^6)/237.170\ldots=75\,894.6\ldots\,\text{V}=75.8946\ldots\,\text{kV}$ , which is $76\,\text{kV}$ to two significant figures.	6
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03.1	$40\,\text{A}$ ; $100\,\text{kW}$ lost; total cable resistance $=62.5\,\Omega$ .	Convert prefixes and use $I=P/V$ : $I=(8.00\times10^6)/(200\times10^3)=40\,\text{A}$ . The cable loss is $8.00-7.90=0.10\,\text{MW}=100\,000\,\text{W}$ . Rearrange $P_{\text{loss}}=I^2R$ to give $R=P_{\text{loss}}/I^2=100\,000/40^2=62.5\,\Omega$ .	5
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04.1	$60\,\text{A}$ ; $180\,\text{kW}$ ; $1.296\times10^9\,\text{J}$ wasted in $2.0$ hours; $2.0\%$ of the transferred power is wasted.	The current is $I=P/V=(9.0\times10^6)/(150\times10^3)=60\,\text{A}$ . Cable-heating power is $I^2R=(60)^2(50)=180\,000\,\text{W}=180\,\text{kW}$ . Convert $2.0$ hours to $7200\,\text{s}$ , so wasted energy is $E=Pt=(180\,000)(7200)=1.296\times10^9\,\text{J}$ . The percentage loss is $(180\,000/(9.0\times10^6))\times100=2.0\%$ .	6
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05.1	A delivers $9.80\,\text{MW}$ and loses $2.0\%$ ; B delivers $3.80\,\text{MW}$ and loses $5.0\%$ . Route A is more efficient. The routes lose the same power in megawatts, but that loss is a larger fraction of B's smaller input power.	Subtract cable loss from input power: route A delivers $10.0-0.20=9.80\,\text{MW}$ and route B delivers $4.0-0.20=3.80\,\text{MW}$ . Route A loses $(0.20/10.0)\times100=2.0\%$ , whereas route B loses $(0.20/4.0)\times100=5.0\%$ . The smaller percentage loss makes A more efficient. Equal absolute losses do not mean equal efficiency when the input powers differ.	5
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4.2.5.1 · Static charge (physics only)

Tier 1 · Easy

Mark scheme for 4.2.5.1 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	X becomes negatively charged Y becomes positively charged	X has gained negative electrons, so it becomes negative. Y has lost the same electrons and is left with an equal positive charge.	2
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02.1	Electrons moved from the jumper to the balloon. The jumper lost exactly the electrons the balloon gained, so the two charges are equal in size and opposite in sign.	Only electrons transfer between the insulating materials. Because the balloon is stated to become negative, it gained electrons; the jumper lost the same number of electrons, leaving it with an equal positive charge.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.2.5.1 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	The second strip is negatively charged The repulsion is a non-contact force	Repulsion occurs only between charges of the same type. Since the known strip is negative, the other charged strip must also be negative. The strips exert the force while separated, so it is non-contact.	3
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02.1	The strip is negatively charged. Repulsion of the negatively charged probe is conclusive because an uncharged object is attracted by either sign of charge.	A charged object repels only another object with the same sign, so repelling the negative probe identifies the strip as negative. Attraction of the uncharged suspended ball is not conclusive because either a positive or a negative charged object can attract an uncharged object.	3
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03.1	A is negative, B is negative and C is positive.	C repels the positive strip D, so C is positive. Because B is known to be charged and attracts positive C, B is negative. A repels negative B, so A is also negative.	3
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Tier 3 · Hard

Mark scheme for 4.2.5.1 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Electrons transfer from the cloth to the rod, leaving the rod negative and the cloth equally positive; when the charged rod approaches the metal, electrons move rapidly through the air gap, producing a spark and reducing the charge difference.	Rubbing transfers electrons onto the insulating rod, where charge can remain because the material does not conduct it away. The cloth loses those electrons. Near the metal object, the large electrical effect across the small gap makes charge move suddenly through the air. This rapid transfer is the spark and partially discharges the rod.	5
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02.1	A becomes positive and B becomes negative, so they attract. Removing some excess electrons makes B less negatively charged, so the attractive force becomes weaker while the signs remain opposite.	Sheet A loses negative electrons and is left positive; sheet B gains them and becomes negative. Opposite charges attract through a non-contact force. The spark transfers away some of B's excess electrons, reducing the amount of negative charge on B. Since B is stated to remain negative, attraction continues but with a weaker electrical effect and therefore a smaller force.	5
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03.1	The rod touched by the bare hand discharges as excess electrons move from the rod through the person's body to earth. The rod held only by the insulating handle remains negatively charged because the handle prevents an easy conducting path to earth.	The metal rod and the body conduct charge. Touching the rod therefore earths it, allowing its stated excess electrons to flow from the rod through the hand and body into Earth until the imbalance is removed. An insulating handle does not let charge flow readily, so it isolates the second rod from earth and its excess electrons remain on the rod.	5
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04.1	Rub two rods that gain electrons with the same cloth and suspend one: they should repel without touching. Rub a different rod that loses electrons and bring it near the suspended negative rod: they should attract without touching. Electron gain makes a rod negative, electron loss makes it positive, like charges repel, and unlike charges attract.	Suspend one insulating rod by a thread so it can move freely. Rub it and a second rod made from the same material with the same type of cloth, using a combination known to transfer electrons onto the rods. Bring the second rod close without touching and observe repulsion, evidence that both rods carry the same negative charge. Charge a different insulating rod using a combination in which it loses electrons, making it positive. Bring it near the suspended negative rod and observe attraction. In both cases movement before contact demonstrates a non-contact force.	6
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05.1	A is positive by six groups, B is negative by four groups, and C is negative by two groups. A and B attract, while B and C repel. The six-group positive charge on A is balanced by a total of six negative groups on B and C, so charge is conserved.	A loses six groups of negative electrons, so it is positive by six groups. B first gains six groups and then loses two, leaving it negative by four groups. C gains the two groups lost by B, so it is negative by two groups. Opposite charges on A and B attract; like negative charges on B and C repel. The final negative total is four plus two groups, equal to the six-group positive charge left on A.	6
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4.2.5.2 · Electric fields (physics only)

Tier 1 · Easy

Mark scheme for 4.2.5.2 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Symmetrical radial field lines perpendicular to the sphere's surface, with arrowheads pointing away from the sphere.	Space several straight radial lines evenly around the sphere. Put arrows on every line pointing outwards because the sphere is positive; the lines must not cross.	2
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02.1	The arrows should point inward, towards the negatively charged sphere. The field lines should be closest together near the sphere because the field is strongest there.	Field arrows show the force direction on a positive test charge, which is towards a negative sphere. Closer field-line spacing represents greater field strength, so the lines must be most closely spaced next to the sphere.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.2.5.2 Tier 2 · Standard
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01.1	The force is away from the sphere in both positions and is stronger at the nearer position.	The sphere's field points outwards, so a positive test charge is repelled along that direction. The electric field is strongest close to the charged sphere and weaker farther away, so the nearer force is larger.	3
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02.1	The force at $3.0\,\text{cm}$ should be between $0.46\,\text{N}$ and $1.8\,\text{N}$ ; the field and force weaken as distance increases.	$3.0\,\text{cm}$ lies between $2.0$ and $4.0\,\text{cm}$ , and the measured force falls throughout the table as distance grows. Therefore its force should lie strictly between the two neighbouring values. The evidence supports a field that is strongest close to the charged sphere and weaker farther away; no inverse-square calculation is required.	3
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03.1	The field points left at P and right at Q and R. P and Q are equally far from the symmetric isolated sphere, so the field strength is equal at those points. A negative test charge at Q is forced left, towards the positive sphere.	The field around an isolated positive sphere is radial and points outwards, which fixes the three field directions. The isolated sphere has the same field pattern in every radial direction, and P and Q are the same distance from its centre, so their field strengths are equal. Field direction is defined for a positive test charge, so a negative test charge experiences force in the opposite direction, towards the positive sphere.	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.2.5.2 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	The dome's field acts on charges in the other sphere without contact; decreasing the gap increases the field and force, and a sufficiently strong field causes charge to cross the air gap as a spark.	The charged dome creates an electric field throughout the surrounding space. Charges in the earthed sphere are therefore acted on even before the objects touch, which explains the non-contact force. Moving the objects closer makes the field in the gap stronger. Eventually charges can move through the air between the objects; the rapid transfer is observed as a spark and reduces the charge separation.	5
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02.1	B is positive and pushes the test charge left more strongly than A pushes it right. Moving P closer to B strengthens B's leftward effect, so the leftward resultant increases.	A positive sphere repels the positive test charge, producing the stated rightward force. For B on the right to produce a leftward force, it must also be positive and repel the test charge away from itself. The observed resultant is leftward, so B's field effect at P is larger than A's opposing effect. Decreasing the distance to B strengthens its field and force, so the leftward resultant becomes larger.	5
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03.1	A field line shows the force direction on a positive test charge. Crossing lines would assign two force directions at one point, which is impossible. An isolated spherical charge distribution is symmetric, so there is no preferred sideways direction; the field follows lines through the sphere's centre, outwards for positive charge and inwards for negative charge.	At any point, the tangent to a field line gives one unique direction for the force on a positive test charge. If two lines crossed, their two tangents would predict two simultaneous force directions. Around an isolated charged sphere, spherical symmetry leaves no preferred tangential direction, so the field direction must lie along the line joining each point to the centre. These centre-lines form the radial pattern, with arrow direction set by the sign of the charge.	4
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