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AQA A-level Physics revision notes

Electricity

Section 3.5
Year 1
Year 1: this is the AS subject content the exam board publishes, which is what most schools teach in Year 12.
6 specification points

Notes and three levels of exam-style practice for each registered specification point in this section.

Checked against AQA 7408 section 3.5

Checked against AQA 7408 section 3.5. Review basis: the qualification registry sourced from the AQA A-level Physics (7408) specification; registry verification recorded 11 July 2026.

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3.5.1.1

Basics of electricity

Notes
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A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Electric current is the rate of flow of charge, I=ΔQ/ΔtI=\Delta Q/\Delta t, measured in amperes. Potential difference is the work done per unit charge, V=W/QV=W/Q, so one volt is one joule per coulomb.
  • Resistance is defined as R=V/IR=V/I and is measured in ohms; this definition remains valid even for a component whose resistance changes with current or temperature.
  • Conventional current describes positive charge flow, while electrons in a metal move oppositely.
  • In practical circuits an ammeter is connected in series to measure current and a voltmeter in parallel across a component to measure potential difference.
  • Examiners expect units, correct meter placement and a distinction between charge transferred, its rate of flow and energy transferred per coulomb.
An ammeter in series and a voltmeter connected in parallel across a component.
Worked example

A component transfers 24J24\,\text{J} when 4.0C4.0\,\text{C} passes through it. The current is 0.75A0.75\,\text{A}. Determine its potential difference and resistance.

  1. 1.Calculate potential difference: V=W/Q=24/4.0=6.0VV=W/Q=24/4.0=6.0\,\text{V}.
  2. 2.Use the measured current in R=V/IR=V/I.
  3. 3.R=6.0/0.75=8.0ΩR=6.0/0.75=8.0\,\Omega.

Answer: The potential difference is 6.0V6.0\,\text{V} and the resistance is 8.0Ω8.0\,\Omega.

Common mistakes

  • Don't treat charge in coulombs as though it were current in amperes.
  • Don't connect an ammeter in parallel or a voltmeter in series.
  • Don't use V=WQV=WQ instead of work done per unit charge.

Exam tip

Define each quantity in words before substituting into I=ΔQ/ΔtI=\Delta Q/\Delta t, V=W/QV=W/Q or R=V/IR=V/I.

Tier 1 · Easy

ORIGINAL

A charge of 3.2C3.2\,\text{C} passes a point in 0.80s0.80\,\text{s}. Calculate the current.

[1 mark]

Total for this question: 1

Evidence from answers you checked

Checked automatically against the model answer once you submit.

Tier 2 · Standard

ORIGINAL

A component transfers 18J18\,\text{J} of energy when 3.0C3.0\,\text{C} passes through it. The current is 0.75A0.75\,\text{A}. Determine the potential difference and the resistance of the component.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A heater operates at 24V24\,\text{V} and carries a current of 2.0A2.0\,\text{A} for 150s150\,\text{s}. Determine the charge transferred, the energy transferred and the resistance of the heater.

[5 marks]

Total for this question: 5

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3.5.1.2

Current-voltage characteristics

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • An ohmic conductor has IVI\propto V only when physical conditions, especially temperature, remain constant, so its current–voltage characteristic is a straight line through the origin.
  • A filament lamp heats as current increases; greater lattice vibration raises its resistance, so an II-against-VV graph becomes less steep at larger magnitudes.
  • A semiconductor diode carries negligible reverse current and little forward current until its turn-on region, after which forward current rises rapidly.
  • Either current or potential difference may be placed on the horizontal axis, so gradient meaning must be checked: on an II-against-VV graph the gradient is conductance.
  • Unless stated otherwise, ammeters are ideal with zero resistance and voltmeters are ideal with infinite resistance.
Typical current–voltage characteristics for an ohmic conductor, filament lamp and semiconductor diode.
Worked example

A point on an ohmic conductor’s II-against-VV graph is V=6.0VV=6.0\,\text{V}, I=0.40AI=0.40\,\text{A}. Determine the resistance and the graph gradient.

  1. 1.Use R=V/I=6.0/0.40=15ΩR=V/I=6.0/0.40=15\,\Omega.
  2. 2.For current on the vertical axis, gradient =I/V=0.40/6.0=I/V=0.40/6.0.
  3. 3.Evaluate the conductance: 0.0667S0.0667\,\text{S}.

Answer: The resistance is 15Ω15\,\Omega and the gradient is 0.0667S0.0667\,\text{S}.

Common mistakes

  • Don't call the gradient of an II-against-VV graph resistance rather than conductance.
  • Don't claim that a filament lamp has constant resistance as it heats.
  • Don't draw substantial reverse current for a diode at the potential differences considered.

Exam tip

Name both axes before using a characteristic graph, because swapping them reverses the gradient interpretation.

Tier 1 · Easy

ORIGINAL

State the condition under which a conductor obeys Ohm's law.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

Describe the current-voltage characteristic of a semiconductor diode for forward and reverse potential differences.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A filament lamp and a fixed metal resistor carry the same small current. Explain why, as the potential difference across each component is increased, the lamp's current-voltage graph curves while the resistor's graph remains approximately straight.

[5 marks]

Total for this question: 5

3.5.1.3

Resistivity

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • For a uniform wire, resistivity is ρ=RA/L\rho=RA/L, a material property measured in Ωm\Omega\,\text{m}. Cross-sectional area for a circular wire is A=πd2/4A=\pi d^2/4, so diameter must be converted to metres before squaring.
  • Metal resistance increases with temperature, while the resistance of a negative-temperature-coefficient thermistor decreases; NTC thermistors are temperature sensors whose resistance–temperature graph slopes downward.
  • A superconductor has zero resistivity at and below its material-dependent critical temperature, enabling strong magnetic fields and reduced transmission losses.
  • Required practical 5 determines wire resistivity using a micrometer, ammeter and voltmeter.
  • Examiners expect repeated diameter measurements, a suitable range of lengths or readings, low current to limit heating, and a gradient or repeated calculation with consistent SI units.
A resistivity experiment measuring current through a known wire length and potential difference across it.
Worked example

A 1.50m1.50\,\text{m} wire has resistance 3.2Ω3.2\,\Omega and diameter 0.40mm0.40\,\text{mm}. Determine its resistivity.

  1. 1.Convert radius: r=0.20mm=2.0×104mr=0.20\,\text{mm}=2.0\times10^{-4}\,\text{m}.
  2. 2.Calculate area: A=πr2=1.26×107m2A=\pi r^2=1.26\times10^{-7}\,\text{m}^2.
  3. 3.Use ρ=RA/L=3.2(1.26×107)/1.50=2.7×107Ωm\rho=RA/L=3.2(1.26\times10^{-7})/1.50=2.7\times10^{-7}\,\Omega\,\text{m}.

Answer: ρ=2.7×107Ωm\rho=2.7\times10^{-7}\,\Omega\,\text{m}.

Common mistakes

  • Don't substitute wire diameter where cross-sectional area is required.
  • Don't square a diameter still expressed in millimetres.
  • Don't let the test wire heat significantly, changing the resistance during the practical.

Exam tip

For required-practical answers, include repeated micrometer readings and explain how a low current limits temperature change.

Tier 1 · Easy

ORIGINAL

A wire of length 1.5m1.5\,\text{m} and cross-sectional area 0.20mm20.20\,\text{mm}^2 has resistance 4.2Ω4.2\,\Omega. Calculate its resistivity.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

An NTC thermistor is used as a temperature sensor. Explain the microscopic change that causes its resistance to fall when its temperature rises.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A 1.80m1.80\,\text{m} wire has diameter 0.400mm0.400\,\text{mm}. A potential difference of 2.40V2.40\,\text{V} produces a current of 0.800A0.800\,\text{A}. Determine the wire's resistivity. Explain why replacing a transmission cable by a material operating below its superconducting critical temperature reduces energy loss.

[5 marks]

Total for this question: 5

3.5.1.4

Circuits

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Series resistances add, RT=R1+R2+R_{\text{T}}=R_1+R_2+\cdots, whereas parallel conductances add, 1/RT=1/R1+1/R2+1/R_{\text{T}}=1/R_1+1/R_2+\cdots. Conservation of charge means current is the same through series components and currents into a junction equal currents out.
  • Conservation of energy means potential rises and drops around a complete loop balance.
  • Electrical energy and power follow E=IVtE=IVt and P=IV=I2R=V2/RP=IV=I^2R=V^2/R.
  • Cells in series have additive emfs; identical cells in parallel retain the emf of one cell and have a smaller effective internal resistance.
  • Examiners expect networks to be reduced in a logical order, with branch currents and component potential differences found only after applying the series, parallel, junction and loop relationships.
A resistor in series with two resistors in parallel, illustrating a mixed network.
Worked example

A 4.0Ω4.0\,\Omega resistor is in series with 6.0Ω6.0\,\Omega and 3.0Ω3.0\,\Omega resistors in parallel across 18V18\,\text{V}. Determine the supply current.

  1. 1.Reduce the parallel pair: Rp=(1/6.0+1/3.0)1=2.0ΩR_p=(1/6.0+1/3.0)^{-1}=2.0\,\Omega.
  2. 2.Add the series resistance: RT=4.0+2.0=6.0ΩR_{\text{T}}=4.0+2.0=6.0\,\Omega.
  3. 3.Use I=V/RT=18/6.0=3.0AI=V/R_{\text{T}}=18/6.0=3.0\,\text{A}.

Answer: The supply current is 3.0A3.0\,\text{A}.

Common mistakes

  • Don't add the resistance values directly for a parallel branch.
  • Don't assign the supply potential difference to one series component.
  • Don't use a branch current as though it were the total current before the junction.

Exam tip

Redraw each simplified network after combining one series or parallel group; this makes the next current or potential difference explicit.

Tier 1 · Easy

ORIGINAL

A 4.0Ω4.0\,\Omega resistor and a 6.0Ω6.0\,\Omega resistor are connected in series. State their total resistance.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

A 12Ω12\,\Omega resistor and a 6.0Ω6.0\,\Omega resistor are connected in parallel across an 18V18\,\text{V} supply. Determine the total resistance, the supply current and the total power.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A 3.0Ω3.0\,\Omega resistor is in series with a parallel pair of 6.0Ω6.0\,\Omega and 3.0Ω3.0\,\Omega resistors. The network is connected to a 20V20\,\text{V} supply. Calculate the energy transferred by the 6.0Ω6.0\,\Omega resistor in 90s90\,\text{s}.

[5 marks]

Total for this question: 5

3.5.1.5

Potential divider

Notes
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Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • A potential divider uses series components to provide a constant or variable fraction of a supply potential difference. For two unloaded resistors, the output across R2R_2 is Vout=VinR2/(R1+R2)V_{\text{out}}=V_{\text{in}}R_2/(R_1+R_2).
  • A variable resistor changes this fraction continuously. Sensor dividers may contain a light-dependent resistor, whose resistance falls as light intensity increases, or an NTC thermistor, whose resistance falls as temperature increases.
  • The sensor’s position determines whether output rises or falls with the stimulus: output across a lower sensor falls when its resistance falls.
  • The potentiometer as a measuring instrument is not required.
  • Examiners expect the output terminals to be identified, the correct resistance placed in the numerator and a circuit designed to achieve the stated response.
A two-resistor potential divider with the output taken across the lower resistor.
Worked example

A 2.0kΩ2.0\,\text{k}\Omega upper resistor and 6.0kΩ6.0\,\text{k}\Omega lower resistor form an unloaded divider across 12V12\,\text{V}. Determine the output across the lower resistor.

  1. 1.Identify the lower resistance as the output resistance, R2=6.0kΩR_2=6.0\,\text{k}\Omega.
  2. 2.Use Vout=12[6.0/(2.0+6.0)]V_{\text{out}}=12[6.0/(2.0+6.0)].
  3. 3.Evaluate the fraction: Vout=9.0VV_{\text{out}}=9.0\,\text{V}.

Answer: The output potential difference is 9.0V9.0\,\text{V}.

Common mistakes

  • Don't place the upper resistance in the numerator when output is taken across the lower component.
  • Don't say an LDR’s resistance rises when light intensity increases.
  • Don't predict the sensor response without first identifying whether output is across the sensor or fixed resistor.

Exam tip

Mark the two output terminals on the circuit before choosing the resistance for the numerator.

Tier 1 · Easy

ORIGINAL

Two equal resistors form an unloaded potential divider across a 12V12\,\text{V} supply. State the potential difference across either resistor.

[1 mark]

Total for this question: 1

Evidence from answers you checked

Checked automatically against the model answer once you submit.

Tier 2 · Standard

ORIGINAL

An LDR is the lower component of a potential divider and a 3.0kΩ3.0\,\text{k}\Omega fixed resistor is the upper component. The supply is 9.0V9.0\,\text{V}. Calculate the output across the LDR when its resistance is 9.0kΩ9.0\,\text{k}\Omega in darkness and 1.0kΩ1.0\,\text{k}\Omega in bright light.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

An NTC thermistor is the lower component of an unloaded potential divider. The upper resistor is 4.70kΩ4.70\,\text{k}\Omega and the supply is 12.0V12.0\,\text{V}. A controller switches when the output across the thermistor is 4.00V4.00\,\text{V}. Determine the thermistor resistance and divider current at switching. State how the output changes if the thermistor then becomes warmer.

[5 marks]

Total for this question: 5

3.5.1.6

Electromotive force and internal resistance

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Electromotive force is energy supplied by a source per unit charge, ε=E/Q\varepsilon=E/Q; it is not a mechanical force. A real source has internal resistance rr.
  • When it supplies current II to external resistance RR, ε=I(R+r)\varepsilon=I(R+r) and terminal potential difference is V=IR=εIrV=IR=\varepsilon-Ir. The term IrIr is the lost volts across the internal resistance, so terminal potential difference decreases as current increases.
  • On a graph of VV against II, the intercept is ε\varepsilon and the gradient is r-r.
  • Required practical 6 finds emf and internal resistance by measuring terminal potential difference over a range of currents.
  • Examiners expect a best-fit line, the negative gradient magnitude for rr, and recognition that terminal potential difference equals emf only when current is zero.
A terminal-potential-difference against current graph with intercept equal to emf and gradient equal to negative internal resistance.
Worked example

A cell has terminal potential difference 5.4V5.4\,\text{V} at 0.50A0.50\,\text{A} and 4.2V4.2\,\text{V} at 2.0A2.0\,\text{A}. Determine its internal resistance and emf.

  1. 1.Find the gradient: (4.25.4)/(2.00.50)=0.80Ω(4.2-5.4)/(2.0-0.50)=-0.80\,\Omega.
  2. 2.Use gradient =r=-r, so r=0.80Ωr=0.80\,\Omega.
  3. 3.Substitute one point into ε=V+Ir\varepsilon=V+Ir: ε=5.4+0.50(0.80)=5.8V\varepsilon=5.4+0.50(0.80)=5.8\,\text{V}.

Answer: The internal resistance is 0.80Ω0.80\,\Omega and the emf is 5.8V5.8\,\text{V}.

Common mistakes

  • Don't report a negative internal resistance instead of taking the magnitude of the negative gradient.
  • Don't use terminal potential difference as equal to emf while current is flowing.
  • Don't add IrIr to the terminal potential difference with inconsistent units or current values.

Exam tip

On a VVII graph, label the intercept ε\varepsilon and state explicitly that rr is the magnitude of the gradient.

Tier 1 · Easy

ORIGINAL

A cell of internal resistance 0.50Ω0.50\,\Omega supplies a 2.5Ω2.5\,\Omega resistor with a current of 2.0A2.0\,\text{A}. Calculate the emf of the cell.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A battery has emf 9.0V9.0\,\text{V} and internal resistance 0.80Ω0.80\,\Omega. It is connected to a 4.2Ω4.2\,\Omega load. Determine the current, terminal potential difference and lost volts.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

Measurements for a cell give terminal potential differences of 5.40V5.40\,\text{V} at 0.50A0.50\,\text{A} and 4.20V4.20\,\text{V} at 2.00A2.00\,\text{A}. Determine the emf and internal resistance. The cell is then connected to a 2.10Ω2.10\,\Omega resistor. Calculate the new current and terminal potential difference.

[5 marks]

Total for this question: 5

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