3.5 Electricity — revision question pack

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

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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Answer all questions in the spaces provided.

3.5.1.1 · Basics of electricity

Explanation

Electric current is the rate of flow of charge, $I=\Delta Q/\Delta t$ , measured in amperes. Potential difference is the work done per unit charge, $V=W/Q$ , so one volt is one joule per coulomb.
Resistance is defined as $R=V/I$ 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 $24\,\text{J}$ when $4.0\,\text{C}$ passes through it. The current is $0.75\,\text{A}$ . Determine its potential difference and resistance.

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

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

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 $V=WQ$ instead of work done per unit charge.

Exam tip

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

Tier 1 · Easy

A charge of $3.2\,\text{C}$ passes a point in $0.80\,\text{s}$ . Calculate the current.
[1 mark]
Total for this question: 1
A charging cable carries a current of $0.75\,\text{A}$ for $80\,\text{s}$ . Calculate the charge passing through the cable.
[1 mark]
Total for this question: 1

Tier 2 · Standard

A component transfers $18\,\text{J}$ of energy when $3.0\,\text{C}$ passes through it. The current is $0.75\,\text{A}$ . Determine the potential difference and the resistance of the component.
[3 marks]
Total for this question: 3
A steady current of $0.420\,\text{A}$ flows through a metal wire for $36.0\,\text{s}$ . Determine the number of electrons that pass a cross-section of the wire. Use $e=1.60\times10^{-19}\,\text{C}$ .
[3 marks]
Total for this question: 3
A device receives $3.60\,\text{kJ}$ of energy over $8.00\,\text{min}$ while connected across a potential difference of $15.0\,\text{V}$ . Determine the mean number of electrons passing through the device per second. Use $e=1.60\times10^{-19}\,\text{C}$ .
[4 marks]
Total for this question: 4

Tier 3 · Hard

A heater operates at $24\,\text{V}$ and carries a current of $2.0\,\text{A}$ for $150\,\text{s}$ . Determine the charge transferred, the energy transferred and the resistance of the heater.
[5 marks]
Total for this question: 5
In an electron beam, $2.50\times10^{15}$ electrons pass through a potential difference of $6.50\times10^2\,\text{V}$ in $4.00\,\text{ms}$ . Determine the charge transferred, the beam current, the energy transferred and the average power transferred. Use $e=1.60\times10^{-19}\,\text{C}$ .
[5 marks]
Total for this question: 5
During each operating cycle, a component first carries $0.800\,\text{A}$ at $6.00\,\text{V}$ for $45.0\,\text{s}$ , then carries $2.40\,\text{A}$ at $3.00\,\text{V}$ for a time $t$ . The energy transferred per unit charge over one complete cycle is $4.50\,\text{J C}^{-1}$ . Determine $t$ and the total number of electrons passing through the component during $25$ cycles. Use $e=1.60\times10^{-19}\,\text{C}$ .
[6 marks]
Total for this question: 6
A sensor draws a current that rises uniformly from $0.180\,\text{A}$ to $0.620\,\text{A}$ during $35.0\,\text{s}$ . It then draws no current for $12.0\,\text{s}$ before drawing a steady $0.440\,\text{A}$ for $28.0\,\text{s}$ . The potential difference across the sensor is $7.50\,\text{V}$ whenever current flows. Determine the total charge and energy transferred, the mean current over the complete $75.0\,\text{s}$ interval and the number of electrons transferred. Use $e=1.60\times10^{-19}\,\text{C}$ .
[5 marks]
Total for this question: 5
A pulsed component transfers $1.68\,\text{kJ}$ while a charge of $2.40\times10^3\,\text{C}$ passes through it. The transfer occurs in $4.00\times10^2$ identical pulses, each lasting $3.00\,\text{ms}$ . Determine the potential difference across the component, the charge in one pulse, the current during a pulse and the component resistance while a pulse is present.
[4 marks]
Total for this question: 4

3.5.1.2 · Current-voltage characteristics

Explanation

An ohmic conductor has $I\propto V$ 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 $I$ -against- $V$ 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 $I$ -against- $V$ 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 $I$ -against- $V$ graph is $V=6.0\,\text{V}$ , $I=0.40\,\text{A}$ . Determine the resistance and the graph gradient.

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

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

Common mistakes

Don't call the gradient of an $I$ -against- $V$ 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

State the condition under which a conductor obeys Ohm's law.
[1 mark]
Total for this question: 1
State the resistances assigned to an ideal ammeter and an ideal voltmeter.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Describe the current-voltage characteristic of a semiconductor diode for forward and reverse potential differences.
[3 marks]
Total for this question: 3
A filament lamp carries $0.32\,\text{A}$ at $4.0\,\text{V}$ and $0.50\,\text{A}$ at $8.0\,\text{V}$ . Determine the factor by which its resistance increases between these operating points.
[3 marks]
Total for this question: 3
Explain how a student should obtain current-voltage data for an ohmic metal resistor over both polarities while keeping its temperature as constant as possible.
[4 marks]
Total for this question: 4

Tier 3 · Hard

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
A diode has a potential difference of $0.720\,\text{V}$ across it when its forward current is $35.0\,\text{mA}$ . It is connected in series with a $1.20\times10^2\,\Omega$ resistor. Determine the supply potential difference, the power dissipated by the diode and its resistance at this operating point. State what happens to the current if the diode is reversed, and explain why the calculated resistance cannot be used at every point on its characteristic.
[5 marks]
Total for this question: 5
A diode is connected in series with a resistor $R$ to a supply that can vary from $4.80\,\text{V}$ to $5.20\,\text{V}$ . The diode current must remain between $18.0\,\text{mA}$ and $22.0\,\text{mA}$ . Its forward potential difference is $0.680\,\text{V}$ at $18.0\,\text{mA}$ and $0.720\,\text{V}$ at $22.0\,\text{mA}$ . Determine the range of values of $R$ that meets both limits and deduce whether a $220\,\Omega$ resistor is suitable.
[6 marks]
Total for this question: 6
A non-ideal voltmeter of resistance $3.00\,\text{k}\Omega$ is connected across a filament lamp. An ammeter placed before the current divides reads $0.415\,\text{A}$ when the voltmeter reads $6.00\,\text{V}$ , and $0.190\,\text{A}$ when it reads $2.00\,\text{V}$ . Correct for the current in the voltmeter and determine the lamp resistance at each operating point. Use the results to explain what the measurements show about the lamp.
[5 marks]
Total for this question: 5
Three components have the following currents when the potential difference across each is set in turn to $-1.20\,\text{V}$ , $-0.600\,\text{V}$ , $+0.600\,\text{V}$ and $+1.20\,\text{V}$ . Component X: $-0.120\,\text{A}$ , $-0.0600\,\text{A}$ , $+0.0600\,\text{A}$ , $+0.120\,\text{A}$ . Component Y: $-0.240\,\text{A}$ , $-0.150\,\text{A}$ , $+0.150\,\text{A}$ , $+0.240\,\text{A}$ . Component Z: $-2\times10^{-6}\,\text{A}$ , $-1\times10^{-6}\,\text{A}$ , $+0.006\,\text{A}$ , $+0.180\,\text{A}$ . Identify the fixed resistor, filament lamp and semiconductor diode, justifying each choice from the data. The three components are then connected in parallel with the polarity that gives $+1.20\,\text{V}$ across each. Determine the supply current and total power transferred.
[6 marks]
Total for this question: 6

3.5.1.3 · Resistivity

Explanation

For a uniform wire, resistivity is $\rho=RA/L$ , a material property measured in $\Omega\,\text{m}$ . Cross-sectional area for a circular wire is $A=\pi d^2/4$ , 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 $1.50\,\text{m}$ wire has resistance $3.2\,\Omega$ and diameter $0.40\,\text{mm}$ . Determine its resistivity.

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

Answer: $\rho=2.7\times10^{-7}\,\Omega\,\text{m}$ .

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

A wire of length $1.5\,\text{m}$ and cross-sectional area $0.20\,\text{mm}^2$ has resistance $4.2\,\Omega$ . Calculate its resistivity.
[2 marks]
Total for this question: 2
State the defining electrical property of a superconductor and one application of this property.
[2 marks]
Total for this question: 2

Tier 2 · Standard

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
Wire A is $1.6\,\text{m}$ long and $0.30\,\text{mm}$ in diameter. Wire B, made from the same material, is $2.4\,\text{m}$ long and $0.45\,\text{mm}$ in diameter. Determine the ratio $R_{\text{B}}/R_{\text{A}}$ .
[3 marks]
Total for this question: 3
A uniform wire has resistance $3.20\,\Omega$ . It is stretched without changing its volume until its length is $1.60$ times its original length. The material and temperature are unchanged. Determine the new resistance.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A $1.80\,\text{m}$ wire has diameter $0.400\,\text{mm}$ . A potential difference of $2.40\,\text{V}$ produces a current of $0.800\,\text{A}$ . 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
A uniform wire has length $12.0\,\text{m}$ , resistance $0.850\,\Omega$ and resistivity $1.72\times10^{-8}\,\Omega\,\text{m}$ . Determine its diameter. A micrometer measures diameter with uncertainty $\pm0.002\,\text{mm}$ ; calculate the percentage uncertainty in the cross-sectional area, and explain why readings should be repeated at several positions and orientations.
[5 marks]
Total for this question: 5
A wire of diameter $(0.250\pm0.003)\,\text{mm}$ is tested without appreciable heating. Its measured resistance is $(2.18\pm0.02)\,\Omega$ at length $0.400\,\text{m}$ and $(6.22\pm0.02)\,\Omega$ at length $1.20\,\text{m}$ . Assume the extra resistance of the contacts is constant and length uncertainty is negligible. Determine the contact resistance and the wire resistivity with its absolute uncertainty. For this calculation, take the uncertainty in a difference as the sum of the absolute uncertainties and add percentage uncertainties in multiplied quantities.
[6 marks]
Total for this question: 6
A wire of uniform diameter $0.360\,\text{mm}$ consists of a $0.800\,\text{m}$ section with resistivity $1.10\times10^{-6}\,\Omega\,\text{m}$ joined in series to a $1.60\,\text{m}$ section with resistivity $4.80\times10^{-7}\,\Omega\,\text{m}$ . A current of $0.250\,\text{A}$ passes through the wire. Determine the resistance of each section, the potential difference across the complete wire and the power dissipated in the first section.
[5 marks]
Total for this question: 5
A coil is made from a uniform wire of mass $32.5\,\text{g}$ , diameter $0.280\,\text{mm}$ and resistance $14.8\,\Omega$ . The wire material has density $8.90\times10^3\,\text{kg m}^{-3}$ . Determine the wire length and its resistivity. A replacement wire is made from the same material with the same length but must have resistance $9.50\,\Omega$ . Determine the required diameter.
[6 marks]
Total for this question: 6

3.5.1.4 · Circuits

Explanation

Series resistances add, $R_{\text{T}}=R_1+R_2+\cdots$ , whereas parallel conductances add, $1/R_{\text{T}}=1/R_1+1/R_2+\cdots$ . 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 $E=IVt$ and $P=IV=I^2R=V^2/R$ .
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 $4.0\,\Omega$ resistor is in series with $6.0\,\Omega$ and $3.0\,\Omega$ resistors in parallel across $18\,\text{V}$ . Determine the supply current.

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

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

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

A $4.0\,\Omega$ resistor and a $6.0\,\Omega$ resistor are connected in series. State their total resistance.
[1 mark]
Total for this question: 1
Three identical cells, each with emf $1.5\,\text{V}$ and internal resistance $0.60\,\Omega$ , are connected in parallel with the same polarity. State the emf and calculate the effective internal resistance of the combination.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A $12\,\Omega$ resistor and a $6.0\,\Omega$ resistor are connected in parallel across an $18\,\text{V}$ supply. Determine the total resistance, the supply current and the total power.
[3 marks]
Total for this question: 3
A current of $2.60\,\text{A}$ reaches a junction and divides between two resistors. One branch carries $0.850\,\text{A}$ . The potential difference across both branches is $7.00\,\text{V}$ . Determine the resistance of the other branch.
[3 marks]
Total for this question: 3
A $12.0\,\text{V}$ cell and a $3.00\,\text{V}$ cell are connected in opposition in a single loop with $2.00\,\Omega$ and $4.00\,\Omega$ resistors in series. Both cells have negligible internal resistance. Determine the current, its direction and the power dissipated by the $4.00\,\Omega$ resistor.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A $3.0\,\Omega$ resistor is in series with a parallel pair of $6.0\,\Omega$ and $3.0\,\Omega$ resistors. The network is connected to a $20\,\text{V}$ supply. Calculate the energy transferred by the $6.0\,\Omega$ resistor in $90\,\text{s}$ .
[5 marks]
Total for this question: 5
A $24\,\text{V}$ supply is connected across two parallel branches. One branch contains an $8.0\,\Omega$ resistor. The other contains a $5.0\,\Omega$ resistor in series with an unknown resistor. The supply current is $4.2\,\text{A}$ . Determine the unknown resistance and the power dissipated by it.
[5 marks]
Total for this question: 5
A $30.0\,\text{V}$ supply of negligible internal resistance is connected to a $6.00\,\Omega$ resistor in series with a parallel combination of a $12.0\,\Omega$ resistor and an unknown resistor $R$ . The power dissipated in the $12.0\,\Omega$ resistor is $27.0\,\text{W}$ . Determine the potential difference across the parallel combination, the current in each resistor, $R$ , and the energy transferred by the supply in $5.00\,\text{min}$ .
[6 marks]
Total for this question: 6
Lamp A is rated $6.0\,\text{V},\ 2.4\,\text{W}$ and lamp B is rated $6.0\,\text{V},\ 1.2\,\text{W}$ . Assume that each lamp keeps the resistance calculated from its rating. The lamps are first connected in series across a $12\,\text{V}$ supply. Determine the power in each lamp and identify which lamp exceeds its rating. A resistor is then connected in parallel with lamp B so that both lamps operate at their rated potential difference. Determine the required resistance and the current from the supply.
[6 marks]
Total for this question: 6
Two resistors are connected first in series and then in parallel across the same $22.0\,\text{V}$ supply. The total power in the parallel arrangement is $6.25$ times the total power in the series arrangement. The series arrangement dissipates $8.80\,\text{W}$ . Determine the two resistance values, the power in each resistor when they are in parallel and the energy transferred by the parallel arrangement in $90.0\,\text{s}$ .
[6 marks]
Total for this question: 6

3.5.1.5 · Potential divider

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 $R_2$ is $V_{\text{out}}=V_{\text{in}}R_2/(R_1+R_2)$ .
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 $2.0\,\text{k}\Omega$ upper resistor and $6.0\,\text{k}\Omega$ lower resistor form an unloaded divider across $12\,\text{V}$ . Determine the output across the lower resistor.

1.Identify the lower resistance as the output resistance, $R_2=6.0\,\text{k}\Omega$ .
2.Use $V_{\text{out}}=12[6.0/(2.0+6.0)]$ .
3.Evaluate the fraction: $V_{\text{out}}=9.0\,\text{V}$ .

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

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

Two equal resistors form an unloaded potential divider across a $12\,\text{V}$ supply. State the potential difference across either resistor.
[1 mark]
Total for this question: 1
The full track of a variable resistor is connected across a $6.0\,\text{V}$ supply. The output is taken between the slider and the lower end of the track. State the full range of output potential difference.
[2 marks]
Total for this question: 2

Tier 2 · Standard

An LDR is the lower component of a potential divider and a $3.0\,\text{k}\Omega$ fixed resistor is the upper component. The supply is $9.0\,\text{V}$ . Calculate the output across the LDR when its resistance is $9.0\,\text{k}\Omega$ in darkness and $1.0\,\text{k}\Omega$ in bright light.
[3 marks]
Total for this question: 3
An unloaded potential divider has an $8.2\,\text{k}\Omega$ upper resistor and a $4.7\,\text{k}\Omega$ lower resistor across a $12.0\,\text{V}$ supply. Determine the output potential difference across the lower resistor and the current drawn from the supply.
[3 marks]
Total for this question: 3
The full $12.0\,\text{k}\Omega$ track of a variable resistor is connected across a $9.00\,\text{V}$ supply. The unloaded output between the slider and the lower end is $5.40\,\text{V}$ . Determine the resistances of the track below and above the slider.
[3 marks]
Total for this question: 3

Tier 3 · Hard

An NTC thermistor is the lower component of an unloaded potential divider. The upper resistor is $4.70\,\text{k}\Omega$ and the supply is $12.0\,\text{V}$ . A controller switches when the output across the thermistor is $4.00\,\text{V}$ . 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
An NTC thermistor is the upper component of a potential divider across a $15.0\,\text{V}$ supply. The output is taken across the fixed lower resistor. The thermistor resistance is $3.3\,\text{k}\Omega$ at $60\,^{\circ}\text{C}$ and $10\,\text{k}\Omega$ at $20\,^{\circ}\text{C}$ . Determine the fixed resistance required for a $6.0\,\text{V}$ output at $60\,^{\circ}\text{C}$ and the output at $20\,^{\circ}\text{C}$ . Explain why cooling reduces the output.
[5 marks]
Total for this question: 5
An LDR is the lower component of an unloaded potential divider across a $12.0\,\text{V}$ supply. Its resistance is $18.0\,\text{k}\Omega$ in darkness and $2.00\,\text{k}\Omega$ in bright light. Determine the range of values of the fixed upper resistance for which the output is greater than $8.00\,\text{V}$ in darkness but less than $3.00\,\text{V}$ in bright light.
[6 marks]
Total for this question: 6
An unloaded divider has a fixed resistor above an LDR. The LDR resistance is $15.0\,\text{k}\Omega$ in one light level and $2.50\,\text{k}\Omega$ in a brighter light. The corresponding outputs measured across the LDR are $10.0\,\text{V}$ and $3.75\,\text{V}$ . Determine the fixed resistance, the supply potential difference and the current drawn at each light level.
[5 marks]
Total for this question: 5
An NTC thermistor of resistance $8.20\,\text{k}\Omega$ is the lower component of a divider whose upper resistor is $4.70\,\text{k}\Omega$ . The supply is $12.0\,\text{V}$ . A controller of input resistance $15.0\,\text{k}\Omega$ is connected across the thermistor, so it loads the divider. Determine the effective resistance of the lower branch and the controller input potential difference. Calculate the unloaded output and hence the percentage by which loading reduces the output. When warm, the thermistor resistance is $3.30\,\text{k}\Omega$ . Determine whether the loaded output then falls below the controller's $4.50\,\text{V}$ switching level.
[6 marks]
Total for this question: 6

3.5.1.6 · Electromotive force and internal resistance

Explanation

Electromotive force is energy supplied by a source per unit charge, $\varepsilon=E/Q$ ; it is not a mechanical force. A real source has internal resistance $r$ .
When it supplies current $I$ to external resistance $R$ , $\varepsilon=I(R+r)$ and terminal potential difference is $V=IR=\varepsilon-Ir$ . The term $Ir$ is the lost volts across the internal resistance, so terminal potential difference decreases as current increases.
On a graph of $V$ against $I$ , the intercept is $\varepsilon$ and the gradient is $-r$ .
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 $r$ , 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 $5.4\,\text{V}$ at $0.50\,\text{A}$ and $4.2\,\text{V}$ at $2.0\,\text{A}$ . Determine its internal resistance and emf.

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

Answer: The internal resistance is $0.80\,\Omega$ and the emf is $5.8\,\text{V}$ .

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 $Ir$ to the terminal potential difference with inconsistent units or current values.

Exam tip

On a $V$ – $I$ graph, label the intercept $\varepsilon$ and state explicitly that $r$ is the magnitude of the gradient.

Tier 1 · Easy

A cell of internal resistance $0.50\,\Omega$ supplies a $2.5\,\Omega$ resistor with a current of $2.0\,\text{A}$ . Calculate the emf of the cell.
[2 marks]
Total for this question: 2
Explain why the terminal potential difference of a cell equals its emf when no current is drawn.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A battery has emf $9.0\,\text{V}$ and internal resistance $0.80\,\Omega$ . It is connected to a $4.2\,\Omega$ load. Determine the current, terminal potential difference and lost volts.
[3 marks]
Total for this question: 3
A battery has an open-circuit potential difference of $12.0\,\text{V}$ . When it supplies $2.40\,\text{A}$ , its terminal potential difference is $10.8\,\text{V}$ . Determine its internal resistance and the power dissipated inside the battery.
[3 marks]
Total for this question: 3
A voltmeter used to investigate a cell reads $0.080\,\text{V}$ too high at every current. The uncorrected best-fit relation is $V_{\text{read}}=1.62-0.350I$ , with SI units. Determine the corrected emf, the internal resistance and the terminal potential difference when $I=0.800\,\text{A}$ .
[4 marks]
Total for this question: 4

Tier 3 · Hard

Measurements for a cell give terminal potential differences of $5.40\,\text{V}$ at $0.50\,\text{A}$ and $4.20\,\text{V}$ at $2.00\,\text{A}$ . Determine the emf and internal resistance. The cell is then connected to a $2.10\,\Omega$ resistor. Calculate the new current and terminal potential difference.
[5 marks]
Total for this question: 5
A cell of emf $18.0\,\text{V}$ and internal resistance $1.20\,\Omega$ is connected to a $6.80\,\Omega$ load for $40.0\,\text{s}$ . Determine the terminal potential difference, the energy transferred to the load, the energy dissipated inside the cell and the efficiency of energy transfer to the load.
[5 marks]
Total for this question: 5
Two identical cells each have emf $1.50\,\text{V}$ and internal resistance $0.800\,\Omega$ . They may be connected either in series or in parallel across the same load resistance $R$ . Derive the value of $R$ for which the power transferred to the load is the same in both arrangements. Hence determine which arrangement transfers more power to a $0.300\,\Omega$ load, supporting your conclusion with calculated powers.
[6 marks]
Total for this question: 6
A battery consists of six identical cells in series, each of emf $1.55\,\text{V}$ and internal resistance $0.180\,\Omega$ . One cell has accidentally been connected with its polarity reversed. The battery is connected to a $3.60\,\Omega$ load. Determine the net emf and the current. The reversed cell is then corrected. Determine the new current and load power, and calculate the factor by which the load power increases.
[5 marks]
Total for this question: 5
A source of emf $14.0\,\text{V}$ and internal resistance $0.800\,\Omega$ charges a rechargeable cell of emf $9.00\,\text{V}$ and internal resistance $0.400\,\Omega$ through a $2.80\,\Omega$ resistor. The emfs oppose. Determine the charging current, the terminal potential difference of each cell, the rate at which energy is stored chemically in the rechargeable cell, the total rate of heating in both cells and the efficiency of transfer from the source emf to stored chemical energy.
[6 marks]
Total for this question: 6

Answer key

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

3.5.1.1 · Basics of electricity

Tier 1 · Easy

Mark scheme for 3.5.1.1 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$4.0\,\text{A}$	Use $I=\dfrac{\Delta Q}{\Delta t}$ . Hence $I=\dfrac{3.2}{0.80}=4.0\,\text{A}$ .	1
02.1	$60\,\text{C}$	Rearrange $I=\Delta Q/\Delta t$ to give $\Delta Q=I\Delta t$ . Therefore $\Delta Q=0.75\times80=60\,\text{C}$ .	1

Tier 2 · Standard

Mark scheme for 3.5.1.1 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$6.0\,\text{V}$ $8.0\,\Omega$	First use $V=\dfrac{W}{Q}=\dfrac{18}{3.0}=6.0\,\text{V}$ . Then $R=\dfrac{V}{I}=\dfrac{6.0}{0.75}=8.0\,\Omega$ .	3
02.1	$9.45\times10^{19}$ electrons	The charge transferred is $Q=It=0.420\times36.0=15.12\,\text{C}$ . Each electron has charge magnitude $e$ , so $N=Q/e=15.12/(1.60\times10^{-19})=9.45\times10^{19}$ electrons to three significant figures.	3
03.1	$3.13\times10^{18}\,\text{electrons s}^{-1}$ (accept $3.12\times10^{18}\,\text{electrons s}^{-1}$ )	The charge transferred is $Q=W/V=3600/15.0=240\,\text{C}$ . The time is $8.00\times60=480\,\text{s}$ , so the mean current is $I=Q/t=240/480=0.500\,\text{A}$ . Therefore the number passing per second is $I/e=0.500/(1.60\times10^{-19})=3.13\times10^{18}\,\text{electrons s}^{-1}$ to three significant figures.	4

Tier 3 · Hard

Mark scheme for 3.5.1.1 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$3.0\times10^2\,\text{C}$ $7.2\times10^3\,\text{J}$ $12\,\Omega$	The charge is $Q=It=2.0\times150=3.0\times10^2\,\text{C}$ . The energy is $W=VQ=24\times300=7.2\times10^3\,\text{J}$ . The resistance is $R=\dfrac{V}{I}=\dfrac{24}{2.0}=12\,\Omega$ .	5
02.1	$4.00\times10^{-4}\,\text{C}$ $0.100\,\text{A}$ $0.260\,\text{J}$ $65.0\,\text{W}$	The charge magnitude is $Q=Ne=(2.50\times10^{15})(1.60\times10^{-19})=4.00\times10^{-4}\,\text{C}$ . The time is $4.00\times10^{-3}\,\text{s}$ , so $I=Q/t=0.100\,\text{A}$ . The energy is $W=VQ=650(4.00\times10^{-4})=0.260\,\text{J}$ . Hence the average power is $P=W/t=0.260/(4.00\times10^{-3})=65.0\,\text{W}$ , with each result given to three significant figures.	5
03.1	$15.0\,\text{s}$ $1.13\times10^{22}$ electrons (accept $1.12\times10^{22}$ electrons)	In the first stage, $Q_1=It=0.800(45.0)=36.0\,\text{C}$ and $W_1=VQ_1=6.00(36.0)=216\,\text{J}$ . In the second stage, $Q_2=2.40t$ and $W_2=3.00(2.40t)=7.20t$ . Since energy transferred per unit charge is $4.50\,\text{J C}^{-1}$ , $(216+7.20t)/(36.0+2.40t)=4.50$ . Hence $216+7.20t=162+10.8t$ , giving $t=15.0\,\text{s}$ . One cycle transfers $36.0+2.40(15.0)=72.0\,\text{C}$ , so $25$ cycles transfer $1800\,\text{C}$ . The number of electrons is $N=Q/e=1800/(1.60\times10^{-19})=1.125\times10^{22}$ , or $1.13\times10^{22}$ to three significant figures.	6
04.1	$26.3\,\text{C}$ $197\,\text{J}$ $0.351\,\text{A}$ $1.65\times10^{20}$ electrons	During the uniform rise, the mean current is $(0.180+0.620)/2=0.400\,\text{A}$ , so the charge is $0.400(35.0)=14.0\,\text{C}$ . The final stage transfers $0.440(28.0)=12.32\,\text{C}$ , giving $Q=14.0+12.32=26.32\,\text{C}=26.3\,\text{C}$ . The energy is $W=VQ=7.50(26.32)=197.4\,\text{J}=197\,\text{J}$ . The mean current over the whole interval, including the zero-current stage, is $26.32/75.0=0.3509\,\text{A}=0.351\,\text{A}$ . Finally, $N=Q/e=26.32/(1.60\times10^{-19})=1.645\times10^{20}$ , or $1.65\times10^{20}$ (accept $1.64\times10^{20}$ if the charge is rounded first) electrons.	5
05.1	$0.700\,\text{V}$ $6.00\,\text{C}$ $2.00\times10^3\,\text{A}$ $3.50\times10^{-4}\,\Omega$	The potential difference is the energy transferred per unit charge: $V=W/Q=1680/(2.40\times10^3)=0.700\,\text{V}$ . Each pulse transfers $Q_{\text{pulse}}=(2.40\times10^3)/(4.00\times10^2)=6.00\,\text{C}$ . During a pulse, $I=Q_{\text{pulse}}/t=6.00/(3.00\times10^{-3})=2.00\times10^3\,\text{A}$ . The pulse resistance is $R=V/I=0.700/(2.00\times10^3)=3.50\times10^{-4}\,\Omega$ .	4

3.5.1.2 · Current-voltage characteristics

Tier 1 · Easy

Mark scheme for 3.5.1.2 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	The current is directly proportional to the potential difference when physical conditions, including temperature, are constant.	Award the statement $I\propto V$ together with the requirement that physical conditions, particularly temperature, remain constant.	1
02.1	An ideal ammeter has zero resistance and an ideal voltmeter has infinite resistance.	State both idealisations: $R_{\text{ammeter}}=0$ and $R_{\text{voltmeter}}\to\infty$ .	2

Tier 2 · Standard

Mark scheme for 3.5.1.2 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	There is negligible reverse current; the forward current is also small at first, then rises rapidly after the turn-on region.	For reverse bias, state that the current is negligible. For small forward potential differences, the current remains small. Beyond the turn-on region, a small further increase in potential difference produces a large increase in current.	3
02.1	$1.3$	At $4.0\,\text{V}$ , $R_1=V/I=4.0/0.32=12.5\,\Omega$ . At $8.0\,\text{V}$ , $R_2=8.0/0.50=16.0\,\Omega$ . The increase factor is $R_2/R_1=16.0/12.5=1.28$ , which is $1.3$ to two significant figures.	3
03.1	Connect an ammeter in series and a voltmeter in parallel with the resistor; vary the current using a variable resistor or variable supply, reverse the supply for the opposite polarity, and use small currents or switch off between readings to limit heating.	The ammeter must measure the current through the resistor, so it is connected in series. The voltmeter must measure the potential difference across the resistor, so it is connected in parallel. Take several paired readings while changing the current with a variable resistor or variable supply, then reverse the supply connections to obtain negative values. Keep currents small and switch off between readings so that heating does not change the resistor's temperature and resistance.	4

Tier 3 · Hard

Mark scheme for 3.5.1.2 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	The lamp filament becomes hotter, increasing lattice vibrations and its resistance, so current rises less rapidly; the fixed resistor stays approximately at constant temperature and remains ohmic.	Increasing the potential difference increases the current and therefore the power dissipated in the lamp. Its filament temperature rises. Greater lattice vibrations cause more frequent scattering of conduction electrons, so the filament resistance increases. Consequently the current rises by progressively smaller amounts for equal potential-difference increases, producing a curved graph. The fixed resistor is assumed to remain at approximately constant temperature, so its resistance is constant and $I\propto V$ gives a straight line through the origin.	5
02.1	$4.92\,\text{V}$ $2.52\times10^{-2}\,\text{W}$ $20.6\,\Omega$ The reverse current is negligible. A diode is non-ohmic, so the ratio $V/I$ changes along its curved characteristic.	The resistor potential difference is $V_R=IR=(35.0\times10^{-3})(1.20\times10^2)=4.20\,\text{V}$ , so the supply potential difference is $4.20+0.720=4.92\,\text{V}$ . The power dissipated by the diode is $P=IV=(35.0\times10^{-3})(0.720)=2.52\times10^{-2}\,\text{W}$ and its operating-point resistance is $R=V/I=0.720/0.0350=20.6\,\Omega$ . Reversing the diode gives only a negligible current. Because its characteristic is not a straight line through the origin, $V/I$ is not constant and the operating-point resistance cannot predict the current elsewhere.	5
03.1	$204\,\Omega\le R\le228\,\Omega$ ; a $220\,\Omega$ resistor is suitable.	The minimum current occurs at the minimum supply potential difference. Requiring at least $18.0\,\text{mA}$ gives $R\le(4.80-0.680)/(18.0\times10^{-3})=228.9\,\Omega$ . The maximum current occurs at the maximum supply potential difference. Requiring no more than $22.0\,\text{mA}$ gives $R\ge(5.20-0.720)/(22.0\times10^{-3})=203.6\,\Omega$ . Therefore the inward-rounded whole-ohm range is $204\,\Omega\le R\le228\,\Omega$ . Since $220\,\Omega$ lies within this interval, it is suitable.	6
04.1	$14.5\,\Omega$ at $6.00\,\text{V}$ $10.6\,\Omega$ at $2.00\,\text{V}$ The lamp is non-ohmic: its resistance is greater at the higher-current operating point because its filament is hotter.	At $6.00\,\text{V}$ the voltmeter current is $6.00/(3.00\times10^3)=2.00\,\text{mA}$ , so the lamp current is $0.415-0.00200=0.413\,\text{A}$ and $R=6.00/0.413=14.5\,\Omega$ . At $2.00\,\text{V}$ the voltmeter current is $2.00/(3.00\times10^3)=0.667\,\text{mA}$ , so the lamp current is $0.190-0.000667=0.18933\,\text{A}$ and $R=2.00/0.18933=10.6\,\Omega$ . The resistance increases by a factor $14.5278/10.5634=1.38$ . The larger current raises the filament temperature, increasing lattice vibrations and electron scattering, so the lamp does not obey Ohm's law across these operating points.	5
05.1	X is the fixed resistor: $I/V$ is constant and the characteristic is symmetric. Y is the filament lamp: its resistance increases from $4.00\,\Omega$ to $5.00\,\Omega$ as the magnitude of the potential difference rises. Z is the diode: it has negligible reverse current and a much larger forward current beyond its turn-on region. $0.540\,\text{A}$ $0.648\,\text{W}$	For X, $V/I=1.20/0.120=0.600/0.0600=10.0\,\Omega$ , with the same behaviour in both polarities, so X is the ohmic fixed resistor. For Y, $R=0.600/0.150=4.00\,\Omega$ at the smaller magnitude but $R=1.20/0.240=5.00\,\Omega$ at the larger magnitude. This symmetric increase is caused by heating of a filament, so Y is the lamp. Z has currents of only $1$ – $2\,\text{mA}$ in reverse, but its forward current rises from $6\,\text{mA}$ to $180\,\text{mA}$ , so Z is the diode. In parallel at $+1.20\,\text{V}$ , $I_{\text{total}}=0.120+0.240+0.180=0.540\,\text{A}$ . Hence $P=VI=1.20(0.540)=0.648\,\text{W}$ .	6

3.5.1.3 · Resistivity

Tier 1 · Easy

Mark scheme for 3.5.1.3 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$5.6\times10^{-7}\,\Omega\,\text{m}$	Convert the area: $0.20\,\text{mm}^2=0.20\times10^{-6}\,\text{m}^2=2.0\times10^{-7}\,\text{m}^2$ . Then $\rho=\dfrac{RA}{L}=\dfrac{4.2\times2.0\times10^{-7}}{1.5}=5.6\times10^{-7}\,\Omega\,\text{m}$ .	2
02.1	At or below its critical temperature a superconductor has zero resistivity; this can reduce transmission losses or enable strong electromagnets.	Give the defining condition, $\rho=0$ at and below the critical temperature, followed by one valid application such as loss-free power transmission or producing a strong magnetic field.	2

Tier 2 · Standard

Mark scheme for 3.5.1.3 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	Heating releases more charge carriers in the semiconductor; the increased carrier density outweighs increased scattering, so resistance decreases.	A higher temperature provides energy that releases more charge carriers in the semiconductor. This increases the number density of mobile carriers. Although collisions may also become more frequent, the carrier-density increase dominates, so conductivity rises and resistance falls.	3
02.1	$0.67$	For the same material, $R\propto L/A$ and $A\propto d^2$ . Hence $R_{\text{B}}/R_{\text{A}}=(2.4/1.6)(0.30/0.45)^2=0.67$ to two significant figures.	3
03.1	$8.19\,\Omega$	Constant volume gives $A'L'=AL$ . Since $L'=1.60L$ , the new area is $A'=A/1.60$ . The resistivity is unchanged because the material and temperature are unchanged. Therefore $R'/R=(L'/L)(A/A')=(1.60)(1.60)=2.56$ , so $R'=2.56(3.20)=8.19\,\Omega$ to three significant figures.	3

Tier 3 · Hard

Mark scheme for 3.5.1.3 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$2.09\times10^{-7}\,\Omega\,\text{m}$ ; below the critical temperature the resistivity and hence the resistive power loss are zero.	The resistance is $R=\dfrac{V}{I}=\dfrac{2.40}{0.800}=3.00\,\Omega$ . The radius is $0.200\,\text{mm}=2.00\times10^{-4}\,\text{m}$ , so $A=\pi r^2=1.257\times10^{-7}\,\text{m}^2$ . Hence $\rho=\dfrac{RA}{L}=\dfrac{3.00\times1.257\times10^{-7}}{1.80}=2.09\times10^{-7}\,\Omega\,\text{m}$ . Below the critical temperature a superconductor has $\rho=0$ , so the cable resistance and the resistive loss $P=I^2R$ are zero.	5
02.1	$0.556\,\text{mm}$ $0.72\%$ Repeated readings reveal diameter variation and reduce random uncertainty when a mean is taken.	From $R=\rho L/A$ , $A=\rho L/R=(1.72\times10^{-8})(12.0)/0.850=2.43\times10^{-7}\,\text{m}^2$ . Since $A=\pi d^2/4$ , $d=2\sqrt{A/\pi}=5.56\times10^{-4}\,\text{m}=0.556\,\text{mm}$ . Because $A\propto d^2$ , its percentage uncertainty is $2(0.002/0.556)\times100=0.72\%$ . Measurements at different positions and orientations test whether the wire is uniform and circular; averaging them reduces random uncertainty.	5
03.1	$0.160\,\Omega$ $(2.48\pm0.08)\times10^{-7}\,\Omega\,\text{m}$	The resistance per unit length is $(6.22-2.18)/(1.20-0.400)=4.04/0.800=5.05\,\Omega\,\text{m}^{-1}$ . The intercept, which is the contact resistance, is $2.18-5.05(0.400)=0.160\,\Omega$ . The wire area is $A=\pi d^2/4=\pi(0.250\times10^{-3})^2/4=4.91\times10^{-8}\,\text{m}^2$ , so $\rho=(R/L)A=5.05(4.91\times10^{-8})=2.48\times10^{-7}\,\Omega\,\text{m}$ . The percentage uncertainty in the resistance difference is $(0.02+0.02)/4.04\times100=0.990\%$ , and that in area is $2(0.003/0.250)\times100=2.40\%$ . Their sum is $3.39\%$ , giving an absolute uncertainty of $0.084\times10^{-7}\,\Omega\,\text{m}$ , reported as $(2.48\pm0.08)\times10^{-7}\,\Omega\,\text{m}$ .	6
04.1	$8.65\,\Omega$ and $7.55\,\Omega$ $4.05\,\text{V}$ $0.540\,\text{W}$	The cross-sectional area is $A=\pi d^2/4=\pi(0.360\times10^{-3})^2/4=1.01788\times10^{-7}\,\text{m}^2$ . Hence $R_1=\rho_1L_1/A=(1.10\times10^{-6})(0.800)/(1.01788\times10^{-7})=8.645\,\Omega$ and $R_2=(4.80\times10^{-7})(1.60)/(1.01788\times10^{-7})=7.545\,\Omega$ . The total potential difference is $V=I(R_1+R_2)=0.250(8.645+7.545)=4.0476\,\text{V}=4.05\,\text{V}$ . The first section dissipates $P_1=I^2R_1=(0.250)^2(8.645)=0.540\,\text{W}$ .	5
05.1	$59.3\,\text{m}$ $1.54\times10^{-8}\,\Omega\,\text{m}$ $0.349\,\text{mm}$	The area is $A=\pi(0.280\times10^{-3})^2/4=6.1575\times10^{-8}\,\text{m}^2$ . The wire volume is $m/\rho_{\text{density}}=0.0325/(8.90\times10^3)=3.6517\times10^{-6}\,\text{m}^3$ , so $L=V/A=3.6517\times10^{-6}/(6.1575\times10^{-8})=59.304\,\text{m}=59.3\,\text{m}$ . Its resistivity is $\rho=RA/L=14.8(6.1575\times10^{-8})/59.304=1.5367\times10^{-8}\,\Omega\,\text{m}$ . For fixed material and length, $R\propto1/d^2$ , so $d_{\text{new}}=0.280\sqrt{14.8/9.50}=0.349\,\text{mm}$ .	6

3.5.1.4 · Circuits

Tier 1 · Easy

Mark scheme for 3.5.1.4 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$10\,\Omega$	For series resistors, $R_{\text{T}}=R_1+R_2=4.0+6.0=10.0\,\Omega$ .	1
02.1	$1.5\,\text{V}$ $0.20\,\Omega$	Identical cells connected in parallel have the same emf as one cell, so the combination has emf $1.5\,\text{V}$ . Their identical internal resistances are in parallel, giving $r_{\text{effective}}=0.60/3=0.20\,\Omega$ .	2

Tier 2 · Standard

Mark scheme for 3.5.1.4 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$4.0\,\Omega$ $4.5\,\text{A}$ $81\,\text{W}$	$\dfrac{1}{R_{\text{T}}}=\dfrac{1}{12}+\dfrac{1}{6}=\dfrac{3}{12}$ , so $R_{\text{T}}=4.0\,\Omega$ . The supply current is $I=\dfrac{V}{R_{\text{T}}}=\dfrac{18}{4.0}=4.5\,\text{A}$ . Hence $P=VI=18\times4.5=81\,\text{W}$ .	3
02.1	$4.00\,\Omega$	Conservation of charge gives the other branch current as $2.60-0.850=1.75\,\text{A}$ . Parallel branches have the same potential difference, so $R=V/I=7.00/1.75=4.00\,\Omega$ .	3
03.1	$1.50\,\text{A}$ in the direction driven by the $12.0\,\text{V}$ cell $9.00\,\text{W}$	The cells oppose, so the resultant emf is $12.0-3.00=9.00\,\text{V}$ in the direction of the larger-emf cell. The total resistance is $2.00+4.00=6.00\,\Omega$ , giving $I=9.00/6.00=1.50\,\text{A}$ . The power in the $4.00\,\Omega$ resistor is $P=I^2R=(1.50)^2(4.00)=9.00\,\text{W}$ .	4

Tier 3 · Hard

Mark scheme for 3.5.1.4 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$9.6\times10^2\,\text{J}$	The parallel resistance is $R_{\parallel}=\left(\dfrac{1}{6.0}+\dfrac{1}{3.0}\right)^{-1}=2.0\,\Omega$ . The total resistance is $3.0+2.0=5.0\,\Omega$ , so the supply current is $20/5.0=4.0\,\text{A}$ . The potential difference across the series $3.0\,\Omega$ resistor is $4.0\times3.0=12\,\text{V}$ , leaving $8.0\,\text{V}$ across the parallel pair. The current in the $6.0\,\Omega$ resistor is $8.0/6.0=1.33\,\text{A}$ . Therefore $E=IVt=1.33\times8.0\times90=9.6\times10^2\,\text{J}$ .	5
02.1	$15\,\Omega$ $22\,\text{W}$	The current in the $8.0\,\Omega$ branch is $24/8.0=3.0\,\text{A}$ . Conservation of charge gives $4.2-3.0=1.2\,\text{A}$ in the other branch. Its total resistance is $24/1.2=20\,\Omega$ , so the unknown resistance is $20-5.0=15\,\Omega$ . Its power dissipation is $P=I^2R=(1.2)^2(15)=21.6\,\text{W}$ , which is $22\,\text{W}$ to two significant figures.	5
03.1	$18.0\,\text{V}$ $2.00\,\text{A}$ in the $6.00\,\Omega$ resistor, $1.50\,\text{A}$ in the $12.0\,\Omega$ resistor and $0.500\,\text{A}$ in $R$ $R=36.0\,\Omega$ $18.0\,\text{kJ}$	For the $12.0\,\Omega$ resistor, $P=V^2/R$ , so the potential difference across the parallel combination is $V=\sqrt{PR}=\sqrt{27.0(12.0)}=18.0\,\text{V}$ . The potential difference across the series resistor is $30.0-18.0=12.0\,\text{V}$ , giving a current of $12.0/6.00=2.00\,\text{A}$ . The current in the $12.0\,\Omega$ resistor is $18.0/12.0=1.50\,\text{A}$ , so conservation of charge gives $2.00-1.50=0.500\,\text{A}$ in $R$ . Hence $R=18.0/0.500=36.0\,\Omega$ . In $5.00\,\text{min}=300\,\text{s}$ , the supply transfers $E=VIt=30.0(2.00)(300)=1.80\times10^4\,\text{J}=18.0\,\text{kJ}$ .	6
04.1	$1.1\,\text{W}$ in lamp A and $2.1\,\text{W}$ in lamp B Lamp B exceeds its $1.2\,\text{W}$ rating. $30\,\Omega$ $0.400\,\text{A}$	From the ratings, $R_A=V^2/P=6.0^2/2.4=15\,\Omega$ and $R_B=6.0^2/1.2=30\,\Omega$ . In series, $I=12/(15+30)=0.2667\,\text{A}$ . Therefore $P_A=I^2R_A=(0.2667)^2(15)=1.07\,\text{W}=1.1\,\text{W}$ and $P_B=(0.2667)^2(30)=2.13\,\text{W}=2.1\,\text{W}$ , so B exceeds its rating. For each lamp to have $6.0\,\text{V}$ , lamp A must be in series with a parallel section whose equivalent resistance is $15\,\Omega$ . Since lamp B is $30\,\Omega$ , its parallel resistor must also be $30\,\Omega$ , giving $15\,\Omega$ for the parallel pair. The total resistance is then $30\,\Omega$ , so the supply current is $12/30=0.400\,\text{A}$ .	6
05.1	$11.0\,\Omega$ and $44.0\,\Omega$ $44.0\,\text{W}$ in the $11.0\,\Omega$ resistor and $11.0\,\text{W}$ in the $44.0\,\Omega$ resistor $4.95\times10^3\,\text{J}$	The series resistance is $R_1+R_2=V^2/P=22.0^2/8.80=55.0\,\Omega$ . Let $R_2=kR_1$ . The power ratio is $P_{\parallel}/P_{\text{series}}=(R_1+R_2)^2/(R_1R_2)=(1+k)^2/k=6.25$ . Thus $k^2-4.25k+1=0$ , giving reciprocal roots $k=4.00$ or $0.250$ ; the two resistances are therefore in the ratio $4:1$ . With a sum of $55.0\,\Omega$ , they are $44.0\,\Omega$ and $11.0\,\Omega$ . In parallel each receives $22.0\,\text{V}$ , so the powers are $22.0^2/11.0=44.0\,\text{W}$ and $22.0^2/44.0=11.0\,\text{W}$ . The total energy in $90.0\,\text{s}$ is $(44.0+11.0)(90.0)=4.95\times10^3\,\text{J}$ .	6

3.5.1.5 · Potential divider

Tier 1 · Easy

Mark scheme for 3.5.1.5 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$6.0\,\text{V}$	Equal series resistors share the supply potential difference equally, so $V=12/2=6.0\,\text{V}$ .	1
02.1	$0\,\text{V}$ to $6.0\,\text{V}$	At one end the slider selects zero resistance below the output, giving $0\,\text{V}$ . At the other end it selects the full supply potential difference, giving $6.0\,\text{V}$ .	2

Tier 2 · Standard

Mark scheme for 3.5.1.5 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$6.8\,\text{V}$ in darkness $2.3\,\text{V}$ in bright light	Use $V_{\text{out}}=9.0\dfrac{R_{\text{LDR}}}{3.0+R_{\text{LDR}}}$ with resistances in $\text{k}\Omega$ . In darkness, $V_{\text{out}}=9.0\times9.0/12.0=6.75\,\text{V}$ , which is $6.8\,\text{V}$ to two significant figures. In bright light, $V_{\text{out}}=9.0\times1.0/4.0=2.25\,\text{V}$ , which is $2.3\,\text{V}$ to two significant figures.	3
02.1	$4.4\,\text{V}$ $0.93\,\text{mA}$	The total resistance is $8.2+4.7=12.9\,\text{k}\Omega$ , so the supply current is $I=12.0/(12.9\times10^3)=0.93\,\text{mA}$ . The output is $V_{\text{out}}=I(4.7\times10^3)=4.4\,\text{V}$ , with both results given to two significant figures.	3
03.1	$7.20\,\text{k}\Omega$ below $4.80\,\text{k}\Omega$ above	The track current is $I=9.00/(12.0\times10^3)=0.750\,\text{mA}$ . The resistance below the slider is $R_{\text{below}}=V_{\text{out}}/I=5.40/(0.750\times10^{-3})=7.20\,\text{k}\Omega$ . The remaining resistance above the slider is $12.0-7.20=4.80\,\text{k}\Omega$ .	3

Tier 3 · Hard

Mark scheme for 3.5.1.5 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$2.35\,\text{k}\Omega$ $1.70\,\text{mA}$ The output decreases as the thermistor becomes warmer.	Let the thermistor resistance be $R$ . Then $4.00=12.0\dfrac{R}{4700+R}$ . Hence $4(4700+R)=12R$ , so $R=2350\,\Omega=2.35\,\text{k}\Omega$ . The total resistance is $4700+2350=7050\,\Omega$ , giving $I=12.0/7050=1.70\times10^{-3}\,\text{A}=1.70\,\text{mA}$ . Warming an NTC thermistor lowers $R$ , so the fraction $R/(4700+R)$ and therefore the output decrease.	5
02.1	$2.2\,\text{k}\Omega$ $2.7\,\text{V}$ Cooling increases the NTC thermistor resistance, so the fixed resistor receives a smaller fraction of the supply potential difference.	At $60\,^{\circ}\text{C}$ , let the fixed lower resistance be $R$ . Then $6.0=15.0R/(3300+R)$ , giving $R=2200\,\Omega=2.2\,\text{k}\Omega$ . At $20\,^{\circ}\text{C}$ , $V_{\text{out}}=15.0\times2.2/(10+2.2)=2.7\,\text{V}$ to two significant figures. Cooling an NTC thermistor raises its resistance, increasing the upper share of the total resistance and reducing the fraction of the supply across the fixed lower resistor.	5
03.1	$6.00\,\text{k}\Omega<R_{\text{fixed}}<9.00\,\text{k}\Omega$	In darkness, $V_{\text{out}}=12.0(18.0)/(R_{\text{fixed}}+18.0)$ with resistances in $\text{k}\Omega$ . Requiring this to exceed $8.00\,\text{V}$ gives $216>8R_{\text{fixed}}+144$ , so $R_{\text{fixed}}<9.00\,\text{k}\Omega$ . In bright light, $V_{\text{out}}=12.0(2.00)/(R_{\text{fixed}}+2.00)$ . Requiring this to be less than $3.00\,\text{V}$ gives $24.0<3R_{\text{fixed}}+6.00$ , so $R_{\text{fixed}}>6.00\,\text{k}\Omega$ . Both conditions are met when $6.00\,\text{k}\Omega<R_{\text{fixed}}<9.00\,\text{k}\Omega$ .	6
04.1	$7.50\,\text{k}\Omega$ $15.0\,\text{V}$ $0.667\,\text{mA}$ and $1.50\,\text{mA}$	Let the fixed upper resistance be $R$ in $\text{k}\Omega$ and the supply be $V_s$ . The readings give $10.0=V_s[15.0/(R+15.0)]$ and $3.75=V_s[2.50/(R+2.50)]$ . Hence $V_s=(2/3)(R+15.0)=1.50(R+2.50)$ . Solving gives $R=7.50\,\text{k}\Omega$ and $V_s=15.0\,\text{V}$ . The currents are $15.0/(22.5\times10^3)=0.6667\,\text{mA}$ and $15.0/(10.0\times10^3)=1.50\,\text{mA}$ .	5
05.1	$5.30\,\text{k}\Omega$ $6.36\,\text{V}$ $7.63\,\text{V}$ unloaded; loading reduces the output by $16.6\%$ $4.38\,\text{V}$ when warm, so the controller switches	The controller and thermistor are in parallel, so the effective lower resistance is $R_{\text{lower}}=(8.20\times15.0)/(8.20+15.0)=5.3017\,\text{k}\Omega=5.30\,\text{k}\Omega$ . The loaded output is $V_{\text{out}}=12.0[5.3017/(4.70+5.3017)]=6.3607\,\text{V}=6.36\,\text{V}$ . Without the controller, $V_{\text{out}}=12.0[8.20/(4.70+8.20)]=7.6279\,\text{V}=7.63\,\text{V}$ . The reduction is $(7.6279-6.3607)/7.6279\times100=16.6\%$ . When warm, the lower resistance is $(3.30\times15.0)/(3.30+15.0)=2.7049\,\text{k}\Omega$ , so the loaded output is $12.0[2.7049/(4.70+2.7049)]=4.3834\,\text{V}=4.38\,\text{V}$ . This is below $4.50\,\text{V}$ , so the controller switches.	6

3.5.1.6 · Electromotive force and internal resistance

Tier 1 · Easy

Mark scheme for 3.5.1.6 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$6.0\,\text{V}$	Use $\varepsilon=I(R+r)=2.0(2.5+0.50)=6.0\,\text{V}$ .	2
02.1	Using $V=\varepsilon-Ir$ , the terminal potential difference equals the emf because the lost volts $Ir$ are zero.	Use $V=\varepsilon-Ir$ . With $I=0$ , there is no potential difference across the internal resistance, so $V=\varepsilon$ .	2

Tier 2 · Standard

Mark scheme for 3.5.1.6 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$1.8\,\text{A}$ $7.6\,\text{V}$ $1.4\,\text{V}$	$I=\dfrac{\varepsilon}{R+r}=\dfrac{9.0}{4.2+0.80}=1.8\,\text{A}$ . The terminal potential difference is $V=IR=1.8\times4.2=7.56\,\text{V}\approx7.6\,\text{V}$ . The lost volts are $Ir=1.8\times0.80=1.44\,\text{V}\approx1.4\,\text{V}$ , and $7.56+1.44=9.0\,\text{V}$ checks the result.	3
02.1	$0.500\,\Omega$ $2.88\,\text{W}$	The open-circuit reading gives $\varepsilon=12.0\,\text{V}$ . The lost volts are $12.0-10.8=1.2\,\text{V}$ , so $r=1.2/2.40=0.500\,\Omega$ . The internal power is $P=I^2r=(2.40)^2(0.500)=2.88\,\text{W}$ .	3
03.1	$1.54\,\text{V}$ $0.350\,\Omega$ $1.26\,\text{V}$	Subtracting the zero error from every voltage gives $V=V_{\text{read}}-0.080=1.54-0.350I$ . Comparing this with $V=\varepsilon-Ir$ gives $\varepsilon=1.54\,\text{V}$ and $r=0.350\,\Omega$ ; the constant voltage offset changes the intercept but not the gradient. At $I=0.800\,\text{A}$ , $V=1.54-0.350(0.800)=1.26\,\text{V}$ .	4

Tier 3 · Hard

Mark scheme for 3.5.1.6 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$5.80\,\text{V}$ $0.800\,\Omega$ $2.00\,\text{A}$ $4.20\,\text{V}$	From $V=\varepsilon-Ir$ , the gradient of a $V$ against $I$ graph is $-r$ . Thus $-r=\dfrac{4.20-5.40}{2.00-0.50}=-0.800\,\Omega$ , so $r=0.800\,\Omega$ . Using the first reading, $\varepsilon=V+Ir=5.40+0.50\times0.800=5.80\,\text{V}$ . With $R=2.10\,\Omega$ , $I=\dfrac{5.80}{2.10+0.800}=2.00\,\text{A}$ . Hence $V=IR=2.00\times2.10=4.20\,\text{V}$ .	5
02.1	$15.3\,\text{V}$ $1.38\times10^3\,\text{J}$ $243\,\text{J}$ $85.0\%$	The current is $I=18.0/(6.80+1.20)=2.25\,\text{A}$ , so the terminal potential difference is $V=IR=2.25(6.80)=15.3\,\text{V}$ . The load energy is $I^2Rt=(2.25)^2(6.80)(40.0)=1.38\times10^3\,\text{J}$ . The internal energy loss is $I^2rt=(2.25)^2(1.20)(40.0)=243\,\text{J}$ . The efficiency is $R/(R+r)=6.80/8.00=0.850$ , or $85.0\%$ .	5
03.1	$R=0.800\,\Omega$ $0.748\,\text{W}$ in series $1.38\,\text{W}$ in parallel; parallel transfers more power.	In series, the total emf is $2\varepsilon$ and internal resistance is $2r$ , so $P_{\text{s}}=4\varepsilon^2R/(R+2r)^2$ . In parallel, the emf is $\varepsilon$ and internal resistance is $r/2$ , so $P_{\text{p}}=\varepsilon^2R/(R+r/2)^2$ . Equating the powers and taking positive square roots gives $2/(R+2r)=1/(R+r/2)$ , so $2R+r=R+2r$ and therefore $R=r=0.800\,\Omega$ . For $R=0.300\,\Omega$ , the series current is $3.00/(0.300+1.60)=1.58\,\text{A}$ , giving $P_{\text{s}}=I^2R=0.748\,\text{W}$ . The parallel current is $1.50/(0.300+0.400)=2.14\,\text{A}$ , giving $P_{\text{p}}=1.38\,\text{W}$ . The parallel arrangement transfers more power.	6
04.1	$6.20\,\text{V}$ $1.32\,\text{A}$ $1.99\,\text{A}$ and $14.2\,\text{W}$ after correction $2.25$ times	Five emfs act in one direction and one opposes them, so the net emf is $(5-1)(1.55)=6.20\,\text{V}$ . All six internal resistances remain in series, giving $r=6(0.180)=1.080\,\Omega$ . Thus $I=6.20/(3.60+1.080)=1.3248\,\text{A}=1.32\,\text{A}$ . Internal heating occurs at $I^2r=(1.3248)^2(1.080)=1.8955\,\text{W}=1.90\,\text{W}$ . After correction, $\varepsilon=6(1.55)=9.30\,\text{V}$ , so $I=9.30/4.680=1.9872\,\text{A}=1.99\,\text{A}$ . The load power is $I^2R=(1.9872)^2(3.60)=14.216\,\text{W}=14.2\,\text{W}$ . Before correction it was $(1.3248)^2(3.60)=6.318\,\text{W}$ , so the increase factor is $14.216/6.318=2.25$ .	5
05.1	$1.25\,\text{A}$ $13.0\,\text{V}$ for the source and $9.50\,\text{V}$ for the cell being charged $11.3\,\text{W}$ $1.88\,\text{W}$ $64.3\%$	The resultant emf is $14.0-9.00=5.00\,\text{V}$ and the total resistance is $2.80+0.800+0.400=4.00\,\Omega$ , so $I=5.00/4.00=1.25\,\text{A}$ . The source supplies current, giving terminal potential difference $14.0-I(0.800)=13.0\,\text{V}$ . Current enters the positive terminal of the rechargeable cell, so its terminal potential difference is $9.00+I(0.400)=9.50\,\text{V}$ . Chemical energy is stored at rate $\varepsilon_{\text{charge}}I=9.00(1.25)=11.25\,\text{W}=11.3\,\text{W}$ . The two internal resistances dissipate $I^2(0.800+0.400)=(1.25)^2(1.20)=1.875\,\text{W}=1.88\,\text{W}$ . The source emf supplies $14.0(1.25)=17.5\,\text{W}$ , so the storage efficiency is $11.25/17.5=0.643$ , or $64.3\%$ .	6