Topic 4: Electricity and Magnetism -- Cambridge Challenge Level
1 hour 15 minutes
80
7
75:00
0625 / 0972
⚡ Cambridge Challenge Level
These questions match real Cambridge IGCSE difficulty. Scoring 50%+ is a solid B, 60%+ is an A, 70%+ is A*. Don't worry if this feels harder -- that's the point!
Instructions
Answer all questions in the spaces provided.
Show all working for calculations. Answers without working may not receive full marks.
You may use a calculator.
Take g = 10 N/kg where needed.
Answer all questions first. When finished, click "Submit Exam" at the bottom of Question 7.
Your answers will be automatically graded when you submit the exam. The model answers will be shown for review.
Question Navigation
Question 1 -- National Grid Power Transmission
Total: 12 marks
The National Grid in England and Wales transmits electrical power from generators at power stations to homes and businesses. A gas-fired power station near Bristol generates 500 MW of electrical power at 25 kV. Step-up transformers increase the voltage to 400 kV for transmission along overhead cables. The cables have a total resistance of 4.0 Ω. Step-down transformers near Cardiff reduce the voltage to 230 V for domestic use.
(a)[2]
State the type of current (a.c. or d.c.) that must be used with transformers. Explain why.
Model Answer -- 1(a)
AC / alternating current [1]
Because a changing magnetic field is needed in the core to induce an e.m.f. in the secondary coil / DC produces a constant field which does not induce an e.m.f. [1]
⚠ If you missed marks here: If you only wrote "a.c." you earned 1 of 2 — the second mark needs the reason: a transformer only works because a CHANGING current makes a CHANGING magnetic field in the core, which induces an e.m.f. in the secondary coil. Saying "d.c. doesn't work" without explaining that a constant field induces no e.m.f. misses the explanation mark.
Mark 1 -- AC alternating current stated (1 mark)
Mark 2 -- changing magnetic field needed to induce emf in secondary coil (1 mark)
(b)[3]
Calculate the current in the transmission cables when 500 MW is transmitted at 400 kV. Then calculate the power lost as heat in the cables.
Model Answer -- 1(b)
I = P / V = 500,000,000 / 400,000 = 1250 A [1]
Ploss = I²R = 1250² × 4.0 [1]
= 6,250,000 W = 6.25 MW [1]
⚠ If you missed marks here: If your current came out as 1.25 A you divided 500 by 400 without converting — put everything in base units first (500,000,000 W ÷ 400,000 V = 1250 A). For the loss, the classic slip is forgetting to square the current: 1250 × 4.0 = 5000 W is wrong; it must be 1250² × 4.0 = 6.25 MW.
Mark 1 -- current calculated 1250 A using P divided by V (1 mark)
Mark 2 -- power loss formula P equals I squared R substituted correctly (1 mark)
Mark 3 -- power lost 6250000 W or 6.25 MW correctly calculated (1 mark)
(c)[3]
Calculate the current and power loss if the same 500 MW were transmitted at the generator voltage of 25 kV instead.
Model Answer -- 1(c)
I = 500,000,000 / 25,000 = 20,000 A [1]
Ploss = 20,000² × 4.0 = 1,600,000,000 W [1]
= 1600 MW, which exceeds the power generated, showing transmission at low voltage is impractical [1]
⚠ If you missed marks here: Same conversion trap as (b): 500,000,000 ÷ 25,000 = 20,000 A. If you got both numbers right but stopped there, you still lost the third mark — you must point out that 1600 MW of loss is MORE than the 500 MW being generated, which is exactly why transmitting at 25 kV is impractical.
Mark 1 -- current calculated 20000 A (1 mark)
Mark 2 -- power loss calculated 1600000000 W or 1600 MW (1 mark)
Mark 3 -- recognised loss exceeds generated power making it impractical (1 mark)
(d)[2]
Explain why electrical power is transmitted at high voltage.
Model Answer -- 1(d)
Higher voltage means lower current for the same power (since P = IV) [1]
Lower current means less power wasted as heat in the cables (since P = I²R) [1]
⚠ If you missed marks here: Jumping straight to "less energy is lost" loses a mark — the chain must be explicit: high voltage → LOWER CURRENT for the same power (P = IV) → less heat wasted in the cables (P = I²R). Saying "less voltage is lost" or "the electricity travels faster" scores zero.
Mark 1 -- higher voltage lower current same power (1 mark)
Mark 2 -- lower current less power wasted heat cables (1 mark)
(e)[2]
The step-down transformer near Cardiff has 8000 turns on the primary coil. Calculate the number of turns needed on the secondary coil to produce 230 V from 400 kV.
Model Answer -- 1(e)
Ns/Np = Vs/Vp -- transformer equation stated or used [1]
Ns = 8000 × 230 / 400,000 = 4.6 turns (accept 4 or 5 as a whole number) [1]
⚠ If you missed marks here: If you got about 14 million turns you flipped the ratio — for a step-DOWN transformer the secondary must have FEWER turns, so Ns = 8000 × 230/400,000 = 4.6. If you got 4600, you used 400 instead of 400,000 — convert kV to V before substituting. A quick sanity check on "more or fewer turns?" catches both errors.
Mark 1 -- transformer equation Ns over Np equals Vs over Vp stated (1 mark)
Mark 2 -- secondary turns calculated as 4.6 or rounded to 5 turns (1 mark)
Question 2 -- Hospital MRI Electromagnet
Total: 11 marks
A hospital in Edinburgh uses a Magnetic Resonance Imaging (MRI) scanner. The scanner contains a large solenoid (coil) that produces a very strong, uniform magnetic field. The solenoid has 50,000 turns of superconducting wire wound around a cylindrical former of length 2.0 m. The current in the coil is 200 A.
(a)[2]
The diagram shows a solenoid. Draw arrows on the diagram to show the direction of the magnetic field inside and outside the solenoid. Describe the direction of the field lines.
Model Answer -- 2(a)
Arrows inside the solenoid pointing from S to N (parallel, uniform, all in the same direction) [1]
Arrows outside showing the field returning from N pole back to S pole (curved paths) [1]
⚠ If you missed marks here: Two common slips: drawing the inside arrows from N to S (inside the solenoid the field runs S → N; it is outside that it goes N → S), and drawing curved or uneven lines inside — the field inside a solenoid is uniform, so the lines must be straight, parallel and evenly spaced.
Mark 1 -- field inside solenoid parallel uniform from S to N (1 mark)
Mark 2 -- field outside returning from N pole back to S pole (1 mark)
(b)[3]
State three ways the magnetic field strength inside the solenoid could be increased.
Model Answer -- 2(b)
Increase the current [1]
Increase the number of turns [1]
Decrease the length of the solenoid (increase turns per unit length) [1]
Use a soft iron core [1]
(Any three of the above for 3 marks)
⚠ If you missed marks here: Vague answers like "use a bigger battery" or "increase the voltage" don't score — the examiner wants the direct factors: more CURRENT, more TURNS (or a shorter solenoid, i.e. more turns per metre), or a soft IRON core. Each of your three must name a specific physical change to the coil or current.
Mark 1 -- increase the current (1 mark)
Mark 2 -- increase number of turns or decrease length (1 mark)
Mark 3 -- soft iron core or another valid method (1 mark)
(c)[2]
The solenoid uses superconducting wire that is cooled to −269°C using liquid helium. Explain the advantage of using a superconductor for the coil.
Model Answer -- 2(c)
A superconductor has zero resistance [1]
So no energy is wasted as heat / no power loss (P = I²R = 0) / the large current can flow continuously without needing a continuous power supply [1]
⚠ If you missed marks here: "Low resistance" or "conducts better" is not enough — a superconductor has ZERO resistance, and that exact word matters. The second mark needs the consequence: no energy wasted as heat (P = I²R = 0), so the huge 200 A current can flow continuously without heating the coil.
Mark 1 -- superconductor has zero resistance (1 mark)
Mark 2 -- no energy wasted heat no power loss current flows continuously (1 mark)
(d)[2]
A patient with a metal hip implant is told they cannot have an MRI scan. Explain why, in terms of magnetic forces.
Model Answer -- 2(d)
The strong magnetic field would exert a force on the metal implant [1]
The force could move or damage the implant causing injury / metal would be attracted into the field / metal could heat up due to induced currents [1]
⚠ If you missed marks here: "Magnets attract metal" on its own is only half the answer. You need both steps: the strong field EXERTS A FORCE on the implant [mark 1], AND that force could move or twist the implant inside the patient's body, causing injury (or induced currents could heat it) [mark 2]. The question says "in terms of magnetic forces" — so the word force must appear.
Mark 1 -- magnetic field exerts force on metal implant (1 mark)
Mark 2 -- force could move damage implant injury attracted heating (1 mark)
(e)[2]
Explain why the MRI room is specially shielded and why hospital staff must remove metal objects (keys, pens, etc.) before entering.
Model Answer -- 2(e)
The magnetic field extends beyond the scanner / is very strong [1]
Metal objects would be attracted with great force / could become dangerous projectiles / could damage equipment [1]
⚠ If you missed marks here: Two separate ideas are needed: (1) the magnetic field EXTENDS BEYOND the scanner into the room — that's why shielding and the whole-room rule exist — and (2) loose metal objects like keys would be accelerated toward the magnet as dangerous projectiles. Giving only one of these caps you at 1 mark.
Mark 1 -- magnetic field extends beyond scanner very strong (1 mark)
Mark 2 -- metal objects attracted great force dangerous projectiles damage (1 mark)
Question 3 -- Wind Farm Generator Design
Total: 12 marks
A wind farm off the coast of Aberdeen, Scotland, has 50 turbines. Each turbine drives an AC generator. The generator consists of a rectangular coil with 800 turns rotating in a magnetic field of strength 0.5 T. Each coil has an area of 0.2 m². The coil rotates at 25 revolutions per second.
(a)[2]
State the name of the rule used to determine the direction of the induced e.m.f. in the coil, and state what each finger represents.
Model Answer -- 3(a)
Fleming's right-hand rule [1]
thuMb = Motion of conductor, First finger = Field direction, seCond finger = Current (induced) direction [1]
⚠ If you missed marks here: This is a GENERATOR, so it's Fleming's RIGHT-hand rule — writing "left-hand rule" (the motor rule) loses the first mark instantly. For the second mark all three assignments must be correct: thuMb = Motion, First finger = Field, seCond finger = induced Current. Swapping field and current is the usual slip.
Mark 1 -- Fleming right hand rule stated (1 mark)
Mark 2 -- thumb motion first finger field second finger current correctly identified (1 mark)
(b)[2]
Explain why slip rings are used instead of a split-ring commutator in the AC generator.
Model Answer -- 3(b)
Slip rings allow continuous contact as the coil rotates [1]
They maintain the alternating current output without reversing it each half turn / a commutator would produce d.c. which is not wanted [1]
⚠ If you missed marks here: "They keep contact while the coil spins" is only 1 mark — the second mark needs the contrast: a split-ring commutator swaps the connections every half turn and would turn the output into d.c., whereas slip rings keep each end of the coil permanently connected so the natural ALTERNATING e.m.f. reaches the circuit unchanged.
Mark 2 -- maintain alternating current output commutator would produce DC (1 mark)
(c)[3]
The graph shows how the e.m.f. varies with time for one complete rotation at 25 revolutions per second.
(i) What is the time period of the output? [1 mark]
(ii) On the graph, sketch the output if the coil were rotated at 50 revolutions per second instead. Describe the changes. [2 marks]
Model Answer -- 3(c)
(i) T = 1/f = 1/25 = 0.04 s [1]
(ii) Frequency doubled: time period halved to 0.02 s (wave compressed horizontally) [1]
(ii) Amplitude (peak e.m.f.) doubled (wave stretched vertically) [1]
⚠ If you missed marks here: For (i), T = 1/f = 1/25 = 0.04 s — writing 25 s means you gave the frequency, not the period. In (ii) nearly everyone remembers the period halves to 0.02 s but forgets the second change: spinning twice as fast cuts the flux twice as quickly, so the PEAK E.M.F. DOUBLES too. Your sketch needs both a squashed wave AND taller peaks.
Mark 1 -- time period calculated 0.04 seconds (1 mark)
Mark 2 -- frequency doubled time period halved 0.02 seconds (1 mark)
Mark 3 -- amplitude peak emf doubled (1 mark)
(d)[3]
State three changes that would increase the peak e.m.f. produced by the generator.
Model Answer -- 3(d)
Increase the speed of rotation [1]
Use a stronger magnet / increase the magnetic field strength [1]
Increase the number of turns on the coil [1]
Increase the area of the coil [1]
(Any three of the above for 3 marks)
⚠ If you missed marks here: "Increase the current" is a MOTOR answer and scores nothing here — a generator produces its own current. The valid changes are: rotate faster, stronger magnetic field, more turns, or bigger coil area. And "more wind" is too vague — you must translate it into "faster rotation of the coil".
Mark 1 -- increase speed rotation (1 mark)
Mark 2 -- stronger magnet increase magnetic field strength (1 mark)
Mark 3 -- increase number turns or increase area coil (1 mark)
(e)[2]
The wind farm has a total power output of 100 MW. A step-up transformer increases the voltage from 33 kV to 275 kV for transmission. Calculate the current in the transmission cables, assuming the transformer is ideal.
Model Answer -- 3(e)
P = IV, so I = P / V [1]
I = 100,000,000 / 275,000 = 363.6 A ≈ 364 A [1]
⚠ If you missed marks here: The current IN THE TRANSMISSION CABLES uses the output voltage: 100,000,000 ÷ 275,000 ≈ 364 A. If you got about 3030 A you divided by 33 kV (the input side of the transformer) instead of 275 kV. If your answer was out by a factor of 1000, you forgot to convert MW to W or kV to V.
Mark 1 -- formula P equals IV rearranged correctly (1 mark)
Mark 2 -- current calculated 364 A approximately (1 mark)
Question 4 -- London Underground Train Motor
Total: 11 marks
The London Underground uses DC electric motors to drive its trains. Each motor consists of a rectangular coil carrying a current in a radial magnetic field. The trains are powered by a 630 V DC supply from a conductor rail.
(a)[2]
State Fleming's left-hand rule and identify what each finger represents.
Model Answer -- 4(a)
First finger = Field (N to S), seCond finger = Current (conventional), thuMb = Motion / force / thrust [1]
The force is perpendicular to both the field and the current [1]
⚠ If you missed marks here: A motor uses the LEFT hand (right hand is for generators — easy to mix up after Q3). Check the finger assignments: First finger = Field (N to S), seCond finger = conventional Current, thuMb = Motion/force. The second mark is often dropped: you must also state that the force is PERPENDICULAR to both the field and the current.
Mark 1 -- first finger field second finger current thumb motion force correctly identified (1 mark)
Mark 2 -- force perpendicular to both field and current (1 mark)
(b)[2]
Explain the function of the split-ring commutator in the DC motor.
Model Answer -- 4(b)
It reverses the direction of current in the coil every half turn [1]
This ensures the force/torque on the coil is always in the same direction, maintaining continuous rotation [1]
⚠ If you missed marks here: "It keeps the motor spinning" is the effect, not the mechanism, and scores at most 1. You need the how: the commutator REVERSES the current in the coil EVERY HALF TURN, so the force (torque) on the coil stays in the same rotational direction. Both the reversal and the constant-direction force are needed for full marks.
Mark 1 -- reverses direction current coil every half turn (1 mark)
Mark 2 -- force torque always same direction continuous rotation (1 mark)
(c)[3]
A motor on a District Line train draws 200 A from the 630 V supply. Calculate:
(i) the power consumed by the motor [1 mark]
(ii) the energy transferred in 5 minutes [2 marks]
Model Answer -- 4(c)
(i) P = IV = 200 × 630 = 126,000 W = 126 kW [1]
(ii) E = Pt = 126,000 × (5 × 60) = 126,000 × 300 [1]
(ii) = 37,800,000 J = 37.8 MJ [1]
⚠ If you missed marks here: (i) is P = IV = 200 × 630 = 126,000 W. The classic slip in (ii) is substituting t = 5: E = 126,000 × 5 = 630,000 J is wrong because E = Pt needs time in SECONDS — 5 minutes = 300 s, giving 37.8 MJ (60 times bigger). The time conversion carries its own mark.
Mark 1 -- power calculated 126000 W or 126 kW (1 mark)
Mark 2 -- time converted to 300 seconds and substituted into E equals Pt (1 mark)
Mark 3 -- energy calculated 37800000 J or 37.8 MJ (1 mark)
(d)[2]
State two ways to increase the speed of the motor.
Model Answer -- 4(d)
Increase the current [1]
Increase the number of turns on the coil [1]
Use a stronger magnet [1]
(Any two of the above for 2 marks)
⚠ If you missed marks here: Your two answers must increase the electromagnetic FORCE on the coil: more current, more turns, or a stronger magnet. Answers like "reduce friction", "oil the axle" or "make the coil lighter" don't score — and "increase the voltage" is risky unless you link it to increasing the current.
Mark 1 -- increase current or stronger magnet (1 mark)
Mark 2 -- increase number turns or another valid method (1 mark)
(e)[2]
When the train brakes, the motors are used as generators (regenerative braking). The kinetic energy of the train is converted back to electrical energy. Explain how the motor acts as a generator in terms of electromagnetic induction.
Model Answer -- 4(e)
The rotating coil/armature continues to move/spin in the magnetic field [1]
This changing magnetic flux induces an e.m.f./current that can be fed back to the power supply / the kinetic energy does work against the back-e.m.f. [1]
⚠ If you missed marks here: "It works in reverse" or "kinetic energy becomes electrical energy" just restates the question and scores nothing. The mechanism is what earns marks: the coil KEEPS ROTATING in the magnetic field, so it cuts field lines / the flux through it keeps CHANGING, and that change INDUCES an e.m.f. which drives current back to the supply.
Mark 1 -- rotating coil armature continues spinning magnetic field (1 mark)
Mark 2 -- changing flux induces emf current fed back power supply (1 mark)
Question 5 -- Smart Home Circuit Analysis
Total: 12 marks
A smart home in Cambridge has the following devices connected to a 230 V mains supply: a smart thermostat control unit (15 W), an LED lighting system (4 lamps, each 8 W, connected in parallel), and a smart speaker (10 W). All devices are connected in parallel to the mains supply. The electricity tariff is 30p per kWh.
(a)[2]
Calculate the total power consumed when all devices are operating.
Model Answer -- 5(a)
LED total = 4 × 8 = 32 W [1]
Total = 15 + 32 + 10 = 57 W [1]
⚠ If you missed marks here: If you got 33 W you counted only ONE LED lamp — there are four in parallel, so the lamp bank is 4 × 8 = 32 W, then 15 + 32 + 10 = 57 W. Show the 32 W step explicitly: it carries its own mark.
Mark 1 -- LED total calculated 32 W (1 mark)
Mark 2 -- total power 57 W correctly summed (1 mark)
(b)[2]
Calculate the total current drawn from the supply.
Model Answer -- 5(b)
I = P / V = 57 / 230 [1]
= 0.248 A ≈ 0.25 A [1]
⚠ If you missed marks here: I = P/V = 57/230 = 0.248 A. If you got about 13,000 you multiplied 57 × 230 instead of dividing; if you got about 4, you divided the wrong way round (230/57). Sanity check: a handful of low-power smart devices should draw well under 1 A from the mains.
Mark 1 -- formula I equals P over V used correctly (1 mark)
Mark 2 -- current calculated 0.248 or 0.25 A (1 mark)
(c)[2]
State the most appropriate fuse rating for this circuit. Explain your choice.
Model Answer -- 5(c)
3 A fuse [1]
Because the operating current (0.25 A) is well below 3 A, and 3 A is the lowest standard fuse rating above the operating current [1]
⚠ If you missed marks here: 13 A is the classic wrong answer — a 13 A fuse would let a dangerous fault current flow for far too long in a 0.25 A circuit. The rule the examiner wants: choose the LOWEST standard rating (3 A, 5 A, 13 A) that is ABOVE the normal operating current, and say so explicitly for the second mark.
Mark 1 -- 3 A fuse stated (1 mark)
Mark 2 -- lowest standard fuse rating above operating current explained (1 mark)
(d)[3]
Calculate the cost of running all devices for 30 days, 24 hours per day.
Model Answer -- 5(d)
Energy = 0.057 kW × 24 h × 30 = 41.04 kWh [1]
Cost = 41.04 × 30p = 1231.2 p [1]
= £12.31 [1]
⚠ If you missed marks here: The kilowatt conversion is the killer: kWh needs power in kW, so use 0.057 kW, not 57 — energy = 0.057 × 24 × 30 = 41.04 kWh. Using 57 gives a cost of about £12,312, a thousand times too big (which should ring alarm bells for three small gadgets). And don't forget the final step: 1231.2p = £12.31 earns its own mark.
Mark 1 -- energy calculated 41.04 kWh (1 mark)
Mark 2 -- cost calculated 1231 pence (1 mark)
Mark 3 -- converted to pounds 12.31 (1 mark)
(e)[3]
One of the 4 LED lamps fails (goes open circuit). State and explain the effect on:
(i) the other LED lamps [1 mark]
(ii) the total current from the supply [2 marks]
Model Answer -- 5(e)
(i) The other three LED lamps continue to work normally / at the same brightness, because they are in parallel so each has the full 230 V across it [1]
(ii) Total current decreases [1]
(ii) Because total power decreases by 8 W / one parallel branch has been removed, reducing total power from 57 W to 49 W [1]
⚠ If you missed marks here: If you said the other lamps go out or get dimmer, you were thinking SERIES — in parallel, each lamp has the full 230 V across it regardless of the others, so the remaining three are completely unaffected. For (ii), "current stays the same" is wrong: one 8 W branch has gone, so total power drops from 57 W to 49 W and the supply current falls.
Mark 1 -- other lamps continue working normally same brightness parallel (1 mark)
Mark 2 -- total current decreases (1 mark)
Mark 3 -- total power decreases one parallel branch removed (1 mark)
Question 6 -- Electric Vehicle Charging
Total: 12 marks
An electric vehicle (EV) charging station in a motorway services on the M4 near Swindon provides Level 2 charging at 7.4 kW (230 V, 32 A) and rapid DC charging at 50 kW. A Tesla Model 3 has a 75 kWh battery. The electricity costs 45p per kWh at this station.
(a)[2]
Calculate the time taken to fully charge the 75 kWh battery using the 7.4 kW Level 2 charger. Give your answer in hours.
Model Answer -- 6(a)
t = E / P = 75 / 7.4 [1]
= 10.1 hours [1]
⚠ If you missed marks here: Because the energy is in kWh and the power in kW, dividing gives hours directly: t = 75/7.4 = 10.1 h — no joule conversion needed. If you got 555 you multiplied instead of dividing; if your answer was thousands of hours you probably converted to joules and slipped a factor of 3600 somewhere.
Mark 1 -- formula t equals E over P used correctly (1 mark)
Mark 2 -- time calculated 10.1 hours (1 mark)
(b)[2]
Calculate the charge (in coulombs) that flows through the Level 2 charger cable in 1 hour at 32 A.
Model Answer -- 6(b)
Q = It = 32 × 3600 [1]
= 115,200 C [1]
⚠ If you missed marks here: If you got 32 C you substituted t = 1 — Q = It only works with time in SECONDS, so 1 hour must become 3600 s, giving 32 × 3600 = 115,200 C. The 3600 conversion is worth a mark on its own, so always write it down.
Mark 1 -- formula Q equals It time converted 3600 seconds (1 mark)
Mark 2 -- charge calculated 115200 coulombs (1 mark)
(c)[3]
The Level 2 charger is 90% efficient. Calculate the energy wasted as heat when fully charging the battery, and the actual energy drawn from the mains.
Model Answer -- 6(c)
Efficiency = useful output / total input, so input = 75 / 0.9 = 83.33 kWh [1]
⚠ If you missed marks here: The classic error is multiplying by 0.9: input = 75 × 0.9 = 67.5 kWh is WRONG, and gives wasted = 7.5 kWh — if that's your answer, this is exactly what happened. The 75 kWh going into the battery is the useful OUTPUT, so the mains input must be BIGGER: 75 / 0.9 = 83.33 kWh, and wasted = 83.33 − 75 = 8.33 kWh.
Mark 1 -- total input energy calculated 83.33 kWh using efficiency formula (1 mark)
Mark 2 -- energy wasted calculated 8.33 kWh (1 mark)
Mark 3 -- clear working shown for efficiency calculation (1 mark)
(d)[3]
Calculate the cost to fully charge the battery using the Level 2 charger, accounting for the 90% efficiency.
Model Answer -- 6(d)
Total energy drawn = 83.33 kWh (from part c) [1]
Cost = 83.33 × 45p = 3750 p [1]
= £37.50 [1]
⚠ If you missed marks here: If you got £33.75 you charged for only 75 kWh and forgot the efficiency — you pay for EVERYTHING drawn from the mains, i.e. 83.33 kWh from part (c), so cost = 83.33 × 45p = 3750p. Then keep pence-to-pounds as its own final step: £37.50 earns the last mark.
Mark 1 -- total energy drawn from mains 83.33 kWh identified (1 mark)
Mark 2 -- cost calculated 3750 pence (1 mark)
Mark 3 -- converted to pounds 37.50 (1 mark)
(e)[2]
The rapid DC charger operates at 50 kW. Explain why a thicker cable is needed for the rapid charger compared to the Level 2 charger.
Model Answer -- 6(e)
The rapid charger delivers more power, requiring a higher current [1]
Thicker cable has lower resistance, reducing heating and power loss in the cable / thicker cable can safely carry the higher current without overheating [1]
⚠ If you missed marks here: Two physics links are needed: 50 kW means a much HIGHER CURRENT than the Level 2 charger's 32 A, and a thicker cable has LOWER RESISTANCE so it heats up less (P = I²R). Saying "it needs to carry more electricity" without mentioning resistance or heating only earns the first mark.
Mark 1 -- rapid charger more power higher current (1 mark)
Mark 2 -- thicker cable lower resistance less heating safely carry current (1 mark)
Question 7 -- Laboratory Resistance Experiment
Total: 10 marks
A student at a school in Oxford investigates how current varies with potential difference for two components: a fixed resistor and a filament lamp. She sets up a circuit with a variable power supply, an ammeter in series, and a voltmeter in parallel with the component.
(a)[2]
Draw or describe the circuit diagram the student should use. Include the variable power supply, ammeter, component under test, and voltmeter. State how the ammeter and voltmeter should be connected.
Model Answer -- 7(a)
Ammeter connected in series with the component [1]
Voltmeter connected in parallel across the component [1]
⚠ If you missed marks here: The classic error is swapping the meters. Remember why they go where they go: the AMMETER must carry the same current as the component, so it sits in SERIES; the VOLTMETER measures the p.d. between the two ends of the component, so it connects in PARALLEL across it. State both placements explicitly — each is a separate mark.
Mark 1 -- ammeter connected series with component (1 mark)
Mark 2 -- voltmeter connected parallel across component (1 mark)
(b)[3]
The student obtains the following results for the fixed resistor:
V (V)
0.0
1.0
2.0
3.0
4.0
5.0
I (A)
0.00
0.10
0.20
0.30
0.40
0.50
Calculate the resistance of the resistor and state whether it obeys Ohm's law. Justify your answer.
Model Answer -- 7(b)
R = V / I = 1.0 / 0.10 = 10 Ω (or any correct pair from the table) [1]
It obeys Ohm's law [1]
Because current is directly proportional to p.d. / the graph would be a straight line through the origin / resistance is constant at all voltages [1]
⚠ If you missed marks here: R = V/I from any pair in the table, e.g. 1.0/0.10 = 10 Ω. The justification mark is the one usually lost: "it obeys Ohm's law" alone isn't a reason — you must say current is DIRECTLY PROPORTIONAL to p.d., i.e. a straight line THROUGH THE ORIGIN (or R = 10 Ω for every pair). "The graph is linear" without mentioning the origin is incomplete.
Mark 1 -- resistance calculated 10 ohms using V over I (1 mark)
Mark 2 -- stated obeys Ohm law (1 mark)
Mark 3 -- current directly proportional to pd straight line through origin constant resistance (1 mark)
(c)[3]
For the filament lamp, the student obtains:
V (V)
0.0
1.0
2.0
3.0
4.0
5.0
I (A)
0.00
0.15
0.25
0.32
0.36
0.39
Explain why the filament lamp does not obey Ohm's law.
Model Answer -- 7(c)
As voltage increases, current increases less than expected / the I-V graph is not a straight line / is a curve [1]
The filament heats up as more current flows [1]
Higher temperature increases the resistance of the metal filament, so each additional volt produces a smaller increase in current [1]
⚠ If you missed marks here: "The graph is a curve" only earns the first mark — the explanation chain is where most marks go missing: more current makes the filament HOT, and the higher TEMPERATURE INCREASES the resistance of the metal, so each extra volt produces a smaller rise in current (0.15 A for the first volt but only 0.03 A for the last). Skip the temperature–resistance link and you're capped at 1.
Mark 1 -- graph not straight line curve current increases less than expected (1 mark)
Mark 2 -- filament heats up as current flows temperature increases (1 mark)
Mark 3 -- higher temperature increases resistance smaller increase current per volt (1 mark)
(d)[2]
Calculate the resistance of the filament lamp at 2.0 V and at 5.0 V. Comment on the values.
Model Answer -- 7(d)
At 2.0 V: R = V / I = 2.0 / 0.25 = 8.0 Ω [1]
At 5.0 V: R = 5.0 / 0.39 = 12.8 Ω; resistance increases with voltage/temperature [1]
⚠ If you missed marks here: R = V/I gives 2.0/0.25 = 8.0 Ω and 5.0/0.39 = 12.8 Ω. If you got 0.125 and 0.078 you inverted the formula to I/V. And don't stop at the numbers — the question says "comment", so the second mark needs the observation that the resistance INCREASES as the filament gets hotter at higher voltage.
Mark 1 -- resistance at 2.0 V calculated 8.0 ohms (1 mark)
Mark 2 -- resistance at 5.0 V calculated 12.8 ohms increases with temperature (1 mark)
When you have finished answering all questions, click Submit to see the model answers.
Self-Assessment
Tick all mark checkboxes you have earned, then click "Calculate Grade" below.