9 What actually comes up

Three years of real questions, part by part, with the setup and values from the paper and the grade band each part sat at according to that year's marking schedule. Start with the pattern table — it is the fastest way to decide what to revise. The circuit diagrams are in the papers, linked beside each year.

Across 3 years

9 questions · 37 parts 10 Achieved 27 Merit 18 Excellence

The pattern — what to prepare

Topic	Appears in	Grade level	What to practise
Capacitor energy and Q = CV	2023, 2025	Merit–Excellence	All three energy forms, and the unit conversions (nF, µF)
RC time constant τ = RC	Every year	Achieved–Excellence	τ from R and C, then Q or V at t = τ and t = 5τ
Charging / discharging curve	2023, 2025	Merit–Excellence	Plotting four correct points; asymptote at Q_max; label τ
Internal resistance and terminal voltage	2024, 2025	Achieved–Excellence	Why V_terminal falls as r rises; Kirchhoff loop with r
Induced EMF, ε = −L(ΔI/Δt)	Every year	Achieved–Excellence	Direction from Lenz; the spark when a switch opens
Eddy currents / Lenz in a real object	2024	Merit–Excellence	Magnet through a ring: flux → induced current → opposing field
Transformers	2023	Achieved–Excellence	AC → changing flux → induced EMF; why the iron core; why DC fails
Reactance X_L = 2πfL and X_C = 1/(2πfC)	Every year	Achieved–Merit	Which rises with f and which falls — and say so in words
Impedance Z and phase angle φ	Every year	Merit–Excellence	Z = √(R² + (X_L−X_C)²); φ, and which leads
Effect of an iron core on L	2024, 2025	Merit–Excellence	Core → L up → X_L up → Z up → I down. One chain, four links
Resonance f₀ = 1/(2π√(LC))	Every year	Achieved–Excellence	X_L = X_C, Z = R, I maximum — and why the lamp is brightest
Power transmission at high voltage	2023	Merit–Excellence	P_loss = I²R; why raising V cuts loss

not tried got it to review 2025 Mereana and the caravan 13 parts

The 2025 paper ↗ Marking schedule ↗

Question 1 — DC circuits

Mereana has a caravan with a simple battery-powered lighting circuit of total resistance 6.00 Ω. The battery has an EMF of 12.6 V and an internal resistance of 0.100 Ω.

R = 6.00 Ω
ε = 12.6 V
r = 0.100 Ω

Not tried yet: a When the switch is closed the current from the battery is 2.07 A. Show that the terminal voltage of the battery is 12.4 V.Worked solution → Achieved
Not tried yet: b As the battery is used over time its internal resistance increases. Explain the effect this has on the terminal voltage.Worked solution → Merit
Not tried yet: c (i) A 90.0 F capacitor is fully charged by a 4.20 V battery, then discharged for 2500 s through a 6.00 Ω resistor. Sketch current against time for 2500 s, showing the initial current and the current after one time constant.Worked solution → Achieved
Not tried yet: c (ii) Explain, using physics principles, why the graph has the shape you drew.Worked solution → Merit to Excellence
Not tried yet: d Six 90.0 F capacitors are charged individually to 4.20 V. Three are connected in series with a 6.00 Ω resistor (Circuit A); three in parallel with a 6.00 Ω resistor (Circuit B). For each circuit, explain how the current varies over time compared with the single capacitor in part (c).Worked solution → Merit to Excellence

Question 2 — inductors

Mereana connects an old coil to her caravan battery. The coil is modelled as a 0.600 H inductor in series with a 2.40 Ω resistor; the 12.6 V battery has an internal resistance of 0.100 Ω.

L = 0.600 H
R(coil) = 2.40 Ω
ε = 12.6 V
r = 0.100 Ω

Not tried yet: a State the size of the induced EMF across the inductor when the switch is first closed.Worked solution → Achieved
Not tried yet: b Calculate the size of the induced EMF across the inductor 0.240 s after the switch is closed.Worked solution → Merit
Not tried yet: c When the switch is opened, a small spark is seen where it breaks the circuit. Explain why the spark is produced.Worked solution → Merit to Excellence
Not tried yet: d Maximum energy is stored in the inductor when a steady current flows. The inductor is replaced by a capacitor storing an equal amount of energy. Calculate the capacitance required, when fully charged by the 12.6 V battery.Worked solution → Merit to Excellence

Question 3 — AC inductors and capacitors

The 0.600 H coil (an ideal inductor plus a 2.40 Ω resistor) is connected in series with a 68.0 Ω resistor to a variable-frequency AC supply set to 25.0 Hz and 9.00 V RMS.

L = 0.600 H
R(coil) = 2.40 Ω
R = 68.0 Ω
V RMS = 9.00 V
f = 25.0 Hz

Not tried yet: a Show that the reactance of the inductor is 94.2 Ω.Worked solution → Achieved
Not tried yet: b An iron nail is inserted inside the coil. Explain what effect this has on the RMS voltage across the 68.0 Ω resistor.Worked solution → Merit
Not tried yet: c The nail is removed and a 1.20 × 10⁻⁴ F capacitor is added in series with the coil and the resistor. As the frequency is reduced, the voltage across the 68.0 Ω resistor rises to a maximum at one particular frequency. Explain why it reaches a maximum at that one frequency. No calculations required.Worked solution → Merit to Excellence
Not tried yet: d The supply frequency is reduced further, to 16.0 Hz. Calculate the RMS current in the circuit. A phasor diagram may assist.Worked solution → Merit to Excellence

2024 Stefan, Harry and Rachel 12 parts

The 2024 paper ↗ Marking schedule ↗

Question 1 — DC and capacitors

Stefan investigates new and used car batteries. His car has a new battery with an EMF of 12.0 V and an internal resistance of 0.0500 Ω, connected to a 7.00 Ω load.

ε = 12.0 V
r = 0.0500 Ω
R = 7.00 Ω

Not tried yet: a Explain the meaning of the term EMF.Worked solution → Achieved
Not tried yet: b Stefan watches the voltmeter, which is across the battery’s terminals, while opening and closing the switch to the load. Explain why the readings are the same or different. Values and calculations are not required.Worked solution → Merit
Not tried yet: c After long use the internal resistance rises to 0.500 Ω. The discharged battery is charged by a 24.0 V DC supply with a 15.5 Ω resistor in series. Calculate the terminal voltage across the car battery. Begin by calculating the circuit current while it is charging.Worked solution → Merit to Excellence
Not tried yet: d A 12.0 V DC supply is connected through a switch to a 0.140 F capacitor, two lamps and a resistor. Lamp A is in series with the capacitor and Lamp B in series with the resistor, and the two branches are in parallel across the supply. The resistor and each lamp have a resistance of 10.0 Ω. Give an in-depth explanation of the brightness of lamps A and B — as soon as the switch closes, and ten seconds later.Worked solution → Merit to Excellence

Question 2 — electromagnetic induction

Harry is experimenting with coils and magnets. Each part gives its own values.

Not tried yet: a An average EMF of 250 µV is induced in a coil when the current through it changes from 10.0 A to 6.00 A in 0.400 s. Calculate the self-inductance of the coil.Worked solution → Achieved
Not tried yet: b The coil has 25 turns and an area of 1.60 × 10⁻⁴ m². It is moved into a magnetic field of flux density 1.80 Wb m⁻², with its plane perpendicular to the field lines, taking 0.300 s. Calculate the EMF induced in the coil.Worked solution → Merit
Not tried yet: c Harry drops a bar magnet through a metal ring. Explain why its acceleration is not equal to g = 9.81 m s⁻². Include what happens as the magnet enters the ring and as it leaves.Worked solution → Merit to Excellence
Not tried yet: d A 0.600 H inductor and a lamp of resistance 2.00 Ω are connected in series to a 12.0 V battery through a switch. Calculate the current through the lamp after 0.600 s. Begin by calculating the time constant.Worked solution → Merit to Excellence

Question 3 — alternating current

Rachel connects a 70.0 mH inductor with a soft iron core and a 40.0 Ω bulb in series to a 110 V RMS, 60.0 Hz supply.

L = 70.0 mH
R = 40.0 Ω
V RMS = 110 V
f = 60.0 Hz

Not tried yet: a Calculate the peak voltage of the power supply.Worked solution → Achieved
Not tried yet: b Calculate the phase difference between the supply voltage and the circuit current, and state which one leads. Begin by showing that the reactance of the inductor is 26.4 Ω.Worked solution → Merit
Not tried yet: c Rachel slowly removes the iron core from the inductor. Explain what she would observe in the brightness of the lamp.Worked solution → Merit to Excellence
Not tried yet: d The iron core is reinserted and a capacitor is added to the circuit to achieve resonance. Explain why the reactance of the capacitor needs to be 26.4 Ω. Begin by describing what happens at resonance.Worked solution → Merit to Excellence

2023 Kate's camera, transformer and AC circuit 12 parts

The 2023 paper ↗ Marking schedule ↗

Question 1 — capacitors

Kate investigates a capacitor found in a camera, labelled 185 nF (1.85 × 10⁻⁷ F). The camera also contains a 1.50 V “AA” battery.

C = 185 nF
V = 1.50 V

Not tried yet: a Show that the energy stored by the capacitor, fully charged by the battery, is 2.08 × 10⁻⁷ J.Worked solution → Merit
Not tried yet: b In the charging circuit the battery has an internal resistance of 0.270 Ω; assume the rest of the circuit has none. Sketch how the charge on the plates varies with time once the switch closes, plotting at least four points and showing the time constant and the maximum charge.Worked solution → Merit to Excellence
Not tried yet: c The rolled-up plates act like two metal rectangles 3.2 × 10⁻² m × 1.83 m, with a dielectric of relative permittivity 2.10 between them. Calculate the distance between the metal rectangles.Worked solution → Achieved
Not tried yet: d Charged from the 1.5 V battery the bulb barely glows. Kate finds wiring that lets the capacitor be charged to 200 V instead. Explain how this gives a much more powerful flash — comparing the stored energy, the initial current, and the brightness.Worked solution → Merit to Excellence

Question 2 — transformers and inductors

Kate’s school has a demonstration transformer. She connects the 12 000-turn primary coil to the mains supply at 240 V RMS. The secondary has 600 turns.

N(p) = 12 000
V(p) = 240 V RMS
N(s) = 600

Not tried yet: a Calculate the RMS voltage she would measure across the 600-turn secondary coil.Worked solution → Achieved
Not tried yet: b The two coils are held by a ring of laminated soft iron running through the core of each. Explain why an AC voltage in the primary produces an AC voltage in the secondary, and why the coils are wrapped around an iron ring.Worked solution → Merit to Excellence
Not tried yet: c The 12 000-turn primary is connected instead to a 12 V DC battery and a 12 V car headlamp bulb, the cores still linked with iron. Explain why the headlamp bulb only comes on after a slight delay.Worked solution → Merit
Not tried yet: d A power station generates 50 kW. The transmission line carries it at 220 kV and has a resistance of 4.00 Ω per kilometre. Calculate the power lost as heat over 300 km, and comment on whether it is significant compared with transmitting at 25 kV.Worked solution → Merit to Excellence

Question 3 — alternating current

Kate builds a series circuit with a signal generator at 200 Hz, an AC ammeter, a lamp of 15.0 Ω and a 1.20 µF capacitor.

f = 200 Hz
R = 15.0 Ω
C = 1.20 µF

Not tried yet: a Show that the capacitive reactance is 663 Ω, and hence determine the impedance of the circuit.Worked solution → Merit
Not tried yet: b Kate raises the frequency from 200 Hz to 20 kHz, then to 200 kHz. Give an in-depth explanation of what she observes at each frequency compared with part (a) — the impedance, the RMS current, and the brightness of the lamp.Worked solution → Merit to Excellence
Not tried yet: c A 0.200 H inductor is added in series with the capacitor. At 2000 Hz the lamp is off; as the frequency is slowly reduced it suddenly glows brightly, then goes off again at lower frequencies. Calculate the frequency at which the bulb glows brightest.Worked solution → Achieved
Not tried yet: d The 0.200 H inductor from part (c) is still in series with the capacitor and the lamp. Explain how the inductor affects the impedance of the circuit, and why there is one frequency at which the impedance equals the 15.0 Ω resistance, making the lamp glow brightly.Worked solution → Merit to Excellence

Read the last column as a to-do list. Every row marked Every year is worth more revision time than anything marked with a single year — and the Excellence rows are where a Merit becomes an Excellence.

Practice for this standard covers the whole topic — flashcards and quiz.