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Check each mark point you believe you earned, then calculate your grade.
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Take g = 10 N/kg where needed.
Question 1
Magnetic Fields and Plotting Experiments -- 11 marks
(a)[2 marks]
Define the term magnetic field.
Model Answer -- Part (a)
A region in which a magnetic pole / magnetic material experiences a force [1]
Field direction is from N to S [1]
⚠ If you missed marks here: Answering "the area around a magnet" is circular and earns nothing — a definition has to say what happens there: a region in which a magnetic pole, or a piece of magnetic material, experiences a force. Mark 2 is the direction convention, that the field points from N to S outside the magnet; leaving direction out of a two-mark definition costs half the part.
(b)[3 marks]
Describe an experiment to plot the magnetic field pattern around a bar magnet using a plotting compass.
Model Answer -- Part (b)
Place the bar magnet on a piece of paper [1]
Place a plotting compass near one pole and mark the position where the compass needle points [1]
Move the compass so its tail is at the previous mark, repeat to trace complete field lines from N to S [1]
⚠ If you missed marks here: Describing iron filings sprinkled on paper answers a different experiment and scores none of these three marks — this part names a plotting compass. The method marks come from the sequence: magnet on paper, compass by one pole with a dot marked where the needle points, then the compass moved so its tail sits on that dot and the process repeated. Writing only "move the compass round the magnet and join the dots" misses the tail-to-previous-mark step.
(c)[3 marks]
Draw the magnetic field pattern when two bar magnets are placed side by side with unlike poles facing each other (N facing S).
Model Answer -- Part (c)
Your diagram should look like this:
Field lines running from N pole of one magnet to S pole of the other [1]
Lines curve smoothly between the magnets [1]
Correct direction shown with arrows [1]
⚠ If you missed marks here: Drawing each magnet with its own separate loops, lines leaving its N and curving back to its own S, loses mark 1. With unlike poles facing, the lines cross the gap from the N of one magnet into the S of the other, curving smoothly and never touching each other. A pattern showing the lines pushed apart with a gap in the middle is the N-facing-N case, not this one, and arrows must run N → S.
(d)[3 marks]
Explain the difference between a permanent magnet and a temporary magnet (induced magnet). Give one example of each.
Model Answer -- Part (d)
A permanent magnet produces its own magnetic field / retains its magnetism [1]
A temporary (induced) magnet is only magnetised when placed in a magnetic field / loses magnetism when removed [1]
Examples: permanent = bar magnet / compass needle; temporary = soft iron near a magnet / electromagnet core [1]
⚠ If you missed marks here: Saying "a permanent magnet lasts for ever and a temporary one does not" is too loose for the marks. Mark 1 is that a permanent magnet keeps its own field; mark 2 is that an induced magnet is magnetised only while it sits in another magnet’s field and loses it on removal. Mark 3 is pure recall — give an example of each (steel bar magnet or compass needle; soft iron near a magnet) and do not swap them round: steel is the permanent-magnet material, soft iron the temporary one.
Question 2
Current Electricity and Resistance Experiment -- 12 marks
(a)[3 marks]
Draw a circuit diagram to show how an ammeter and a voltmeter should be connected to measure the resistance of a resistor. Use standard circuit symbols.
Model Answer -- Part (a)
Battery, switch and ammeter in series with the resistor [1]
Voltmeter connected in parallel across the resistor [1]
Correct circuit symbols used [1]
⚠ If you missed marks here: Swapping the meters is the error that costs two marks: an ammeter placed across the resistor short-circuits it, and a voltmeter in series lets almost no current through. The ammeter goes in series with the resistor, the voltmeter in parallel across the resistor only — not across the battery. Mark 3 needs standard symbols: A and V in circles, the resistor a plain rectangle, the cell a long line and a short line.
(b) (i)[1 mark]
State the equation used to calculate resistance from the ammeter and voltmeter readings.
Model Answer -- Part (b)(i)
R = V / I
Correct equation R = V/I stated [1]
⚠ If you missed marks here: Turning the equation over, R = I/V, scores nothing. Quoting V = IR is accepted only if you go on to make R the subject, since the part asks how resistance is obtained from the two meter readings. Adding units to the symbols is not required and does not earn anything.
(b) (ii)[3 marks]
A student records the following results:
Voltage V / V
2.0
4.0
6.0
8.0
10.0
Current I / A
0.10
0.20
0.30
0.40
0.50
Calculate the resistance.
Model Answer -- Part (b)(ii)
Use any pair of readings, e.g. R = V/I = 2.0/0.10 [1]
R = 2.0 / 0.10 = 20 Ω
R = 20 ohm [1]
Note that resistance is constant / the same for all readings (showing the resistor obeys Ohm's law) [1]
⚠ If you missed marks here: Dividing the wrong way gives 0.10 / 2.0 = 0.05 Ω instead of 20 Ω. One pair of readings is enough for the first two marks — you do not need to average the table. Mark 3 is the one most often missed: say explicitly that every pair gives the same 20 Ω, so the resistance is constant and the resistor obeys Ohm’s law.
(c)[2 marks]
The student plots a graph of current (y-axis) against voltage (x-axis). Describe and explain the shape of the graph.
Model Answer -- Part (c)
A straight line through the origin [1]
This shows that current is directly proportional to voltage / the resistor obeys Ohm's law / resistance is constant [1]
⚠ If you missed marks here: A bare "straight line" is worth at most one mark; the description must include through the origin, because that is what shows there is no offset. The second mark is the explanation, that current is directly proportional to voltage at constant temperature. Sketching a curve that flattens off belongs to the filament lamp, not to a fixed resistor. With I on the y-axis the gradient is 1/R, not R.
(d)[3 marks]
Explain why the resistance of a metallic conductor increases when its temperature increases.
Model Answer -- Part (d)
As temperature increases, the metal ions/atoms vibrate more vigorously [1]
The free electrons / charge carriers collide more frequently with the vibrating ions [1]
This opposes the flow of current, increasing the resistance [1]
⚠ If you missed marks here: Writing "the electrons move faster so they bump into each other more" is not the mark scheme — it is the metal ions that vibrate more vigorously. The chain is: temperature rises → ions vibrate more → free electrons collide with them more often → the flow of charge is opposed, so resistance rises. Note this is a metal; an NTC thermistor does the opposite, and mixing the two up loses every mark.
Question 3
I-V Characteristics and Component Behaviour -- 11 marks
(a)[3 marks]
Sketch the I-V characteristic graph for a filament lamp. Label the axes.
Model Answer -- Part (a)
Axes correctly labelled: current (I) on y-axis, voltage (V) on x-axis [1]
Curve passes through the origin [1]
Curve shows decreasing gradient as V increases (curves towards the V-axis) [1]
⚠ If you missed marks here: Curving the line upwards, away from the voltage axis, is the wrong way round: as V rises the filament heats, its resistance rises, and the graph bends towards the V-axis with a decreasing gradient. Mark 1 is for labelling the axes, I on the y-axis and V on the x-axis, which an unlabelled sketch throws away immediately, and mark 2 needs the curve through the origin. A straight line through the origin scores only the first two marks.
(b)[2 marks]
Sketch the I-V characteristic graph for a diode.
Model Answer -- Part (b)
Very small / zero current for negative (reverse bias) voltages [1]
Current increases rapidly after the threshold voltage (around 0.6-0.7 V for silicon) in forward bias [1]
⚠ If you missed marks here: A straight line through the origin is a resistor’s characteristic, not a diode’s. Reverse bias must be drawn flat along the axis at essentially zero current, and in forward bias almost no current flows until roughly 0.6–0.7 V, after which the curve rises steeply. Starting the steep rise at 0 V loses the threshold mark.
(c) (i)[2 marks]
State how the resistance of a thermistor (NTC type) changes as the temperature increases.
Model Answer -- Part (c)(i)
Resistance decreases [1]
As temperature increases, more charge carriers are released [1]
⚠ If you missed marks here: Stating that the resistance increases is the behaviour of a metal wire, not of an NTC thermistor — here it decreases. Mark 2 is the reason, that heating releases more charge carriers so more current flows for the same p.d.; answering "it lets more current through" repeats the effect without giving a cause.
(c) (ii)[2 marks]
Give one practical use of a thermistor.
Model Answer -- Part (c)(ii)
Temperature sensor / thermostat [1]
E.g. used in fire alarm, temperature-controlled heating system, electronic thermometer [1]
⚠ If you missed marks here: Vague answers such as "used in electrical circuits" earn nothing. One mark is the role — temperature sensor or thermostat — and one is a named device: fire alarm, electronic thermometer, temperature-controlled heater. A light-operated switch or automatic streetlight is an LDR application and gains no credit here.
(d)[2 marks]
State how the resistance of a light-dependent resistor (LDR) changes as the light intensity increases and give one practical application.
Model Answer -- Part (d)
Resistance decreases as light intensity increases [1]
⚠ If you missed marks here: Bright light means low resistance, and reversing that is the single most common error in this part. Mark 2 needs a named application — streetlights that switch on at dusk, a camera light meter, a burglar alarm beam — because "used to detect light" repeats the question rather than applying it.
Question 4
Circuit Problem with Series/Parallel Combination -- 12 marks
The circuit below shows a 24 V battery connected to three resistors. R1 = 6.0 Ω is connected in series with a parallel combination of R2 = 12 Ω and R3 = 4.0 Ω.
(a) (i)[3 marks]
Calculate the combined resistance of R2 and R3 in parallel.
Model Answer -- Part (a)(i)
1/R_parallel = 1/R2 + 1/R3 = 1/12 + 1/4.0
Correct parallel resistance formula with substitution [1]
1/R_parallel = 1/12 + 3/12 = 4/12
Correct addition of fractions giving 4/12 [1]
R_parallel = 12/4 = 3.0 Ω
R parallel = 3.0 ohm [1]
⚠ If you missed marks here: Adding the two resistances (16 Ω) is the standard error, and stopping at 1/R = 4/12 to write 0.33 Ω is the next commonest. Invert at the end: R = 12/4 = 3.0 Ω. A parallel combination is always smaller than the smaller resistor, so any answer above 4.0 Ω is wrong before the arithmetic is checked.
(a) (ii)[2 marks]
Calculate the total resistance of the circuit.
Model Answer -- Part (a)(ii)
R_total = R1 + R_parallel = 6.0 + 3.0
Correct series addition formula used [1]
R_total = 9.0 Ω
R total = 9.0 ohm [1]
⚠ If you missed marks here: Adding all three resistances, 6.0 + 12 + 4.0 = 22 Ω, ignores the fact that R2 and R3 are in parallel. Only the 3.0 Ω combination is in series with R1, giving 9.0 Ω. Re-using the parallel formula on 6.0 and 3.0 (giving 2.0 Ω) also scores nothing — a series step is a simple sum.
(a) (iii)[2 marks]
Calculate the current drawn from the battery.
Model Answer -- Part (a)(iii)
I = V / R_total = 24 / 9.0
Correct substitution I = V / R = 24 / 9.0 [1]
I = 2.67 A (accept 2.7 A or 8/3 A)
Current = 2.67 A or equivalent [1]
⚠ If you missed marks here: Dividing by the wrong resistance is the trap here: 24 / 6.0 = 4.0 A uses R1 alone, when the battery drives the current through the whole 9.0 Ω. Keep the answer as 2.67 A (or 8/3 A) rather than rounding to 3 A, because parts (b) and (c) reuse this value and an early round-off drags the error through all of them.
(b) (i)[2 marks]
Calculate the potential difference across R1.
Model Answer -- Part (b)(i)
V = IR = 2.67 x 6.0
Correct substitution V = I R [1]
V = 16.0 V
Voltage across R1 = 16.0 V [1]
⚠ If you missed marks here: Using the full 24 V, or dividing the supply evenly between the resistors, both fail. R1 carries the whole circuit current, so V = IR = 2.67 × 6.0 = 16.0 V. If you substituted a parallel-branch resistance you would get 2.67 × 12 = 32 V, which is above the battery voltage and so impossible — a quick check that catches the mistake.
(b) (ii)[1 mark]
State the potential difference across the parallel combination of R2 and R3.
Model Answer -- Part (b)(ii)
V_parallel = 24 - 16.0 = 8.0 V
Voltage across parallel combination = 8.0 V [1]
⚠ If you missed marks here: The expected wrong answer is 24 V, treating the parallel section as though it sat straight across the battery. R1 has already taken 16.0 V, and in a series arrangement the p.d.s add up to the supply, so 24 − 16.0 = 8.0 V is left. No fresh calculation is needed; recomputing with I = 2.67 A and 3.0 Ω should give the same 8.0 V.
(c)[2 marks]
Calculate the power dissipated in R1.
Model Answer -- Part (c)
P = I²R = (2.67)² x 6.0 or P = IV = 2.67 x 16.0
Correct power formula substitution [1]
P = 42.7 W
Power = 42.7 W or equivalent [1]
⚠ If you missed marks here: The values must be R1’s own: it carries 2.67 A and drops 16.0 V, so P = IV = 2.67 × 16.0 = 42.7 W (or I2R = 2.672 × 6.0). Using the battery’s 24 V gives 64 W, which is the power of the whole circuit, not of this resistor. In I2R only the current is squared, not the product with R.
Question 5
Electrical Safety in the Home -- 10 marks
(a)[3 marks]
Draw and label a diagram of the inside of a three-pin plug, showing the three wires correctly connected.
Model Answer -- Part (a)
Live (brown) wire connected to the fuse and live pin [1]
Neutral (blue) wire connected to the neutral pin [1]
Earth (green and yellow) wire connected to the earth pin / correctly positioned at the top [1]
⚠ If you missed marks here: The mark most often lost is the fuse: it belongs in the live (brown) wire between the live pin and the appliance, never in the neutral. Earth, green and yellow, goes to the top pin, which is the longest. All three wires must be labelled with their colours — a correct but unlabelled drawing gains nothing — and red/black/green is the obsolete code.
(b)[2 marks]
Explain the purpose of earthing a metal-cased appliance.
Model Answer -- Part (b)
If a fault causes the live wire to touch the metal case, the current flows to earth through the earth wire [1]
This causes a large current that blows the fuse / trips the circuit breaker, disconnecting the supply and preventing electric shock [1]
⚠ If you missed marks here: Writing "it stops you getting an electric shock" is the outcome, not the mechanism, and scores nothing on its own. Mark 1: if a fault lets the live wire touch the metal case, current flows through the low-resistance earth wire to earth. Mark 2: that current is large enough to blow the fuse or trip the breaker, cutting off the supply so the case cannot remain live. Saying the earth wire "carries the extra electricity away" without the fuse blowing loses mark 2.
(c)[2 marks]
Explain the purpose of a fuse and how it works.
Model Answer -- Part (c)
A fuse is a thin wire that melts / blows when the current exceeds the rated value [1]
This breaks the circuit, protecting the appliance and the user from damage or electric shock [1]
⚠ If you missed marks here: Answering "it protects the appliance" without saying how earns neither mark. A fuse is a thin wire that melts when the current exceeds its rated value, and melting breaks the circuit. Saying it "stops too much voltage" is wrong — a fuse responds to current. Do not describe a circuit breaker instead: that switches off magnetically and can be reset, while a fuse must be replaced.
(d)[3 marks]
Explain what is meant by double insulation. Why do some appliances not need an earth wire?
Model Answer -- Part (d)
Double insulation means the appliance has no exposed metal parts that could become live [1]
The outer casing is made of an insulating material such as plastic [1]
There is no risk of the case becoming live, so an earth connection is not needed [1]
⚠ If you missed marks here: Answering "there are two layers of insulation" is not what the marks are for. The point is that no exposed metal part can become live, because the casing is made of an insulator such as plastic. Mark 3 follows from that: with no metal case that could become live there is nothing for an earth wire to make safe, so it is omitted. Double insulation does not remove the need for a fuse — the two protect against different things.
Question 6
Transformer and Power Transmission -- 12 marks
(a) (i)[2 marks]
Describe the structure of a simple transformer.
Model Answer -- Part (a)(i)
Two coils of insulated wire (primary and secondary) [1]
Both coils are wound on a shared soft iron core [1]
⚠ If you missed marks here: One coil wound on iron is an electromagnet, not a transformer. Both marks need the pair: a primary and a secondary coil, each of insulated wire, wound on the same soft iron core. Dropping "insulated" or naming steel instead of soft iron loses a mark, because steel would stay magnetised and could not follow the alternating field.
(a) (ii)[2 marks]
Explain how a transformer works.
Model Answer -- Part (a)(ii)
An alternating current in the primary coil produces a changing magnetic field in the iron core [1]
This changing field links the secondary coil and induces an alternating e.m.f. (by electromagnetic induction) [1]
⚠ If you missed marks here: Quoting the turns-ratio equation answers a different question — this part wants the mechanism. Alternating current in the primary sets up a continually changing magnetic field in the core, and that changing field through the secondary induces an alternating e.m.f. Saying the current "passes across" to the secondary is wrong: the two coils are never electrically connected. A d.c. supply would give a steady field and no output at all.
(b)[3 marks]
A step-down transformer has 4600 turns on the primary coil and 230 turns on the secondary. The input voltage is 230 V. Calculate the output voltage. If the transformer is ideal, and the output current is 8.0 A, calculate the input current.
Model Answer -- Part (b)
Vs = Vp x Ns/Np = 230 x 230/4600 = 11.5 V
Output voltage = 11.5 V correctly calculated [1]
IpVp = IsVs so Ip = IsVs / Vp
Power conservation equation for ideal transformer stated [1]
Ip = 8.0 x 11.5 / 230 = 0.40 A
Input current = 0.40 A correctly calculated [1]
⚠ If you missed marks here: Inverting the turns ratio gives 230 × 4600/230 = 4600 V, a step-up answer, when the question has already told you it steps down. For the current the ratio works the opposite way to the voltage: the low-voltage side carries the larger current, so the input current 0.40 A must be smaller than the 8.0 A output. An answer of 160 A means you multiplied by the voltage ratio where you should have divided.
(c)[2 marks]
Explain why the core of a transformer is made of soft iron, and state what would happen to the output if the core were removed.
Model Answer -- Part (c)
Soft iron is easily magnetised and demagnetised, so it carries the changing magnetic field from the primary coil to the secondary coil (it links almost all of the field into the secondary) [1]
Without the core, much less of the changing magnetic field would reach the secondary coil, so the induced output voltage would be much smaller [1]
⚠ If you missed marks here: Saying "iron is magnetic" is not enough for mark 1 — the core has to magnetise and demagnetise easily so that it follows the rapidly alternating field and carries almost all of it from the primary to the secondary. Mark 2 needs a specific effect of removing it: far less of the changing field reaches the secondary, so the output voltage is much smaller. "It would stop working completely" overstates it and is not credited.
(d)[3 marks]
A power station generates 500 MW of power. Compare the power loss in transmission cables (total resistance 4.0 Ω) when the power is transmitted at 25 kV versus 400 kV.
Model Answer -- Part (d)
At 25 kV: I = P/V = 500x10⁶ / 25x10³ = 20000 A
Loss = I²R = (20000)² x 4.0 = 1.6x10⁹ W = 1600 MW
Power loss at 25 kV correctly calculated as 1600 MW [1]
At 400 kV: I = P/V = 500x10⁶ / 400x10³ = 1250 A
Loss = I²R = (1250)² x 4.0 = 6.25x10⁶ W = 6.25 MW
Power loss at 400 kV correctly calculated as 6.25 MW [1]
Higher transmission voltage gives much lower current and dramatically lower power losses, so high voltage is essential for efficient transmission [1]
⚠ If you missed marks here: Powers of ten decide this part: 500 MW is 500 × 106 W and 25 kV is 25 × 103 V, giving I = 20 000 A. Slipping a factor of a thousand produces 20 A and a loss of 1600 W instead of 1600 MW. The loss is I2R, not IV or V2/R — and because the current is squared, dropping it by a factor of 16 cuts the loss by 256, from 1600 MW to 6.25 MW.
Question 7
Force on Conductor, DC Motor, and Loudspeaker -- 12 marks
(a)[2 marks]
State the conditions necessary for a current-carrying conductor to experience a force in a magnetic field.
Model Answer -- Part (a)
The conductor must carry a current [1]
The conductor must not be parallel to the magnetic field / must have a component perpendicular to the field [1]
⚠ If you missed marks here: Repeating the stem — "the wire must be in a magnetic field" — earns nothing. The two conditions are that the wire must be carrying a current, and that it must not lie parallel to the field: the force is greatest at 90° to the field and falls to zero when the wire points along it. However strong the magnet, a wire with no current feels no force.
(b) (i)[2 marks]
State Fleming's left-hand rule and what each finger represents.
Model Answer -- Part (b)(i)
Fleming's left-hand rule: hold the left hand with the thumb, first finger and second finger at right angles [1]
First finger = magnetic Field direction, seCond finger = Current direction, thuMb = direction of Motion/Force [1]
⚠ If you missed marks here: Reaching for the right hand gives a force in exactly the opposite direction — the right hand belongs to the generator effect, the left to the motor effect. Mark 1 is the geometry: thumb, first finger and second finger held mutually at right angles. Mark 2 is the labelling: First finger → Field, seCond finger → Current, thuMb → Motion. The current meant is conventional current, from + to −, not the electron flow.
(b) (ii)[2 marks]
A straight wire of length 0.15 m carries a current of 4.0 A at right angles to the field between the poles of a magnet. The wire experiences a force. State two changes the student could make that would increase the size of this force.
Model Answer -- Part (b)(ii)
Increase the current in the wire [1]
Use a stronger magnet / stronger magnetic field (accept: increase the length of wire that lies in the field) [1]
⚠ If you missed marks here: Answering "increase the voltage" is indirect and does not earn the current mark — say increase the current in the wire. The two changes must be genuinely different: a bigger current, a stronger magnet, or more of the wire’s length lying inside the field. "Use a bigger battery" and "turn the current up" are the same idea twice and score once, and turning the wire to lie along the field would reduce the force to zero.
(c)[3 marks]
Describe the structure and operation of a simple DC motor. Explain the role of the split-ring commutator.
Model Answer -- Part (c)
A rectangular coil is placed between the poles of a permanent magnet and connected to a DC supply via a split-ring commutator and brushes [1]
Current in the coil creates forces on opposite sides (one up, one down by Fleming's LHR), producing a turning effect [1]
The split-ring commutator reverses the current direction every half turn, ensuring the coil continues to rotate in the same direction [1]
⚠ If you missed marks here: Saying "the coil spins because of the magnet" earns none of the three marks. Mark 1 is the structure — a coil between the poles of a permanent magnet, fed through a split-ring commutator and brushes. Mark 2 is why it turns: the currents in the two opposite sides run in opposite directions, so the forces act one up and one down, giving a turning effect. Mark 3 is the commutator reversing the current every half turn so the coil keeps going the same way instead of rocking back. Slip rings belong to the a.c. generator.
(d)[3 marks]
Describe how a loudspeaker works.
Model Answer -- Part (d)
A coil of wire (voice coil) is attached to a paper cone and sits in the magnetic field of a permanent magnet [1]
An alternating current in the coil creates a force (by the motor effect) that alternates in direction [1]
This causes the cone to vibrate back and forth, producing sound waves at the frequency of the AC signal [1]
⚠ If you missed marks here: Describing a microphone — sound moving a coil and inducing a current — is this process in reverse and scores nothing. The marks are for the voice coil attached to the cone sitting in a permanent magnet’s field, the alternating current producing a force that alternates in direction by the motor effect, and the cone vibrating back and forth to send out sound waves at the frequency of the signal. "Electrical energy is turned into sound energy" names the outcome without the mechanism.
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