Show all working for calculation questions -- marks are awarded for method.
Use the question navigation buttons to move between questions.
After submitting, use "Show Model Answer" to reveal mark schemes and self-assess.
Check each mark point you believe you earned, then calculate your grade.
A calculator may be used.
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]
(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]
(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)
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]
(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]
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]
(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]
(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]
(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]
(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]
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]
(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]
(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]
(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]
(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]
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]
(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]
(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]
(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]
(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]
(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]
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]
(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]
(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]
(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]
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]
Wound on a shared soft iron core (laminated) [1]
(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]
(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]
(c)[2 marks]
Explain why the iron core of a transformer is laminated.
Model Answer -- Part (c)
Laminations reduce eddy currents in the core [1]
This reduces energy loss / heating in the core, making the transformer more efficient [1]
(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]
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]
(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]
(b) (ii)[2 marks]
A wire of length 0.15 m carries a current of 4.0 A perpendicular to a magnetic field of flux density 0.30 T. Calculate the force on the wire using F = BIl.
Model Answer -- Part (b)(ii)
F = BIl = 0.30 x 4.0 x 0.15
Correct substitution into F = BIl [1]
F = 0.18 N
Force = 0.18 N [1]
(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]
(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]
When you have finished answering all questions, click Submit to see the model answers.
Exam Submitted -- Marking Mode Active
Click "Show Model Answer" on each question to check your work.