HKDSE Physics Practice Paper: Current, Circuits And Electromagnetism

HKDSE Physics Practice Paper: Current, Circuits and Electromagnetism

2026-11-03

HKDSE Physics: Current, Circuits and Electromagnetism

HKDSE Physics practice paper on current, circuits and electromagnetism: 12 questions, 100 marks, 75 minutes, with full answers, marking points and A4 printing.

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Paper details　Level: S6 | Subject: Physics | Time: 75 minutes | Marks: 100 | Questions: 12

Section A (Questions 1 to 4, 26 marks)

Read each question and write the correct answer with an explanation. Show working and units for any calculation.

#	Question	Answer and marking notes
1	A current of 2.5 A flows through a wire for 4 minutes. Find the charge passing a cross section, and the number of electrons this charge represents (elementary charge = 1.6 × 10⁻¹⁹ C).	Q = It = 2.5 × 240 = 600 C. Number of electrons = 600 divided by 1.6 × 10⁻¹⁹ = 3.75 × 10²¹. Marking: formula 1, substitution 1, charge 1, electron count 1.
2	A 12 V battery of negligible internal resistance is connected to a 6 ohm resistor in series with a parallel pair of 12 ohm and 6 ohm. Find the equivalent resistance of the pair, the total current, the voltage across the pair, and the current in the 12 ohm resistor.	Parallel resistance = (12 × 6) divided by (12 + 6) = 4 ohm. Total resistance = 6 + 4 = 10 ohm, total current = 12 divided by 10 = 1.2 A. Voltage across the pair = 1.2 × 4 = 4.8 V. Current in the 12 ohm branch = 4.8 divided by 12 = 0.4 A. Marking: equivalent resistance 2, total current 1, voltage 1, branch current 2.
3	A 2000 W kettle runs on a 220 V supply for 3 minutes. Find the energy used in joules and in kWh, the cost at 1.2 dollars per kWh, the current, and the resistance of the kettle.	Energy = 2000 × 180 = 360 000 J = 0.1 kWh. Cost = 0.1 × 1.2 = 0.12 dollars. Current = 2000 divided by 220 = 9.09 A. Resistance = 220 divided by 9.09 = 24.2 ohm. Marking: energy 2, kWh conversion 1, cost 1, current 2, resistance 2.
4	A copper wire is 2.0 m long with cross sectional area 1.0 × 10⁻⁶ m². The resistivity of copper is 1.7 × 10⁻⁸ ohm metre. Find the resistance, and the new resistance if the length doubles while the area is halved.	R = resistivity × length divided by area = (1.7 × 10⁻⁸ × 2.0) divided by 1.0 × 10⁻⁶ = 0.034 ohm. Doubling the length doubles R and halving the area doubles it again, so the new resistance is 0.136 ohm. Marking: formula 2, substitution 2, first value 2, second value 2.

Section B (Questions 5 to 8, 32 marks)

Answer the following questions with working, units and evidence drawn from the data given.

#	Question	Answer and marking notes
5	Describe how to measure the resistance of a lamp with an ammeter and a voltmeter, and explain why the current voltage graph of a lamp is not a straight line.	Connect the ammeter in series with the lamp and the voltmeter in parallel across it. Vary the supply in steps, record several current and voltage pairs, and compute R = V divided by I for each. The tungsten filament heats up as the current rises, and the resistance of a metal increases with temperature, so the graph curves upward instead of staying straight. Marking: connections 2, procedure 2, calculation 1, temperature explanation 2.
6	A 0.25 m length of wire carries 3.0 A and sits in a uniform magnetic field of 0.40 T at 90 degrees to the wire. Find the magnitude of the magnetic force and state how its direction is found.	F = BIL = 0.40 × 3.0 × 0.25 = 0.30 N. The direction follows Fleming left hand rule: first finger along the field, second finger along the current, and the thumb gives the force. Marking: formula 2, value 2, rule named 1, finger mapping 2.
7	A transformer has 1000 turns on the primary and 50 turns on the secondary, with 220 V applied to the primary. Find the secondary voltage. If the output power is 44 W at 100 per cent efficiency, find both currents. Find the actual output when efficiency falls to 90 per cent.	Secondary voltage = 220 × (50 divided by 1000) = 11 V. Secondary current = 44 divided by 11 = 4.0 A. Primary current = 44 divided by 220 = 0.2 A. At 90 per cent efficiency the output is 44 × 0.90 = 39.6 W. Marking: turns ratio 2, secondary voltage 2, each current 2, efficiency output 2.
8	State Faraday law of electromagnetic induction and Lenz law, and explain why the induced current reverses direction when a magnet leaves the coil rather than entering it.	Faraday law: the induced emf is proportional to the rate of change of magnetic flux. Lenz law: the induced current always opposes the change that produced it. As the magnet enters, flux grows and the induced current makes an opposing field; as it leaves, flux falls and the induced current makes a field in the same direction to sustain the flux. Hence the two currents flow in opposite directions. Marking: Faraday 3, Lenz 3, both cases 1 each.

Section C (Questions 9 to 12, 42 marks)

Write fuller answers using correct scientific terms and support them with examples from Hong Kong or daily life.

#	Question	Answer and marking notes
9	A battery of emf 6.0 V and internal resistance 0.5 ohm is connected to a 2.5 ohm external resistor. Find the current, the terminal voltage, the power wasted inside the battery, and the efficiency.	Current = 6.0 divided by (2.5 + 0.5) = 2.0 A. Terminal voltage = 6.0 – 2.0 × 0.5 = 5.0 V. Power in the internal resistance = 2.0 squared times 0.5 = 2.0 W. Efficiency = 5.0 divided by 6.0 = 83.3 per cent. Marking: current 2, terminal voltage 2, power 2, efficiency 4.
10	A 2200 W appliance is plugged into a 220 V household circuit. Find the operating current and state whether a 5 A, 13 A or 15 A fuse should be used. Explain the function of the earth wire and why it must never carry the operating current.	Current = 2200 divided by 220 = 10 A, so a 13 A fuse is correct: 5 A would blow and 15 A gives too little protection. The earth wire connects the metal casing to the ground, so leakage current flows safely to earth and the risk of shock falls. The earth wire carries no current in normal use, and using it as a return path would make the casing live and destroy that protection. Marking: current 3, fuse choice with reason 3, earth function 2, not a return path 2.
11	Design an experiment to compare the resistance of two lamps. Give the measuring steps, the variables to control, and three sources of error.	Connect the lamp, ammeter and variable supply in series, with the voltmeter across the lamp. Raise the voltage in steps, record current and compute R = V divided by I. Control the lamp model, the ambient temperature and the wiring layout. Errors come from meter resistance, from the filament heating up during readings, and from random reading error. Marking: circuit 2, steps 3, controls 3, three errors 3.
12	High voltage transmission reduces power loss along a cable. Explain why using P = I squared R, and calculate the loss and the percentage lost when 100 MW is transmitted through a cable of resistance 5 ohm at 400 kV and at 25 kV.	With power fixed, a higher voltage means a smaller current, and since loss = I squared R, halving the current cuts the loss to one quarter. At 400 kV the current is 100 × 10⁶ divided by 400 × 10³ = 250 A, so the loss is 250 squared times 5 = 312 500 W, which is 0.3125 MW or 0.31 per cent. At 25 kV the current is 4000 A and the loss is 4000 squared times 5 = 80 000 000 W, which is 80 MW or 80 per cent. Marking: explanation 3, 400 kV calculation 4, 25 kV calculation 4.

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