HKDSE Physics Practice Paper: Electricity, Current, Resistance And Power

HKDSE Physics Practice Paper: Electricity, Current, Resistance and Power

2026-11-03

HKDSE Physics: Electricity

HKDSE Physics paper on electricity: 12 questions, 100 marks, 75 minutes, covering Ohm law, series and parallel networks, power, energy cost, efficiency and mains safety.

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

Section A (Questions 1 to 4, 33 marks)

Read each question and write the correct answer with an explanation.

#	Question	Answer and marking notes
1	Define electric current, state its SI unit, and state the relation between current, charge and time. If 600 C of charge passes a cross-section of a wire in 5 minutes, find the current.	Current is the rate of flow of charge, namely the charge passing a cross-section per unit time; the SI unit is the ampere, and one ampere is one coulomb per second. The relation is current equals charge divided by time. Working: 5 minutes is 300 s, so the current is 600 divided by 300, which is 2.0 A. Marking points: definition 2 marks, unit 1 mark, relation 2 marks, calculation and answer 2 marks. Common weakness: forgetting to convert minutes into seconds, or counting electrons instead of charge.
2	State the law of Ohm and the conditions under which it applies. A resistor has a potential difference of 12 V across it and a current of 3.0 A through it. (a) Find its resistance. (b) Find the power it dissipates.	Law: provided temperature and other physical conditions remain constant, the potential difference across a conductor is proportional to the current through it, so V equals I multiplied by R. It applies to ohmic components such as metal conductors at constant temperature, but not to lamp filaments or diodes, whose resistance changes with temperature or with the direction of current. (a) R equals 12 divided by 3.0, which is 4.0 Ω. (b) P equals 12 multiplied by 3.0, which is 36 W, and I squared multiplied by R, 9 multiplied by 4.0, gives the same 36 W as a check. Marking points: law and condition 3 marks, resistance 2 marks, power 3 marks. Common weakness: quoting the formula without the constant temperature condition.
3	Resistors of 6 Ω and 3 Ω are connected in parallel, and the combination is connected in series with a 4 Ω resistor across a 12 V supply of negligible internal resistance. (a) Find the resistance of the parallel part. (b) Find the total resistance. (c) Find the total current.	(a) Parallel resistance equals 6 multiplied by 3 divided by 6 plus 3, that is 18 divided by 9, which is 2.0 Ω. (b) Total resistance equals 2.0 plus 4.0, which is 6.0 Ω. (c) Total current equals 12 divided by 6.0, which is 2.0 A. Marking points: parallel formula 3 marks, series addition 2 marks, working and unit for the current 4 marks. Common weakness: adding the parallel resistors directly to get 9 Ω, or omitting the 4 Ω resistor.
4	Using the circuit in question 3, (a) find the potential difference across the parallel part, (b) find the current in the 6 Ω resistor to two decimal places, and (c) find the energy dissipated in the 4 Ω resistor in 5 minutes.	(a) The voltage across the parallel part equals 2.0 multiplied by 2.0, which is 4.0 V. (b) The current in the 6 Ω resistor is 4.0 divided by 6, which is 0.67 A, while the other branch carries 1.33 A, and the two add to 2.0 A. (c) Energy equals I squared multiplied by R multiplied by t, that is 4.0 multiplied by 4.0 multiplied by 300, which is 4 800 J. Marking points: 3 marks for each part, with units. Common weakness: treating 12 V as the voltage across the parallel part, or failing to convert minutes into seconds.

Section B (Questions 5 to 8, 32 marks)

Answer the following questions. Show your working and units, and explain with physics concepts.

#	Question	Answer and marking notes
5	Resistors of 10 Ω and 15 Ω are connected in series across a 25 V supply. (a) Find the current. (b) Find the potential difference across each resistor. (c) Find the power of the 15 Ω resistor.	(a) Total resistance equals 10 plus 15, which is 25 Ω, so the current is 25 divided by 25, that is 1.0 A. (b) The voltage across the 10 Ω resistor is 10 V and across the 15 Ω resistor is 15 V, and the two add to the 25 V of the supply. (c) Power equals I squared multiplied by R, that is 1.0 multiplied by 15, which is 15 W. Marking points: current 3 marks, voltages 3 marks, power 2 marks. Common weakness: assuming the voltage is the same everywhere in a series circuit.
6	Resistors of 12 Ω and 4 Ω are connected in parallel across a 12 V supply. (a) Find the equivalent resistance. (b) Find the total current. (c) Find the total power.	(a) Equivalent resistance equals 12 multiplied by 4 divided by 12 plus 4, that is 48 divided by 16, which is 3.0 Ω. (b) Total current equals 12 divided by 3.0, which is 4.0 A, and the branch currents of 1.0 A and 3.0 A add to the same value. (c) Total power equals 12 multiplied by 4.0, which is 48 W. Marking points: equivalent resistance 3 marks, total current 3 marks, total power 2 marks. Common weakness: obtaining an equivalent resistance larger than any branch.
7	An electric water heater rated at 2 000 W is used for 3 hours a day for 30 days. Electricity costs 1.2 dollars per kilowatt-hour. (a) Find the energy used in kilowatt-hours. (b) Find the total cost.	(a) Energy each day equals 2 multiplied by 3, which is 6 kilowatt-hours, so over 30 days it is 180 kilowatt-hours. (b) The cost is 180 multiplied by 1.2, which is 216 dollars. Marking points: power conversion 3 marks, kilowatt-hours 3 marks, cost 2 marks. Common weakness: failing to convert watts into kilowatts, which makes the answer a thousand times too large.
8	An electric motor has an input power of 500 W and delivers a mechanical output of 400 W. (a) Find its efficiency. (b) Find the power lost as heat and sound. (c) Suggest one specific way to raise the efficiency.	(a) Efficiency equals useful output divided by input, multiplied by 100, that is 400 divided by 500, which is 80 per cent. (b) Power lost equals 500 minus 400, which is 100 W. (c) Suitable measures include using thicker copper wire to reduce coil resistance, improving the bearings to reduce friction, and improving cooling so that the coil temperature and resistance do not rise. Marking points: efficiency 3 marks, loss 2 marks, a specific and workable suggestion 3 marks. Common weakness: giving 20 per cent as the efficiency, which is really the fraction lost.

Section C (Questions 9 to 12, 35 marks)

Write fuller answers and support them with specific examples and data.

#	Question	Answer and marking notes
9	An electric kettle rated at 2 200 W is connected to a 220 V mains supply. (a) Find the current when it operates. (b) Should a 3 A, 13 A or 30 A fuse be used? Explain. (c) Explain why a fuse must be fitted in the live wire.	(a) Current equals 2 200 divided by 220, which is 10.0 A. (b) A 13 A fuse should be used: its rating is above the normal current of 10.0 A, so the kettle works normally, yet a fault current above 13 A melts the fuse and breaks the circuit. A 3 A fuse would melt during normal use, while a 30 A fuse would allow a large excess current to flow for a long time and give no protection. (c) With the fuse in the live wire, the kettle is isolated from the high potential once the fuse melts, so a person touching the internal conductors is not at risk; if the fuse were in the neutral wire, the appliance would still be connected to the live wire after melting and a shock hazard would remain. Marking points: current 3 marks, choice with reasons 4 marks, live wire explanation 3 marks. Common weakness: choosing 13 A without comparing the consequences of the three ratings.
10	Two identical lamps are connected (i) in series and (ii) in parallel to the same supply, with resistance assumed constant. (a) Compare the brightness of each lamp in the two cases and explain. (b) Compare the total power of the circuit. (c) State what happens to the other lamp when one lamp burns out in each case.	(a) In parallel each lamp has the full supply voltage across it, whereas in series the supply voltage is shared and each lamp gets only half, so each lamp is brighter in parallel. (b) Let each lamp have resistance R and let the supply be V. In series the total resistance is 2R and the total power is V squared divided by 2R; in parallel the equivalent resistance is R divided by 2 and the total power is 2 times V squared divided by R, four times as large. (c) In series the circuit is broken and the other lamp goes out as well; in parallel the branches are independent and the other lamp keeps working normally. Marking points: part (a) 3 marks, part (b) 3 marks with a comparison, part (c) 3 marks. Common weakness: saying only that parallel is brighter without explaining the sharing of voltage.
11	A uniform metal wire is 2.0 m long, has a cross-sectional area of 1.0 × 10⁻⁶ m² and a resistivity of 1.0 × 10⁻⁶ Ω m. (a) Find its resistance. (b) The wire is stretched to a length of 4.0 m at constant volume; find the new resistance.	(a) Resistance equals resistivity multiplied by length divided by area, that is 1.0 multiplied by 2.0 divided by 1.0, which is 2.0 Ω. (b) At constant volume a doubled length halves the cross-sectional area to 0.5 × 10⁻⁶ m², so the resistance becomes 1.0 multiplied by 4.0 divided by 0.5, which is 8.0 Ω, four times the original value. Marking points: formula and substitution 3 marks, 2.0 Ω 2 marks, new area 1 mark, 8.0 Ω 2 marks. Common weakness: assuming that doubling the length only doubles the resistance and ignoring the halving of area.
12	A 9 W LED lamp and a 60 W filament lamp give similar brightness. (a) Find the energy each uses in 1 000 hours in kilowatt-hours. (b) At 1.2 dollars per kilowatt-hour, find the difference in cost. (c) State one benefit of the LED other than cost.	(a) The LED uses 9 multiplied by 1 000 divided by 1 000, which is 9 kilowatt-hours, while the filament lamp uses 60 kilowatt-hours. (b) The LED costs 9 multiplied by 1.2, which is 10.8 dollars, the filament lamp costs 72 dollars, and the difference is 61.2 dollars. (c) The LED produces less heat, so room temperature rises less and the air conditioner load falls; its life is also usually longer, which reduces replacement and disposal. Marking points: part (a) 3 marks, part (b) 3 marks, part (c) 2 marks. Common weakness: treating the wattage directly as the number of kilowatt-hours.

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