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HKDSE Physics Practice Paper: Electricity, Circuits and Electromagnetic Induction (12 questions, 100 marks, 75 minutes)
HKDSE Physics: Electricity, Circuits and Electromagnetic Induction
Original HKDSE Physics practice paper on electricity, circuits and electromagnetic induction: 12 questions, 100 marks and 75 minutes, with worked answers and marking points.
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Paper details Level: S6 | Subject: Physics | Time: 75 minutes | Marks: 100 | Questions: 12
Section A (32 marks)
Resistance, power and household-circuit questions: show the formula and the unit.
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | In a circuit, a 12 V battery of negligible internal resistance is connected to a 4 Ω resistor in series with a parallel pair of 6 Ω and 3 Ω. (a) Find the equivalent resistance of the parallel part. (b) Find the total resistance and the total current. (c) Find the voltage across the 4 Ω resistor and the currents through the 6 Ω and 3 Ω resistors. |
Answer: (a) 1/R_p = 1/6 + 1/3 = 0.5, so R_p = 2.0 Ω. (b) Total resistance R = 4 + 2.0 = 6.0 Ω; total current I = 12/6.0 = 2.0 A. (c) Voltage across 4 Ω = 2.0 × 4.0 = 8.0 V; across the parallel part = 12 − 8.0 = 4.0 V; current in 6 Ω = 4.0/6 ≈ 0.67 A, in 3 Ω = 4.0/3 ≈ 1.33 A. Marking points: parallel formula 1, R_p value 1, total resistance 1, total current 1, p.d. split 1, each branch current 1, units 1. |
| 2 | A nichrome wire is 2.0 m long, has cross-sectional area 5.0×10⁻⁷ m² and resistivity 1.1×10⁻⁶ Ω m. (a) Find its resistance. (b) If the p.d. across it is 6.0 V, find the current and the power dissipated. (c) After some time the resistance rises slightly. Explain why. |
Answer: (a) R = ρL/A = (1.1×10⁻⁶ × 2.0)/(5.0×10⁻⁷) = 4.4 Ω. (b) I = 6.0/4.4 ≈ 1.36 A; P = VI = 6.0 × 1.36 ≈ 8.2 W. (c) The current heats the metal; lattice vibrations become stronger and electrons collide more often, so resistance rises (positive temperature coefficient of metals). Marking points: formula 1, substitution and value 2, current 1, power 2, temperature and lattice explanation 2. |
| 3 | A 2,000 W water heater is used for 1.5 hours a day for 30 days. (a) Find the energy used in that month in kWh. (b) If electricity costs 1.2 dollars per kWh, find the monthly cost. (c) A 1,200 W model used for 2 hours a day is considered instead. Compare the monthly energy use. |
Answer: (a) 2.0 kW × 1.5 h × 30 = 90 kWh. (b) 90 × 1.2 = 108 dollars. (c) 1.2 kW × 2.0 h × 30 = 72 kWh, which is 18 kWh less than 90 kWh, a reduction of 20%. Marking points: kW conversion 1, time and days 2, cost 2, alternative model energy 2, comparison 1. |
| 4 | The mains socket voltage in Hong Kong is 220 V. A 1,500 W kettle and an 800 W rice cooker are plugged into the same socket circuit protected by a 13 A fuse. (a) Find the current through each appliance when both are on. (b) Find the total current and state whether the fuse will blow. (c) Explain the function of the fuse and why a copper wire must not replace it. |
Answer: (a) Kettle I = 1,500/220 ≈ 6.8 A; rice cooker I = 800/220 ≈ 3.6 A. (b) Total current ≈ 6.8 + 3.6 = 10.4 A, below 13 A, so the fuse will not blow. (c) On overload or short circuit the fuse wire melts and breaks the circuit, preventing overheating and fire. A copper wire has a high melting point and low resistance, so it will not melt in time and gives no protection. Marking points: each current 1, total 1, judgement 1, fuse principle 2, copper danger 2. |
Section B (36 marks)
Field, capacitor, magnetic-force and induction questions: show the steps and give a judgement.
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | Two parallel metal plates are 5.0 mm apart and connected to a 200 V d.c. supply. (a) Find the electric field strength between the plates. (b) A particle carrying 2.0 µC is placed between the plates. Find the electric force on it. (c) Find the work done by the field in moving the particle from the negative to the positive plate, and state the energy change. |
Answer: (a) E = V/d = 200/(5.0×10⁻³) = 4.0×10⁴ V/m. (b) F = qE = 2.0×10⁻⁶ × 4.0×10⁴ = 0.080 N. (c) W = qV = 2.0×10⁻⁶ × 200 = 4.0×10⁻⁴ J; the field does work and the electric potential energy falls by 4.0×10⁻⁴ J, becoming kinetic energy. Marking points: formula and E 3, F 3, W 2, direction of energy change 1. |
| 6 | A 470 µF capacitor is charged fully by a 12 V supply. (a) Find the charge stored. (b) Find the energy stored. (c) It is then discharged through a 10 kΩ resistor. Find the time constant and state the time for the voltage to fall to about 37% of its initial value. |
Answer: (a) Q = CV = 470×10⁻⁶ × 12 = 5.6×10⁻³ C (5.6 mC). (b) E = ½CV² = 0.5 × 470×10⁻⁶ × 144 ≈ 3.4×10⁻² J. (c) τ = RC = 10×10³ × 470×10⁻⁶ = 4.7 s; after one time constant the voltage is about 37% of its initial value. Marking points: Q 2, E 3, τ 3, 37% statement 1. |
| 7 | A straight wire 0.25 m long carries a current of 4.0 A and is placed perpendicular to a uniform magnetic field of 0.30 T. (a) Find the magnetic force on the wire. (b) Find the force if the wire makes 30° with the field. (c) Explain how Fleming left-hand rule gives the direction of the force. |
Answer: (a) F = BIL = 0.30 × 4.0 × 0.25 = 0.30 N. (b) F = BIL sin30° = 0.30 × 0.5 = 0.15 N. (c) In Fleming left-hand rule the first finger points along the field, the second finger along the current, and the thumb gives the force; the three are mutually perpendicular. Marking points: force 3, angle correction 3, rule description 3. |
| 8 | A 200-turn coil of area 5.0×10⁻³ m² per turn lies in a magnetic field perpendicular to its plane. The flux density increases uniformly from 0.10 T to 0.30 T in 0.20 s. The coil resistance is 2.0 Ω. (a) Find the flux change per turn. (b) Find the induced e.m.f. (c) Find the induced current and the power dissipated in the coil. |
Answer: (a) ΔΦ = ΔB × A = 0.20 × 5.0×10⁻³ = 1.0×10⁻³ Wb. (b) ε = NΔΦ/Δt = 200 × 1.0×10⁻³/0.20 = 1.0 V. (c) I = ε/R = 1.0/2.0 = 0.50 A; P = I²R = 0.50² × 2.0 = 0.50 W. Marking points: ΔΦ 2, ε 3, current 2, power 2 (must use I²R or VI). |
Section C (32 marks)
Transformer, transmission and a.c. questions: set out the calculation and explain its physical meaning.
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | A transformer of efficiency 90% has its primary connected to 220 V with 1,100 turns. The secondary supplies 11 V to a 22 W device. (a) Find the number of secondary turns. (b) Find the input power. (c) Find the primary current. |
Answer: (a) N_s/N_p = V_s/V_p, so N_s = 1,100 × 11/220 = 55 turns. (b) Input power = 22/0.90 ≈ 24 W. (c) I_p = 24/220 ≈ 0.11 A. Marking points: turns ratio 2, N_s value 2, efficiency treatment 2, primary current 2. |
| 10 | A transmission cable has a resistance of 0.40 Ω and delivers 100 kW. (a) If transmission is at 400 V, find the current and the power loss in the cable. (b) Repeat for transmission at 20 kV. (c) Hence explain the advantage of high-voltage transmission. |
Answer: (a) I = 100,000/400 = 250 A; P_loss = I²R = 250² × 0.40 = 25,000 W (25 kW, 25% of the input). (b) I = 100,000/20,000 = 5.0 A; P_loss = 5.0² × 0.40 = 10 W. (c) Raising the voltage 50 times cuts the current to 1/50 and the loss by I² to 1/2,500, so high-voltage transmission greatly reduces heating loss in the cable. Marking points: (a) 3, (b) 3, squared relationship 2. |
| 11 | A battery of e.m.f. 6.0 V and internal resistance 1.5 Ω is connected to an external resistor of 4.5 Ω. (a) Find the current. (b) Find the terminal voltage. (c) Find the power dissipated inside the battery and the efficiency of the circuit. |
Answer: (a) I = ε/(R + r) = 6.0/(4.5 + 1.5) = 1.0 A. (b) Terminal voltage V = IR = 1.0 × 4.5 = 4.5 V. (c) P_int = I²r = 1.0² × 1.5 = 1.5 W; efficiency = 4.5/6.0 = 75%. Marking points: (a) 3, (b) 2, internal power 2, efficiency 1. |
| 12 | An a.c. generator has a 100-turn coil of area 0.020 m² per turn rotating at 50 Hz in a uniform field of 0.15 T. (a) Find the angular frequency. (b) Find the peak induced e.m.f. (c) Find the rms e.m.f. and state which value the domestic 220 V rating refers to. |
Answer: (a) ω = 2πf = 2π × 50 ≈ 314 rad/s. (b) ε₀ = NBAω = 100 × 0.15 × 0.020 × 314 ≈ 94 V. (c) ε_rms = ε₀/√2 ≈ 67 V; the domestic 220 V rating is an rms value. Marking points: (a) 2, (b) formula 2 and value 2, (c) 2. |
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