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S5 Physics Practice Paper: Electromagnetic Induction, Transformers and Alternating Current (12 Questions, 100 Marks, 75 Minutes, with Answers)
HKDSE Physics: Electromagnetic Induction, Transformers and Alternating Current
A 12-question, 100-mark, 75-minute physics paper on electromagnetic induction, transformers, eddy currents, rms values and transmission losses, with worked solutions.
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Paper details Level: S5 | Subject: Physics | Time: 75 minutes | Marks: 100 | Questions: 12 | Answers included
Section A: Electromagnetic Induction (32 marks)
Questions 1 to 4
| # | Question | Answer and marking notes |
|---|---|---|
| 1 | A coil of 200 turns, each of area 0.050 m2, lies in a uniform magnetic field whose flux density falls uniformly from 0.20 T to zero in 0.4 s. Find the average induced emf across the coil. | Flux linkage per turn: magnetic flux = B x A = 0.20 x 0.050 = 0.010 Wb (1 mark). By Faraday’s law the average induced emf is N x dPhi/dt = 200 x 0.010 / 0.4 = 5.0 V (3 marks). The emf acts in a direction that opposes the decrease of flux (1 mark). Answer: 5.0 V (1 mark). |
| 2 | A 0.50 m metal rod moves at 4.0 m/s perpendicular to the field lines of a uniform 0.30 T field, and its ends are connected to a 1.5 ohm resistor. Find (a) the induced emf, (b) the current, (c) the power dissipated in the resistor, (d) the external force needed to keep the rod moving at constant speed. | (a) emf = B x l x v = 0.30 x 0.50 x 4.0 = 0.60 V (2 marks). (b) I = emf / R = 0.60 / 1.5 = 0.40 A (2 marks). (c) P = I squared x R = 0.40 x 0.40 x 1.5 = 0.24 W (2 marks). (d) At constant speed the applied force balances the magnetic force, F = B x I x l = 0.30 x 0.40 x 0.50 = 0.060 N (2 marks). |
| 3 | State Lenz’s law and explain why, when the N pole of a magnet is moved towards a coil, the end of the coil facing the magnet behaves as an N pole. | Lenz’s law: the induced current flows in a direction such that its magnetic field opposes the change in flux that produced it (3 marks). As the N pole approaches, the downward flux through the coil increases (2 marks); the induced current produces an upward field to oppose the increase (2 marks), so the end facing the magnet is an N pole and repels the magnet (2 marks). Energy is conserved because work is done against this repulsion. |
| 4 | For the coil in Question 1 (resistance 20 ohm), find the total induced charge passing through the coil during the 0.4 s and explain how this charge depends on the time taken. | Induced charge q = N x flux change / R = 200 x 0.010 / 20 = 0.10 C (4 marks). The formula shows that the charge depends only on the total change of flux and the resistance, not on the time taken (4 marks): a longer time means a smaller current, and the product of current and time stays the same (1 mark). |
Section B: Transformers (36 marks)
Questions 5 to 8
| # | Question | Answer and marking notes |
|---|---|---|
| 5 | An ideal transformer has 1200 turns on its primary, connected to a 240 V a.c. supply, and 60 turns on its secondary. Find (a) the secondary voltage, (b) the primary current when the secondary current is 5.0 A, (c) the power transferred. | (a) Vs / Vp = Ns / Np, so Vs = 240 x 60 / 1200 = 12.0 V (3 marks). (b) For an ideal transformer input power equals output power, so Vp x Ip = Vs x Is and Ip = 5.0 x 12.0 / 240 = 0.25 A (4 marks). (c) P = Vs x Is = 12.0 x 5.0 = 60 W (3 marks). |
| 6 | A transformer of efficiency 90 percent takes 100 W from the supply and delivers 12 V from its secondary. Find the secondary current and the power lost in the transformer. | Output power = 100 x 0.90 = 90.0 W (3 marks). Secondary current = 90.0 / 12 = 7.5 A (3 marks). Power lost = 100 – 90.0 = 10.0 W (3 marks). The loss is mainly joule heating in the coils plus eddy current and hysteresis losses in the core. |
| 7 | Compare the turns ratio of step-up and step-down transformers, and explain why high-voltage transmission reduces the power lost in transmission cables. | A step-up transformer has more secondary turns than primary turns, so the secondary voltage is higher than the primary voltage (2 marks); a step-down transformer has fewer secondary turns (2 marks). Cable loss is I squared R (2 marks). For the same transmitted power, raising the voltage by a factor n reduces the current by n, so the loss falls by a factor of n squared (2 marks). Hence high-voltage transmission greatly reduces cable loss. |
| 8 | Explain how eddy currents are produced in an iron core and how a laminated core reduces the energy loss. | The alternating flux in the core induces circulating currents known as eddy currents (3 marks). Because the iron has low resistance, these currents cause large joule heating and energy loss (3 marks). A laminated core is built from thin sheets separated by insulation, which breaks up the eddy current paths and raises the resistance, so the eddy currents and the loss are much reduced (3 marks). |
Section C: Alternating Current and Transmission (32 marks)
Questions 9 to 12
| # | Question | Answer and marking notes |
|---|---|---|
| 9 | An a.c. supply has a peak voltage of 340 V at 50 Hz. Find (a) the rms voltage, (b) the period. | (a) rms voltage = peak voltage / square root of 2 = 340 / 1.414 = 240.4 V (4 marks). (b) period = 1 / f = 1 / 50 = 0.020 s (4 marks). |
| 10 | A 60 ohm resistor is connected to an a.c. supply of rms value 240 V. Find (a) the rms current, (b) the average power dissipated. | (a) rms current = 240 / 60 = 4.0 A (4 marks). (b) average power = 240 x 4.0 = 960 W, or rms voltage squared divided by R = 240 x 240 / 60 = 960 W (4 marks). |
| 11 | Explain why the average value of an alternating current over a complete cycle is zero, and why the rms value rather than the average value is used to calculate power. | The positive and negative half cycles of an a.c. wave are equal in size but opposite in direction (3 marks), so the average over a complete cycle is zero and cannot be used for power (2 marks). Power depends on the square of the current, so the current is squared, averaged and then square-rooted to give the rms value (2 marks); average power = rms current squared x R (1 mark). |
| 12 | A transmission cable of total resistance 2.0 ohm carries 20 kW. Find the power lost in the cable when the power is transmitted at 400 V and at 4000 V, and comment on the result. | At 400 V the current is 20000 / 400 = 50.0 A and the loss is 50.0 x 50.0 x 2.0 = 5000 W, about 25.0 percent of the transmitted power (3 marks). At 4000 V the current is 20000 / 4000 = 5.0 A and the loss is 5.0 x 5.0 x 2.0 = 50 W, only 0.25 percent (3 marks). A tenfold increase in voltage reduces the current tenfold and the loss a hundredfold, so high-voltage transmission is far more efficient (2 marks). |
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