DSE Physics Practice Paper: Electricity And Magnetism (12 Questions, 50 Marks, 75 Minutes)

DSE Physics Practice Paper: Electricity and Magnetism (12 Questions, 50 Marks, 75 Minutes)

2026-11-02

DSE Physics: Electricity and Magnetism

DSE Physics paper: 12 questions, 50 marks, 75 minutes on circuits, power bills, magnetic force, induction, transformers and Coulomb law. Full mark schemes included.

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Paper details　Level: S6 | Subject: Physics | Time: 75 minutes | Marks: 50 | Questions: 12

Section A (Short Questions)

Each question carries 2 marks. Show the equation and give units.

#	Question	Answer and marking notes
1	A 12 V supply is connected across a 4 ohm resistor. Find (a) the current through the resistor; (b) the power it dissipates.	I = V / R = 12 / 4 = 3 A; P = V I = 12 x 3 = 36 W. [(a) 1 mark, (b) 1 mark]
2	A steady current of 2.5 A flows through a wire for 240 s. (a) Find the charge passing any cross-section. (b) If the current is halved and the time doubled, how does the charge change?	Q = I t = 2.5 x 240 = 600 C. Halving the current and doubling the time leaves the charge unchanged at 600 C. [(a) 1 mark, (b) 1 mark]
3	A 60 W lamp is left on for 5 hours. (a) Find the energy used in kilowatt-hours. (b) Find the cost at 1.2 dollars per kilowatt-hour.	E = P t = 60 x 5 = 300 Wh = 0.3 kWh; cost = 0.3 x 1.2 = 0.36 dollars. [(a) 1 mark, (b) 1 mark]
4	Resistors of 12 ohm and 6 ohm are connected in parallel across a supply. (a) Find the equivalent resistance. (b) If the supply is 12 V, what is the total current?	1 / R = 1 / 12 + 1 / 6, so R = 4 ohm; I = V / R = 12 / 4 = 3 A. [(a) 1 mark, (b) 1 mark]
5	A straight wire of length 0.2 m carrying 5 A is placed at right angles to a uniform magnetic field of flux density 0.4 T. (a) Find the magnitude of the magnetic force. (b) How does the force change if the current is reversed?	F = B I L = 0.4 x 5 x 0.2 = 0.4 N. Reversing the current leaves the magnitude unchanged and reverses the direction (Fleming left-hand rule). [(a) 1 mark, (b) 1 mark]
6	An ideal transformer has 1000 turns in the primary and 50 turns in the secondary, with 220 V across the primary. (a) Find the secondary voltage. (b) Is it a step-up or step-down transformer?	Vs = Vp x Ns / Np = 220 x 50 / 1000 = 11 V. The secondary voltage is lower, so it is a step-down transformer. [(a) 1 mark, (b) 1 mark]

Section B (Multi-step Calculations)

Each question carries 4 marks. Show all working.

#	Question	Answer and marking notes
7	A battery of e.m.f. 6 V and negligible internal resistance is connected in series with resistors of 8 ohm and 12 ohm. (a) Find the current. (b) Find the voltage across the 12 ohm resistor. (c) Find the power dissipated in it.	(a) Total resistance = 8 + 12 = 20 ohm, so I = 6 / 20 = 0.3 A. (b) V = I R = 0.3 x 12 = 3.6 V. (c) P = I^2 R = 0.3^2 x 12 = 1.08 W. [(a) 1, (b) 1, (c) 2 marks]
8	A 2000 W heater raises 2 kg of water from 20 degrees to 80 degrees. The specific heat capacity of water is 4200 J per kg per degree. (a) Find the energy absorbed by the water. (b) Assuming no heat loss, find the time needed. (c) The real time is longer. Explain why.	(a) E = m c dT = 2 x 4200 x 60 = 504000 J. (b) t = E / P = 504000 / 2000 = 252 s. (c) In practice heat is lost to the container and the surrounding air, so the heater is less than 100% efficient and the real time is longer. [(a) 2, (b) 1, (c) 1 marks]
9	A coil of 200 turns and cross-sectional area 0.005 m2 lies perpendicular to a magnetic field whose flux density rises uniformly from 0.20 T to 0.60 T in 0.1 s. (a) Find the change in flux through the coil. (b) Find the induced e.m.f. (c) Which law determines the direction of the induced current?	(a) Change in flux = dB x A = 0.4 x 0.005 = 0.002 Wb. (b) e.m.f. = N x change in flux / time = 200 x 0.002 / 0.1 = 4 V. (c) The direction is given by Lenz law: the induced current produces a field that opposes the increase in flux. [(a) 1, (b) 2, (c) 1 marks]

Section C (Integrated and Essay Questions)

Follow the instructions and support answers with physical principles and calculations.

#	Question	Answer and marking notes
10	A domestic socket circuit runs at 220 V and supplies a 2200 W kettle and a 1100 W hot-water pot connected in parallel. (a) Find the current through each appliance. (b) Find the total current and state whether a 13 A fuse would blow. (c) Find the resistance of each appliance. (d) If both run for 2 hours, find the energy used and the cost at 1.2 dollars per kWh.	(a) Kettle: I = P / V = 2200 / 220 = 10 A; pot: 1100 / 220 = 5 A. (b) In parallel the total current is 15 A, which is larger than 13 A, so the 13 A fuse melts. (c) Kettle: R = V / I = 22 ohm; pot: 44 ohm. (d) Energy = (2.2 + 1.1) x 2 = 6.6 kWh; cost = 6.6 x 1.2 = 7.92 dollars. [(a) 2, (b) 2, (c) 2, (d) 2 marks]
11	A metal rod of length 0.5 m moves at 2 m/s through a horizontal magnetic field of flux density 0.8 T, perpendicular to both the rod and its motion. The ends of the rod are joined by a resistor of 0.4 ohm. (a) Find the induced e.m.f. (b) Find the current and explain how Lenz law fixes its direction. (c) Find the magnetic force on the rod and state its direction. (d) Compare the mechanical power needed to keep the speed constant with the heating power in the resistor, and explain.	(a) e.m.f. = B L v = 0.8 x 0.5 x 2 = 0.8 V. (b) I = e.m.f. / R = 0.8 / 0.4 = 2 A; by Lenz law the induced current produces a force that opposes the motion of the rod. (c) F = B I L = 0.8 x 2 x 0.5 = 0.8 N, opposite to the velocity. (d) Mechanical power = F v = 1.6 W; heating power = I^2 R = 1.6 W. They are equal, because at constant speed all the work done by the applied force becomes electrical energy and then heat, as energy conservation requires. [(a) 2, (b) 2, (c) 2, (d) 3 marks]
12	In a vacuum, two point charges of +4 microcoulomb and +1 microcoulomb are 0.3 m apart. Take the Coulomb constant as 9.0 x 10^9 N m2/C2. (a) Find the magnitude and direction of the electrostatic force between them. (b) Find the resultant field strength at the midpoint of the line joining them, giving magnitude and direction. (c) Find the point on that line where the resultant field is zero.	(a) F = k Q1 Q2 / r^2 = 9.0 x 10^9 x 4.0 x 10^-6 x 1.0 x 10^-6 / 0.3^2 = 0.4 N, repulsive because the charges have the same sign. (b) The midpoint is 0.15 m from each charge: E1 = 1.6 x 10^6 N/C pointing away from the 4 microcoulomb charge and E2 = 4 x 10^5 N/C pointing away from the 1 microcoulomb charge. The two fields oppose, so the resultant is 1.2 x 10^6 N/C towards the smaller charge. (c) At x from the 4 microcoulomb charge, k Q1 / x^2 = k Q2 / (0.3 – x)^2 gives 2 / x = 1 / (0.3 – x), so x = 0.2 m, between the charges and closer to the smaller one. [(a) 3, (b) 4, (c) 2 marks]

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