IGCSE Physics: Electricity and Magnetism (Extended) — Practice Questions (Cambridge 0625)

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs — Cambridge International holds copyright in its own papers. Use these alongside the official past papers available from your board.

Tier note: everything marked (Extended) is Supplement content of the 0625 syllabus. Two parts are Core: the transformer turns ratio, Vₚ/Vₛ = Nₚ/Nₛ, in 5(a) (syllabus 4.5.6 Core 3), and finding a current from P = IV in 6(a) (4.2.5 Core 2). The later parts of those questions (IₚVₚ = IₛVₛ and power loss P = I²R, 4.5.6 Supplement 7–8) and all of questions 1–4 and 7 are Extended only.

After each answer there is a common mistake to avoid.

---

Questions

1. (Extended) A charge of 450 C flows through a torch bulb in 2.5 minutes. (a) Calculate the current in the bulb. (b) State the direction in which the free electrons flow around the circuit, compared with the direction of the conventional current. [3]

2. (Extended) A 12 Ω resistor and a 6.0 Ω resistor are connected in parallel across a 6.0 V battery. Calculate (a) the combined resistance of the two resistors and (b) the current supplied by the battery. [3]

3. (Extended) A piece of metal wire has a resistance of 2.0 Ω. A second wire, made of the same metal, is twice as long and has three times the cross-sectional area. Calculate the resistance of the second wire. [2]

4. (Extended) The north pole of a bar magnet is pushed towards one end of a coil that is connected to a sensitive meter. (a) State the magnetic pole produced at the end of the coil nearest the magnet while the magnet is moving in. (b) Explain, using the idea of energy, why the induced current must produce this pole. [2]

5. A transformer has 1600 turns on its primary coil, which is connected to a 240 V a.c. supply. The output voltage is 12 V. (a) Calculate the number of turns on the secondary coil. (b) (Extended) The current in the secondary coil is 2.0 A. Assuming the transformer is 100% efficient, calculate the current in the primary coil. [3]

6. A power station sends 20 MW of electrical power along transmission cables at 400 kV. The total resistance of the cables is 5.0 Ω. (a) Calculate the current in the cables. (b) (Extended) Calculate the power wasted as heat in the cables. (c) (Extended) Explain why sending the same power at a lower voltage would waste more energy. [3]

7. (Extended) A student draws the magnetic field lines around a bar magnet. (a) State how the drawing shows where the magnetic field is strongest, and where that is. (b) The N poles of two bar magnets are held close together. State what causes the force between them. [3]

---

Answers

1. (Extended) (a) t = 2.5 × 60 = 150 s [1]; I = Q ÷ t = 450 ÷ 150 = 3.0 A [1]. (b) Electrons flow from the negative terminal to the positive terminal, which is opposite to the conventional current (positive to negative) [1].

Common mistake: dividing by 2.5 instead of 150. Current is charge per second, so time must be in seconds.

2. (Extended) (a) 1/R = 1/12 + 1/6.0 = 3/12 [1], so R = 4.0 Ω [1]. (b) I = V ÷ R = 6.0 ÷ 4.0 = 1.5 A [1].

Common mistake: forgetting to turn 1/R back into R, giving 0.25 Ω. A check: the combined resistance of a parallel pair must be less than the smaller resistor.

3. (Extended) Doubling the length doubles the resistance to 4.0 Ω; tripling the cross-sectional area divides it by 3 [1]. R = 2.0 × 2 ÷ 3 = 1.3 Ω [1].

Common mistake: multiplying by 3 for the larger area. A thicker wire has more room for the current, so resistance is inversely proportional to cross-sectional area.

4. (Extended) (a) A north pole [1]. (b) The induced pole repels the approaching magnet, opposing the change that causes it, so work must be done to push the magnet in, and this work is transferred to electrical energy in the coil; if the coil attracted the magnet, electrical energy would be produced without any work being done, which would break the conservation of energy [1].

Common mistake: saying a south pole forms “to attract the magnet”. The induced e.m.f. always opposes the change causing it.

5. (a) Vₚ ÷ Vₛ = Nₚ ÷ Nₛ, so Nₛ = 1600 × 12 ÷ 240 = 80 turns [1]. (b) (Extended) IₚVₚ = IₛVₛ [1], so Iₚ = (2.0 × 12) ÷ 240 = 0.10 A [1].

Common mistake: thinking that a step-down transformer also steps down the current. With 100% efficiency the power is the same on both sides, so the lower voltage side has the larger current.

6. (a) I = P ÷ V = 20 000 000 ÷ 400 000 = 50 A [1]. (b) (Extended) P = I²R = 50² × 5.0 = 12 500 W (12.5 kW) [1]. (c) (Extended) For the same power, a lower voltage needs a larger current, and since the power lost is I²R, the loss rises with the square of the current [1].

Common mistake: using P = V²/R with the 400 kV supply voltage. The 400 kV is not the voltage across the cables, so the loss must be worked out from the current using I²R.

7. (Extended) (a) The field is strongest where the field lines are closest together [1], which is near the poles [1]. (b) The force is due to the interaction between the magnetic fields of the two magnets [1] (here it is a repulsion, because the poles are alike).

Common mistake: drawing more lines to show a stronger magnet but spacing them evenly. Strength is shown by how close together the lines are.

---

Where marks are usually lost

• Leaving time in minutes when using I = Q ÷ t.
• Giving 1/R instead of R for resistors in parallel.
• Getting the effect of cross-sectional area on resistance the wrong way round.
• Saying the induced current helps the magnet move, instead of opposing the change.
• Using the supply voltage in V²/R to find the power lost in transmission cables.