AS Physics: Electricity — Practice Questions

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 — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Electricity revision notes

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Questions

1. Define electric current and the coulomb. [2]

2. Define potential difference. [2]

3. State Ohm’s law and the condition under which it applies. [2]

4. A copper wire of length 2.4 m and cross-sectional area 1.8 × 10⁻⁷ m² has a resistance of 0.23 Ω.

(a) Calculate the resistivity of copper. [3] (b) State the effect on resistance of doubling the length. [1] (c) State the effect on resistance of doubling the diameter. [2]

5. Sketch or describe the I–V characteristic of:

(a) a metallic conductor at constant temperature [1] (b) a filament lamp, with an explanation of its shape [3] (c) a semiconductor diode [2]

6. Explain, using I = nAvq, why a thin section of a wire becomes hotter than the rest when a current flows. [4]

7. A 60 W lamp operates at 12 V.

(a) Calculate the current. [2] (b) Calculate its resistance. [2] (c) Calculate the charge passing in 5.0 minutes. [2]

8. Distinguish between e.m.f. and potential difference, and explain why the terminal p.d. of a battery is always less than its e.m.f. when current flows. [3]

9. A thermistor’s resistance falls as its temperature rises.

(a) Sketch or describe its I–V characteristic. [1] (b) Explain, in terms of charge carriers, why its resistance falls with increasing temperature. [2] (c) Explain why this behaviour is the opposite of a metallic conductor’s. [2]

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10. (a) Define the potential difference across a component. [1]

(b) A wire has length 2.0 m, cross-sectional area 5.0 × 10⁻⁷ m² and is made of a metal with resistivity 1.1 × 10⁻⁶ Ω m. Calculate its resistance. [2]

Answers

1. Current is the rate of flow of charge [1]. One coulomb is the charge passing when a current of one ampere flows for one second [1].

2. The energy transferred from electrical to other forms per unit charge [1] passing between two points; one volt is one joule per coulomb [1].

3. The current is directly proportional to the potential difference across a conductor [1], provided the temperature (and other physical conditions) remain constant [1].

4. (a) ρ = RA ÷ L = (0.23 × 1.8 × 10⁻⁷) ÷ 2.4 [1] [1] = 1.725 × 10⁻⁸ ≈ 1.7 × 10⁻⁸ Ω m (2 s.f., matching the data) [1]. (b) Resistance doubles [1]. (c) Area is proportional to the square of the diameter, so area quadruples [1] and resistance is quartered [1].

5. (a) A straight line through the origin [1]. (b) A curve that flattens as current increases [1]. The current heats the filament [1], increasing lattice vibration so electrons collide more frequently and resistance rises [1]. (c) Almost no current in reverse bias or below about 0.6 V forward [1], then a sharp rise above that threshold [1].

6. The current I is the same throughout a series circuit [1]. In the thin section the cross-sectional area A is smaller, so from I = nAvq the drift velocity v must be greater [1]. The thin section also has a greater resistance R, since R = ρL/A and A is smaller [1]; since the same current I flows through it, the power dissipated there, P = I²R, is greater than in the thicker sections, so it heats up more [1]. (The higher drift speed shows charge is moving faster through the constriction, but the heating itself is best explained by the greater I²R power dissipation from the higher local resistance, not by asserting that faster electrons directly cause more frequent lattice collisions — collision rate also depends on lattice temperature and the material, not drift speed alone.)

7. (a) I = P ÷ V = 60 ÷ 12 [1] = 5.0 A [1]. (b) R = V ÷ I = 12 ÷ 5.0 [1] = 2.4 Ω [1]. (c) Q = It = 5.0 × 300 [1] = 1500 C [1].

8. E.m.f. is the energy converted from other forms to electrical per unit charge, supplied by the source [1]; p.d. is the energy transferred from electrical to other forms per unit charge, across a component [1]. While the battery is discharging — supplying current to an external circuit — some energy per coulomb is transferred to the battery’s own internal resistance, so the terminal p.d. is less than the e.m.f. by the “lost volts” across that internal resistance [1]. (The reverse holds while the battery is being charged: current is driven into its positive terminal, and the applied terminal voltage must then exceed the e.m.f. by Ir.)

9. (a) A curve where resistance falls as current (and temperature) increases — the opposite curvature to a filament lamp [1]. (b) As temperature rises, more charge carriers (electrons) are released and become free to move [1], so for a given p.d. more current can flow, meaning resistance falls [1]. (c) In a metal, the number of charge carriers stays constant with temperature, and it is increased lattice vibration (more frequent collisions) that raises resistance [1]; in a thermistor, the number of charge carriers itself increases with temperature, and this effect dominates over any increase in collisions, so resistance falls instead of rising [1].

10. (a) The energy transferred from electrical to other forms per unit charge passing through the component [1]. (b) R = ρL ÷ A [1] = (1.1 × 10⁻⁶ × 2.0) ÷ (5.0 × 10⁻⁷) = 4.4 Ω [1].

Examiner insight (June 2025): the most common mistake was to define potential difference as the energy needed to “move” a unit charge, instead of the energy transferred per unit charge. Try the real question next: Cambridge International AS & A Level Physics 9702, June 2025, Paper 22, Question 6.

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Where marks are usually lost

• Confusing resistivity with resistance.
• Halving rather than quartering resistance when the diameter doubles.
• Explaining the filament lamp curve without mentioning temperature.
• Omitting the constant-temperature condition from Ohm’s law.
• Using e.m.f. and p.d. interchangeably, or forgetting that internal resistance is what causes the terminal p.d. to be less than the e.m.f.
• Explaining a thermistor’s falling resistance using the same “more lattice collisions” reasoning as a filament lamp — the dominant effect there is the number of charge carriers increasing, not the collision rate.

Metals vs thermistors — same equation, opposite outcome

Both a metal and a thermistor are governed by I = nAvq, but temperature affects the two variables that matter (n, the charge carrier density, and the collision rate) in opposite proportions. In a metal, n is already enormous and essentially fixed, so a temperature rise mainly increases lattice vibration and collision frequency, raising resistance. In a thermistor, by contrast, only a small number of charge carriers are free at room temperature, so a temperature rise releases a large number of additional carriers, and this dominates over any increase in collision frequency, lowering resistance. Recognising which effect dominates for a given material is the key skill this pair of components tests. For the full I–V characteristic table and resistivity treatment, see the Electricity revision notes.