Free Electricity questions for AS Level Physics (9702), part of Waves, electricity & particles. Each question has the answer and the marking points an examiner would look for.
Generate a full Electricity worksheet →All Waves, electricity & particles questionsA wire has a cross-sectional area that changes along its length. Section X has 2 times the cross-sectional area of section Y. There is a constant current in the wire.
What is average drift speed of electrons in Xaverage drift speed of electrons in Y?
Answer: D
A wire of resistivity 4.9e-7 Ω m, length 1.7 m and diameter 0.52 mm. What is its resistance?
Answer: B
How does the resistance of a filament lamp change as the current increases?
Answer: D
A copper wire of cross-sectional area 0.80 mm2 and length 5.0 m carries a current of 1.5 A. Copper has 8.5 × 1028 free electrons per m3 and resistivity 1.7 × 10−8 Ω m.
Calculate the drift speed of the electrons.
Calculate the number of electrons passing a point in 2.0 minutes.
Calculate the resistance of the wire.
A wire of half the diameter carries the same current. State the factor by which the drift speed changes.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 1.38 × 10−4 m s−1 I = nAvq C1 v = 1.38 × 10−4 m s−1 A1 | 2 |
| (b) | 1.13 × 1021 Q = 180 C C1 N = 1.13 × 1021 A1 | 2 |
| (c) | 0.106 Ω R = ρL/A C1 0.106 Ω A1 | 2 |
| (d) | 4 ×4 B1 | 1 |
A wire of length 2.5 m and diameter 0.40 mm is made of a metal of resistivity 1.1 × 10−6 Ω m.
Define resistivity in words, or by stating the meaning of each symbol in an equation.
Calculate the resistance of the wire.
A current of 1.50 A passes through a copper wire with the same cross-section. The number density of free electrons in copper is 8.5 × 1028 m−3. Calculate the mean drift speed of the electrons.
Explain, in terms of electrons and lattice ions, why the resistance of a metal increases with temperature.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | ρ = RA/L resistance × cross-sectional area / length (ρ = RA/L, symbols explained) B1 | 1 |
| (b) | 21.9 Ω R = ρL/A, A = 1.26 × 10−7 m2 C1 21.9 Ω A1 | 2 |
| (c) | 8.78 × 10−4 m s−1 I = nAvq C1 v = 1.50/(8.5 × 1028 × 1.26 × 10−7 × 1.60 × 10−19) C1 8.78 × 10−4 m s−1 A1 | 3 |
| (d) | ions vibrate more; more frequent collisions with electrons lattice ions vibrate with greater amplitude B1 electrons collide with them more often, reducing drift speed for a given p.d. B1 | 2 |
A wire of length 1.5 m and diameter 0.3 mm has resistivity 4.9 × 10−7 Ω m.
Calculate the resistance of the wire.
The wire is stretched to twice its length without change of volume. Determine the new resistance.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 10.4 Ω R = ρL/A C1 10.4 Ω A1 | 2 |
| (b) | 41.6 Ω A halves and L doubles C1 41.6 Ω A1 | 2 |
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Meet Riyaz →Original Cambridge-style practice questions written for mentorriyaz.in. These are not Cambridge past paper questions, and this site is not affiliated with or endorsed by Cambridge University Press & Assessment. Updated 2026-11-04. Paper generator · mentorriyaz.in