Current, I–V characteristics and resistivity (RP5) – A-level Physics exam questions with mark scheme

1Core4 marks

A current of 2.4 mA flows in a wire for 5.0 minutes.

(a) Calculate the charge that passes a point in the wire. [2]

(b) Calculate the number of electrons that pass this point. [1]

(a) $Q = It = 2.4\times10^{-3} \times 300$(1)
(a) 0.72 C(1)
(b) $N = \dfrac{0.72}{1.60\times10^{-19}} = 4.5\times10^{18}$ ecf from (a)(1)
(c) it becomes zero allow falls to zero / no resistance; reject 'very small' or 'decreases'(1)
Answer: (a) 0.72 C (b) $4.5\times10^{18}$ (c) becomes zero
Examiner tip: Convert mA to A and minutes to seconds before using Q = It.

2Standard5 marks

A student determines the resistivity of nichrome. A length of 0.75 m of wire of diameter 0.32 mm has a resistance of 10.1 Ω.

(a) Calculate the resistivity of nichrome. [3]

(b) State why the diameter should be measured at several points along the wire and in different orientations. [1]

(c) The student instead plots a graph of resistance against length for several lengths. Explain why the resistance of the connecting leads and contacts does not affect the value of resistivity found from the gradient. [1]

(a) $A = \pi\left(0.16\times10^{-3}\right)^2 = 8.04\times10^{-8}$ m$^2$(1)
(a) $\rho = \dfrac{RA}{L} = \dfrac{10.1\times8.04\times10^{-8}}{0.75}$(1)
(a) $1.1\times10^{-6}$ Ω m allow 1.08 × 10$^{-6}$; unit required(1)
(b) the wire may not be uniform / may not be perfectly circular, so a mean diameter is more representative allow to reduce the effect of random error(1)
(c) the contact resistance adds a constant amount to every reading, which only shifts the intercept, not the gradient (gradient = $\rho/A$)(1)
Answer: (a) $1.1\times10^{-6}$ Ω m (b) wire not uniform; take a mean (c) constant offset changes the intercept only
Examiner tip: The unit of resistivity is Ω m, not Ω m⁻¹ or Ω/m — a wrong unit loses the final mark.

3Hard6 marks

Describe how a student could obtain the current–voltage characteristic of a filament lamp, including both positive and negative values of p.d., and explain the shape of the graph she would obtain. [6]

circuit: lamp in series with an ammeter, voltmeter in parallel with the lamp(1)
p.d. varied using a potential divider (or variable resistor) connected to the supply(1)
record several pairs of $V$ and $I$ over the range, then reverse the supply / lamp connections to obtain negative values(1)
plot $I$ (y-axis) against $V$ (x-axis) allow axes swapped, with the shape described to match(1)
graph is a curve through the origin with a decreasing gradient (for $I$ against $V$); the same shape in the third quadrant (rotational symmetry about the origin)(1)
as current increases the filament temperature rises(1)
lattice (metal) ions vibrate with larger amplitude, so electrons collide with them more often; resistance ($V/I$) increases(1)

Level	Marks	What it looks like
Level 3	5–6	A full, coherent account of both the method (circuit, varying and reversing the p.d.) and the shape of the graph with its explanation, in a logical order; physics terminology used correctly throughout.
Level 2	3–4	A partial account covering several indicative points with some linking; may have gaps or minor errors in the sequence or terminology.
Level 1	1–2	Isolated, relevant points with little or no linking between them.
Level 0	0	No relevant content.

Answer: Lamp with series ammeter and parallel voltmeter, p.d. varied with a potential divider and reversed; curve through the origin with decreasing gradient, same shape in the negative quadrant, because resistance rises with temperature.
Examiner tip: Resistance is V/I at a point, not the gradient — explain the curve in terms of rising temperature and resistance.

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