Electricity (resistivity, circuits, EMF & internal resistance, potential dividers) – 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.

3Standard4 marks

The diagram shows a circuit containing a battery of emf 6.0 V and negligible internal resistance.

Calculate the current in the 12 Ω resistor. [4]

resistance of parallel combination: $\left(\dfrac{1}{6.0} + \dfrac{1}{12}\right)^{-1} = 4.0$ Ω(1)
total resistance = 8.0 Ω so current from battery = $\dfrac{6.0}{8.0} = 0.75$ A(1)
p.d. across parallel section = $0.75 \times 4.0 = 3.0$ V or by ratio of resistances(1)
current in 12 Ω = $\dfrac{3.0}{12} = 0.25$ A(1)
Answer: 0.25 A
Examiner tip: The p.d. across parallel branches is the same, but the current splits in inverse ratio to resistance — the larger resistor gets the smaller current.

4Standard5 marks

A battery is connected to a variable resistor. When the current is 0.40 A, the terminal p.d. is 5.6 V. When the current is 0.75 A, the terminal p.d. is 4.9 V.

(a) Calculate the internal resistance and the emf of the battery. [3]

(b) In a required practical, a student records several pairs of values of terminal p.d. $V$ and current $I$. Describe how she should use a graph to find the emf and internal resistance. [2]

(a) $\varepsilon = V + Ir$: $\varepsilon = 5.6 + 0.40r$ and $\varepsilon = 4.9 + 0.75r$ allow $r = \frac{\Delta V}{\Delta I} = \frac{5.6-4.9}{0.75-0.40}$(1)
(a) $0.35r = 0.70$ so $r = 2.0$ Ω(1)
(a) $\varepsilon = 5.6 + 0.40 \times 2.0 = 6.4$ V allow 4.9 + 0.75 × 2.0; ecf for their r(1)
(b) plot $V$ (y-axis) against $I$ (x-axis); emf = the y-intercept allow $I$ against $V$ with emf = intercept on the $V$ axis(1)
(b) internal resistance = −gradient (magnitude of gradient) allow for an $I$ against $V$ graph: $r$ = −1/gradient(1)
Answer: (a) $r = 2.0$ Ω, $\varepsilon = 6.4$ V (b) $\varepsilon$ = intercept on V axis; $r$ = −gradient
Examiner tip: Write ε = V + Ir for both readings and solve simultaneously — the emf is the same in both.

5Hard5 marks

The diagram shows a potential divider used as a temperature sensor. The supply has negligible internal resistance. At 20 °C the (negative temperature coefficient, NTC) thermistor has a resistance of 4.7 kΩ.

(a) Calculate the output p.d. $V_{out}$ at 20 °C. [2]

(b) The temperature of the thermistor rises. Explain what happens to $V_{out}$. [2]

(a) $V_{out} = 9.0 \times \dfrac{2.2}{2.2 + 4.7}$(1)
(a) 2.9 V(1)
(b) the resistance of the (NTC) thermistor decreases as temperature rises(1)
(b) the fixed resistor gets a larger share of the supply p.d. (current increases), so $V_{out}$ increases allow p.d. across thermistor decreases so $V_{out} = 9.0 - V_{th}$ increases(1)
(c) 2.2 kΩ (p.d. is shared equally, so resistances are equal)(1)
Answer: (a) 2.9 V (b) $V_{out}$ increases (c) 2.2 kΩ
Examiner tip: Always state the thermistor's resistance change first, then link it to the share of p.d. — the second mark depends on the first.

6Standard5 marks

A battery of emf 12 V and internal resistance 0.50 Ω is connected to a heater of resistance 5.5 Ω.

(a) Calculate the power dissipated in the heater. [3]

(b) Calculate the percentage of the total energy supplied by the battery that is transferred to the heater. [2]

(a) $I = \dfrac{\varepsilon}{R + r} = \dfrac{12}{6.0} = 2.0$ A(1)
(a) $P = I^2R = 2.0^2 \times 5.5$ or $VI$ with $V = 11$ V(1)
(a) 22 W(1)
(b) total power = $\varepsilon I = 12 \times 2.0 = 24$ W(1)
(b) $\dfrac{22}{24}\times100 = 92\%$ allow 91.7%(1)
Answer: (a) 22 W (b) 92%
Examiner tip: The total power supplied by the battery is εI, not VI; the difference, I²r, is wasted inside the battery.

7Hard5 marks

Two 100 kΩ resistors are connected in series to a 12 V supply of negligible internal resistance. A student connects a voltmeter of resistance 200 kΩ across one of the resistors.

(a) Calculate the reading on the voltmeter. [3]

(b) The student expected a reading of 6.0 V. Explain why the reading is lower and how the measurement could be improved. [2]

(a) combined resistance of voltmeter and resistor: $\left(\dfrac{1}{100}+\dfrac{1}{200}\right)^{-1} = 66.7$ kΩ(1)
(a) $V = 12 \times \dfrac{66.7}{66.7 + 100}$(1)
(a) 4.8 V(1)
(b) the voltmeter in parallel reduces the resistance of that section, so it takes a smaller share of the 12 V(1)
(b) use a voltmeter with a much higher resistance (ideally infinite) compared with 100 kΩ, e.g. a digital voltmeter of ~10 MΩ(1)
Answer: (a) 4.8 V (b) voltmeter lowers resistance of the section; use a much higher-resistance voltmeter
Examiner tip: An ideal voltmeter has infinite resistance; real ones only behave ideally when their resistance is much larger than the component they measure.

8Hard6 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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