A cell measured with a biased voltmeter

8 marks · A-level · original practice case

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The situation

A student varies the load on a cell and records terminal p.d. and current. The data are simulated with rounding and small scatter. A best-fit line passes through (0.10 A, 1.43 V) and (0.50 A, 1.03 V). Use these fitted positions for slope/intercept calculations, not adjacent readings. The cell model is V = ε − Ir.

Use the axes and units. The table provides the text alternative.

Simulated cell readings
I / A	V / V
0.10	1.44
0.20	1.32
0.30	1.24
0.40	1.12
0.50	1.04

Part (a)

Determine the internal resistance from the fitted line. [2 marks]

Give the numerical answer to 2 significant figures.

Final value / Ω

Working

Part (b)

Determine the emf inferred from this fitted line. [2 marks]

Give the numerical answer to 3 significant figures.

Final value / V

Working

Part (c)

Explain how a constant positive voltmeter error of 0.04 V affects the inferred emf and internal resistance. [2 marks]

Your explanation

Part (d)

Suggest why readings should be taken quickly and the cell allowed to rest between readings. [2 marks]

Your explanation

Marking rubric and worked answers

Indicative practice marks. Method marks assess valid working; accuracy marks require the associated valid method. Independent points stand on their own. Tick only points supported by your response. Do not award the same explanation twice. These ticks record your judgement, not an examiner’s result.

Part (a) marking points

Final answer: 1.0 Ω. Numerical checks assess the final value only; they do not check working or the number of digits typed.

Use gradient (1.03 − 1.43)/(0.50 − 0.10) = −1.0 V A⁻¹. (1 method mark)
Identify r = −gradient = 1.0 Ω. (1 accuracy mark)

No earlier numerical result is required. Award method credit for valid working and accuracy credit for the stated answer. Accept a defensible equivalent method; wrong physics cannot earn dependent accuracy credit.

Common errors: Internal resistance is minus the gradient, so it is positive.

Part (b) marking points

Final answer: 1.53 V. Numerical checks assess the final value only; they do not check working or the number of digits typed.

Extrapolate using ε = V + Ir. (1 method mark)
Obtain 1.53 V. (1 accuracy mark)

Common errors: This is a fitted voltage at non-zero current, not the intercept.

Part (c) marking points

All p.d. readings and the intercept increase by 0.04 V, so emf is overestimated by 0.04 V. (1 independent mark)
Voltage differences and gradient are unchanged, so inferred r is unchanged. (1 independent mark)

Award each distinct physical point once. Equivalent scientifically correct wording is creditable; a contradictory statement cannot earn the same point.

Part (d) marking points

Current heats the cell or can change its state during the test. (1 independent mark)
Keeping temperature/state approximately stable helps keep ε and r constant. (1 independent mark)

Award each distinct physical point once. Equivalent scientifically correct wording is creditable; a contradictory statement cannot earn the same point.

Acceptable alternatives: A switch opened between readings is a valid suggestion if its purpose is explained.

Report a problem with this case

Related skills

internal resistance practical practice

Scope and reference sources

AQA practical/mathematical skills with selected 3.4/3.5; OCR A Modules 1/2 and selected 3/4. These are shared original exercises; wording, indicative marks and difficulty have not been endorsed by either board.

AQA specification · OCR A specification

Case reference: cell-calibration · version 1.1.0 · seed 0. Choose another case