Powering a flash from a capacitor

10 marks · A-level · original practice case

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The situation

A 2200 μF capacitor is initially charged to 12.0 V. It discharges through an idealised flash circuit modelled as a constant 30.0 Ω resistance. The flash stops when capacitor voltage reaches 5.00 V. Assume capacitance and resistance are constant; 18.0% of the energy transferred from the capacitor becomes visible light.

Flash model
Parameter	Value
Capacitance	2200 μF
Initial voltage	12.0 V
Cut-off voltage	5.00 V
Discharge resistance	30.0 Ω

Part (a)

Calculate the energy emitted as visible light during one flash. [2 marks]

Give the numerical answer to 3 significant figures.

Final value / J

Working

Part (b)

Determine how long the flash lasts. [3 marks]

Give the numerical answer to 3 significant figures.

Final value / s

Working

Part (c)

Calculate the mean electrical power delivered to the resistor while the flash operates. [2 marks]

Give the numerical answer to 3 significant figures.

Final value / W

Working

Part (d)

Explain why doubling capacitance does not double the initial electrical power in this model, although it increases the flash duration. [3 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: 0.0236 J. Numerical checks assess the final value only; they do not check working or the number of digits typed.

Calculate the decrease in stored energy: ½C(12.0² − 5.00²) = 0.1309 J. (1 method mark)
Multiply the decrease by 0.180 to obtain 0.0236 J. (1 accuracy mark)

No earlier numerical result is required. Credit a valid alternative method. Accuracy credit requires a physically valid method.

Part (b) marking points

Final answer: 0.0578 s. Numerical checks assess the final value only; they do not check working or the number of digits typed.

Use exponential discharge V/V₀ = exp(−t/RC). (1 method mark)
Rearrange t = RC ln(12.0/5.00). (1 method mark)
Obtain 0.0578 s. (1 accuracy mark)

No earlier numerical result is required. Credit a valid alternative method. Accuracy credit requires a physically valid method.

Part (c) marking points

Final answer: 2.27 W. Numerical checks assess the final value only; they do not check working or the number of digits typed.

Use energy decrease divided by discharge duration, not the initial power. (1 method mark)
Obtain 2.27 W. (1 accuracy mark)

Credit correct division using the student’s discharge time from part (b), after an arithmetic or prefix error. If part (a) has an arithmetic error, an energy value recovered by dividing the light energy by 0.180 may be carried forward. Reject use of visible-light energy as electrical energy without reversing the efficiency.

Part (d) marking points

Initial power is V₀²/R, independent of capacitance at fixed V₀ and R. (1 independent mark)
The time constant RC doubles, so reaching the same voltage ratio takes twice as long. (1 independent mark)
The energy transferred also doubles; the time dependence remains exponential. (1 independent mark)

Credit equivalent correct explanations. Award each distinct point once; a contradiction invalidates the affected point.

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Related skills

Scope and reference sources

AQA 3.7.4; OCR A 6.1.1–6.1.3. These are shared original exercises; wording, indicative marks and difficulty have not been endorsed by either board.

AQA specification · OCR A specification

Case reference: capacitor-flash-energy · version 1.1.0 · seed 0. Choose another case