AQA AS Level Physics Paper 1, June 2019: Question 2

9 marks · Medium difficulty · Short Answer

Calculate the energy dissipated in a heating element, determine the suitability of a carbon fibre tape for the heating element using resistivity, and calculate the resistance of a resistor in a circuit containing an LED.

Practise this question

Question

Three-part physics question about a heating element and an LED circuit. Question 02.1 asks to calculate energy dissipated given a current of 2.0 A and time of 240 minutes. Question 02.2 asks to deduce if a carbon fibre tape with cross-sectional area 4.9 x 10^-6 m^2 and resistivity 2.0 x 10^-5 ohm meters is suitable for a 0.85 m heating element of 3.7 ohms. Question 02.3 provides a circuit diagram with a battery, switch, LED, and resistor R, alongside an I-V characteristic graph for the LED, asking to calculate the resistance of R.
Question text

02 A battery of emf 7.4 V and negligible internal resistance is used to power a heating

element inside a glove. The heating element has a resistance of 3.7 Ω.

02.1 The designers state that the battery can produce a current of 2.0 A in the heating

element for 240 minutes.

Calculate the energy dissipated in the heating element in this time.

[3 marks]

energy dissipated = J

02.2 The length of the heating element needed is about 0.85 m.

The designer considers using a carbon fibre tape for the heating element.

Table 1 gives information for the carbon fibre tape.

Table 1

m2 Resistivity / Ω m

Cross-sectional area /

4.9 × 10−6 2.0 × 10−5

Deduce whether the carbon fibre tape is suitable for making the heating element for

the glove.

[2 marks]

02.3 A light emitting diode (LED) is used to indicate that the switch in the glove is closed,

as shown in Figure 1. Resistor R limits the current in the LED.

Figure 1

Figure 2 shows part of the characteristic graph for the LED.

*05* Figure 2

The circuit is designed so that the potential difference across the LED is 2.2 V when

the switch is closed.

Calculate the resistance of R.

[4 marks]

resistance = Ω

Mark scheme

Show the mark scheme Mark scheme showing answers for questions 02.1, 02.2, and 02.3. Question 02.1 awards 3 marks for correct power, time conversion, and energy calculation of 2.1 x 10^5 J. Question 02.2 awards 2 marks for using resistivity equation to find length or resistance and making an appropriate conclusion. Question 02.3 awards 4 marks for reading current from graph, finding pd across resistor as 5.2 V, and calculating resistance as 350 ohms.

Question Answers Additional Comments/Guidance Mark

details

V2

P=VI P=I2R P =

Use of or or

R

Use of ∆W=P∆t

OR

02.1 3 AO2.1g

Use of ∆Q=I∆t

Use of W=VQ

2 marks if time not converted to seconds (3600

2.1 × 105 (J)

J)

𝑅𝑅𝑅𝑅

Use of 𝜌𝜌 =

𝐿𝐿

0.91 (m) + appropriate conclusion Allow calculation of R, ρ or A assuming 0.85 m

length, and conclusion for second mark:

02.2 2 5

R = 3.5 Ω

A = 4.6 × 10–6 m2

ρ = 2.1 × 10–5 Ω m

Full marks for correct answer

350 (Ω)

Max 3 from:

14.5 15.5 AO3.1a

Allow to

02.3 15 (mA) read from graph 4

AO2.1h

Conversion to A

pd across resistor = 7.4 – 2.2 = 5.2 V

𝑉𝑉

Use of 𝑅𝑅 =

𝐼𝐼 Do not allow gradient calculation for R.

Total 9

How to answer it

Electric Current, Potential Difference, and Resistance

What this question tests

This question assesses core circuit equations (power, energy, resistance, and resistivity), interpreting non-linear characteristic graphs (specifically for an LED), applying Kirchhoff's second law / potential dividers, and handling unit conversions such as minutes to seconds and milliamperes to amperes.

Part 0.2.1: Energy Dissipated by the Heating Element

Calculate the energy dissipated in the heating element in 240 minutes. [3 marks]

✅ Correct Answer

2.1 × 10⁵ J (or 210 000 J )

💡 Key Knowledge

  • Power formula: P = I²R (or P = VI , P = V² / R ).
  • Energy formula: ΔW = P × Δt or W = VQ where Q = IΔt .
  • Time conversion: Always convert minutes into SI base units (seconds). 240 min = 240 × 60 = 14 400 s .

📐 Step-by-Step Calculation

  1. Find Power (P): P = I²R = (2.0 A)² × 3.7 Ω = 4 × 3.7 = 14.8 W
  2. Convert Time (t): t = 240 min × 60 s/min = 14 400 s
  3. Calculate Energy (ΔW): ΔW = P × t = 14.8 W × 14 400 s = 213 120 J ≈ 2.1 × 10⁵ J

❌ Common Errors & Examiner Comments

  • The Time Trap: Forgetting to multiply by 60 and using t = 240 directly loses 1 mark (giving 3552 J or capped marks). The mark scheme awards 2 marks if time is left in minutes (resulting in 3552 J or equivalent intermediate slip).
  • Significant Figures: Give your final answer to 2 significant figures to match the precision of the input data (2.0 A, 7.4 V, 3.7 Ω).
Mark breakdown: 1 mark for calculating power (or charge/voltage method), 1 mark for correct time conversion / substitution into energy equation, 1 mark for final correct numerical value with unit.

Part 0.2.2: Suitability of Carbon Fibre Tape

Deduce whether the carbon fibre tape is suitable for making the heating element for the glove. [2 marks]

✅ Correct Answer

The required length for the tape is 0.91 m (or calculating required resistance/area). Since this is close to 0.85 m , it is suitable (with valid comparative commentary).

💡 Key Knowledge

Resistivity equation: R = ρL / A , rearranged to solve for length: L = RA / ρ

  • Resistance of element ( R ) = 3.7 Ω
  • Cross-sectional area ( A ) = 4.9 × 10⁻⁶ m²
  • Resistivity ( ρ ) = 2.0 × 10⁻⁵ Ω m

📐 Step-by-Step Calculation

  1. Rearrange formula: L = (R × A) / ρ
  2. Substitute values: L = (3.7 × 4.9 × 10⁻⁶) / (2.0 × 10⁻⁵)
  3. Calculate L: L = 0.9065 m ≈ 0.91 m
  4. Conclusion: Compare 0.91 m with the required 0.85 m . They are close enough, making it suitable (allow minor variations depending on whether R, ρ, or A was checked as the target subject).

🧠 Exam Technique

A "deduce" question requires a calculated numerical value plus an explicit concluding statement explaining why that value makes the tape suitable or unsuitable.

Mark breakdown: 1 mark for correct use/rearrangement of the resistivity equation, 1 mark for calculating the target parameter (e.g., length = 0.91 m) and giving a valid concluding comparison.

Part 0.2.3: Calculating Resistance of Resistor R

Calculate the resistance of R given that the potential difference across the LED is 2.2 V when the switch is closed. [4 marks]

✅ Correct Answer

350 Ω (Acceptable range: 340 Ω to 360 Ω based on graph reading)

💡 Key Knowledge

  • Kirchhoff's Second Law: The total emf across the series loop equals the sum of p.d.s across components: emf = V(LED) + V(R)
  • Graph Reading: Find the current flowing through the LED at V = 2.2 V using Figure 2.
  • Ohm's Law: R = V(R) / I

📐 Step-by-Step Calculation

  1. Read current from graph (Figure 2): At V = 2.2 V , the current is 15 mA ( = 0.015 A ). Note: Allow 14.5 mA to 15.5 mA.
  2. Find p.d. across resistor R: V(R) = emf - V(LED) = 7.4 V - 2.2 V = 5.2 V
  3. Calculate resistance of R: R = V(R) / I = 5.2 V / 0.015 A = 346.67 Ω
  4. Round appropriately: 350 Ω (to 2 significant figures).

❌ Common Errors & Examiner Comments

  • mA to A conversion error: Forgetting to convert 15 mA to 0.015 A is the most frequent place students drop marks here.
  • Incorrect graph reading: Ensure you look strictly at 2.2 V on the horizontal axis and trace up to the curve.
  • Gradient misconception: Examiners explicitly note: Do not allow gradient calculation for R—this is a fixed resistor in series, not an attempt to find dynamic resistance from the LED curve.
Mark breakdown: Max 3 marks from: reading 15 mA from graph, correct unit conversion to A, calculating p.d. across resistor (5.2 V), and using R = V/I. 1 final mark for the correct answer (350 Ω).

Topics

Physics · 3.5 Electricity

Question and mark scheme from the AQA AS Level Physics examination, Paper 1, June 2019. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.