AQA AS Level Physics Paper 1, June 2023: Question 6

13 marks · Medium difficulty · Extended Answer

Analyze battery emf, calculate electron flow, determine LED power output, find resistor value, and deduce LED behavior with internal resistance and an appliance.

Practise this question

Question

A series of exam questions about electrical circuits containing LEDs and resistors powered by a 12.0 V battery. Figure 10 shows a circuit with a battery, a resistor R, and three identical LEDs in series with a current of 44 mA. Figure 11 shows a graph of current in mA against voltage in V for an LED, showing current rising sharply past approximately 2.8 V. Figure 12 shows a circuit diagram with an appliance and switch S connected in parallel across the battery, while the LEDs and resistor remain in series.
Question text

06.1 State what is meant by the emf (electromotive force) of a battery.

[1 mark]

Figure 10 shows the circuit diagram for a battery-powered torch.

The circuit contains three identical light emitting diodes (LEDs) and a resistor R.

The current in the circuit is 44 mA.

Figure 10

06.2 Calculate the number of electrons that pass a point in the circuit in 37 minutes.

[2 marks]

number of electrons =

Figure 11 is the current–voltage characteristic for an LED used in the torch.

Figure 11

06.3 Determine the power output of one LED when the torch is on.

[3 marks]

power output = W

The battery has an emf of 12.0 V and an internal resistance of 1.5 Ω.

06.4 Determine the resistance of R in Figure 10.

[4 marks]

24 resistance = Ω

06.5 Another appliance is connected to the battery as shown in Figure 12.

The current in the battery is 3.5 A when switch S is closed.

Figure 12

*23* Each LED requires a voltage of at least 2.9 V to light.

Deduce whether the LEDs will light when S is closed.

[3 marks]

Mark scheme

Show the mark scheme Mark scheme giving detailed solutions and acceptable alternatives for parts 06.1 through 06.5, showing calculations for charge, electron number, power output from graph readings, internal resistance and terminal potential difference considerations.

Question Answers Additional Comments/Guidance Mark AO

06.1 Amount of chemical energy transferred / converted to Allow: 1 AO1.1a

electrical energy for 1 C of charge (through the battery).

(The emf is) the terminal pd (of the battery)

OR when there is no current in the battery.

Work done in moving 1 C of charge whole way round circuit

06.2 Use of Q = It Substitutes for I and t. 2 AO1.1a

(Q =) 0.044 × 37 × 60 OR AO2.1b

OR (Q =) 0.044 × 2220 OR

(Q=) 97.68 (C)

their 𝑄𝑄 18

(N=) or (N =) 6.25 x 10 x their Q

Use of Q = Ne 1.6 × 10–19

their Q must have supporting work which

identifies it as Q

(N =) 6.1 × 1020

Accept answer correctly rounded to at least 2 sf.

Calculator display = 6.105 x 1020

Accept any of the following pairs of values:

06.3 Read off V = 3.4 V when I = 44 (mA) 3 2 ×

AO1.1a

I V P AO2.1b

(mA) (V) (W)

4 to 6 2.8 0.011 to 0.017

8 to 12 2.9 0.023 to 0.035

12 to 15 3 0.036 to 0.045

20 3.1 0.062

22 to 26 3.2 0.070 to 0.083

40 to 47 3.4 0.136 to 0.160

64 to 72 3.6 0.23 to 0.26

96 to 100 3.8 0.36 to 0.38

Substitutes a voltage between 2.8 V and 3.8 V

Use of P = VI and a corresponding current value from table.

Expect to see a consistent power in range

quoted for that voltage.

Condone POT error in sub for current and its

subsequent power. In MP1 and MP2

OR Uses I = 0.044 (A) and 3 𝑉𝑉 < V <4V and

obtaining a consistent answer.

(P =) 0.15 (W)

Accept answer correctly rounded to at least 2 sf.

MP3 must be 0.15 (W) or 0.150 (W) or 0.1496 (W) .

Alternative

06.4 MP1 4 AO1.1a

MP1 Use of 𝜀𝜀 = 𝐼𝐼(𝑅𝑅 + 𝑜𝑜) by substituting

Use of V = IR: (to find lost volts = Ir =) 0.044 × 1.5 OR 0.066 V for 𝜀𝜀, I and r (where R is external resistance) 3 ×

OR

MP2 AO2.1b

𝜀𝜀 12 (To find total resistance = 3000 or 272.7 (Ω)) Rearrange 𝜀𝜀 = 𝐼𝐼(𝑅𝑅 + 𝑜𝑜) to find R = 271.2 (Ω)

( =) 11

𝐼𝐼 0.044

MP2

(Total resistance – r = R =) 5967 or 271.2 (Ω)

OR

(Pd across R =) 271.2 × 0.044 or 11.9328 V or their R × 0.044

OR

(Total pd across LEDs =) 3 × 3.4 or 10.2 V ECF

OR

(Resistance of an LED =) 3.4 or 77.3 (Ω) ECF

0.044

OR ECF

MP3 Condone POT in any of the working for MP1, MP2

(Total resistance of 3 LEDs =) 3 × 77.3 or 231.9 or 232.438 (Ω) or and MP3.

3 x their resistance of one LED

OR

(Pd across R =) 12 – their total pd across LED – their lost volts or 1.734 (V)

OR

OR

(R of R =) 𝑡𝑡ℎ𝑖𝑖𝑖𝑖𝑖𝑖 𝑝𝑝𝑝𝑝 𝑎𝑎𝑖𝑖𝑖𝑖𝑖𝑖𝑎𝑎𝑎𝑎 𝒖𝒖𝒖𝒖𝒖𝒖𝒖𝒖𝒖𝒖𝒖𝒖𝒖𝒖 𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒖𝒖𝒓𝒓 𝑜𝑜𝑜𝑜 1.734 Condone answers in range 38.5 to 39.7

0044 0044 Ω Ω

. .

Treat use of their V from 06.3 as an ECF e.g.

MP4 An answer of 53 (Ω) gains 4 marks (uses V= 3.2 V)

(R of R =) 39(.4) (Ω) or 38.8 (Ω) Other consistent uses of their V identifiable in 06.3

can achieve 4 marks.

06.5 3 3 ×

MP1

3.5 × 1.5 or 5.25 V AO3.1b

lost volts =

OR Allow between 8 mA and 10 mA for this

Current in LEDs = 8 mA (when V = 2.9 V)

OR read-off.

LEDs require 8.7 V to light

MP2

terminal pd = 6.75 V

MP3

LEDs won’t light:

MP3 gives a valid reason why 6.75 V is

-because terminal pd is less than 8.7 V insufficient.

-because pd across LED is less than 2.5 V, therefore, no current in

LEDs

Needs to state LEDS won’t light for to gain

-because pd across R is zero as the resistance of LED is much

greater the R, therefore, no current in LEDs MP3.

-Resistor R would require a pd of 0.315 V. Therefore, total pd required

= 9.01 V is greater than terminal pd.

-their pd across each LED is below switch-on voltage (of 2.9 V)

Total 13

How to answer it

DC Circuits, Emf, and Non-Ohmic Components Study Guide

What this question tests

This question assesses your understanding of electromotive force (emf), charge quantization, current relationships (Q = It and Q = Ne), graphical analysis of non-ohmic components (LED characteristics), Kirchhoff's laws, internal resistance, and terminal potential difference calculations.

Question 06.1 [1 mark]

Definition of Electromotive Force (Emf)

✅ Correct Answer

The amount of chemical energy transferred/converted to electrical energy per 1 C of charge (passing through the battery), OR the work done in moving 1 C of charge the whole way round a complete circuit.

💡 Key Knowledge

Emf is a measure of energy per unit charge supplied by the source, not a force despite its traditional name. Alternatively, it equals the terminal potential difference of the battery when no current flows.

Question 06.2 [2 marks]

Calculating Number of Electrons

📐 Step-by-Step Calculation

  1. Find total charge (Q): Use Q = I × t
    Convert current to amperes: 44 mA = 0.044 A
    Convert time to seconds: 37 mins = 37 × 60 = 2220 s
    Q = 0.044 × 2220 = 97.68 C
  2. Find number of electrons (N): Use N = Q / e
    Elementary charge e = 1.60 × 10⁻¹⁹ C
    N = 97.68 / (1.60 × 10⁻¹⁹) = 6.105 × 10²⁰

❌ Common Errors

  • Forgetting to convert minutes into seconds (omitting the factor of 60).
  • Failing to convert milliamperes into amperes ( 10⁻³ factor).
Mark breakdown: 1 mark for calculating Q (or substitution into Q = Ne), 1 mark for the final correct value of N (6.1 × 10²⁰).
Question 06.3 [3 marks]

Determining Power Output of One LED

📐 Step-by-Step Calculation

  1. Read off voltage from graph: For I = 44 mA , follow the curve on Figure 11 to find the corresponding potential difference. V = 3.4 V (Accept values between 2.8 V and 3.8 V based on valid graph readings).
  2. Calculate power: Use P = V × I
    P = 3.4 V × 0.044 A = 0.1496 W
  3. Apply sig figs: Round to 2 significant figures: 0.15 W (or 0.150 W).

🧠 Exam Technique

Always double-check graph scale divisions before reading values. Ensure current is converted back to amperes when computing power in watts.

Mark breakdown: 1 mark for reading correct V from graph at 44 mA, 1 mark for substituting into P = VI, 1 mark for correct evaluation to 2+ sf (0.15 W).
Question 06.4 [4 marks]

Determining Resistance of R

📐 Step-by-Step Calculation

  1. Account for internal resistance / total circuit resistance:
    Using ε = I(R + r) or finding lost volts: v = Ir = 0.044 × 1.5 = 0.066 V .
  2. Find total resistance of circuit:
    R_total = ε / I = 12 / 0.044 = 272.7 Ω .
  3. Subtract internal resistance and component drops:
    Total external resistance = 272.7 - 1.5 = 271.2 Ω .
    Subtract resistance of the 3 identical LEDs (each LED resistance = 3.4 V / 0.044 A = 77.3 Ω ; total for 3 = 231.9 Ω ).
  4. Find R:
    R = 271.2 - 231.9 = 39 Ω (or 39.4 Ω depending on intermediate rounding).

❌ Common Errors

  • Forgetting to multiply the single LED voltage or resistance by 3 for the three identical LEDs in series.
  • Omitting internal resistance ( r = 1.5 Ω ) from total loop equations.
Mark breakdown: 1 mark for accounting for internal resistance / lost volts, 1 mark for evaluating total circuit resistance or pd, 1 mark for subtracting LED drops/resistances, 1 mark for final correct value of R (39 Ω or 38.8 Ω).
Question 06.5 [3 marks]

Deducing LED Operation When Switch S is Closed

💡 Key Knowledge

When switch S closes, the extra appliance draws a larger current from the battery ( 3.5 A ), which drastically increases "lost volts" across the internal resistance, pulling down the terminal potential difference available to the rest of the circuit.

✅ Correct Deduction & Marks

  • MP1: Calculate lost volts: 3.5 A × 1.5 Ω = 5.25 V (or note required total voltage for LEDs is 3 × 2.9 = 8.7 V ).
  • MP2: Determine terminal pd: 12.0 V - 5.25 V = 6.75 V .
  • MP3: Conclude clearly that LEDs will not light because the terminal pd (6.75 V) is less than the minimum required voltage (8.7 V) to exceed the threshold voltage of all three LEDs in series.
Mark breakdown: 1 mark for calculating lost volts / total required voltage, 1 mark for calculating terminal pd, 1 mark for stating LEDs won't light with valid supporting numerical justification.

Topics

Physics · 3.5 Electricity

Question and mark scheme from the AQA AS Level Physics examination, Paper 1, June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.