AQA GCSE Combined Science: Trilogy Physics Paper 1 (Foundation), November 2021: Question 3

✅ Correct Answer

Tick the middle box: A circle containing an "X" ( ⊗ ).

💡 Key Knowledge

• Filament Lamp: Circle with a cross inside.
• First symbol (triangle + line with arrows): Light Emitting Diode (LED).
• Third symbol (resistor in circle with arrows): Light Dependent Resistor (LDR).

✅ Correct Answer

Tick: "The flow of electrical charge" (second option).

❌ Common Errors

• "The speed at which charge moves": Incorrect! Speed is rate of displacement, not current.
• "The energy carried by each unit of charge": Incorrect! That is the definition of potential difference (voltage).
• "The number of electrons": Incorrect! Current is the rate of flow of charge.

✅ What to Draw

Draw a single closed rectangular loop connecting all four components in series:

• Battery: Two or more cells connected (long thin line, short thick line, dashed line, long thin line).
• Ammeter: Circle with an A inside.
• Filament Lamp: Circle with an X inside.
• Variable Resistor: Rectangle with an arrow passing diagonally through it.

🧠 Exam Technique

• Always use a ruler for connecting wires.
• Never leave gaps between wire connections and components.
• Never draw components on the corners of the rectangle.
• The order of components in a simple series loop does not matter for this mark, as long as they form one continuous loop.

✅ Correct Answer

Tick: "Decrease the resistance of the variable resistor."

💡 Key Knowledge

From Ohm's Law ( I = V ÷ R ):

• If total circuit resistance decreases, current increases.
• Decreasing potential difference would decrease current.
• Adding an ammeter has virtually zero effect because ideal ammeters have negligible resistance.

📐 Step-by-Step Calculation

1. Identify values: Potential difference, V = 0.75 V ; Current, I = 0.16 A .
2. Substitute into formula: P = 0.75 × 0.16 [1 mark]
3. Calculate result: P = 0.12 W [1 mark]

🧠 Exam Technique

• The unit W (Watts) is already given on the answer line. Always check whether units are provided or if you must write them yourself.
• Always write the substitution step clearly to secure method marks even if a calculator typo occurs.

✅ Correct Equation

charge flow = current × time

or using standard symbols: Q = I × t (also accepted: Q = It )

💡 Units Reminder

• Q (Charge flow): Coulombs (C)
• I (Current): Amperes / Amps (A)
• t (Time): Seconds (s)

📐 Step-by-Step Calculation

1. Convert milliamperes to amperes:
   200 mA = 200 ÷ 1000 = 0.2 A [1 mark]
2. Substitute into formula:
   Q = 0.2 × 15 [1 mark]
3. Calculate final value:
   Q = 3.0 C (or 3 C) [1 mark]

❌ Common Trap: Unit Prefix

The prefix milli- (m) means one-thousandth ( 10⁻³ ).

Students who forget to convert calculate: 200 × 15 = 3000 C .

Examiner note: If you fail to convert, you can still get 2 marks via "error carried forward" (ecf), but converting guarantees all 3 marks!

✅ Calculation & Answer

Divide the larger current by the original current:

320 mA ÷ 200 mA = 1.6

Answer: 1.6 (times greater)

🧠 Exam Technique

• "How many times greater" always means divide, not subtract!
• Subtraction trap: 320 - 200 = 120 (Incorrect).
• Because both values are in mA, units cancel out, so no conversion is needed.

✅ Correct Answer

Tick the fourth box: "260 mA to 380 mA"

💡 How Uncertainty Range Works

The measurement is given as 320 ± 60 mA :

• Minimum value: 320 - 60 = 260 mA
• Maximum value: 320 + 60 = 380 mA
• Total spread (range): 260 mA to 380 mA.