AQA GCSE Physics Physics Paper 1 (Higher), November 2021: Question 7

📐 Step-by-Step Calculation

1. Read current from Figure 9:
   Find Resistance = 36 Ω on the x-axis. Follow up to the line of best fit and read across to the y-axis:
   I = 0.08 A
2. Convert power to standard SI units:
   P = 230 mW = 230 / 1000 = 0.230 W
3. Select formula & substitute:
   P = I × V
   0.230 = 0.08 × V
4. Rearrange and solve for V:
   V = 0.230 / 0.08 = 2.875 V
   (Acceptable range: 2.88 V or 2.9 V)

✅ Mark Scheme Breakdown

• Mark 1: Correct reading of current: I = 0.08 (A) .
• Mark 2: Substitution of power and current into P = I × V (or V = I × R ): 0.230 = 0.08 × V .
• Mark 3: Correct rearrangement: V = 0.230 / 0.08 .
• Mark 4: Correct final answer: 2.875 (V) or 2.88 (V) .

❌ Common Errors

• Unit conversion forgotten: Using 230 instead of 0.230 gives 2875 V , losing conversion marks.
• Misreading the grid: Each small horizontal square is 2 Ω ; each small vertical square is 0.01 A . Reading 0.075 A or 0.09 A causes an immediate error (though error-carried-forward applies).

🧠 Exam Technique

• Draw pencil projection lines directly on the graph from 36 Ω up to the curve and across to the current axis.
• Always write down the unconverted and converted values clearly: P = 230 mW = 0.23 W . This locks in method marks even if a calculator typing error occurs later.

💡 Key Knowledge

If two variables y and x are inversely proportional ( y ∝ 1/x ), their product is always constant:

I × R = constant (k)

Alternatively, whenever resistance doubles, current must halve.

✅ Model Answer (Full 2 Marks)

1. State the relationship rule (1 mark):
"For inverse proportionality, the product of current and resistance must be constant ( I × R = k )."

2. Test using at least 3 pairs from Figure 9 (1 mark):

• Pair 1: 12 Ω × 0.24 A = 2.88
• Pair 2: 24 Ω × 0.12 A = 2.88
• Pair 3: 36 Ω × 0.08 A = 2.88
• Pair 4: 48 Ω × 0.06 A = 2.88

"Since I × R ≈ 2.88 for each pair, the student's conclusion is correct."

❌ Common Errors & Examiner Warnings

• Vague descriptions: Stating only "as resistance goes up, current goes down" scores 0 marks. That only proves a negative correlation, not inverse proportionality!
• Using only one pair of values: Calculating I × R for just one point proves nothing. You must calculate it for three or more pairs to demonstrate constancy.

🧠 Exam Technique

If you don't calculate products, you can get 1 mark by quoting values that double and halve:

"When resistance doubles from 12 Ω to 24 Ω, the current halves from 0.24 A to 0.12 A."

However, calculating the product of 3 coordinate pairs is the safest and clearest way to guarantee both marks.

💡 Circuit Behaviour

In Figure 10, the switch is wired in parallel directly across the branch containing the ammeter and variable resistor, connecting directly from one side of the battery to the other.

A closed switch and ideal wires have effectively zero resistance.

✅ Model Answer

• Effect on current (1 mark):
  The current in the variable resistor would drop to zero (or almost zero).
• Reason (1 mark):
  The closed switch creates a short circuit (or has effectively zero resistance / the potential difference across the variable resistor drops to zero).

❌ Common Errors

• Thinking current increases: Students see a switch closing and assume total circuit current increases, forgetting the question asks specifically about the current in the variable resistor.
• Incomplete explanations: Saying "the current takes the easiest path" without mentioning that the switch branch has virtually zero resistance or creates a short circuit.

🧠 Examiner's Insight

Current divides between parallel paths inversely according to resistance ( I ∝ 1/R ). Because the switch branch has near-zero resistance, almost 100% of the current flows through the switch branch, bypassing the variable resistor completely.