AQA AS Level Physics Paper 2, June 2025: Question 33

Resistance of a Non-Ideal Voltmeter

📋 What this question tests

This question assesses your understanding of Ohm's law (V = IR), Kirchhoff's first law (conservation of charge in parallel branches), the practical implications of a non-ideal voltmeter (which draws current), and how negligible internal resistance ensures a constant terminal potential difference across parallel branches.

Question 33 (Multiple Choice)

AQA AS Physics — Circuit Electricity & Non-Ideal Meters

✅ Correct Answer

Option B: 3.0 kΩ

Mark Scheme Breakdown:
• 1 mark for selecting B (AO2 — application of circuit laws in practical contexts).

💡 Key Knowledge

Ideal vs Non-Ideal Voltmeter: An ideal voltmeter has infinite resistance and draws zero current. A non-ideal voltmeter has a finite resistance and therefore draws current.
Negligible Internal Resistance ( r = 0 ): Terminal p.d. remains constant and equal to the battery's emf ( V = ε ), regardless of current drawn.
Parallel Branches: Voltage across parallel branches is identical. Adding an extra branch does not change the voltage across or current through the original resistor when r = 0 .

📐 Step-by-Step Calculations

1 Find the battery EMF (terminal p.d.):

Initially, only the 2.0 kΩ resistor is connected across the battery:

V = I × R = (6.0 × 10⁻³ A) × (2.0 × 10³ Ω) = 12.0 V

Because internal resistance is negligible, the terminal p.d. remains 12.0 V after the voltmeter is added.

2 Determine current through the voltmeter branch:

When the voltmeter is placed in parallel, the current through the 2.0 kΩ resistor is unchanged at 6.0 mA (since its p.d. is still 12.0 V).

The total current supplied by the battery increases by 4.0 mA:

I_total = 6.0 mA + 4.0 mA = 10.0 mA

By Kirchhoff's First Law (or simply considering the added parallel branch):

I_voltmeter = 4.0 mA = 4.0 × 10⁻³ A

3 Calculate the resistance of the voltmeter (R_V):

R_V = V / I_voltmeter = 12.0 V / (4.0 × 10⁻³ A) = 3000 Ω = 3.0 kΩ

Alternative Parallel Method:
Total resistance R_total = 12 V / 10 mA = 1.2 kΩ .
Using 1 / R_total = 1 / R + 1 / R_V :
1 / R_V = (1 / 1.2) - (1 / 2.0) = (5 / 6) - (3 / 6) = 2 / 6 = 1 / 3 kΩ⁻¹
Therefore, R_V = 3.0 kΩ .

🧠 Exam Technique & Examiner Insights

Watch the preposition: Notice the phrasing "increases by 4.0 mA" vs "increases to 4.0 mA". The wording tells you directly how much extra current the voltmeter branch draws!
Take the direct route: Since the voltmeter is connected directly across the 12.0 V source and draws 4.0 mA, you can bypass equivalent resistance formulas entirely and calculate R = V / I directly for that single branch.
Sanity check: If the voltmeter drew current equal to the resistor (6.0 mA), its resistance would be 2.0 kΩ. Since it draws less current (4.0 mA), its resistance must be greater than 2.0 kΩ. This immediately eliminates Option A!

❌ Common Errors & Pitfalls

Misreading the total current as 4.0 mA: Students who use 4.0 mA as total current get nonsensical negative resistances or Option D (12 V / 4 mA? No, 12 / (6-4) = 6 kΩ).
Assuming internal resistance causes a p.d. drop: The question states negligible internal resistance. If internal resistance were present, terminal p.d. would drop as current increased. Here, V stays strictly at 12.0 V.
Calculation trap (Option C = 3.3 kΩ): Often picked by students making reciprocal addition errors, such as computing (2.0 × 10) / (2.0 + 10) or misapplying the parallel resistance formula.

AQA AS Level Physics Paper 2, June 2025: Question 33

Question

Mark scheme

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