AQA AS Level Physics Paper 2, June 2023: Question 11

1 mark · Medium difficulty · Multiple Choice

Identify the correct equation relating the currents in a parallel circuit containing two resistors of resistance R and 4R connected to a cell.

Practise this question

Question

A multiple-choice question showing a circuit diagram with a cell connected in parallel to two branches: one containing a resistor of resistance R with current I2 flowing downwards, and another containing a resistor of resistance 4R with current I3 flowing downwards. The total current from the cell is I1. Below the diagram are four options A, B, C, and D representing relations between the currents.
Question text

11 A cell with negligible internal resistance is connected to two resistors of

resistance 4R and R.

The currents I1, I2 and I3 in the circuit are shown.

Which equation is correct for this circuit?

[1 mark]

A I1 = 4I2

B I1 = 4I3

C I2 = 4I3

D I3 = 4I1

Mark scheme

Show the mark scheme The mark scheme indicates that the correct answer for question 11 is C, which corresponds to the equation I2 = 4I3.

11 C I2 = 4I3

How to answer it

Analyzing Currents in Parallel Resistor Branches

AQA AS Level Physics • Multiple Choice Question

What this question tests

This question assesses your understanding of parallel circuits, potential difference across parallel branches, and Ohm's Law (I = V / R). You need to relate the inverse proportionality between current and resistance when components share the same potential difference.

Question 1.1

Determining the correct relationship between circuit currents

✅ Correct Answer

Option C ( I₂ = 4I₃ )

💡 Key Knowledge

  • The two branches containing resistor R and resistor 4R are connected in parallel across the same cell.
  • Parallel branches share the exact same potential difference ( V ).
  • By Ohm's Law ( I = V / R ), current is inversely proportional to resistance when V is constant.

📐 Step-by-Step Derivation

  1. Identify that the potential difference V across the branch with resistor R equals the potential difference across the branch with resistor 4R .
  2. Write the expression for current I₂ through resistor R : I₂ = V / R
  3. Write the expression for current I₃ through resistor 4R : I₃ = V / (4R)
  4. Compare the two equations: since I₃ = (1/4) × (V / R) , it follows that I₃ = I₂ / 4 , or rearranging gives I₂ = 4I₃ .

🧠 Exam Technique

Do not waste time calculating numerical values for voltage. Use algebraic proportionality. Since resistance is 4 times smaller in the middle branch ( R vs 4R ), the current must be 4 times larger.

❌ Common Errors

  • Direct Proportionality Trap: Students often mistakenly assume that a larger resistance draws a larger current ( I₃ = 4I₂ , Option D), forgetting that current chooses the path of least resistance.
  • Confusing Branch and Total Current: Mistaking Kirchhoff's first law junctions by equating total current I₁ directly with a single branch multiplier without accounting for parallel splitting.
Mark Scheme Allocation: 1 mark total for selecting option C ( I₂ = 4I₃ ).

Topics

Physics · 3.5 Electricity

Question and mark scheme from the AQA AS Level Physics examination, Paper 2, June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.