AQA A-Level Physics Paper 1, June 2025: Question 29
1 mark · Medium difficulty · Multiple Choice
Determine the total charge that flows through point P in 60 seconds given the variation of two junction currents with time.
Practise this questionQuestion
Question text
29 The diagram shows part of a circuit.
The graph shows how the two currents I1 and I2 vary with time.
What is the total charge that flows through point P in 60 s?
[1 mark]
A 1.44 C
B 0.93 C
C 0.72 C
D 0.42 C
Mark scheme
Show the mark scheme
29 A 1.44 C AO1
How to answer it
Total Charge Flow at a Circuit Junction
This question assesses your ability to apply Kirchhoff’s First Law (conservation of charge) to determine the total current entering a junction, interpret a current–time (I–t) graph to calculate charge flow (area under the graph), and convert prefix units ( mA to A ) correctly under timed conditions.
Question 29 Breakdown
Multiple Choice (1 Mark)
✅ Correct Answer
A: 1.44 C
💡 Key Knowledge
- Kirchhoff’s First Law: The total current entering a junction equals the total current leaving it:
IP = I₁ + I₂ - Charge & Current: Charge is the integral of current with respect to time:
ΔQ = Area under the I–t graph - Unit Conversion: Current is given in milliamperes ( mA ):
1 mA = 1 × 10⁻³ A
📐 Step-by-Step Calculation
There are two efficient methods to solve this question:
Method 1: Elegant Constant Current Insight (Fastest)
- Check current at t = 0 s:
I₁ = 5 mA , I₂ = 19 mA
IP(0) = 5 + 19 = 24 mA - Check current at t = 60 s:
I₁ = 12 mA , I₂ = 12 mA
IP(60) = 12 + 12 = 24 mA - Observe linearity: Both graphs are straight lines with equal and opposite gradients. Therefore, the total current flowing into point P remains constant at 24 mA for the entire 60 s!
- Calculate total charge:
Q = I × t = (24 × 10⁻³ A) × 60 s = 1.44 C
Method 2: Area Under Each Curve (Standard Trapezium Method)
- Charge from I₁ (trapezium 1):
Q₁ = ½ × (5 + 12) × 10⁻³ A × 60 s = 8.5 × 10⁻³ × 60 = 0.51 C - Charge from I₂ (trapezium 2):
Q₂ = ½ × (19 + 12) × 10⁻³ A × 60 s = 15.5 × 10⁻³ × 60 = 0.93 C - Total charge Q:
Q = Q₁ + Q₂ = 0.51 C + 0.93 C = 1.44 C
🧠 Exam Technique & Strategy
- Spot symmetry in multiple-choice questions: Noticing that 5 + 19 = 24 and 12 + 12 = 24 turns a complex dual-trapezium calculation into a single, instant multiplication ( 24 × 60 ).
- Check units on axes immediately: The y-axis is in mA , not A . Always write the × 10⁻³ factor down first to prevent power-of-ten mistakes.
- Inspect junction arrows: Verify whether currents are both entering or if one is leaving. Here, both arrows point into the node, meaning they add together.
❌ Common Distractors & Traps
- Selecting B (0.93 C): This corresponds only to the charge delivered by current I₂ ( 15.5 mA × 60 s ). Students who forget Kirchhoff's Law and only compute the upper trace fall into this trap.
- Selecting C (0.72 C): Obtained by calculating the charge if current stayed at the final intersection value of 12 mA for the whole duration ( 12 mA × 60 s = 0.72 C ).
- Selecting D (0.42 C): Obtained by subtracting the two charges ( 0.93 C − 0.51 C = 0.42 C ) instead of adding them, mistaking current directions.
Topics
Physics · Practical skills · 3.5 Electricity · Data analysis
Question and mark scheme from the AQA A-Level Physics examination, Paper 1, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.