Circuits — AQA A-Level Physics
Teaching note checked against the official specification.
Circuits in AQA A-Level Physics covers the analysis of series and parallel resistor networks, electromotive force (emf), internal resistance, potential dividers, and the conservation of charge and energy in electrical loops. Mastery requires applying Kirchhoff's laws, Ohm's law, and the power equations to solve quantitative problems involving current, potential difference, resistance, and power in both ideal and real circuit scenarios.
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Circuits explained
In any dc circuit, charge is neither created nor destroyed, so the current entering a junction equals the total current leaving it.
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Energy is also conserved: each charge carrier that completes the circuit transfers energy from the source to the components, and the sum of potential differences around any closed loop equals the source emf. For example, in a series circuit the current is the same at every point because charge is conserved, while the p.d.s across components add to the supply p.d. because energy is conserved. In parallel branches, the branch currents sum to the total current, and each branch has the same p.d. Applying these principles lets you analyse unknown currents and p.d.s without memorising every circuit configuration.
Your focus
- State the principle of conservation of charge at a junction in a dc circuit.
- Apply conservation of charge to calculate an unknown current in a series or parallel circuit.
- Apply conservation of energy to calculate an unknown potential difference in a closed loop.
Which command word is this?
You must use mathematical operations to find a numerical answer. Show all working, include units, and round to an appropriate number of significant figures. Marks are awarded for correct substitution and final answer.
Circuits exam tips
Topic Overview
This topic explores the fundamental principles governing electrical circuits, including Ohm's law, Kirchhoff's laws, and the behaviour of components in series and parallel arrangements. It extends to real-world applications such as potential dividers, internal resistance, and power dissipation, forming the bedrock for understanding more complex electronic systems.
Circuits are essential in A-Level Physics because they link theoretical concepts of charge, energy, and potential difference to practical measurement and problem-solving. Mastery of this topic is crucial for exams and provides a foundation for further study in fields like electronics, engineering, and renewable energy.
Key Concepts
- →Ohm's law: For an ohmic conductor at constant temperature, current is directly proportional to potential difference (V = IR).
- →Kirchhoff's laws: The sum of currents entering a junction equals the sum leaving (conservation of charge); the sum of emfs equals the sum of pds around any closed loop (conservation of energy).
- →Series and parallel circuits: In series, current is constant and pds add; in parallel, pd is constant and currents add. Total resistance rules differ accordingly.
- →Electromotive force (emf) and internal resistance: The emf is the energy transferred per unit charge by the source; internal resistance causes a drop in terminal pd when current flows (ε = I(R + r)).
- →Potential dividers: Circuits that split a supply voltage in proportion to resistance, often used with sensors like thermistors and LDRs to produce variable output pds.
Marking Points
- State that electric charge is conserved at a junction: the sum of currents entering equals the sum of currents leaving.
- Apply conservation of charge to series and parallel circuits to deduce unknown currents.
- State that energy is conserved in a closed loop: the sum of potential differences around the loop equals the source emf.
- Apply conservation of energy to series circuits to deduce unknown potential differences.
- Use the relationship between energy transferred, charge and potential difference, W = QV, to support energy conservation arguments.
- Recognise that in a parallel circuit the p.d. across each branch is the same because each branch connects the same two nodes.
Examiner Tips
- 💡When asked to find an unknown current, identify a junction and write an equation based on charge conservation before substituting numbers.
- 💡When asked to find an unknown p.d., trace a closed loop and write an equation based on energy conservation, taking care with the direction of travel.
- 💡Show the conservation principle explicitly in your working; a correct numerical answer without the principle may not gain full credit.
- 💡Always define your symbols and state the equations you use before substituting numbers. This secures method marks even if arithmetic errors occur.
- 💡When drawing circuit diagrams, use standard symbols and label all components, currents, and potential differences clearly. Ambiguous diagrams can lose marks.
- 💡For 'show that' questions, work through the calculation step-by-step and ensure your final answer matches the given value to the required precision. Do not just state the answer.
Common Mistakes
- Thinking current is used up as it passes through components. Correction: current is the rate of flow of charge; charge is conserved, so the same current enters and leaves a series component.
- Adding branch currents incorrectly in a parallel circuit, for example treating total current as the difference between branch currents. Correction: the total current is the sum of the branch currents.
- Assuming energy is used up in the sense that charge loses energy permanently. Correction: energy is transferred from the source to components; the total energy transferred per unit charge around a closed loop is conserved.
- Students often think current is 'used up' in a circuit. In reality, current is conserved; energy is transferred, causing potential difference across components.
- Many believe that adding resistors in parallel increases total resistance. Actually, adding parallel resistors provides more paths for current, decreasing total resistance.
- Some confuse emf with terminal potential difference. Emf is the total energy per unit charge supplied, while terminal pd is the energy per unit charge available to the external circuit, reduced by internal resistance.
Revision Plan
- 1Day 1-2: Review definitions of current, potential difference, resistance, and emf. Practice using I = ΔQ/Δt and V = W/Q with numerical examples.
- 2Day 3-4: Master series and parallel circuit rules. Solve problems calculating total resistance, current, and pd in various combinations.
- 3Day 5-6: Study Kirchhoff's laws and apply them to multi-loop circuits. Practice deriving equations for unknown currents and pds.
- 4Day 7-8: Learn about internal resistance and terminal pd. Solve problems involving ε = I(R + r) and V = ε - Ir.
- 5Day 9-10: Explore potential dividers, including with thermistors and LDRs. Practice calculating output voltage and interpreting graphs.
- 6Day 11-12: Complete past paper questions on circuits, focusing on structured 6-mark questions and calculations. Review mark schemes for examiner expectations.
Exam Question Types
- 📋Calculation questions: Often ask for current, pd, resistance, or power in series/parallel circuits. Advice: Write down the relevant equation, substitute values with units, and give the answer to an appropriate number of significant figures.
- 📋Circuit analysis with Kirchhoff's laws: May present a multi-loop circuit and ask to calculate unknown currents or pds. Advice: Label currents and pds clearly, apply junction and loop rules systematically, and solve simultaneous equations.
- 📋Practical-based questions: Involve interpreting I-V graphs for ohmic and non-ohmic conductors, or explaining the effect of internal resistance. Advice: Describe the shape of graphs and relate to physical principles, using correct terminology.
- 📋Potential divider design: Ask to calculate output voltage or explain how a thermistor/LDR changes output. Advice: Use the potential divider equation Vout = Vin × (R2/(R1+R2)) and explain how resistance changes with temperature or light intensity.
Command Word Expectations (AQA)
You must use mathematical operations to find a numerical answer. Show all working, include units, and round to an appropriate number of significant figures. Marks are awarded for correct substitution and final answer.
You must provide a detailed justification that links cause and effect, using correct physics terminology. Typically requires a sequence of logical steps, often worth 2-3 marks. Avoid vague statements; be specific about physical principles.
You must give a detailed account of a process, pattern, or observation without necessarily explaining why. Use precise language and include key features. Often worth 2-3 marks, with marks for each correct point.
How Students Lose Marks (Examiner Pitfalls)
Step-by-Step Worked Solutions
Question: A battery of emf 9.0 V and internal resistance 0.50 Ω is connected to a resistor of resistance 4.0 Ω. Calculate the current in the circuit and the terminal potential difference across the battery.
- 1.Step 1: Identify given facts: emf ε = 9.0 V, internal resistance r = 0.50 Ω, external resistance R = 4.0 Ω.
- 2.Step 2: Apply the equation for a real battery: ε = I(R + r). Rearrange to find current: I = ε / (R + r).
- 3.Step 3: Substitute values: I = 9.0 / (4.0 + 0.50) = 9.0 / 4.5 = 2.0 A.
- 4.Step 4: Calculate terminal pd using V = IR: V = 2.0 × 4.0 = 8.0 V. Alternatively, V = ε - Ir = 9.0 - (2.0 × 0.50) = 8.0 V.
- 5.Step 5: State final answers with units: current = 2.0 A, terminal pd = 8.0 V.
Question: Two resistors, R1 = 6.0 Ω and R2 = 3.0 Ω, are connected in parallel across a 12 V supply of negligible internal resistance. Calculate the total resistance, the current through each resistor, and the total current drawn from the supply.
- 1.Step 1: Identify given facts: R1 = 6.0 Ω, R2 = 3.0 Ω, V = 12 V (same across parallel branches).
- 2.Step 2: Calculate total resistance using 1/RT = 1/R1 + 1/R2 = 1/6 + 1/3 = 1/6 + 2/6 = 3/6 = 1/2. Therefore RT = 2.0 Ω.
- 3.Step 3: Calculate current through R1: I1 = V / R1 = 12 / 6.0 = 2.0 A.
- 4.Step 4: Calculate current through R2: I2 = V / R2 = 12 / 3.0 = 4.0 A.
- 5.Step 5: Total current I = I1 + I2 = 2.0 + 4.0 = 6.0 A. Check with I = V / RT = 12 / 2.0 = 6.0 A.
- 6.Step 6: State final answers with units: RT = 2.0 Ω, I1 = 2.0 A, I2 = 4.0 A, total current = 6.0 A.