Basics of electricity — AQA A-Level Physics
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The basics of electricity in AQA A-Level Physics covers electric current as the rate of flow of charge, potential difference as energy transferred per unit charge, and resistance as the opposition to current. It underpins Ohm's law, I-V characteristics, resistivity, and power calculations, forming the foundation for all later circuit analysis and electromagnetism topics.
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Basics of electricity explained
Electric current is the rate of flow of charge, I = ΔQ/Δt, so one ampere is one coulomb per second.
Read the full explanation
In a metal, electrons drift through the lattice; in an electrolyte, positive and negative ions move in opposite directions. Potential difference is the work done per unit charge, V = W/Q, so one volt is one joule per coulomb. When charge Q moves through a component with potential difference V, the work done is W = QV. For example, a 12 V battery driving 2.0 A for 5.0 s transfers charge Q = It = 10 C and energy W = QV = 120 J. These definitions underpin circuit analysis and energy transfer calculations.
Your focus
- State and use the equation I = ΔQ/Δt to calculate current, charge or time.
- State and use the equation V = W/Q to calculate potential difference, work done or charge.
- Describe the direction of conventional current and the motion of charge carriers in metals and electrolytes.
Which command word is this?
Give a precise, formal statement of the quantity in terms of energy, charge, or time. For example, 'The volt is the potential difference across a component when one joule of energy is transferred per coulomb of charge passing through it.' No explanation or example is required.
Basics of electricity exam tips
Topic Overview
This topic introduces the fundamental quantities of electric current, potential difference, and resistance, and the relationships between them. You will learn how to define these quantities in terms of charge and energy, apply Ohm's law, and analyse series and parallel circuits. It also covers the I-V characteristics of ohmic and non-ohmic components, including filament lamps, diodes, and thermistors.
Understanding these basics is essential for all later work on electricity, including resistivity, power, electromotive force, and internal resistance. It also underpins practical skills such as circuit building, measurement, and graph analysis. Mastery of this topic is crucial for success in AQA A-Level Physics exams and provides a foundation for studying electromagnetism and electronics.
Key Concepts
- →Electric current is the rate of flow of charge, I = ΔQ/Δt, measured in amperes (A). In metals, current is carried by electrons; in electrolytes, by ions.
- →Potential difference (p.d.) is the energy transferred from electrical energy to other forms per unit charge, V = W/Q. Electromotive force (e.m.f.) is the energy transferred from other forms to electrical energy per unit charge.
- →Resistance is the opposition to current, R = V/I, measured in ohms (Ω). Ohm's law states that for an ohmic conductor at constant temperature, current is directly proportional to potential difference.
- →Resistivity is a property of a material, defined by ρ = RA/L, and depends on temperature. Resistance of a component depends on its dimensions and material.
- →In series circuits, current is the same at all points, and the total p.d. is the sum of individual p.d.s. In parallel circuits, p.d. is the same across each branch, and total current is the sum of branch currents.
Marking Points
- States current as rate of flow of charge: I = ΔQ/Δt, with Q in coulombs, t in seconds, I in amperes.
- Explains that one ampere equals one coulomb per second.
- Defines potential difference as work done per unit charge: V = W/Q, with W in joules, Q in coulombs, V in volts.
- Explains that one volt equals one joule per coulomb.
- Applies W = QV to calculate energy transferred when charge moves through a potential difference.
- Describes charge carriers in metals as electrons and in electrolytes as ions, and notes conventional current direction.
- Uses consistent units and prefixes when substituting into I = ΔQ/Δt or V = W/Q.
Examiner Tips
- 💡Write the defining equation before substituting numbers; this makes the physics clear and helps secure method marks.
- 💡Check units carefully: convert mA to A, minutes to seconds, and kJ to J before calculating.
- 💡When explaining definitions, use the phrase 'per unit charge' or 'per second' precisely, as these are the key ideas being tested.
- 💡For energy transfer questions, identify Q first using I = ΔQ/Δt, then use W = QV.
- 💡Always define quantities in terms of energy or charge per unit something, e.g., 'potential difference is energy transferred per unit charge'. Avoid vague statements like 'voltage pushes current'.
- 💡When drawing or interpreting I-V graphs, label axes correctly (I on y-axis, V on x-axis) and state whether the component is ohmic or non-ohmic. For non-ohmic components, explain the physical reason for the shape, such as temperature increase or threshold voltage.
- 💡In calculations, show all steps clearly, including rearrangement of equations and substitution of values with units. Round final answers to an appropriate number of significant figures, usually 2 or 3, and always include units.
Common Mistakes
- Confusing current with charge: current is the rate of flow of charge, not the total charge. Correct by writing I = ΔQ/Δt and checking units.
- Reversing the definition of potential difference as charge per unit work done. Correct by stating V = W/Q and checking that the unit is J C⁻¹.
- Using the wrong time interval or forgetting to convert minutes to seconds. Correct by converting all times to seconds before calculating charge or current.
- Assuming conventional current direction is the same as electron flow in metals. Correct by stating that conventional current is opposite to electron drift.
- Students often think that current is 'used up' as it goes around a circuit. In reality, current is conserved; the same number of charge carriers per second flows through each component in series. Energy is transferred, not current.
- Many believe that a battery always provides a constant potential difference regardless of the current drawn. In fact, due to internal resistance, the terminal p.d. decreases as current increases, following V = ε - Ir.
- Students frequently confuse resistance and resistivity. Resistance depends on the dimensions of the component, while resistivity is a material property independent of shape and size.
Revision Plan
- 1Start by revising the definitions of current, potential difference, and resistance, ensuring you can state them in words and equations. Use flashcards to memorise the precise wording.
- 2Practice calculations using I = ΔQ/Δt, V = W/Q, R = V/I, and resistivity ρ = RA/L. Work through examples from textbooks and past papers, checking units and significant figures.
- 3Draw and interpret I-V graphs for a resistor, filament lamp, and diode. For each, describe the shape and explain the physical reasons, linking to temperature and threshold voltage.
- 4Build and test simple circuits to measure current and potential difference, using a variable resistor to investigate I-V characteristics. Record data and plot graphs to reinforce practical skills.
- 5Complete past exam questions on this topic, focusing on definitions, calculations, and explanations. Mark your answers using the mark scheme and note recurring mistakes.
Exam Question Types
- 📋Definition and explanation questions: e.g., 'Define the volt' or 'Explain why the resistance of a filament lamp increases as current increases'. These require precise wording and clear physics reasoning.
- 📋Calculation questions: e.g., using R = V/I, ρ = RA/L, or ε = I(R + r). Show all working and pay attention to unit conversions.
- 📋Graph analysis: e.g., sketching or interpreting I-V graphs for different components. Label axes, describe the shape, and explain the behaviour in terms of resistance changes.
- 📋Practical-based questions: e.g., describing how to measure the I-V characteristics of a component, including circuit diagrams and safety precautions.
Command Word Expectations (AQA)
Give a precise, formal statement of the quantity in terms of energy, charge, or time. For example, 'The volt is the potential difference across a component when one joule of energy is transferred per coulomb of charge passing through it.' No explanation or example is required.
Provide a detailed account of why something happens, using correct physics terminology and linking cause and effect. For example, 'Explain why the resistance of a thermistor decreases as temperature increases' requires stating that more charge carriers are released, so current increases for a given p.d., hence resistance decreases.
Use the given data and appropriate equations to find a numerical answer. Show all steps, including rearrangement and substitution, and give the final answer with correct units and significant figures. No explanation of the physics is needed unless asked.
How Students Lose Marks (Examiner Pitfalls)
Step-by-Step Worked Solutions
Question: A copper wire of length 2.5 m and cross-sectional area 1.2 × 10^-6 m^2 carries a current of 3.0 A. Calculate the potential difference across the wire. The resistivity of copper is 1.7 × 10^-8 Ω m.
- 1.Step 1: Identify given facts: length L = 2.5 m, area A = 1.2 × 10^-6 m^2, current I = 3.0 A, resistivity ρ = 1.7 × 10^-8 Ω m.
- 2.Step 2: Apply the resistivity formula: R = ρL / A. Substitute values: R = (1.7 × 10^-8 × 2.5) / (1.2 × 10^-6) = 4.25 × 10^-8 / 1.2 × 10^-6 = 0.0354 Ω.
- 3.Step 3: Use Ohm's law V = IR to find potential difference: V = 3.0 × 0.0354 = 0.106 V.
- 4.Step 4: State final answer with units: V = 0.11 V (2 s.f.).
Question: A battery of e.m.f. 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 of the battery.
- 1.Step 1: Identify given facts: e.m.f. ε = 9.0 V, internal resistance r = 0.50 Ω, external resistance R = 4.0 Ω.
- 2.Step 2: Apply the equation for e.m.f.: ε = I(R + r). Rearrange to find current: I = ε / (R + r) = 9.0 / (4.0 + 0.50) = 9.0 / 4.5 = 2.0 A.
- 3.Step 3: Calculate terminal p.d. using V = IR: V = 2.0 × 4.0 = 8.0 V. Alternatively, V = ε - Ir = 9.0 - (2.0 × 0.50) = 8.0 V.
- 4.Step 4: State final answers with units: current = 2.0 A, terminal p.d. = 8.0 V.