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    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.

    Read the Basics of electricity study guideFull revision notes for AQA A-Level Physics

    Your focus

    1. State and use the equation I = ΔQ/Δt to calculate current, charge or time.
    2. State and use the equation V = W/Q to calculate potential difference, work done or charge.
    3. 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
    1. 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.
    2. 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.
    3. 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.
    4. 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.
    5. 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)
    Define

    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.

    Explain

    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.

    Calculate

    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)
    Pitfall: Students often confuse potential difference (p.d.) with electromotive force (e.m.f.) and incorrectly state that p.d. is 'used up' by a component, rather than energy being transferred from electrical to other forms.
    ❌ Weak Answer (Loses Marks):The current goes through the resistor and the voltage is used up, so there is less voltage after the resistor.
    Example improved answer:The potential difference across the resistor is the energy transferred from electrical potential energy to thermal energy per unit charge passing through it. The e.m.f. of the supply is the energy transferred from chemical to electrical energy per unit charge. In a series circuit, the sum of the potential differences across components equals the e.m.f. of the supply, in accordance with conservation of energy.
    Examiner Tip: Always define p.d. and e.m.f. in terms of energy transfer per unit charge, and use the correct terminology: 'energy is transferred' not 'voltage is used up'. Refer to conservation of energy when explaining series circuits.
    Pitfall: When sketching I-V graphs for a filament lamp, students often draw a straight line through the origin or forget to label the axes with current on the y-axis and p.d. on the x-axis. They also fail to explain the non-ohmic behaviour in terms of temperature increasing resistance.
    ❌ Weak Answer (Loses Marks):The graph is a curve because the lamp is not an ohmic conductor.
    Example improved answer:As the potential difference across the filament lamp increases, the current increases, causing the temperature of the filament to rise. At higher temperatures, the lattice ions vibrate with greater amplitude, leading to more frequent collisions with charge carriers (electrons). This increases the resistance, so the current increases at a decreasing rate, producing a curve that flattens off. The graph is not a straight line through the origin, so the lamp is non-ohmic.
    Examiner Tip: Always state the physical reason for resistance change: increased lattice vibrations cause more collisions. Ensure axes are correctly labelled (I on y-axis, V on x-axis) and that you describe the shape as a curve that becomes less steep.
    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. 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. 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. 3.Step 3: Use Ohm's law V = IR to find potential difference: V = 3.0 × 0.0354 = 0.106 V.
    4. 4.Step 4: State final answer with units: V = 0.11 V (2 s.f.).
    Final Answer: The potential difference across the wire is 0.11 V.

    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. 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. 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. 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. 4.Step 4: State final answers with units: current = 2.0 A, terminal p.d. = 8.0 V.
    Final Answer: The current in the circuit is 2.0 A and the terminal potential difference is 8.0 V.
    Active Recall Memory Test
    State the equation linking current, charge, and time, and define the coulomb.
    Key Fact: I = ΔQ/Δt. The coulomb is the charge transferred by a current of 1 ampere in 1 second.
    What is the difference between potential difference and electromotive force?
    Key Fact: Potential difference is the energy transferred from electrical energy to other forms per unit charge. Electromotive force is the energy transferred from other forms to electrical energy per unit charge.
    Sketch the I-V graph for a silicon diode and explain its shape.
    Key Fact: The graph shows zero current for negative p.d. and for p.d. below about 0.6 V in forward bias. Above 0.6 V, current increases rapidly. This is because the diode only conducts when the forward p.d. exceeds the threshold voltage, which provides enough energy to overcome the potential barrier at the p-n junction.
    Define resistivity and state its unit.
    Key Fact: Resistivity is defined by ρ = RA/L, where R is resistance, A is cross-sectional area, and L is length. It is a property of a material and is measured in ohm-metres (Ω m).
    Frequently Asked Questions
    What is the difference between current and voltage?
    Current is the rate of flow of electric charge, measured in amperes (A), and is the same at all points in a series circuit. Voltage, or potential difference, is the energy transferred per unit charge between two points, measured in volts (V). Current is the flow of charge, while voltage is the driving force that causes the flow. In equations, current is I = ΔQ/Δt, and voltage is V = W/Q.
    Why does the resistance of a filament lamp increase as the current increases?
    As current increases, the temperature of the filament increases. At higher temperatures, the metal lattice ions vibrate with greater amplitude, which increases the frequency of collisions between the charge carriers (electrons) and the ions. This makes it harder for electrons to flow, so resistance increases. This is why the I-V graph for a filament lamp is a curve that flattens off at higher potential differences.
    What is Ohm's law and when does it apply?
    Ohm's law states that the current through a conductor is directly proportional to the potential difference across it, provided physical conditions (such as temperature) remain constant. It applies to ohmic conductors, such as a resistor at constant temperature, where the I-V graph is a straight line through the origin. It does not apply to non-ohmic components like filament lamps or diodes.
    How do you calculate the total resistance in a series and parallel circuit?
    In a series circuit, the total resistance is the sum of individual resistances: R_total = R1 + R2 + ... In a parallel circuit, the reciprocal of the total resistance is the sum of the reciprocals of individual resistances: 1/R_total = 1/R1 + 1/R2 + ... For two resistors in parallel, this can be simplified to R_total = (R1 × R2) / (R1 + R2). Always check that your answer is less than the smallest individual resistance in parallel.
    What is internal resistance and how does it affect terminal potential difference?
    Internal resistance is the resistance within a power source, such as a battery, that opposes the flow of current. It causes some energy to be dissipated inside the source, so the terminal potential difference (the p.d. across the external circuit) is less than the e.m.f. The relationship is given by ε = I(R + r) or V = ε - Ir, where ε is e.m.f., I is current, R is external resistance, r is internal resistance, and V is terminal p.d. As current increases, the terminal p.d. decreases.
    What is the difference between resistance and resistivity?
    Resistance is a property of a specific component and depends on its dimensions (length and cross-sectional area) and the material it is made from. It is measured in ohms (Ω). Resistivity is a property of the material itself, independent of shape or size, and is defined by ρ = RA/L. It is measured in ohm-metres (Ω m). For example, copper has a low resistivity, making it a good conductor, while nichrome has a higher resistivity, making it suitable for heating elements.