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GCSE/Physics/CCEA

U2.1Electricity — current, voltage, resistance, IV characteristics, circuits

Notes

Electricity

Charge and current

Electric charge is measured in coulombs C. Electrons carry a charge of −1.6 × 10⁻¹⁹ C.

Electric current (I) is the rate of flow of charge: I = Q / t (A = C/s). In metals, current is the flow of electrons. Conventional current flows from + to − (opposite to electron flow).

Voltage (potential difference)

Potential difference (p.d.) or voltage (V) is the energy transferred per unit charge: V = W / Q (volts = J/C).

A 12 V battery transfers 12 J of energy for every coulomb of charge that passes through it.

Electromotive force (e.m.f.) is the energy supplied per unit charge by a source (cell, battery, generator).

Resistance

Ohm's law: V = IR (for ohmic conductors at constant temperature). Resistance = voltage / current: R = V / I (Ω).

Ohmic conductors (e.g. metal wire at constant temperature): I ∝ V → straight line through origin on IV graph. Non-ohmic devices:

  • Filament bulb: resistance increases with temperature → IV graph curves (gradient decreases at high I).
  • Diode: only allows current in one direction → IV graph shows near-zero current for reverse voltage and rapid rise for forward voltage above ~0.7 V.
  • Thermistor: resistance decreases as temperature increases (negative temperature coefficient, NTC). Used in temperature sensors and thermostats.
  • LDR (light-dependent resistor): resistance decreases as light intensity increases. Used in automatic lighting.

Series circuits

  • Same current through every component.
  • Total voltage = sum of individual voltages: V_total = V₁ + V₂ + …
  • Total resistance: R_total = R₁ + R₂ + … (increases with more resistors).

Parallel circuits

  • Same voltage across each branch.
  • Total current = sum of branch currents: I_total = I₁ + I₂ + …
  • Total resistance is less than the smallest individual resistance: 1/R_total = 1/R₁ + 1/R₂ + …
  • Adding more branches in parallel decreases overall resistance and increases total current from the supply.

Electrical power

P = IV = I²R = V²/R (W = J/s)

Energy: E = Pt = IVt (J)

For electricity bills, energy is measured in kilowatt-hours: 1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J.

Safety and fuses

A fuse contains a thin wire that melts if current exceeds a rated value, breaking the circuit. Choose the fuse with the lowest rating that exceeds normal operating current. Circuit breakers (RCCBs) and earth wires protect against electric shock and fire.

Common mistakes

  1. Confusing voltage and current — voltage is the "push" (energy per charge); current is the "flow" (charge per second).
  2. Adding resistances in parallel directly — use the reciprocal formula.
  3. Forgetting to convert kW to W and hours to seconds in energy calculations.
  4. IV graph for filament bulb — the curve is because resistance increases with temperature, not because it breaks Ohm's law in principle.

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Practice questions

Try each before peeking at the worked solution.

  1. Question 16 marks

    Ohm's law and resistance

    CCEA Unit 2 Paper

    A resistor has a current of 0.4 A when a voltage of 12 V is applied across it.

    (a) Calculate the resistance of the resistor. (2 marks)
    (b) Calculate the power dissipated. (2 marks)
    (c) Calculate the charge that flows in 30 s. (2 marks)

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  2. Question 27 marks

    Series and parallel circuits

    CCEA Unit 2 Paper

    Two resistors, R₁ = 10 Ω and R₂ = 15 Ω, are connected to a 12 V supply.

    (a) Calculate the total resistance when connected in series. (1 mark)
    (b) Calculate the current from the supply in series. (2 marks)
    (c) Calculate the total resistance when connected in parallel. (2 marks)
    (d) Calculate the total current from the supply in parallel. (2 marks)

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  3. Question 36 marks

    Describe IV characteristics

    CCEA Unit 2 Paper

    (a) Sketch IV characteristic graphs for: (i) an ohmic resistor, (ii) a filament lamp, (iii) a diode. (3 marks)
    (b) Explain why a filament lamp is not ohmic. (3 marks)

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  4. Question 46 marks

    Electrical energy and cost

    CCEA Unit 2 Paper

    A 2.5 kW electric kettle is used for 4 minutes. The cost of electricity is 28 p per kWh.

    (a) Calculate the energy transferred in joules. (3 marks)
    (b) Calculate the energy in kWh. (1 mark)
    (c) Calculate the cost of using the kettle. (2 marks)

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  5. Question 57 marks

    Thermistor in a circuit — application

    CCEA Unit 2 Paper

    A thermistor is connected in series with a fixed resistor (4700 Ω) and a 9 V battery.

    (a) Describe how the resistance of a thermistor changes with temperature. (1 mark)
    (b) When the thermistor is cold, its resistance is 14,300 Ω. Calculate the current in the circuit. (3 marks)
    (c) When the thermistor warms up, explain what happens to the current and the voltage across the fixed resistor. (3 marks)

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Flashcards

U2.1 — Electricity — current, voltage, resistance, IV characteristics, circuits

8-card SR deck for CCEA Physics topic U2.1

8 cards · spaced repetition (SM-2)