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Capacitor Charge and Discharge (A level)

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For A level physics (AQA, OCR A and Edexcel). Charge and discharge a capacitor through a resistor in the animated explorer: change R, C and the emf and watch the Q–t, V–t and I–t graphs draw themselves. Then practise reading time constants from graphs, the exponential equations, Q = CV and energy, and combining capacitors.

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Worked solution

    Key facts

    Q = CV    C = Aε₀εᵣ / d

    E = ½QV = ½CV² = ½Q²/C

    Discharging through R:

    Q = Q₀e−t/RC   (V and I decay in the same way)

    Charging from a cell of emf ε:

    Q = Q₀(1 − e−t/RC)   V = ε(1 − e−t/RC)

    Time constant RC (in s): the time to fall to 0.37 of the start value when discharging, or to reach 0.63 of the final value when charging. From a graph of ln V against t, the gradient is −1/RC.

    T½ = ln 2 × RC ≈ 0.69RC

    The area under an I–t graph is the charge that flows. The area under a Q–V graph is the energy stored. When charging from a cell, the cell supplies Qε, the capacitor stores ½Qε and the other half is dissipated in the resistance.

    series: 1/C = 1/C₁ + 1/C₂ + …    parallel: C = C₁ + C₂ + …

    ε₀ = 8.85 × 10−12 F m−1. Prefixes: µ = 10−6, k = 103, so kΩ × µF = ms.

    Calculated answers within 2% are marked correct. Times read from a graph are marked correct within half a small square. Gradients within 5%, and areas within 6%, are marked correct.

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