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Capacitors

Capacitor calculations and graph work

  • 3 things to know
  • Unit A2 2

Overview

Recall questions combining charge, energy and time constant for a stated capacitor, what to plot for a gradient equal to the capacitance and what for the time constant, reading a time constant and an initial voltage off a logarithmic plot, comparing the energy stored in series and in parallel arrangements, and the half-life of a discharge as 0.69 CR.

What you need to know

Answer each one before you reveal it.

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Numbers

A 47 μF capacitor is charged to 24 V. Calculate the charge and the energy stored.

Key term

What is plotted to obtain a straight line whose gradient is the capacitance?

Key term

What is plotted to obtain a straight line whose gradient gives the time constant?

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    A 47 μF capacitor is charged to 24 V. Calculate the charge and the energy stored.

    Model answer

    Q = CV = 47 × 10⁻⁶ × 24 = 1.13 × 10⁻³ C, and E = ½QV = 0.5 × 1.13 × 10⁻³ × 24 = 1.4 × 10⁻² J.

  2. 2

    What is plotted to obtain a straight line whose gradient is the capacitance?

    Model answer

    Charge stored on the vertical axis against potential difference on the horizontal axis.

  3. 3

    How is the energy stored obtained from a charge against potential difference graph?

    Model answer

    It is the area between the line and the horizontal axis up to the potential difference to which the capacitor is charged.

  4. 4

    What is plotted to obtain a straight line whose gradient gives the time constant?

    Model answer

    The natural logarithm of the potential difference on the vertical axis against time on the horizontal axis.

  5. 5

    A discharge graph of ln V against t has a gradient of −0.25 s⁻¹ and an intercept of 2.30. Find the time constant and the initial voltage.

    Model answer

    The time constant is 1 ÷ 0.25 = 4.0 s, and the initial voltage is e^(2.30), which is 10 V.

  6. 6

    Two capacitors of 100 μF and 100 μF are charged in parallel to 20 V. Calculate the total energy stored.

    Model answer

    The total capacitance is 200 μF, so E = 0.5 × 200 × 10⁻⁶ × 20² = 0.040 J.

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