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Capacitance

Charging at constant current and measuring capacitance

  • 3 things to know
  • A2 Unit 4 Section A

Overview

Recall questions on Charging at constant current and measuring capacitance.

What you need to know

Answer each one before you reveal it.

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Key term

Why does the charging current fall if the series resistance is left unaltered?

Numbers

A capacitor is charged at a constant current of 50 μA and the potential difference across it rises at 0.25 V s⁻¹. Calculate its capacitance.

Numbers

A capacitor charged at a steady current of 20 μA for 30 s reaches a potential difference of 4.0 V. Calculate its capacitance.

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    How can a capacitor be charged at a constant current in the laboratory?

    Model answer

    A variable resistor in series with the capacitor is adjusted continuously to hold the ammeter reading fixed as the potential difference across the capacitor rises, or a constant-current source is used instead.

  2. 2

    Why does the charging current fall if the series resistance is left unaltered?

    Model answer

    As charge collects on the plates the potential difference across the capacitor rises and opposes the supply, so less potential difference is left across the resistor and the current falls.

  3. 3

    What is the shape of a graph of potential difference against time for a capacitor charged at a constant current?

    Model answer

    It is a straight line through the origin, because the charge increases at a steady rate and the potential difference is proportional to the charge.

  4. 4

    What does the gradient of a graph of potential difference against time equal for a capacitor charged at constant current, and how is the capacitance found from it?

    Model answer

    The gradient equals the charging current divided by the capacitance, so the capacitance is the current divided by the gradient.

  5. 5

    A capacitor is charged at a constant current of 50 μA and the potential difference across it rises at 0.25 V s⁻¹. Calculate its capacitance.

    Model answer

    The capacitance is 50 × 10⁻⁶ ÷ 0.25 = 2.0 × 10⁻⁴ F. In more usual units that is 200 μF.

  6. 6

    A capacitor is charged at a steady current of 20 μA for 30 s. Calculate the charge stored.

    Model answer

    Using Q = It gives 20 × 10⁻⁶ × 30 = 6.0 × 10⁻⁴ C.

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