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Electric and Magnetic Fields

Investigating capacitor discharge and time-constant calculations

  • 1 thing to know
  • Papers 1 and 3

Overview

Recall questions on recording a discharge, why a large resistance and a high-resistance voltmeter are needed, and finding the time constant and half-time.

What you need to know

Answer each one before you reveal it.

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Key term

Two identical capacitors discharge through different resistors. Which discharges more quickly and why?

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    Why should a large resistance and capacitance be chosen when recording a discharge by hand or with a slow data logger?

    Model answer

    Because a large time constant makes the decay slow enough for a good number of readings to be taken before the capacitor is discharged.

  2. 2

    Why must the instrument measuring the potential difference across a discharging capacitor have a very high resistance?

    Model answer

    Because a low-resistance meter would provide an additional discharge path, making the capacitor discharge faster and giving a time constant that is systematically too small.

  3. 3

    In an investigation of capacitor discharge, why is it good practice to repeat the discharge and average the timings?

    Model answer

    Repeating and averaging reduces the effect of random errors, particularly in reaction time when readings are taken by hand.

  4. 4

    A 47 μF capacitor discharges through a 100 kΩ resistor. What is the time constant?

    Model answer

    RC = 1.00 × 10⁵ × 47 × 10⁻⁶ = 4.7 s.

  5. 5

    A capacitor is charged to 6.0 V and then discharged. What is the potential difference across it after one time constant?

    Model answer

    V = V₀e⁻¹ = 6.0 × 0.368 = 2.2 V.

  6. 6

    For a resistor-capacitor circuit with a time constant of 4.7 s, how long does the charge take to fall to half its initial value?

    Model answer

    The time is t = RC ln 2 = 4.7 × 0.693 = 3.3 s.

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