Edexcel A-Level Physics 9PH0
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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.
Questions from this subtopic
Each one with the model answer underneath.
- 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
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
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
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
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
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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