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Capacitor discharge practical questions

Calculate capacitor discharge, interpret log-linear graphs and find capacitance and time constants.

Practise

Capacitance from gradient

Use the supplied natural-log line and R = 40 kΩ to find capacitance in farads. Give 3 significant figures.

Practical setup and assumptions

A capacitor initially at V₀ is disconnected from its charging supply and discharges through one resistor R. The voltmeter is ideal except in the separately stated loading task.

Assume R and C remain constant and there is no leakage. The readings follow V = V₀ exp(−t/RC). Use the table values without rounding intermediate results.

Readings for this dataset
Same-discharge readings
t / sV / V
08
24.8522452777
42.94303552937
61.78504128119
81.08268226589

The linked tasks use the same readings. These are generated practice data: treat them as exact unless an uncertainty is stated. Use the stated uncertainty rather than estimating it from scatter.

Natural log of voltage ratio against timePractice readings with a given straight line; use the line shown for calculations. Horizontal axis: Time / s, 0 to 10. Vertical axis: ln(V/V₀) / 1, -2.5 to 0.5. Exact values are also listed below the graph.0246810-2.5-2-1.5-1-0.500.5Time / sln(V/V₀) / 112345
Read the axes and units. The values are also available in the table below.
Larger graph
Natural log of voltage ratio against timePractice readings with a given straight line; use the line shown for calculations. Horizontal axis: Time / s, 0 to 10. Vertical axis: ln(V/V₀) / 1, -2.5 to 0.5. Exact values are also listed below the graph.0246810-2.5-2-1.5-1-0.500.5Time / sln(V/V₀) / 112345
Graph values
Coordinates
PointTime / sln(V/V₀) / 1
100
22-0.5
34-1
46-1.5
58-2

Use decimals or e notation, e.g. 4.2e-3. Enter the unit separately.

Hint

Gradient m = −1/(RC), so C = −1/(mR). Convert R to Ω.

Worked solution
  1. m = −1/4 s⁻¹; R = 40000 Ω.
  2. C = −1/[m × 40000] F.
  3. Answer: 1.00 × 10-4 F (3 significant figures).
Next task in this dataset
Work through this dataset
  1. Convert R and C
  2. Voltage during discharge
  3. Charge remaining
  4. Discharge current magnitude
  5. Energy remaining
  6. Natural-log voltage ratio
  7. Choose log-linear axes
  8. Signed log-linear gradient
  9. Capacitance from gradient (current)
  10. Time to halve voltage
  11. Capacitance uncertainty
  12. Voltmeter loading

These tasks keep the same readings. Another question starts a new dataset. Your answers are checked separately.

Another question
Question type: Capacitance from gradient
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Discharge through a resistor

With the supply disconnected, V = V₀ exp(−t/RC). Convert kΩ to Ω and μF to F before finding RC. After one time constant, voltage is V₀/e. The time to halve is RC ln(2).

Charge, current and energy

Q = CV, current magnitude through the discharge resistor is V/R, and stored energy is ½CV². Voltage and charge have the same decay time constant; energy falls with voltage squared.

Use a natural-log graph

ln(V/V₀) is dimensionless. Plotted against time, it has gradient −1/(RC) and zero intercept. Base-10 logs give a different gradient. Capacitance is −1/(gradient × R), with R in Ω.

Check voltmeter loading

A finite-resistance voltmeter across the capacitor supplies a parallel discharge path. The effective resistance and time constant become smaller. Inferring C using nominal R then gives a value that is too low. Other tasks assume an ideal voltmeter, constant R/C and no leakage; they cover discharge only.

A-level relevance

AQA 7408: 3.7.4.1, 3.7.4.3–4; required practical 9 (selected); OCR A H556: 6.1.1–3 (selected). Tasks cover the selected skills listed above.

AQA specification · OCR A specification · AQA practical handbook · OCR practical skills handbook

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