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Capacitor discharge practical questions

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

Practise

Voltage during discharge

Initially V₀ = 8 V; RC = 16 s. Calculate V at t = 16 s. 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
84.8522452777
162.94303552937
241.78504128119
321.08268226589

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.

Capacitor voltage during dischargeSynthetic discharge readings and supplied exponential model curve; use the table for calculations. Horizontal axis: Time / s, 0 to 40. Vertical axis: Voltage / V, 0 to 9.6. Exact values are also listed below the graph.081624324001.63.24.86.489.6Time / sVoltage / V12345
Read the axes and units. The values are also available in the table below.
Larger graph
Capacitor voltage during dischargeSynthetic discharge readings and supplied exponential model curve; use the table for calculations. Horizontal axis: Time / s, 0 to 40. Vertical axis: Voltage / V, 0 to 9.6. Exact values are also listed below the graph.081624324001.63.24.86.489.6Time / sVoltage / V12345
Graph values
Coordinates
PointTime / sVoltage / V
108
284.8522452777
3162.94303552937
4241.78504128119
5321.08268226589

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

Hint

Use V = V₀ exp(−t/RC); this is discharge, so the exponent is negative.

Worked solution
  1. V = 8 × exp(−16/16) V.
  2. After one time constant, V/V₀ = 1/e, not 1/2.
  3. Answer: 2.94 V (3 significant figures).
Next task in this dataset
Work through this dataset
  1. Convert R and C
  2. Voltage during discharge (current)
  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
  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: Mixed practice
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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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