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A Level · A2 · data analysis

A Level extension · Question 12

A Level extension · Original GioPhysics question with a detailed, mark-by-mark answer guide.

Demand
discriminating
Marks
15
Topics
2
Answer
Complete
A charged capacitor discharging through a resistor, with a voltmeter across itCVSR = 100 kΩ
Fig. 12.1A circuit with three branches between an upper and a lower horizontal wire. On the left is a capacitor labelled C; in the middle is a voltmeter connected permanently across it, with junction dots where its branch joins each of the two wires; on the right is a resistor labelled R = 100 kΩ. An open switch S lies in the upper wire between the voltmeter branch and the resistor branch, so that closing it connects the resistor across the capacitor.
data analysis15 marks

A student investigates the discharge of a capacitor of capacitance C through a resistor of resistance R = 100 kΩ. The potential difference V across the capacitor is recorded at intervals, and ln(V/V) is calculated. The student expects V = V₀e^(−t/RC).

The student's processed results
t / s0.010.020.030.040.0
V / V9.006.054.072.731.84
ln (V / V)2.1971.8001.4041.0040.610
  1. (a)

    Explain Explain why a graph of ln V against t should be a straight line if the discharge is exponential, and state what its gradient and intercept represent.

    4 marks
  2. (b)

    Determine Use the first and last data points to determine the gradient of the line, and hence determine the capacitance C.

    4 marks
  3. (c)

    Determine The resistance is quoted as 100 ± 2 kΩ, and the uncertainty in the gradient is ±1.5%. Determine the percentage uncertainty in C, and state the value of C with its absolute uncertainty.

    4 marks
  4. (d)

    Suggest The voltmeter used has a finite resistance connected across the capacitor throughout the experiment. Suggest how this affects the value obtained for C, and suggest an improvement.

    3 marks
Ready to self-mark?Reveal the detailed answer guide
Answer overviewKey answer: taking natural logarithms of V = V₀e^(−t/RC) gives ln V = ln V₀ − t/(RC) this is of the form y = mx + c with ln V as y and t as x, so the points lie on a straight line if the model holds the gradient is −1/(RC)
01

(a)

4 marks

Explain Explain why a graph of ln V against t should be a straight line if the discharge is exponential, and state what its gradient and intercept represent.

How to approach it

Read the data before explaining it. Quote the relevant values or trend, show the comparison or calculation, and then connect that numerical evidence to the physical conclusion—including uncertainty or anomalies when they matter.

  1. 1

    taking natural logarithms of V = V₀e^(−t/RC) gives ln V = ln V₀ − t/(RC)

  2. 2

    this is of the form y = mx + c with ln V as y and t as x, so the points lie on a straight line if the model holds

  3. 3

    the gradient is −1/(RC)

  4. 4

    the intercept on the ln V axis is ln V₀

02

(b)

4 marks

Determine Use the first and last data points to determine the gradient of the line, and hence determine the capacitance C.

How to approach it

Read the data before explaining it. Quote the relevant values or trend, show the comparison or calculation, and then connect that numerical evidence to the physical conclusion—including uncertainty or anomalies when they matter.

  1. 1

    gradient = (0.610 − 2.197) / (40.0 − 0.0) = −1.587 / 40.0

  2. 2

    gradient = −0.0397 s⁻¹

  3. 3

    RC = −1/gradient = 25.2 s

  4. 4

    C = 25.2 / 100 × 10³ = 2.5 × 10⁻⁴ F (250 μF)

03

(c)

4 marks

Determine The resistance is quoted as 100 ± 2 kΩ, and the uncertainty in the gradient is ±1.5%. Determine the percentage uncertainty in C, and state the value of C with its absolute uncertainty.

How to approach it

Read the data before explaining it. Quote the relevant values or trend, show the comparison or calculation, and then connect that numerical evidence to the physical conclusion—including uncertainty or anomalies when they matter.

  1. 1

    percentage uncertainty in R = (2/100) × 100 = 2%

  2. 2

    C is found by dividing by the gradient and by R, so the percentage uncertainties add: 1.5% + 2% = 3.5%

  3. 3

    absolute uncertainty = 0.035 × 250 = 8.75 ≈ 9 μF

  4. 4

    C = 250 ± 9 μF

04

(d)

3 marks

Suggest The voltmeter used has a finite resistance connected across the capacitor throughout the experiment. Suggest how this affects the value obtained for C, and suggest an improvement.

How to approach it

Read the data before explaining it. Quote the relevant values or trend, show the comparison or calculation, and then connect that numerical evidence to the physical conclusion—including uncertainty or anomalies when they matter.

  1. 1

    the voltmeter provides a second discharge path in parallel with R, so the effective resistance is less than 100 kΩ

  2. 2

    the capacitor therefore discharges faster than the model assumes; the measured time constant is too small, and since C is calculated using R = 100 kΩ, the value obtained for C is too small

  3. 3

    improvement: use a voltmeter (or digital multimeter) of much higher resistance, or a data logger with a high-impedance input

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