Capacitor discharge: when a good straight line hides a systematic error
A student must determine the capacitance of an unmarked capacitor by discharging it through a carbon resistor marked 100 kΩ ± 5%. He charges the capacitor to 6.00 V, then connects it across the resistor and records the p.d. every 10 s with a digital voltmeter on its 20 V range; the specification states an input resistance of 1 MΩ ± 2%. Readings: 0 s, 6.00 V; 10 s, 4.75 V; 20 s, 3.76 V; 30 s, 2.97 V; 40 s, 2.35 V; 50 s, 1.86 V. He plots ln V against t, obtains a straight line of gradient −0.0234 s⁻¹, and calculates C = 1/(100 × 10³ × 0.0234) = 4.3 × 10⁻⁴ F. He quotes C = 430 µF ± 5%, the tolerance of the resistor. His partner says the graph is 'a good straight line, so the method must be sound'.
1. Check the gradient against the readings, then determine the true capacitance, taking the voltmeter into account. Compare the resulting error with the ±5% the student quotes.
Read the worked answer
Check. ln 6.00 = 1.792 and ln 1.86 = 0.620, so the gradient over 50 s is (0.620 − 1.792)/50 = −0.0234 s⁻¹, matching the plot. The time constant is τ = 1/0.0234 = 42.7 s.
Student's value. C = τ/R = 42.7/1.00 × 10⁵ = 4.27 × 10⁻⁴ F ≈ 430 µF.
Correct value. The voltmeter is connected across the capacitor throughout, so it discharges through the resistor and the meter in parallel: R_eff = (100 kΩ × 1000 kΩ)/(1100 kΩ) = 90.9 kΩ. C = 42.7/9.09 × 10⁴ = 4.70 × 10⁻⁴ F ≈ 470 µF.
Comparison. The 5% resistor tolerance and 2% meter tolerance combine in quadrature to about 5.4%, i.e. roughly ±23 µF. The meter loading shifts the result by 40 µF - about 10%, twice the quoted uncertainty, and it is systematic, always in the same direction. The quoted figure 430 ± 23 µF does not even contain the true value, so the uncertainty statement is both too small and of the wrong kind.
2. Explain why a straight-line graph does not validate the method here, and describe two changes that would remove or correct the systematic error.
Read the worked answer
Why linearity proves nothing. The theory predicts V = V₀e^(−t/R_effC) whatever R_eff is. Any parallel resistance simply changes the time constant; the decay remains a clean exponential, so ln V against t remains perfectly straight. Linearity tests only that there is a single time constant, not that the time constant is the one the student thinks he is measuring. The partner's inference is invalid.
Removal (choose either):
- Use a measuring instrument whose resistance is negligible in parallel with 100 kΩ - a 10 MΩ electrometer or data-logger input gives an effective 99 kΩ, a 1% effect, and the residual can be corrected. Alternatively measure the meter's resistance directly and calculate R_eff, as above.
- Swap the roles: pass the discharge current through a low-resistance ammeter (a few ohms) and measure current instead of p.d. across the capacitor, or use a known reference capacitor to calibrate the whole arrangement.
A cross-check is also worth adding: extrapolating the line back to t = 0 should give ln V₀ = 1.792. If the intercept is low, the switching moment was missed and the timing has a zero error, which is a second, independent systematic error.
3. The student wants to reduce the random uncertainty in C. Using the readings, decide which measurement limits the precision and what he should change, and write down the resulting expression for the fractional uncertainty in C.
Read the worked answer
Where the random error sits. The voltage is read to ±0.01 V on the 20 V range, i.e. 0.17% of 6.00 V, which is negligible. The timing of each point is limited by reaction time, perhaps ±0.3 s in 10 s, i.e. 3%, but the gradient depends on the spread of the points, not on one point. The run covers 0 to 50 s, only about 1.2τ. Over that short baseline a small error in any single point tilts the line appreciably, and the last points (V = 1.86 V) are hardest to time because he is waiting for a slowly falling value to cross a mark.
Changes: extend the run to at least 3τ–4τ (about 130–170 s, down to roughly 0.2 V) so the line spans a wide range of ln V; use a data logger sampling every 1 s so timing is exact; and use a larger initial p.d. (e.g. 10 V on the 20 V range) to keep the voltage resolution a small fraction of V₀.
Uncertainty expression: since C = τ/R_eff, ΔC/C = Δτ/τ + ΔR/R + ΔR_V/(R + R_V), with the last term being the loading correction (about 100 × 2%/1100 ≈ 0.2%). The gradient itself comes from a least-squares fit, so Δτ/τ is obtained from the standard error of the gradient, not from a single reading.