Log-Linear Fit of a Capacitor Discharge: Is the Fourth Point Real?
A technician discharges a 20 µF capacitor from 10.00 V through a 150 kΩ resistor and logs the voltage with a 4½-digit DMM once per second: at t = 0, 1, 2, 3, 4, 5 s the readings are 10.00, 7.17, 5.13, 3.68, 3.10 and 1.89 V. A repeat run reproduces every reading within 0.01 V except the t = 4 s value, which repeats at 3.09–3.11 V. The nominal time constant is RC = 3.00 s. Before these numbers go into a calibration report, the engineer must decide whether the fourth point is a valid datum, extract the time constant from the data that remain, and state how far the model can be trusted when extrapolated to the 1.00 V crossing.
1. Linearise V(t) = V₀e^(−t/τ) and test the t = 4 s reading. Should it be discarded? Support the decision numerically.
Read the worked answer
Taking ln V gives 2.3026, 1.9699, 1.6351, 1.3029, 1.1314, 0.6366 at t = 0…5 s. A straight line is expected with slope −1/τ.
Fitting the five uncontested points (t = 0, 1, 2, 3, 5) by least squares: t̄ = 2.2 s, ȳ = 1.5694, Σ(t−t̄)(y−ȳ) = −4.932, Σ(t−t̄)² = 14.8, so slope = −0.33324 and intercept = 2.3025.
Residuals with this fit: +0.0001, +0.0006, −0.0010, +0.0001, +0.0002 - all under 0.001 in log (≈0.1 % in volts). The t = 4 s point is predicted at ln V = 2.3025 − 4(0.33324) = 0.9696 → V = 2.64 V, but reads 3.10 V: residual +0.1618, about 17.6 % high and roughly 200× larger than any other residual.
Because the error repeats (3.09–3.11 V) it is not random noise - it is systematic, e.g. a range change, a mis-keyed timestamp, or a probe contact at that step. Discard it and document why; keeping it would bias τ to ≈3.13 s (4 % high).
2. Report the best estimate of τ and V₀ from the retained data, and comment on the agreement with the nominal RC.
Read the worked answer
τ = 1/0.33324 = 3.00 s and V₀ = e^2.3025 = 10.00 V. The fit therefore reproduces the nominal 150 kΩ × 20 µF = 3.00 s to better than 0.1 %.
Uncertainty: the largest retained residual is 0.001 in ln V over a 5 s lever arm, so the slope is good to roughly ±0.0005 s⁻¹, i.e. τ = 3.00 ± 0.01 s. Quoting more digits would be dishonest given 3-significant-figure meter readings.
One systematic check worth noting: a 10 MΩ DMM input loads the 150 kΩ source, giving an effective 147.8 kΩ and τ = 2.96 s - about 1.5 % low. The measured 3.00 s is if anything slightly high, so meter loading is not masking a larger leakage or dielectric-absorption effect. If the design had demanded τ known to 0.1 %, this loading term could not be ignored.
3. Predict the time at which V falls to 1.00 V and state the cautions that belong with that number.
Read the worked answer
Set 1.00 = 10.00·e^(−t/3.00): t = τ·ln 10 = 3.00 × 2.302585 = 6.91 s (call it 6.9 s).
Cautions:
- The data stop at 5 s (V = 1.89 V), so 6.91 s is a modest extrapolation - about 2.3 time constants out - and inherits any error in τ directly (a 1 % error in τ is a 0.07 s error here).
- The exponential model assumes a fixed R and C. Leakage paths, dielectric absorption in the capacitor, and the meter's 10 MΩ input all change the effective τ, and their relative weight grows as V gets small: at 1 V the leakage current needed to matter is only nanoamps.
- The result is a prediction for this network, not a general calibration constant; it should be re-verified with a measurement near the threshold rather than shipped as a measured value.