Infer a decay without confusing the meter and leakage
A capacitor discharging through leakage resistance Rl and parallel meter resistance Rm has equivalent resistance Req = RlRm/(Rl + Rm).
Learning outcome: Compare inferred leakage with and without the known meter. Predict a second meter’s time constant while holding the leakage fixed, and identify the shared parameter. Write a baseline-offset counterexample showing why a straight two-point log fit is not always a valid physical inference.
Starting recap
Integrate current or resistance, preserve node topology and derive transient state with a signed energy account.
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For C = 2 microfarad, Rl = 3 Mohm and Rm = 2 Mohm, Req = 1.2 Mohm and tau = 2.4 s. Starting at 4 V, supplied readings at 1 and 3 s are about 2.63696 and 1.14602 V. Their ratio recovers tau; the known meter then gives Rl = 3 Mohm.
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What does one ideal exponential decay identify when both parallel resistances are unknown?
Experiment
Compare inferred leakage with and without the known meter. Predict a second meter’s time constant while holding the leakage fixed, and identify the shared parameter. Write a baseline-offset counterexample showing why a straight two-point log fit is not always a valid physical inference.
Distinguish leakage from meter loading
Worked teaching model · synthetic values · Supplied capacitor 2 microfarad initially at 4 V, discharging through leakage and a parallel meter. Exact synthetic observations have no additive voltage baseline. One measured time constant identifies equivalent resistance, not two unknown resistances separately; meter resistance is stipulated known for the inverse step.
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Meter-loaded discharge
Axes rescale independently. Curves join sampled coordinates; a joining segment is not additional measured evidence. Read physical units rather than screen angles or apparent slopes.
Meter-loaded discharge: numerical coordinates
| time (s) | voltage (V) |
|---|---|
| 0 | 4 |
| 0.166667 | 3.73165 |
| 0.333333 | 3.4813 |
| 0.5 | 3.24775 |
| 0.666667 | 3.02986 |
| 0.833333 | 2.82659 |
| 1 | 2.63696 |
| 1.16667 | 2.46005 |
| 1.33333 | 2.29501 |
| 1.5 | 2.14105 |
| 1.66667 | 1.99741 |
| 1.83333 | 1.86341 |
| 2 | 1.73839 |
| 2.16667 | 1.62177 |
| 2.33333 | 1.51297 |
| 2.5 | 1.41146 |
| 2.66667 | 1.31677 |
| 2.83333 | 1.22843 |
| 3 | 1.14602 |
| 3.16667 | 1.06914 |
| 3.33333 | 0.997409 |
| 3.5 | 0.930495 |
| 3.66667 | 0.86807 |
| 3.83333 | 0.809832 |
| 4 | 0.755502 |
| 4.16667 | 0.704817 |
| 4.33333 | 0.657532 |
| 4.5 | 0.61342 |
| 4.66667 | 0.572267 |
| 4.83333 | 0.533874 |
| 5 | 0.498058 |
| 5.16667 | 0.464644 |
| 5.33333 | 0.433472 |
| 5.5 | 0.404391 |
| 5.66667 | 0.377261 |
| 5.83333 | 0.351952 |
| 6 | 0.32834 |
| 6.16667 | 0.306312 |
| 6.33333 | 0.285762 |
| 6.5 | 0.266591 |
| 6.66667 | 0.248706 |
| 6.83333 | 0.232021 |
| 7 | 0.216455 |
| 7.16667 | 0.201934 |
| 7.33333 | 0.188386 |
| 7.5 | 0.175748 |
| 7.66667 | 0.163957 |
| 7.83333 | 0.152958 |
| 8 | 0.142696 |
| 8.16667 | 0.133123 |
| 8.33333 | 0.124192 |
| 8.5 | 0.11586 |
| 8.66667 | 0.108087 |
| 8.83333 | 0.100836 |
| 9 | 0.094071 |
| 9.16667 | 0.0877599 |
| 9.33333 | 0.0818723 |
| 9.5 | 0.0763797 |
| 9.66667 | 0.0712555 |
| 9.83333 | 0.0664751 |
| 10 | 0.0620154 |
| Quantity | Value | Unit |
|---|---|---|
| True equivalent resistance | 1.2 | Mohm |
| Measured exponential time constant | 2.4 | s |
| Supplied voltage at 1 s | 2.63696 | V |
| Supplied voltage at 3 s | 1.14602 | V |
| Two-point fitted time constant | 2.4 | s |
| Leakage inferred with known meter | 3 | Mohm |
| Naive leakage if meter ignored | 1.2 | Mohm |
One decay identifies the parallel combination; a known meter resistance is additional information, not something learned from the same slope.
With zero additive baseline, ln[V(t1)/V(t2)] = (t2−t1)/tau. Then 1/Rleak = C/tau − 1/Rmeter must be positive for a finite passive leakage model. A zero result gives an infinite-leakage-resistance limit; a negative result rejects the assumed parameters. Unknown voltage baseline invalidates this simple two-reading logarithm.
Derive before checking. Retain contrasting captures and explain one assumption that could fail; calculated evidence is not actual laboratory work.
One decay identifies the parallel combination; a known meter resistance is additional information, not something learned from the same slope.
With zero additive baseline, ln[V(t1)/V(t2)] = (t2−t1)/tau. Then 1/Rleak = C/tau − 1/Rmeter must be positive for a finite passive leakage model. A zero result gives an infinite-leakage-resistance limit; a negative result rejects the assumed parameters. Unknown voltage baseline invalidates this simple two-reading logarithm.
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Which source, topology, initial condition or instrument assumption would invalidate this inference?
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