Topic 9 of 9
Infer a time constant from data
A discharge record can reveal a time constant through its exponential shape or a straight-line logarithmic plot. The area under its current-time graph gives charge transferred, with the sign set by the chosen current reference.
For a capacitor discharging through fixed R, let V0 be its initial p.d. and τ = RC. With positive current defined towards the initially positive plate, V = V0e-t/τ and I = -V/R. The capacitor voltage falls while current is negative and approaches zero. The switching diagrams identify the corresponding paths.
Make the exponential relationship linear
Divide by V0 before taking the natural logarithm:
ln(V/V0) = -t/τ
The logarithm's input is a positive, dimensionless ratio. Plot ln(V/V0) vertically against time t horizontally. The ideal straight line has gradient -1/τ, with unit s-1, and zero intercept. Its gradient gives τ; a known discharge resistance then gives C = τ/R.
A raw ln(V) label would conceal the required reference unit. A zero or negative ratio cannot be processed by LN. In an actual record, preserve such readings and investigate zero offset, noise, polarity and the usable range; do not silently delete them or invent positive replacements.
Signed area during only the first four seconds
Dots and straight edges use the rounded supplied readings; the dashed curve uses the exact model. Signed area is negative. The four trapezia overestimate the magnitude by about 2.09%.
A logarithmic transform gives a straight-line model
Fit the transformed points with a free intercept. The small departures from the exact model come from rounding these generated readings, not experimental scatter.
Software exercise with generated values
Enter the data and preserve its units
Use a blank sheet. The supplied model has R = 200 kΩ, C = 10 µF and V0 = 10 V, giving τ = 2.0 s. Enter the following rounded generated voltages in A2:B6 under A1 = Time / s and B1 = Capacitor voltage / V. These are model values, not a performed experiment.
| Row | A: time / s | B: voltage / V |
|---|---|---|
| 2 | 0 | 10.00 |
| 3 | 1 | 6.07 |
| 4 | 2 | 3.68 |
| 5 | 3 | 2.23 |
| 6 | 4 | 1.35 |
Put labels in K1:N1 for resistance / Ω, capacitance / F, initial voltage / V and time constant / s. Enter 200000 in K2, 1.00E-5 in L2 and 10 in M2. In N2 enter =K2*L2, which gives 2.0 s. Keep units in headers and cells numeric. Formatting can display fewer decimals without replacing the stored values.
Add derived headers C1 = V/V0, D1 = ln(V/V0), E1 = Model voltage / V and F1 = Signed current / A. Use these first-row formulas and copy each through row 6:
| Cell | Formula |
|---|---|
| C2 | =B2/$M$2 |
| D2 | =LN(C2) |
| E2 | =$M$2*EXP(-A2/$N$2) |
| F2 | =-B2/$K$2 |
The dollar signs keep the supplied constants fixed as each row changes. EXP evaluates an exponential; it does not multiply its input by an approximate value of e. Its exponent -t/τ is dimensionless. The motion-data method explains numeric entry, copying formulas and XY plotting if you need a reminder.
Fit the log plot and infer capacitance
Make a numeric XY plot with time from A2:A6 horizontally and both B2:B6 and E2:E6 vertically to compare rounded values with the exact voltage model. Make a second XY plot of D2:D6 against A2:A6. Label its axes time / s and ln(V/V0), and fit a straight line with a free intercept.
Enter the following result formulas. Put descriptive labels beside the cells so a number retains its meaning:
| Cell and quantity | Formula |
|---|---|
| K5: gradient / s-1 | =SLOPE(D2:D6,A2:A6) |
| K6: intercept | =INTERCEPT(D2:D6,A2:A6) |
| K7: fitted τ / s | =-1/K5 |
| K8: inferred C / F | =K7/$K$2 |
Compare the fit with the supplied model
The gradient is about -0.5006318 s-1, with intercept +0.0008710. Hence τfit = 1.99748 s and Cfit = 9.9874 × 10-6 F, or about 9.99 µF.
The exact model would give gradient -0.5000000 s-1 and zero intercept. These small departures follow from rounding the generated input voltages; they do not represent measured scatter. A real unknown-capacitance investigation would additionally need reliable R, timing, voltage and loading information.
Compare current area with the change in charge
Use G1 = Interval / s and H1 = Estimated charge change / C. Enter =A3-A2 in G2 and =(F2+F3)*G2/2 in H2. Fill both formulas through row 5 only. Row 6 is the last supplied point, so it has no following interval; keep G6/H6 blank.
Each interval formula averages its two current endpoints and multiplies by its duration, giving the signed trapezium area. Calculate the totals and comparisons:
| Cell and quantity | Formula |
|---|---|
| M5: trapezium sum / C | =SUM(H2:H5) |
| M6: rounded-endpoint change / C | =$L$2*(B6-B2) |
| M7: exact-model change / C | =$L$2*$M$2*(EXP(-A6/$N$2)-1) |
| M8: relative magnitude excess | =M5/M7-1 |
Format M8 as a percentage. It compares two negative changes, so a positive result indicates that the coarse area has a greater magnitude.
Compare the area results and explain the difference
The four signed interval areas are -40.175, -24.375, -14.775 and -8.950 µC. Their sum is -88.275 µC. From the rounded endpoints, C(Vfinal - Vinitial) = -86.50 µC; the exact model gives -86.4665 µC.
The trapezium result overestimates the exact magnitude by about 2.09%. The positive current magnitude is a convex exponential curve, so the coarse chords lie above it. Smaller time intervals improve the curved-area approximation. Rounding contributes a separate small discrepancy.
Charge conservation has not failed: exact signed current area equals ΔQ. The positive amount leaving the initially positive plate is the magnitude of that negative change. At 4 s discharge is still incomplete, so this interval does not transfer the full initial 100 µC.
Turn a real record into a defensible time constant
Record the actual switch event and time zero, sample times, resistor value, initial capacitor voltage and capacitor information. Connect the voltage sensor across the capacitor's B/C terminals. Check its zero, polarity, range, resolution and response time; sample often enough to resolve the change over the expected time constant. A range ending below the initial 10 V would not capture this supplied example.
Use the real circuit's discharge path, check component ratings and capacitor polarity where applicable, and repeat the timing where useful. Retain the original record and document any excluded interval or correction. The measurement-record procedure explains how to enter or import actual readings while keeping them distinct from generated examples.
The simple exponential assumes negligible loading. A finite-resistance voltmeter across the capacitor is also a discharge path. For a 10 MΩ meter across the supplied 200 kΩ resistor, using 1 MΩ = 106 Ω:
= 196078 Ω
τ = ReffectiveC ≈ 1.9608 s
This is about 1.96% shorter than 2.0 s. Dividing that measured time constant by 200 kΩ while ignoring the meter would underestimate capacitance. In a charging circuit, the same finite parallel meter can also lower the final capacitor voltage. Check the actual connections before choosing the ideal formula.