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University Physics II · Kirchhoff's Rules and RC Circuits · 7.8

RC Design & Data Studio

Eight problems, from a comparator threshold to a timing window no nominal value can meet. Each one asks which resistances are really in the loop while the capacitor charges, and holds you to three significant figures.

Open the lab ↓Practice & worked examples ↓
01

Build the model

Connect the measurement to the mechanism.

Timing thresholds, sensor smoothing, component tolerances, exponential fitting, residuals, and choosing a defensible circuit model. Treat this course-map statement as a claim to test rather than an invitation to import a familiar equation. In Kirchhoff's Rules and RC Circuits, begin from time-constant inference, then state the system, observable, assumptions, and evidence before calculating.

Simple definition
Timing thresholds, sensor smoothing, component tolerances, exponential fitting, residuals, and choosing a defensible circuit model.
Example
A strong response uses energy-versus-time plots and states where the model stops being reliable.
01

The subsection's claim

Timing thresholds, sensor smoothing, component tolerances, exponential fitting, residuals, and choosing a defensible circuit model.

02

How to work with it

Start from time-constant inference. Then writing independent conservation equations and testing transient solutions at zero and infinite time. Select an equation only after its variables and assumptions match the stated system.

03

What evidence would decide

Which exponential model and time constant are supported by capacitor-voltage data? Useful evidence includes charging and discharging traces, fitted parameters, residuals, uncertainty, initial-condition checks, and component comparison.

04

Keep the boundary visible

This GioPhysics course map is an adaptable learning sequence, not academic credit, accreditation, or a universal university syllabus. Departments may redistribute weeks, laboratory hours, optics, or the modern-physics survey to match local requirements. This GioPhysics course map is an adaptable learning sequence, not academic credit, accreditation, or a universal university syllabus. Departments may redistribute weeks, laboratory hours, optics, or the modern-physics survey to match local requirements. Thermal physics appears as an unnumbered institutional extension: some universities assess it within Physics II, while others teach it in a separate course, so include the thermal extensions only where the local syllabus requires them. A result should be checked against units, signs, limiting cases, and the conditions under which its model was derived.

02

Change one variable at a time

Make the relationship visible.

Interactive model
1.00 MΩ
10 MΩ

Drag P down to 1 MΩ with R at 1 MΩ: the logged trace falls twice as fast as the dashed one, and τ/R reports half the true capacitance. Push P to 40 MΩ with R at 0.40 MΩ and the two curves lie on top of each other, with τ/R reporting 2.18 μF for an error of only −1.0%.

Interactive physics modelA 2.20 μF capacitor charged to 9.00 V discharges through a series resistor R = 1.00 MΩ while a meter input or leakage path P = 10 MΩ sits across it. The solid curve is the trace that actually gets logged: it falls with the fitted time constant 2.00 s, set by the parallel combination R ∥ P = 0.909 MΩ, and the marker drops to the time axis where the trace has fallen to 3.31 V — one time constant in. The dashed curve is the slower fall the same capacitor would show with P removed. Dividing the fitted τ by R alone therefore reports C = 2.00 μF against a true 2.20 μF, an error of −9.1 %.logged (solid): τ = 2.00 sP removed (dashed): τ = 2.20 strue C = 2.20 μFτ/R → C = 2.00 μF (−9.1 %)9.00 V3.31 V06 st = τtime / s

FITTED τ2.00 s

R ∥ P0.909 MΩ

C REPORTED AS τ/R2.00 μF

ERROR IN C-9.1 %

Live interpretationFITTED τ: 2.00 s. R ∥ P: 0.909 MΩ. C REPORTED AS τ/R: 2.00 μF. ERROR IN C: −9.1 %

03

Catch the common trap

Explain before calculating.

What should a RC design and data studio solution make visible?

Choose an answer to test the model.

04

Practice & worked examples

Reason from the model, then test the result.

01 · EasyA timing stage charges an initially uncharged capacitor from a 12.0 V rail through a series resistor, and a comparator trips when the capacitor voltage reaches 8.00 V. The resistor is 47.0 kΩ and the capacitor is 2.20 μF, both at nominal value. Find the time constant, the delay before the comparator trips, and the charging current at the instant of the trip.
  1. τ = RC = (47.0 × 10³)(2.20 × 10⁻⁶) = 0.1034 s, so 103 ms.
  2. Charging from zero, Vcap(t) = E(1 − e(−t/τ)). Setting Vcap = 8.00 V with E = 12.0 V gives e(−t/τ) = 1 − 8.00/12.0 = 1/3.
  3. t = τ ln 3 = (0.1034)(1.09861) = 0.113597 s = 114 ms. The threshold sits at exactly two-thirds of the rail, so the delay is ln 3 time constants whatever the rail voltage — it is a ratio threshold, not an absolute one.
  4. The series resistor carries the whole loop current, driven by the part of the rail the capacitor has not yet taken: I = (E − Vcap)/R = (12.0 − 8.00)/(47.0 × 10³) = 8.5106 × 10⁻⁵ A = 85.1 μA.
  5. Model check: the initial current is E/R = 255 μA, so the current has fallen to one third at the same moment the remaining voltage gap has. Both decay with the same τ, as a single exponential requires.

Answerτ = 103 ms, trip delay t = 114 ms, charging current at the trip I = 85.1 μA.

02 · EasyA logic input registers HIGH at 2.40 V and is driven by an RC delay from a 5.00 V rail with a 4.70 μF capacitor. Choose the resistor that puts the transition at 250 ms, then take the nearest E24 value (… 75, 82, 91 kΩ …) and find the delay it actually gives and the percentage error.
  1. Vcap = E(1 − e(−t/RC)) = 2.40 V with E = 5.00 V gives e(−t/RC) = 1 − 0.480 = 0.520.
  2. So t/RC = ln(1/0.520) = ln 1.92308 = 0.653926. The threshold ratio alone fixes how many time constants the delay is worth; the rail cancels out of the design.
  3. RC = 0.250/0.653926 = 0.382306 s, so R = 0.382306/(4.70 × 10⁻⁶) = 8.1342 × 10⁴ Ω = 81.3 kΩ.
  4. The nearest E24 value is 82 kΩ — 81.3 sits 0.7 from 82 and 6.3 from 75 — giving RC = (82.0 × 10³)(4.70 × 10⁻⁶) = 0.38540 s.
  5. t = (0.38540)(0.653926) = 0.252023 s = 0.252 s, which is 0.252023/0.250 − 1 = 0.809% long.
  6. Note what this error is not: it is the deliberate, known cost of using a preferred value, and it is small beside the component tolerances stacked on top of it in the next problem.

AnswerR = 81.3 kΩ ideal; with the E24 82 kΩ part the delay is 0.252 s, 0.809% long.

03 · MediumTake the delay circuit of problem 2 into production: R = 82.0 kΩ ±5% and C = 4.70 μF ±20%, with the rail and threshold held exact so the delay stays 0.653926RC. Find the nominal delay, the worst-case shortest and longest delays, and the root-sum-square spread. Then find the capacitor tolerance that would bring the worst case within ±6% of nominal.
  1. Nominal: t = 0.653926 × (82.0 × 10³)(4.70 × 10⁻⁶) = 0.252023 s. The delay is proportional to RC, so each tolerance enters as a multiplying factor and the threshold ratio never appears.
  2. Worst case long, both parts high: ×(1.05)(1.20) = ×1.2600, so t = (0.252023)(1.2600) = 0.317549 s = 0.318 s.
  3. Worst case short, both parts low: ×(0.95)(0.80) = ×0.7600, so t = (0.252023)(0.7600) = 0.191538 s = 0.192 s.
  4. Worst case assumes the two errors conspire. Treating them as independent, the fractional spreads add in quadrature: √(0.050² + 0.200²) = 0.2062, so ±20.6%, or 0.252 ± 0.0520 s.
  5. For a ±6% worst case with a ±5% resistor, the capacitor tolerance b must satisfy (1.05)(1 + b) ≤ 1.06 and (0.95)(1 − b) ≥ 0.94. The first binds, giving b ≤ 0.952%.
  6. So even a ±1% capacitor misses, at (1.05)(1.01) = +6.05%. The capacitor dominates every version of this budget; tightening the resistor alone buys almost nothing, which is why timing circuits are trimmed rather than specified.

Answert = 0.252 s nominal; worst case 0.192 s to 0.318 s; RSS ±20.6%, i.e. ±0.0520 s; needs C held to ±0.952%.

04 · MediumA capacitor marked 4.70 μF is discharged through a 100.0 kΩ resistor and the capacitor voltage is logged as (t/s, V/V): (0, 9.00), (0.200, 5.46), (0.400, 3.31), (0.600, 2.01), (0.800, 1.22). Fit ln V against t by least squares to find the time constant and the initial voltage, then find the capacitance and say what the residuals show.
  1. Take logs: ln V = ln V₀ − t/τ, a straight line of slope −1/τ. The values are 2.19722, 1.69745, 1.19695, 0.69813, 0.19885.
  2. The times are evenly spaced, so t̄ = 0.400 s and Σ(t − t̄)² = 0.400 s². The mean log is ȳ = 1.19772.
  3. Σ(t − t̄)(ln V − ȳ) = −0.999212, so the slope is −0.999212/0.400 = −2.49803 s⁻¹ and τ = 1/2.49803 = 0.400 s.
  4. Intercept: ȳ + (2.49803)(0.400) = 2.19693, so V₀ = e2.19693 = 9.00 V — fitted from all five points, not assumed from the first one.
  5. C = τ/R = 0.400315/(1.000 × 10⁵) = 4.00 × 10⁻⁶ F. That is 4.00 μF against a 4.70 μF marking, 14.8% low: inside a ±20% electrolytic tolerance, but worth recording as the part's real value.
  6. Residuals in ln V are +2.9 × 10⁻⁴, +1.2 × 10⁻⁴, −7.7 × 10⁻⁴, +1.9 × 10⁻⁵, +3.4 × 10⁻⁴ — at most 2.6 mV once converted back to volts, the size of the rounding in a three-figure reading, with no run of one sign. One exponential is enough.

Answerτ = 0.400 s, V₀ = 9.00 V, C = 4.00 μF — 14.8% below the 4.70 μF marking; residuals are rounding-sized and unpatterned.

05 · MediumA sensor output carries switching noise near 1.00 kHz and feeds a single RC low-pass filter built with a 10.0 kΩ resistor. A step in the sensor reading must settle to within 1.00% of its final value in 20.0 ms. Find the largest usable time constant, choose the largest E12 capacitor that meets it, then find the actual 1% settling time, the −3 dB frequency, the 10–90% rise time, and the gain at 1.00 kHz.
  1. The step response approaches its final value as e(−t/τ), so settling to 1.00% needs e(−t/τ) = 0.0100, that is t = τ ln 100 = 4.60517τ.
  2. τ ≤ 0.0200/4.60517 = 4.34294 × 10⁻³ s, so C ≤ 4.34294 × 10⁻³/(10.0 × 10³) = 434 nF. The largest E12 value at or below that is 390 nF.
  3. τ = (10.0 × 10³)(390 × 10⁻⁹) = 3.90 ms and the 1% settling time is (3.90)(4.60517) = 18.0 ms, inside the budget. The 10.0 kΩ must stay small against the following stage's input resistance, or that divider shifts the final value as well as τ.
  4. −3 dB frequency: fc = 1/(2πτ) = 1/(2π × 3.90 × 10⁻³) = 40.8 Hz.
  5. 10–90% rise time: t = τ ln 9 = (3.90 × 10⁻³)(2.19722) = 8.57 × 10⁻³ s = 8.57 ms.
  6. At 1.00 kHz, f/fc = 24.50 and |H| = 1/√(1 + 24.50²) = 0.0408, a factor of 24.5 down, or −27.8 dB. Smoothing and speed trade directly: any τ that rejects more noise also lengthens the settling.

Answerτ ≤ 4.34 ms; take C = 390 nF, giving τ = 3.90 ms, 1% settling 18.0 ms, fc = 40.8 Hz, rise time 8.57 ms, |H| = 0.0408 at 1.00 kHz.

06 · MediumA 2.20 μF capacitor is discharged through a 1.00 MΩ resistor while a digital multimeter of 10.0 MΩ input resistance sits across the capacitor recording the trace. The fitted time constant is 2.00 s. Find the capacitance you would report by dividing by the 1.00 MΩ resistor alone, the error that makes, the correct capacitance, and the meter resistance needed to keep the error below 1.00%.
  1. The meter is across the capacitor, so its input resistance is in parallel with the discharge resistor: the capacitor sees Rpar = (1.00)(10.0)/(1.00 + 10.0) = 0.909091 MΩ.
  2. Reporting C = τ/R = 2.00/(1.00 × 10⁶) = 2.00 × 10⁻⁶ F = 2.00 μF, which is (2.00 − 2.20)/2.20 = −9.09% against the true value.
  3. The defensible model divides by the resistance the capacitor actually discharges through: C = τ/Rpar = 2.00/(9.09091 × 10⁵) = 2.20 × 10⁻⁶ F = 2.20 μF.
  4. Equivalently C = τ(1/R + 1/Rmeter) = 2.00(1.00 × 10⁻⁶ + 1.00 × 10⁻⁷) = 2.20 μF. Conductances add, which is the cleaner way to see what the meter contributes.
  5. The measured constant is short by the factor Rmeter/(R + Rmeter), so 1.00% accuracy needs Rmeter/(R + Rmeter) ≥ 0.990, that is Rmeter ≥ 99R = 99.0 MΩ.
  6. With megohm-scale resistors that rules out an ordinary 10 MΩ meter. Use a unity-gain buffer or a ×10 probe, or keep R small enough that the meter is negligible — and either way, quote which resistance the fit was divided by.

AnswerNaive C = 2.00 μF, 9.09% low; corrected C = 2.20 μF; 1.00% accuracy needs a meter of at least 99.0 MΩ.

07 · HardA 10.00 V step is applied through a 220 kΩ resistor to a capacitor that has an internal leakage resistance across it. A plot of ln(10.00 − V) against t curves instead of running straight, and the trace settles at 9.40 V rather than 10.00 V. Refitting against the true final value, ln(9.40 − V) against t, gives a straight line of time constant 0.310 s. Find the leakage resistance, the capacitance, the capacitance an ideal-capacitor fit would have reported, and the self-discharge time constant.
  1. At long times no current flows into the capacitor, so the series resistor and the leakage resistance act as a divider: Vfinal/E = Rleak/(R + Rleak) = 9.40/10.00 = 0.940.
  2. Rleak = 0.940R/0.060 = 15.667 × 220 × 10³ = 3.4467 × 10⁶ Ω = 3.45 MΩ. The curvature was the signature: ln(10.00 − V) flattens toward ln 0.60 = −0.511 instead of falling without limit.
  3. During the transient the capacitor charges through both resistances in parallel: Rpar = R Rleak/(R + Rleak) = R(Vfinal/E) = (220 × 10³)(0.940) = 2.068 × 10⁵ Ω.
  4. C = τ/Rpar = 0.310/(2.068 × 10⁵) = 1.4990 × 10⁻⁶ F = 1.50 μF.
  5. Assuming an ideal capacitor gives C = τ/R = 0.310/(2.20 × 10⁵) = 1.41 μF, low by 6.00%. Unlike random scatter, this error does not shrink when you log more points — it is a wrong model, not a noisy one.
  6. Self-discharge: Rleak C = (3.4467 × 10⁶)(1.4990 × 10⁻⁶) = 5.17 s. That is only 16.7 times the charging constant, so any run lasting seconds must carry the leakage term explicitly rather than treat it as a slow background.

AnswerRleak = 3.45 MΩ, C = 1.50 μF; the ideal-capacitor fit gives 1.41 μF, 6.00% low; self-discharge constant Rleak C = 5.17 s.

08 · HardA monostable must hold a relay closed for between 1.80 s and 2.20 s. It releases when a capacitor charging through a resistor from a fixed rail reaches half that rail. The capacitor is 10.0 μF with ±10% tolerance plus a further ±4% drift over the operating temperature range; the resistor is ±1%. Size the resistor for a 2.00 s nominal hold, test the specification against the tolerances, and find what the capacitor would have to be.
  1. A half-rail threshold gives t = τ ln 2, so τ = 2.00/0.693147 = 2.88539 s and R = 2.88539/(10.0 × 10⁻⁶) = 2.885 × 10⁵ Ω = 289 kΩ.
  2. Tolerance and drift are separate fractional deviations that add, so the capacitor spans 0.860 to 1.140 of nominal and the resistor 0.990 to 1.010. The product RC therefore spans ×0.8514 to ×1.1514.
  3. Worst case: t = (2.00)(1.1514) = 2.30 s and t = (2.00)(0.8514) = 1.70 s. Both violate the specification, so the design fails before a single part is soldered.
  4. Re-centring the nominal cannot rescue it. The parts span a ratio 1.1514/0.8514 = 1.3524 while the specification allows only 2.20/1.80 = 1.2222, and no choice of nominal fits a wider interval inside a narrower one.
  5. Solve for the capacitor variation b that just fits: 1.80 × 1.01(1 + b) ≤ 2.20 × 0.99(1 − b) gives 3.996b ≤ 0.360, so b ≤ 9.01%. Holding the nominal at exactly 2.00 s tightens it to 1.01(1 + b) ≤ 1.10, that is b ≤ 8.91%.
  6. The fix is a component, not a formula: a ±5% film capacitor with ±2% temperature drift clears the window with margin, while the ±10% electrolytic cannot meet it without trimming R on every unit.

AnswerR = 289 kΩ; worst case 1.70 s to 2.30 s, outside the 1.80–2.20 s window; C must hold ±8.91% at a 2.00 s nominal, ±9.01% if re-centred.