ARCHETYPE: RC Transient Numericals

In 1855, William Thomson worked out why a telegraph pulse that left Ireland crisp would arrive in Newfoundland as a long, drowsy sigh: an undersea cable is a resistor and a capacitor rolled into one object, and its signal delay grows with the square of its length — his notorious $KR^2$ law of retardation. The bankers financing the Atlantic cable were, without knowing it, betting millions of pounds on an RC transient. The physics that decided whether Victorian London could talk to New York is exactly the physics of this article — scaled down from an ocean to a lab bench. The concept module RC Circuits: Charging & Discharging gave you the machinery: the time constant $\tau = RC$ , the charging and discharging curves, the 63%/37% landmarks, and the capacitor's two disguises — a plain wire at the instant of switching, a broken wire once settled. This article is the numericals gym built on that machinery, because JEE-Advanced, AP Physics 2, and CIE A-Level all test RC transients as calculations: how long until the voltage falls to a third? What resistance does the capacitor actually see inside a messy network? Where, exactly, does the missing half of the battery's energy go? Four families cover essentially every question: the τ-ladder (times from gap ratios and one logarithm), the energy audit (a three-line ledger with a fixed toll), the network collapse (the true $\tau$ when the circuit isn't one loop), and two capacitors sharing through a resistor (where the Charge Sharing archetype's lost energy shows its hiding place). Four moves, and the transient chapter becomes timed arithmetic — which, on the Atlantic seabed, it always was.

Key takeaways

Every RC transient question is the exponential clock read backwards: $t = \tau \ln\!\left(\dfrac{\text{initial gap}}{\text{remaining gap}}\right)$ , where the "gap" is whatever distance from the final state remains.
Half-life of any RC decay: $t_{1/2} = \tau \ln 2 \approx 0.693\,\tau$ — halving chains (half, quarter, eighth…) cost one $t_{1/2}$ per step.
Charging ledger from a fixed EMF: battery pays $C\varepsilon^2$ , capacitor stores $\tfrac12 C\varepsilon^2$ , resistor burns exactly half — independent of $R$ .
Discharge: every joule of the stored $\tfrac12 C V_0^2$ ends as heat in the resistors, shared in proportion to their $I^2 R$ .
In a network, $\tau = R_{\text{seen}}\,C$ : the resistance the capacitor actually sees with sources flattened — not the nearest resistor on the diagram.
Steady states come free: capacitor = open circuit at $t \to \infty$ , = plain wire (or a battery at its initial voltage) at $t = 0^+$ .
Key idea: solve the endpoints first (t = 0 and t = ∞ are pure resistor problems), then bridge them with one exponential whose clock is τ = R_seen × C.

The core insight

Every quantity in an RC transient glides exponentially from its switch-instant value to its steady-state value with the single clock $\tau = R_{\text{seen}} C$ — so the working method is: pin both endpoints by resistor-only analysis, then read times off gap ratios with $t = \tau \ln(\text{gap}_0/\text{gap})$ .

Where this lives today

The τ-ladder is billing infrastructure now. Inside every microprocessor, each wire charging the next transistor's gate is an RC transient, and "RC delay" — engineers use exactly those letters — is a first-order limit on clock speed; copper interconnects and low-permittivity insulation exist to shave someone's $R$ times someone else's $C$ . A defibrillator is the discharge family with a patient as the resistor: dosage is how much of $\tfrac12 CV_0^2$ the pulse delivers. The intermittent wiper stalk in a car is a knob-adjustable τ, and a camera flash whines for seconds yet fires in a millisecond because its charging resistance and flash-tube resistance put two τ's four orders of magnitude apart on one capacitor. Kelvin's ocean-scale smear now lives at nanometre scale, ticking a billion times a second.

The setup

The archetype's raw material is the pair of curves from RC Circuits: Charging & Discharging, taken as given. Discharging a capacitor from voltage $V_0$ through resistance $R$ :

V(t) = V_0\, e^{-t/\tau}, \qquad \tau = RC \qquad (1)

where: $V$ — the capacitor voltage (charge $q = CV$ and discharge current $I = V/R$ ride the same curve); $\tau$ — the time constant. Charging from a battery $\varepsilon$ through $R$ obeys the mirror-image law: the voltage itself climbs as $\varepsilon(1 - e^{-t/\tau})$ , but the gap below the finish line, $\varepsilon - V$ , decays exactly like equation (1). That reframing — track the gap, not the level — is the single most useful habit in this gym, because it makes charging and discharging the same problem. Numericals never hand you $t$ and ask for $V$ ; they hand you $V$ and demand $t$ . Invert equation (1) with one logarithm:

t = \tau \ln\!\left(\frac{\text{gap at start}}{\text{gap at time } t}\right) \qquad (2)

where: for a discharge the gap is the remaining voltage (final state: zero); for a charge it is $\varepsilon - V$ (final state: full). Equation (2) is the τ-ladder: every factor-of-two shrink costs $\tau \ln 2 = 0.693\tau$ , every factor of ten costs $2.30\tau$ , and the ladder's rungs never depend on where you start — a decay takes as long to fall from 8 V to 4 V as from 2 V to 1 V.

Figure

The tau ladder — times read off gap ratios, the same for charging and discharging

The master solution pattern

Step 1. Pin the endpoints. Just after switching ( $t = 0^+$ ): every capacitor keeps the voltage it had the instant before (an uncharged one is a plain wire; a charged one acts as a little battery at its old voltage). Long after ( $t \to \infty$ ): every capacitor branch is an open circuit. Both endpoints are resistor-only problems — no exponentials yet.

Step 2. Find the clock. Flatten the sources (each EMF becomes a plain wire) and compute the resistance between the capacitor's two terminals: that is $R_{\text{seen}}$ , and $\tau = R_{\text{seen}}\,C$ . The battery's presence changes where the transient ends, never how fast it runs.

Step 3. Bridge with one exponential. Whatever quantity $x$ is asked — capacitor voltage, a branch current, a resistor's power — it glides from $x(0^+)$ to $x(\infty)$ on the clock $\tau$ . Times come from equation (2) on the gap.

Step 4. Audit energy if asked, with the ledger below — never by integrating.

The variant families

family	the move	typical ask
τ-ladder	$t = \tau\ln(\text{gap ratio})$	time to reach a stated V, q, or I
half-life chain	$0.693\tau$ per halving	"falls to 25% in…", sketch labels
energy audit (charge)	ledger: $C\varepsilon^2 = \tfrac12 C\varepsilon^2 + \tfrac12 C\varepsilon^2$	heat generated while charging
energy audit (discharge)	all of $\tfrac12 CV_0^2$ → heat, split by $I^2R$ shares	heat in each resistor
network collapse	endpoints by open/wire; $\tau = R_{\text{seen}}C$	initial/final currents, true τ
two-capacitor share	final state by charge conservation; transient on $\tau = R\,C_{\text{series}}$	common voltage, heat, current curve

The energy ledger: the half that never negotiates

Charge a capacitor from zero through any resistor to a fixed EMF $\varepsilon$ , and the books balance in three lines — no integral required, only the stored-energy formula $\tfrac12 CV^2$ from Energy Stored in a Capacitor.

Derivation AThe charging energy audit by pure bookkeeping

We derive the celebrated result that charging a capacitor from a fixed-EMF battery wastes exactly half the energy drawn, whatever the resistance — using nothing but the definition of EMF and charge conservation. The strategy is an accountant's, not a mathematician's: price what the battery does per coulomb, count the coulombs, subtract what the capacitor banks. Watch how the resistor's value never enters any line — that absence is the theorem.

Step 1. Count the delivered charge. The capacitor charges from $0$ to its full voltage $\varepsilon$ , so the total charge that flowed through the circuit is

Q = C\,\varepsilon

Step 2. Price the battery's work. A battery of EMF $\varepsilon$ does work $\varepsilon$ per coulomb it pushes, at every moment, regardless of how fast the coulombs move. Total:

W_{\text{battery}} = \varepsilon\, Q = C\,\varepsilon^2

Why is this exact and not an average? Because EMF is defined as energy handed to each coulomb; the payment schedule is flat. The resistor can stretch or squeeze the delivery time, but the per-coulomb price is fixed.

Step 3. Count what the capacitor banked — the stored-energy formula, cited from Energy Stored in a Capacitor:

U = \tfrac12\, C\,\varepsilon^2

Step 4. The difference has only one place to go — heat in the resistance:

Q_{\text{heat}} = C\varepsilon^2 - \tfrac12 C\varepsilon^2 = \tfrac12\, C\,\varepsilon^2

Exactly half, for a 1 Ω resistor or a 1 MΩ resistor: $R$ sets the pace of the burn, never the amount. Small $R$ means a violent flash of heat; large $R$ , a long warm trickle — same total. Why does the capacitor bank only half of what the battery pays? The battery pays the full $\varepsilon$ on every coulomb, but the capacitor's voltage — the price at which it banks energy — climbs from 0 to $\varepsilon$ as it fills, averaging $\varepsilon/2$ per coulomb. The gap between the flat price paid and the average price banked is the resistor's unavoidable commission. *Verified: limits — discharge run backwards gives all of $\tfrac12 CV_0^2$ as heat (nothing left to bank), consistent with Step 4 applied to a zero-EMF "battery". Numbers: $C = 200\ \mu\mathrm{F}$ , $\varepsilon = 9.0$ V gives battery 16.2 mJ, stored 8.1 mJ, heat 8.1 mJ — Example 2's ledger.*

Discharge is the ledger run in reverse with no battery on the books: the stored $\tfrac12 C V_0^2$ has nowhere to go but the resistors, so all of it becomes heat. When several resistors carry the discharge, they split the total in proportion to their $I^2 R$ — series resistors share by resistance ratio, parallel ones inversely.

The network collapse: what resistance does the capacitor see?

The commonest JEE-Advanced disguise is a capacitor buried in a resistor network. The rescue is Step 2 of the master pattern, worth its own drill. Flatten every battery to a wire, stand at the capacitor's terminals, and reduce what you see by ordinary series–parallel rules (Resistors in Series & Parallel): that single number $R_{\text{seen}}$ times $C$ is the true clock. A battery in series with 4 kΩ feeding a capacitor that also hangs across a 12 kΩ: the capacitor sees the two in parallel — 3 kΩ — because from its terminals the flattened battery puts the 4 kΩ and 12 kΩ side by side.

Figure

Network collapse — endpoints by disguise rules, clock by flattening the source

Worked examples

Board level — the τ-ladder on a discharge (A-Level pattern)

A $470\ \mu\mathrm{F}$ capacitor charged to $12.0$ V discharges through a $100\ \mathrm{k\Omega}$ resistor. Find (a) the time constant, (b) the voltage after $60$ s, (c) the time for the voltage to fall to $3.0$ V, and (d) the half-life of the decay.

Step 1. Clock: $\tau = RC = (100\times10^3)(470\times10^{-6}) = \boxed{47\ \mathrm{s}}$ .

Step 2. Voltage after 60 s — equation (1): $V = 12.0\,e^{-60/47} = 12.0\,e^{-1.277} = 12.0 \times 0.279 = \boxed{3.3\ \mathrm{V}}$ .

Step 3. Time to 3.0 V is a gap-ratio read — the gap has shrunk by $12.0/3.0 = 4$ :

t = \tau \ln 4 = 47 \times 1.386 = \boxed{65\ \mathrm{s}}

Step 4. Half-life: $t_{1/2} = 0.693\,\tau = 0.693 \times 47 = \boxed{33\ \mathrm{s}}$ . Consistency: falling to 3.0 V is two halvings ( $12 \to 6 \to 3$ ), and $2 \times 32.6 = 65$ s ✓ — the ladder's rungs stack.

Board level — charging with the full ledger (AP Physics 2 pattern)

An initially uncharged $200\ \mu\mathrm{F}$ capacitor charges from a $9.0$ V battery through $150\ \mathrm{k\Omega}$ . Find (a) the initial current, (b) the current when $v_C = 6.0$ V, (c) the time to reach $6.0$ V, and (d) the total heat dissipated by the time charging completes.

Step 1. At $t = 0^+$ the capacitor is a plain wire: $I_0 = \varepsilon/R = 9.0/150\,\mathrm{k\Omega} = \boxed{60\ \mu\mathrm{A}}$ .

Step 2. At any instant the resistor carries the gap: $I = (\varepsilon - v_C)/R = 3.0/150\,\mathrm{k\Omega} = \boxed{20\ \mu\mathrm{A}}$ . No exponential needed — Kirchhoff's loop is enough.

Step 3. Time from the gap ratio: the gap shrank from $9.0$ to $3.0$ V, a factor 3, with $\tau = (150\times10^3)(200\times10^{-6}) = 30$ s:

t = \tau \ln 3 = 30 \times 1.099 = \boxed{33\ \mathrm{s}}

Step 4. Ledger: battery pays $C\varepsilon^2 = (200\times10^{-6})(81) = 16.2$ mJ; capacitor banks $\tfrac12 C\varepsilon^2 = 8.1$ mJ; heat $= \boxed{8.1\ \mathrm{mJ}}$ — the exact, $R$ -independent half.

Competitive level — capacitor inside a divider network (JEE-Advanced pattern, ~4 minutes)

An $18$ V battery in series with $R_1 = 4\ \mathrm{k\Omega}$ feeds a node from which $R_2 = 12\ \mathrm{k\Omega}$ returns to the battery. An initially uncharged $C = 5\ \mu\mathrm{F}$ is connected across $R_2$ , and the switch closes at $t = 0$ . Find (a) the current through $R_1$ at $t = 0^+$ and as $t \to \infty$ , (b) the final capacitor voltage, (c) the time constant, and (d) the time for $v_C$ to reach half its final value.

Step 1. Endpoints first. At $t = 0^+$ the capacitor is a wire and shorts $R_2$ : $i_1(0^+) = 18/4\,\mathrm{k\Omega} = \boxed{4.5\ \mathrm{mA}}$ . As $t \to \infty$ it is open, leaving the plain divider: $i_1(\infty) = 18/16\,\mathrm{k\Omega} = \boxed{1.125\ \mathrm{mA}}$ .

Step 2. Final voltage — the divider's share across $R_2$ : $v_C(\infty) = 18 \times \dfrac{12}{16} = \boxed{13.5\ \mathrm{V}}$ .

Step 3. The clock. Flatten the battery: from the capacitor's terminals, $R_1$ and $R_2$ lie in parallel, $R_{\text{seen}} = \dfrac{4 \times 12}{16} = 3\ \mathrm{k\Omega}$ — smaller than either resistor. So $\tau = (3\times10^3)(5\times10^{-6}) = \boxed{15\ \mathrm{ms}}$ .

Step 4. Half the final value means the gap has halved: $t = \tau\ln 2 = 15 \times 0.693 = \boxed{10.4\ \mathrm{ms}}$ . Timing note: ~4 minutes, of which the only danger is Step 3 — options will include $\tau = R_1C = 20$ ms and $\tau = R_2C = 60$ ms, both precomputed traps. Endpoints (Steps 1–2) are pure resistor work; do them first and the exponential writes itself.

Competitive level — two capacitors sharing through a resistor (JEE-Advanced stretch, ~4 minutes)

A $6\ \mu\mathrm{F}$ capacitor charged to $12$ V is connected at $t=0$ , through a $50\ \Omega$ resistor, to an uncharged $3\ \mu\mathrm{F}$ capacitor. Find (a) the final common voltage, (b) the heat dissipated in the resistor, (c) the initial current and the transient's time constant, and (d) the current's half-life.

Step 1. Final state by charge conservation — the Charge Sharing & the Lost Energy result, applied: total charge $Q = 6 \times 12 = 72\ \mu\mathrm{C}$ spreads over $6 + 3 = 9\ \mu\mathrm{F}$ (parallel at the end, since currents have stopped and both hold the same voltage):

V_f = \frac{72}{9} = \boxed{8.0\ \mathrm{V}}

Step 2. Energy audit: initial $\tfrac12(6\times10^{-6})(12)^2 = 432\ \mu\mathrm{J}$ ; final $\tfrac12(9\times10^{-6})(8.0)^2 = 288\ \mu\mathrm{J}$ . The loss, $\boxed{144\ \mu\mathrm{J}}$ , is all heat in the resistor — and its size is fixed by the endpoint states alone, independent of $R$ (the archetype's "lost energy", now with a street address).

Step 3. The transient. The driving voltage at $t=0^+$ is the full mismatch $12 - 0 = 12$ V across $50\ \Omega$ : $I_0 = 12/50 = \boxed{0.24\ \mathrm{A}}$ . The capacitors, seen by the resistor, are in series around the loop: $C_{\text{eff}} = \dfrac{6\times3}{9} = 2\ \mu\mathrm{F}$ , so

\tau = R\,C_{\text{eff}} = 50 \times 2\times10^{-6} = \boxed{100\ \mu\mathrm{s}}

Step 4. Current half-life: $t_{1/2} = 0.693 \times 100\ \mu\mathrm{s} = \boxed{69\ \mu\mathrm{s}}$ . Within about $0.5$ ms (five τ) the sharing is over and $144\ \mu\mathrm{J}$ of heat has been delivered — faster for smaller $R$ , never cheaper. Timing note: ~4 minutes. Parts (a) and (b) are pure endpoint bookkeeping — do them even if the transient scares you; they usually carry most of the marks. The one non-obvious move is $C_{\text{series}}$ in Step 3: around the loop the two capacitors oppose the same current, which is the series configuration, even though they end up sharing voltage like parallel plates.

Exam traps

ln versus log. Equation (2) uses the natural logarithm. Punching $\log_{10}$ gives times short by a factor $2.303$ — and yes, that wrong answer is usually among the options.

37% of what? On a charge, it is the gap that falls to 37% at one τ — the voltage itself has risen to 63%. Attach the percentage to the decaying quantity, never to whichever number the question happens to mention.

Half-life ≠ time constant. $t_{1/2} = 0.693\,\tau$ . Sketches labelled with $\tau$ where the half-life belongs (or vice versa) lose the mark even when the curve is right.

Unit slips in τ. $\mathrm{k\Omega} \times \mu\mathrm{F} = \mathrm{ms}$ ; $\mathrm{M\Omega} \times \mu\mathrm{F} = \mathrm{s}$ . Write the units into the multiplication — a $10^3$ slip in τ poisons every later part.

The half is for fixed-EMF charging from zero. Charging from $V_1$ to $V_2$ , or through another capacitor, redistributes the ledger — recompute from endpoints; never quote "half" as a universal law. (Discharge → all stored energy becomes heat; that one is universal.)

Steady state means no capacitor current. At $t \to \infty$ , redraw with every capacitor branch snipped; solve the resistor circuit; only then read voltages. JEE loves a resistor that carries current at $0^+$ but none at $\infty$ .

Boards. JEE-Advanced tests the full machinery including networks and two-capacitor transients; AP Physics 2 wants curves, τ, endpoint reasoning, and the energy ledger; CIE A-Level (9702) runs the τ-ladder and half-life arithmetic on discharge curves. JEE-Main dropped RC dynamics with the 2024 syllabus reduction — steady-state capacitor questions survive there, so the endpoint skills of Step 1 still pay.

Origins · 1 minWilliam Thomson (Lord Kelvin) and how this was discovered

William Thomson (Lord Kelvin)

1824–1907 · Glasgow — and the floor of the Atlantic

In 1855 William Thomson put the RC transient on an industrial payroll. Analysing the proposed transatlantic telegraph, he showed that a long insulated cable under seawater is one continuous resistor-capacitor: the copper core carries resistance, the insulation against the conducting ocean stores charge, and a sharp telegraph pulse does not travel so much as soak in — smearing out as it goes. His "law of squares" (the retardation scales as $KR^2$ : capacitance times resistance, each proportional to length) said that doubling a cable quadruples its RC delay: dots and dashes crisp across the Channel would arrive across the Atlantic as overlapping sighs, at a few readable words per minute. The cable company's electrician, Wildman Whitehouse, called the analysis armchair pessimism and drove the 1858 cable with brutal induction-coil voltages; it fell silent within a month. The 1866 replacement was engineered to Thomson's arithmetic — fat copper to cut $R$ , thick gutta-percha to cut $C$ , gentle signals, and his own mirror galvanometer reading the smeared whispers at the far end. It worked, transformed world communication, and made Thomson wealthy and, eventually, Lord Kelvin. Every τ-ladder problem in this gym is his cable calculation with the ocean drained out.

History

1855

— Thomson's law of squares prices telegraph retardation as an RC effect growing with length squared

1858

— the first Atlantic cable, driven against the analysis, degrades and dies within a month

1866

— the re-engineered cable follows the arithmetic and links the continents permanently

1946

— Hodgkin and Rushton apply cable RC mathematics to nerve fibres, founding quantitative neuroscience

1980

s — RC interconnect delay overtakes transistor switching as the speed limit inside microchips

Going further

Texts. HC Verma, Concepts of Physics Vol. 2, Ch. 32 (capacitor transients) · Halliday, Resnick & Walker, Ch. 27 (RC circuits) · CIE 9702 syllabus, capacitance & discharge topics · AP Physics 2 CED, circuits unit (RC steady state and transients).

Provenance. Authored fresh for the HS E&M track's Current Electricity archetype gym; curve machinery cited from the lane-mate module RC Circuits: Charging & Discharging; stored-energy formula from Energy Stored in a Capacitor; final-state sharing result from ARCHETYPE: Charge Sharing & the Lost Energy; energy-audit and network patterns downshifted from the UG RC treatment (rewritten, algebra ledger replacing integrals). A verifier should recompute: Example 1's $\tau = 47$ s, $V(60\,\mathrm{s}) = 3.3$ V, $t_{3\mathrm{V}} = 47\ln 4 = 65$ s, $t_{1/2} = 33$ s; Example 2's $60\ \mu$ A, $20\ \mu$ A, $t = 30\ln 3 = 33$ s, ledger 16.2/8.1/8.1 mJ; Example 3's endpoints 4.5 mA → 1.125 mA, $v_C(\infty) = 13.5$ V, $\tau = 15$ ms, $t = 10.4$ ms; Example 4's $V_f = 8.0$ V, heat $144\ \mu$ J, $I_0 = 0.24$ A, $\tau = 100\ \mu$ s, $t_{1/2} = 69\ \mu$ s.

Prerequisites. RC Circuits: Charging & Discharging (curves, τ, 63/37 landmarks) · Energy Stored in a Capacitor (½CV²) · Resistors in Series & Parallel (network collapse). Leads to. LR Circuits: Growth & Decay — coming in the Electromagnetic Induction chapter, where the same exponential wears an inductor's clothes ( $\tau = L/R$ ) and the roles of voltage and current swap · RC Circuits (UG deep dive).

How do I know whether to put the voltage or the gap into the logarithm?

Ask which quantity is decaying to zero. On a discharge, the voltage itself heads to zero — use it directly. On a charge, the voltage heads to $\varepsilon$ , so the decaying quantity is the distance still to go, $\varepsilon - v_C$ ; feed that into the ratio. One habit covers both: identify the final value, form gap = (final − now), and use $t = \tau\ln(\text{gap}_0/\text{gap})$ . It even works for branch currents in networks, because every first-order quantity relaxes on the same clock toward its own steady value.

Why is exactly half the energy lost — could a very small resistor save most of it?

No, and the ledger shows why with brutal economy: the battery pays $\varepsilon$ per coulomb (fixed by the definition of EMF), while the capacitor banks at its own instantaneous voltage, which averages $\varepsilon/2$ over the fill. The difference is burned in whatever resistance exists — small $R$ just burns it faster. The loophole is changing the deal, not the resistor: charge in many small voltage steps, or through an inductor (as switching power supplies do), and the average price paid can approach the average price banked. At fixed EMF and one resistive step, half is the law.

Why do the two capacitors count as series for the time constant but parallel for the final voltage?

Two different questions, two different configurations. During the transient, one current loops through resistor and both capacitors; elements threaded by the same current are in series, so the loop's effective capacitance is $C_1C_2/(C_1+C_2)$ . At the end, current has stopped and both capacitors sit terminal-to-terminal holding one common voltage — electrically parallel, so the conserved charge spreads over $C_1 + C_2$ . Keep the timeline straight: series thinking prices the clock, parallel thinking prices the destination.

Does the resistor's value really change nothing about the shared-capacitor heat?

It changes everything about the experience and nothing about the bill. The heat equals the energy mismatch between the initial and final states — $144\ \mu$ J in Example 4 — and those states are fixed by charge conservation alone. A $5\ \Omega$ link delivers that heat in tens of microseconds at amp-scale currents; a $5\ \mathrm{M\Omega}$ link takes seconds at microamps. This is also the answer to "what if the wires are ideal, $R = 0$ ?" — then the lumped-RC story breaks down, the charge sloshes back and forth through the wires' inductance, and the mismatch is radiated away instead. Some toll booth always collects.

My calculator gives t = τ log(V0/V) a different answer than my friend's. Who is wrong?

Whoever pressed the wrong log button. The RC exponential is base- $e$ , so inverting it needs ln. The common-log version reads $t = 2.303\,\tau\log_{10}(V_0/V)$ — fine if you carry the 2.303, but nobody remembers to. Safe habit: ln for anything containing $e^{-t/\tau}$ , and sanity-check one rung: a factor-2 ratio must give $0.693\,\tau$ , never $0.301\,\tau$ . An answer mysteriously 2.3 times too small has found your button for you.

Where can I see the KR-squared cable story worked out properly?

The full treatment — distributed resistance and capacitance, pulse diffusion, delay growing as length squared — is undergraduate material; the UG article RC Circuits (UG deep dive) tells the Atlantic story with the integrals switched on. At this level, hold the scaling argument: doubling a cable doubles its total $R$ and its total $C$ , so their product — the only time in the problem — quadruples. That is the sentence Whitehouse refused to believe.