RC Circuits

Every circuit you have met so far in this chapter answered instantly: close the switch, and Ohm's law announces the current before your finger leaves the button. Add one capacitor and the circuit acquires a history. Close the switch and the current starts large and dies away; the capacitor's voltage creeps upward, fast at first, then slower and slower, never quite arriving. Ask "what is the current?" and the honest answer is now "when?" This article is about the single number that governs that story: the time constant $\tau = RC$ . It is the first appearance in this track of a theme that will follow you through inductors, oscillations, and AC circuits — that circuits with energy-storage elements obey differential equations, and their behaviour unfolds in time with a rhythm set by the product of their parts. The exponential curve you will derive here is, without exaggeration, one of the most-reused solutions in all of applied physics: it times intermittent windshield wipers and camera flashes, paces artificial pacemakers, sets the speed limit of every logic gate on every chip, and once determined how fast Victorian bankers could learn the price of cotton across the Atlantic — the story that made Lord Kelvin rich and nearly broke the first transatlantic cable. You already own every tool this needs: capacitance from the start of the chapter, Kirchhoff's loop rule, and one honest integration. What you get in return is the ability to answer when — and a precise accounting of a famous energy scandal: charging a capacitor through a resistor always wastes exactly half the battery's work, no matter what the resistor is.

Key takeaways

A resistor–capacitor loop obeys a first-order differential equation; its solutions are exponentials with time constant $\tau = RC$ — the only combination of $R$ and $C$ with units of time.
Charging toward final charge $Q_f = C\varepsilon$ : $q(t) = Q_f\,(1 - e^{-t/\tau})$ ; the current starts at $\varepsilon/R$ (capacitor acts like a wire) and decays to zero (capacitor acts like an open circuit).
The 63% ladder: after $1\tau$ , 63% of the way there; $2\tau$ , 86%; $3\tau$ , 95%; $5\tau$ , 99.3% — "about five time constants" is the practical meaning of done.
Discharge: $q(t) = Q_0\,e^{-t/\tau}$ ; the dissipated energy $\int i^2R\,dt = \tfrac12 C V_0^2$ exactly, independent of $R$ — the resistor sets the pace, not the price.
Charging through any resistor stores $\tfrac12 C\varepsilon^2$ but draws $C\varepsilon^2$ from the battery: exactly half the energy is always lost — the resolved form of the two-capacitor paradox from the energy article.
Any network with one capacitor is an RC circuit in disguise: $\tau = R_{\text{Th}}C$ , with $R_{\text{Th}}$ the Thévenin resistance the capacitor sees.
Key idea: a capacitor cannot change its voltage instantly — so every RC circuit relaxes exponentially toward its steady state, with the pace set once and for all by $\tau = RC$ .

The core insight

Charge on a capacitor in a resistive circuit cannot jump; it relaxes exponentially toward equilibrium with time constant $\tau = RC$ , and every voltage and current in the circuit inherits that same exponential clock.

Where this lives today

The RC time constant is the metronome of the information age, mostly in the role of villain. Every wire on every chip charges the tiny capacitance of the next gate through its own resistance; that "RC delay" — engineers use exactly this phrase — is what caps processor clock speeds, and a large share of semiconductor process innovation (copper interconnects replacing aluminium, low-k dielectrics replacing silica) has been a two-decade campaign to shrink someone's $R$ times someone else's $C$ . The same arithmetic, benign now, debounces every pushbutton (a keypress "rings" mechanically for milliseconds; an RC filter waits it out), sets the sampling window of touchscreens, and paces the intermittent wiper stalk in your car with a knob that is literally a variable $R$ in a $\tau = RC$ . In the body, the neuron's membrane is a leaky capacitor (~1 μF/cm²) behind the axon's series resistance, and its RC time constant (~10 ms) sets how fast synaptic inputs can be summed — Kelvin's cable mathematics, applied by Hodgkin and colleagues, remains the quantitative foundation of neural dynamics. And in the lab, the first thing an experimentalist checks when a fast signal arrives rounded is the oscilloscope probe's RC compensation — the exponential of this article, on screen, several times a day.

The physical picture

Think of filling a bicycle tyre with a hand pump. The first strokes are easy — the tyre is empty, nothing pushes back. As pressure builds, each stroke fights harder against the air already inside, and progress slows. The tyre approaches its final pressure asymptotically: the closer you get, the weaker the net push, the slower the filling. Replace tyre with capacitor, pump with battery, and the thin connecting hose with a resistor, and you have the RC charging circuit exactly. The capacitor's growing voltage opposes the battery, the net driving voltage $\varepsilon - v_C$ shrinks as charge accumulates, and since that net voltage is what drives the current through the resistor, the process throttles itself. Self-throttling growth is the signature of the exponential.

Figure

Charging an RC circuit: the capacitor voltage climbs as the current dies — both on the same clock τ = RC

Two boundary facts anchor everything, and they are worth carrying as instincts. At the instant of switching, a capacitor behaves like a plain wire. Its charge — and therefore its voltage $q/C$ — cannot change in zero time (that would take infinite current), so if it starts uncharged it momentarily holds zero volts and the circuit sees it as a short circuit: the initial current is $\varepsilon/R$ , as if the capacitor were not there. After a long time, a capacitor behaves like a break in the wire. Once fully charged, no more charge flows onto its plates; the steady-state current through any capacitor branch is exactly zero. Between these two limits — wire at $t=0$ , open circuit at $t\to\infty$ — the exponential is simply the smooth, self-throttled crossing. The same two instincts, run backwards, cover discharge: a charged capacitor dumped across a resistor starts by driving current $V_0/R$ (it is the battery now), and both its voltage and the current bleed away exponentially together. Nothing else in this article is new physics — it is the loop rule plus the definition of capacitance, watched in time.

The equation and its clock

Walk the charging loop once with Kirchhoff's loop rule (the rules article of this chapter): battery gain $\varepsilon$ , resistor drop $iR$ , capacitor drop $q/C$ . The sum vanishes,

\varepsilon - iR - \frac{q}{C} = 0 \qquad (1)

where: $q$ — the charge on the capacitor's positive plate; $i = dq/dt$ — the current delivering it. Equation (1) is not yet an answer; it is a differential equation wearing circuit clothes, because $i$ is the rate of change of $q$ . It says the current at every instant is proportional to how far the capacitor still has to go: $i = (\varepsilon - q/C)/R$ . Far from full — big drive, big current. Nearly full — small drive, trickle. This is the tyre-pump sentence written in symbols. Before solving it, extract the clock. The only way to build a time from an $R$ and a $C$ is their product,

\tau = RC \qquad (2)

where: $\tau$ (tau) — the time constant, in seconds when $R$ is in ohms and $C$ in farads (check: $\Omega\cdot\text{F} = \tfrac{\text{V}}{\text{A}}\cdot\tfrac{\text{C}}{\text{V}} = \tfrac{\text{C}}{\text{C/s}} = \text{s}$ ). Big resistor, slow delivery; big capacitor, big tank to fill — both stretch the process, and the physics depends on them only through the product. A 1 kΩ resistor with a 1000 μF capacitor and a 1 MΩ resistor with a 1 μF capacitor tell identical stories, each one second per chapter.

Derivation ACharging: solving the RC loop honestly

We want $q(t)$ from equation (1) with the capacitor initially uncharged, $q(0) = 0$ — derived by actual integration, not quoted, because this is the track's first differential equation of motion for a circuit and the method (separate, integrate, impose the initial condition) is the reusable skill. The route: rewrite (1) as a statement about the shortfall from full charge, watch the shortfall obey the simplest decay law there is, and integrate. Watch two things: where the initial condition enters, and why the current is allowed to jump at $t=0$ even though the charge is not.

Step 1. Substitute $i = dq/dt$ into (1) and isolate the derivative:

\frac{dq}{dt} = \frac{\varepsilon C - q}{RC}

Define the final charge $Q_f = \varepsilon C$ — the charge at which the capacitor's voltage balances the battery and the drive dies. The equation reads: the rate of filling is proportional to the remaining shortfall $Q_f - q$ .

Step 2. Separate the variables — all the $q$ dependence left, all the $t$ dependence right:

\frac{dq}{Q_f - q} = \frac{dt}{RC}

Why separation? Because the equation is autonomous and first-order: the rate depends only on the present state, so present state and time can be disentangled onto opposite sides and integrated independently. This one move solves every first-order relaxation problem in this track.

Step 3. Integrate both sides from the start ( $q = 0$ at $t = 0$ ) to a general moment:

-\ln\!\left(\frac{Q_f - q}{Q_f}\right) = \frac{t}{RC}

Exponentiating and solving for $q$ :

q(t) = Q_f\left(1 - e^{-t/\tau}\right), \qquad \tau = RC

Step 4. Differentiate for the current and divide by $C$ for the capacitor voltage:

i(t) = \frac{\varepsilon}{R}\,e^{-t/\tau}, \qquad v_C(t) = \varepsilon\left(1 - e^{-t/\tau}\right)

The current starts at $\varepsilon/R$ — the uncharged capacitor is momentarily invisible — and decays on the same clock the charge climbs. Why may $i$ jump while $q$ may not? The capacitor's energy lives in its charge (voltage); changing $q$ instantly would demand infinite current through $R$ , hence infinite power — forbidden. But $i$ itself is not stored anywhere in this circuit: closing the switch rearranges the driving voltage instantly, and the current follows instantly. (When we meet inductors in the next article, the roles reverse — current becomes the quantity that cannot jump.) Verified: at $t=\tau$ , $1 - e^{-1} = 0.632$ — the 63% landmark; $t=2\tau$ : 86.5%; $3\tau$ : 95.0%; $5\tau$ : 99.33% ("practically done"). Numerically integrating $dq/dt = (Q_f - q)/\tau$ by small steps reproduces the closed form to five figures; the solution satisfies (1) identically on substitution.

The exponential's landmarks deserve one paragraph of respect, because engineers speak this dialect fluently. After one time constant the gap to the finish has shrunk to $e^{-1} = 37\%$ of what it was; after each further $\tau$ , to 37% of that. Hence the ladder 63% – 86% – 95% – 99.3% at $\tau, 2\tau, 3\tau, 5\tau$ : the process never mathematically ends, but "five tau" is the working definition of finished across electronics, and "one tau" is where you check your oscilloscope cursor. The same geometry gives a drafting trick: the tangent drawn at any point of the curve reaches the final value exactly $\tau$ later — the curve always "thinks" it will finish in one more time constant, and is always wrong by the same factor. Discharge is the same mathematics with the battery deleted: the loop rule gives $iR + q/C = 0$ with $i = dq/dt$ , hence $q(t) = Q_0\,e^{-t/\tau}$ — and now everything decays together, charge, voltage, and current, each losing 63% of itself per time constant.

Where the energy goes: the exact half

Charging looks innocent, but audit the books and a scandal appears. The battery moves total charge $Q_f = C\varepsilon$ through its full EMF, doing work $W_{\text{bat}} = \varepsilon Q_f = C\varepsilon^2$ . The capacitor ends up storing — from the energy article of this chapter — only $\tfrac12 C\varepsilon^2$ . Half the energy is missing, and the resistor took it. What is remarkable is not the loss but its exactness: the missing half does not depend on $R$ at all.

Go deeper · Derivation B · 2 minThe dissipation integrals: half is not negotiable

Derivation BThe dissipation integrals: half is not negotiable

We want two results computed honestly: the total heat in the resistor during a full discharge (claim: exactly the stored energy $\tfrac12 CV_0^2$ , no matter what $R$ is), and during a full charge (claim: exactly half the battery's work, again independent of $R$ ). Both follow by integrating Joule heating $P = i^2R$ over the whole exponential — a two-line integral each — and both explain, at last, where the energy vanished in the two-capacitor paradox of the Capacitor Energy and Dielectrics article, which promised the mechanism to this article by name. Watch $R$ cancel: it sets how fast the heat is released, never how much.

Step 1. Discharge from $V_0$ : the current is $i(t) = (V_0/R)\,e^{-t/\tau}$ . Integrate the Joule power:

E_{\text{heat}} = \int_0^\infty i^2 R\,dt = \frac{V_0^2}{R}\int_0^\infty e^{-2t/\tau}dt = \frac{V_0^2}{R}\cdot\frac{\tau}{2} = \frac{V_0^2}{R}\cdot\frac{RC}{2} = \tfrac12 C V_0^2

Why did $R$ cancel? The peak power scales as $1/R$ , but the burn lasts a time $\propto R$ ; the product is fixed. A small resistor releases the energy as a violent flash, a large one as a long warm sigh — same total heat, because the source of the heat is the capacitor's fixed stockpile, and all of it must pass through the resistor on the way out.

Step 2. Charging: the current is the same exponential $i(t) = (\varepsilon/R)e^{-t/\tau}$ , so by the identical integral the heat is $\tfrac12 C\varepsilon^2$ . The battery's work is $\varepsilon\int_0^\infty i\,dt = \varepsilon Q_f = C\varepsilon^2$ . The ledger:

\underbrace{C\varepsilon^2}_{\text{battery}} = \underbrace{\tfrac12 C\varepsilon^2}_{\text{stored}} + \underbrace{\tfrac12 C\varepsilon^2}_{\text{heat}}

Exactly half, for any $R$ whatsoever — a 50% energy efficiency written into the topology itself.

Step 3. This is the two-capacitor paradox resolved as promised: connect a charged capacitor to an uncharged one and the "missing" energy is this same resistive toll, paid in whatever resistance the connecting wires have — and if the wires were truly resistanceless, the charge would slosh as an oscillation and radiate the difference away instead (the LC article picks up that thread). The energy was never mislaid; it was spent in transit. Why can't a cleverer $R(t)$ or a different resistor beat the half? For this fixed-EMF, start-from-zero topology it cannot: the battery's work per coulomb is $\varepsilon$ always, while the capacitor's average voltage over the charge delivery runs from 0 to $\varepsilon$ linearly in $q$ , averaging $\varepsilon/2$ . The loss is the gap between the price paid and the value banked, coulomb by coulomb. (Charging in many small voltage steps, or through an inductor — a switching converter's trick — changes the topology and can approach lossless charging; that is precisely why your phone charger contains inductors.) Verified: $\int_0^\infty e^{-2t/\tau}dt = \tau/2$ directly; both energy integrals recomputed numerically for $R = 10\ \Omega$ and $R = 10\ \text{M}\Omega$ with $C = 100\ \mu\text{F}$ , $V_0 = 10$ V give 5.000 mJ each — matching $\tfrac12CV_0^2$ to machine precision and confirming $R$ -independence.

Figure

The charging ledger: the battery pays a full rectangle, the capacitor banks the triangle, the resistor burns the difference — half, for any R

One capacitor, any network: the Thévenin clock

Real circuits rarely hand you a clean battery–resistor–capacitor loop. The capacitor may sit behind a divider, feed several branches, or share the board with a dozen resistors. The rescue is the equivalent-source theorem from the Series and Parallel Networks article: seen from the capacitor's two terminals, everything else — sources and resistors together — collapses to one EMF $V_{\text{Th}}$ behind one resistance $R_{\text{Th}}$ . The capacitor cannot tell the difference, so its entire behaviour is the simple loop's, with

\tau = R_{\text{Th}}\,C, \qquad v_C(\infty) = V_{\text{Th}} \qquad (3)

where: $R_{\text{Th}}$ — the resistance looking back into the network with sources quenched; $V_{\text{Th}}$ — the open-circuit voltage where the capacitor connects. This single sentence turns every one-capacitor problem on any exam into the same three-step ritual: find the final voltage (capacitor as open circuit), find $R_{\text{Th}}$ (sources quenched), write the exponential between the initial and final values. The general form worth memorizing, valid for any quantity $x(t)$ in a first-order circuit — capacitor voltage, resistor current, anything linear in them:

x(t) = x_\infty + \left(x_0 - x_\infty\right)e^{-t/\tau} \qquad (4)

where: $x_0$ — the value just after switching (found from "capacitor voltage cannot jump"); $x_\infty$ — the steady-state value (capacitor open). Start where you must, end where you're headed, glide exponentially between: equation (4) is the whole of first-order circuit theory in one line, and it works unchanged for the RL circuits of the next article with $\tau = L/R$ .

Limiting cases

case	result	meaning
$t \ll \tau$	$q \approx (\varepsilon/R)\,t\,$ : linear ramp	too early for feedback — constant-current filling
$t = \tau$	63.2% charged; $i$ down to 36.8%	the universal landmark
$t \gtrsim 5\tau$	within 1% of final	the engineer's "done"
$R \to 0$	$\tau \to 0$ : instant charge, but $\tfrac12C\varepsilon^2$ still burned	the half survives even as the flash becomes violent
$C \to \infty$	$\tau \to \infty$ , $v_C$ frozen	a huge capacitor is a temporary battery — the decoupling trick
$t=0^+$ instinct	capacitor = wire	initial currents by pure resistor analysis
$t\to\infty$ instinct	capacitor = open circuit	final state by pure resistor analysis
network, one C	$\tau = R_{\text{Th}}C$ , eq. (4)	every one-capacitor circuit is the simple loop in disguise

Common trapAfter one time constant the capacitor is charged — and other misreadings of τ

One $\tau$ leaves you at 63%, not done — a capacitor "charged for one time constant" still has 37% of its climb ahead, and timing designs that treat $\tau$ as a finish line run fast by a factor of three to five. The twin error is dimensional: students asked for a time constant often multiply the wrong pair or report $1/RC$ ; always run the units ( $\Omega \cdot \text{F} = \text{s}$ ) and sanity-check against the curve — bigger $R$ or bigger $C$ must mean slower. And keep the two non-jump rules straight between this article and the next: it is the capacitor's voltage that cannot change instantly (current may leap freely), while an inductor's current is the protected quantity. Mixing those up inverts every $t=0^+$ analysis you will ever do.

Worked examples

1 · Camera flash: charge slowly, spend instantly (confidence builder)

A photoflash charges $C = 120\ \mu\text{F}$ through $R = 10\ \text{k}\Omega$ from a 300 V converter, then dumps it through the flash tube (effective resistance $\sim 5\ \Omega$ ) when you press the shutter. Find the charging time constant, the "ready" time (95%), the stored energy, and the flash duration scale.

Step 1. Charging clock: $\tau = RC = (10^4)(1.2\times10^{-4}) = 1.2$ s. Ready at $3\tau \approx 3.6$ s — the familiar few-second whine between flashes.

Step 2. Stored energy at full charge: $U = \tfrac12CV^2 = \tfrac12(1.2\times10^{-4})(300)^2 = 5.4$ J.

Step 3. Discharge clock through the tube: $\tau_{\text{flash}} = (5)(1.2\times10^{-4}) = 0.6$ ms. The same 5.4 J that dribbled in over seconds floods out in under a millisecond — power amplification from ~1.5 W (average charging) to ~9 kW (peak flash) with no moving parts, purely by the asymmetry of the two resistances.

Step 4. Audit: charging through $10\ \text{k}\Omega$ also burned 5.4 J in the resistor (the exact half). The flash you see cost the battery 10.8 J.

Answer: $\tau = 1.2$ s, ready in ≈3.6 s; stores 5.4 J; dumps it with $\tau_{\text{flash}} = 0.6$ ms at ~kW peak power.

2 · One capacitor in a real network (exam level)

In the circuit below, a 12 V battery feeds $R_1 = 6\ \text{k}\Omega$ ; from the junction, $R_2 = 3\ \text{k}\Omega$ returns to the battery's negative terminal, and $C = 100\ \mu\text{F}$ (initially uncharged) is connected across $R_2$ . The switch closes at $t=0$ . Find $v_C(t)$ , the initial and final currents through $R_1$ , and the time for $v_C$ to reach 3.0 V.

Step 1. Final state (capacitor = open circuit): the divider carries $12/(6+3) = 1.33$ mA, so $v_C(\infty) = V_{\text{Th}} = 12 \times \tfrac{3}{9} = 4.0$ V.

Step 2. Thévenin resistance seen by $C$ (battery quenched to a wire): $R_1$ and $R_2$ in parallel, $R_{\text{Th}} = \dfrac{(6)(3)}{9} = 2.0\ \text{k}\Omega$ . Clock: $\tau = R_{\text{Th}}C = (2\times10^3)(10^{-4}) = 0.20$ s.

Step 3. Assemble by equation (4) with $v_0 = 0$ : $v_C(t) = 4.0\,(1 - e^{-t/0.20\,\text{s}})$ V.

Step 4. Currents through $R_1$ : at $t=0^+$ the capacitor is a wire, shorting $R_2$ , so $i_1(0^+) = 12/6\,\text{k}\Omega = 2.0$ mA; at $t\to\infty$ it is open, $i_1(\infty) = 12/9\,\text{k}\Omega = 1.33$ mA. (Note the current stepped down as the capacitor filled — equation (4) applies to $i_1$ too, same $\tau$ .)

Step 5. Time to 3.0 V, i.e. three-quarters of the way: $1 - e^{-t/\tau} = 0.75 \Rightarrow t = \tau\ln 4 = 0.20 \times 1.386 = 0.28$ s.

Answer: $v_C(t) = 4.0(1 - e^{-t/0.2\text{ s}})$ V; $i_1$ steps from 2.0 mA to 1.33 mA; $v_C = 3.0$ V at $t = 0.28$ s.

3 · Pacing a heart (design pass)

An implanted pacemaker times its pulses by charging a capacitor to 63% of a reference and firing. For a pulse every 0.80 s using $C = 4.7\ \mu\text{F}$ , what $R$ does the timing network need?

Step 1. Firing at 63% means firing at exactly $t = \tau$ . Demand $\tau = RC = 0.80$ s.

Step 2. $R = \tau/C = 0.80/(4.7\times10^{-6}) = 170\ \text{k}\Omega$ .

Step 3. Design reading: the comparator threshold is the 63% landmark — chips like the classic 555 timer bake the $1-e^{-1}$ arithmetic into silicon so that the period is $\ln$ -free for the designer. Precision then rests entirely on how stable $R$ and $C$ are, which is why medical and instrumentation timing uses film capacitors and metal-film resistors rather than electrolytics.

Answer: $R \approx 170\ \text{k}\Omega$ — and the 63% threshold is the time constant, exploited as a design primitive.

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

William Thomson (Lord Kelvin)

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

Thomson met the RC circuit at industrial scale before the world had a name for it. When financiers proposed a telegraph cable across the Atlantic in the 1850s, he worked out the physics of a signal in a long insulated conductor: the cable's copper core is a distributed resistance, its insulation against the surrounding seawater a distributed capacitance, and a telegraph pulse does not travel down such a line so much as diffuse — smearing out with a delay that grows as the square of the length, his 1855 "law of squares." Doubling a cable doubles both its total $R$ and its total $C$ , quadrupling the RC delay: dots and dashes that were crisp across the English Channel would arrive across the Atlantic as long, overlapping sighs. The cable's promoter, the self-taught surgeon-turned-projector Wildman Whitehouse, disputed the mathematics and drove the 1858 cable with five-kilofoot induction coils; it worked wretchedly for three weeks (Queen Victoria's 99-word inaugural telegram took sixteen hours) and then the insulation failed. The 1866 replacement was engineered to Thomson's arithmetic instead — thicker copper to cut $R$ , thicker gutta-percha to cut $C$ , and his mirror galvanometer reading microampere whispers at the far end. It made telegraphy transoceanic and Thomson wealthy and famous; the knighthood that eventually became "Lord Kelvin" traces directly to winning an argument about a time constant.

History

1745

— Leyden jar stores charge; "discharge through a conductor" becomes the first transient experiment

1827

— Ohm's law prices the resistor; with Faraday's capacitance work the RC pair's two halves exist separately

1853

— Thomson computes the oscillatory and non-oscillatory discharge of a Leyden jar — the first honest circuit differential equation

1855

— Thomson's law of squares: telegraph delay grows as length², the RC time constant at continental scale

1858

— First Atlantic cable fails in weeks, vindicating the analysis its builders ignored; the 1866 cable follows the mathematics and works

1946

— Hodgkin and Rushton measure nerve fibres as RC cables — Kelvin's telegraph equations reborn as neuroscience's "cable theory"

Going further

References. Halliday, Resnick & Walker, Fundamentals of Physics, ch. 27 (RC circuits) · Purcell & Morin, Electricity and Magnetism, 3rd ed., §4.11 · Horowitz & Hill, The Art of Electronics, 3rd ed., ch. 1 (RC timing, debouncing, and the practitioner's τ folklore) · Hodgkin & Rushton (1946), the membrane cable paper — Kelvin's mathematics in biology.

Provenance. Authored fresh for the SuperSTEM E&M track's circuits chapter (recon-§d gap heal); sources: v3 circuit pilots (rewritten to UG register) + legacy wiki circuit articles (narrative) — no AllQFT base exists for this topic. A verifier should recompute: the separation-of-variables solution and the 63/86/95/99.3 ladder; both dissipation integrals ( $\int_0^\infty e^{-2t/\tau}dt = \tau/2$ giving $\tfrac12CV^2$ independent of $R$ ); example 1's $\tau = 1.2$ s, 5.4 J, 0.6 ms; example 2's $V_{\text{Th}} = 4.0$ V, $R_{\text{Th}} = 2.0$ kΩ, $\tau = 0.20$ s, $t_{3\text{V}} = \tau\ln 4 = 0.28$ s; example 3's $R = 170$ kΩ.

Builds on: Capacitance · Capacitor Energy and Dielectrics · Series and Parallel Networks (Thévenin) · Kirchhoff's Rules Leads to: RL Circuits (the same equation with the roles swapped) · LC and RLC Oscillations · AC Circuits and Phasors (where $1/\omega C$ makes the clock continuous)

Why exactly can't the capacitor's voltage change instantly?

Voltage on a capacitor is charge divided by capacitance, and charge arrives only by current flowing for some time: $\Delta q = \int i\,dt$ . To change $q$ in zero time you would need infinite current, which the series resistance forbids (it would also mean infinite power). So $v_C$ is continuous always — that is the boundary condition you impose at every switching event. The current, by contrast, is not stored in anything in an RC circuit, so it may — and does — jump discontinuously the moment a switch changes the topology.

Does a capacitor ever actually finish charging?

Mathematically, never — the exponential approaches its asymptote without touching it. Physically, yes, and sooner than the mathematics admits: once the remaining gap is smaller than the thermal voltage noise (microvolts), "not yet arrived" has no measurable meaning. The engineering convention is that $5\tau$ (99.3%) is complete, and precision work states its own tolerance ("settled to 0.01% = $9.2\tau$ "). The never-arriving exponential is an idealization in the same spirit as the frictionless plane — exact language for a process whose tail disappears into noise.

Where does the "lost" half of the energy actually go if I use superconducting wires?

With truly zero resistance the lossless-looking circuit becomes an LC oscillator: the wires always have some inductance, so the charge overshoots, sloshes back and forth, and the energy difference is carried away as electromagnetic radiation and (in real wires) eventually as heat in whatever small resistance exists. You cannot dodge the toll by shrinking $R$ — you only change the collector. The LC and RLC Oscillations article of this chapter makes the sloshing quantitative; switching power converters dodge the half legitimately, by charging through an inductor so the "average voltage gap" argument no longer applies.

Is the 63% number special, or just a convention?

It is forced, not chosen: $1 - e^{-1} = 0.632\ldots$ is what any self-throttling process (rate proportional to remaining gap) achieves in one natural unit of its own clock. The same 63% appears in radioactive-decay problems (there it is 37% remaining), in thermal equilibration, in charging inductors, in first-order chemical kinetics — anywhere the differential equation $\dot x = (x_\infty - x)/\tau$ rules. Engineers standardized on quoting $\tau$ precisely because the fraction is universal; only the clock differs from system to system.

How do I find the initial conditions in a circuit that was already running before the switch flipped?

Two-step ritual. First, analyse the old configuration in steady state (every capacitor an open circuit) to find each capacitor's voltage just before switching. Second, carry those voltages across the switching instant unchanged — they cannot jump — and compute everything else (currents, resistor voltages) fresh in the new topology with the capacitors momentarily acting as fixed-voltage sources. The values you get are the $x_0$ of equation (4); the new steady state gives $x_\infty$ ; the new Thévenin resistance gives $\tau$ . Three numbers, one exponential, done.

Why does my oscilloscope show a rounded edge when the function generator says square wave?

Because the probe, cable, and scope input form an RC divider with the generator's own output resistance — you are watching this article's exponential, applied to every rising and falling edge. A ×10 probe with its compensation trimmer mis-set makes it worse (the trimmer exists precisely to match two RC time constants inside the probe). Rule of thumb: your measurement chain must have $\tau$ several times shorter than the fastest edge you care about, or you are measuring your instrument, not your signal.

Do the RC formulas work when the resistor and capacitor aren't literal components — like a nerve fibre or a chip interconnect?

Yes, with one honest upgrade. When resistance and capacitance are distributed along a length (cable insulation, axon membrane, on-chip wire) rather than lumped in two components, the single exponential becomes a diffusion equation — Kelvin's telegraph mathematics — and the delay grows as length². But every short segment still obeys $\tau = rc\,\Delta x^2$ locally, and engineers routinely approximate distributed lines as ladders of small lumped RC sections (the Elmore delay of chip design is exactly this). The lumped circuit of this article is the honest first term of that story, and often all you need.