RC Circuits: Charging & Discharging

The RC time constant is the metronome of the information age — usually cast as the villain. Every microscopic wire on every chip must charge the tiny input capacitance of the next transistor through its own resistance, and that RC delay (engineers use exactly this phrase) is what ultimately caps clock speeds; two decades of semiconductor innovation amount to a campaign to shrink someone's $R$ multiplied by someone else's $C$. The same arithmetic, wearing friendlier clothes, debounces every pushbutton, sets the timing of intermittent wipers (the stalk knob is literally a variable $R$ inside a $\tau = RC$), and paces touchscreen sampling. Biology got there first. A neuron's membrane is a leaky capacitor fed through the resistance of its ion channels, with a time constant around 10 ms — that RC clock sets how quickly a neuron can integrate its inputs, and it is why neural signalling tops out in the hundreds of hertz while a processor runs a billion times faster. Same equation, nine orders of magnitude apart.

Fill a bicycle tyre with a hand pump. The first strokes are easy — nothing inside pushes back. As pressure builds, every stroke fights the air already stored, and progress slows. Now relabel: pump = battery, tyre = capacitor, the narrow hose between = resistor. The capacitor's growing voltage opposes the battery, the net push shrinks as charge accumulates, and since that net push drives the current, the filling throttles itself. Self-throttling growth — rate proportional to what's left to do — is exactly what produces an exponential curve. Run the movie backwards for discharge: a charged capacitor across a resistor is its own battery. It starts by driving $V_0/R$; but every coulomb that leaves lowers its voltage, which lowers the current, which slows the leaving. The decay eats 63% of whatever remains in each time constant — fast, then slow, then a long asymptotic fade.

Two boundary facts anchor everything — engrave them as instincts before any formula. At the instant of switching, a capacitor's voltage cannot jump. Voltage is charge divided by capacitance, and charge only arrives by current flowing for some time — changing $q$ in zero time would need infinite current, which the series resistor forbids. So an uncharged capacitor holds zero volts at $t = 0^+$, dropping nothing: the circuit sees it as a plain wire, and the first-instant current is found by ignoring it. After a long time, a capacitor is an open switch. Once its voltage has risen to block any further net push, charge stops arriving; the steady current through any capacitor branch is exactly zero. Wire at the start, gap at the end — the exponential is just the smooth glide between two circuits you already know how to solve.

First, the clock. The only way to build a quantity with units of time out of a resistance and a capacitance is to multiply them:

where: $\tau$ (tau) — the time constant, in seconds when $R$ is in ohms and $C$ in farads. The units check is worth reproducing in an exam: $\Omega \times \mathrm{F} = \dfrac{\mathrm{V}}{\mathrm{A}} \times \dfrac{\mathrm{C}}{\mathrm{V}} = \dfrac{\mathrm{C}}{\mathrm{C/s}} = \mathrm{s}$. Physically, a bigger $R$ means charge is delivered more slowly, and a bigger $C$ means a bigger tank to fill — both stretch the process, and the circuit cares only about their product. A $1\ \mathrm{k\Omega}$ resistor with a $1000\ \mathrm{\mu F}$ capacitor and a $1\ \mathrm{M\Omega}$ resistor with a $1\ \mathrm{\mu F}$ capacitor tell the identical one-second story. Charging a capacitor of final charge $Q_0 = C\varepsilon$ from empty, through resistance $R$ from a battery of emf $\varepsilon$, the charge follows

where: $Q_0 = C\varepsilon$ — the final (asymptotic) charge; $e \approx 2.718$ — the exponential base your calculator carries. The voltage tracks the charge, $v_C = q/C = \varepsilon(1 - e^{-t/\tau})$, and the current is the decaying partner $i(t) = \dfrac{\varepsilon}{R}e^{-t/\tau}$: it starts at the full Ohm's-law value $\varepsilon/R$ (the capacitor is a wire at that instant) and dies to zero (open switch). At $t = \tau$ the bracket is $1 - e^{-1} = 0.632$ — the 63% landmark. Discharging from initial charge $Q_0$ (initial voltage $V_0 = Q_0/C$) through a resistor, everything decays together:

where: the voltage $v_C = V_0 e^{-t/\tau}$ and the current magnitude $i = (V_0/R)e^{-t/\tau}$ obey the same law. One reading matters for graphs: during discharge the capacitor drives the current instead of receiving it, so the current through the circuit flows in the opposite direction to the charging current — plotted on the same axes, the discharge current is the charging current flipped below the time axis. At $t = \tau$, exactly $e^{-1} = 36.8\%$ remains: the 37% landmark, the discharge twin of 63%.

The $5\tau$ row is the engineer's contract: the exponential never mathematically finishes, but within one percent is "done". A question saying "fully charged" is invoking this row — and licensing the open-switch doctrine.

Exams test the pictures as hard as the algebra. Charging: $q$ and $v_C$ are the same rising curve (differing by the factor $C$), steepest at the start, 63% at $\tau$, flattening toward the asymptote; the current is the mirror story, falling from $\varepsilon/R$. Discharging: charge, voltage and current magnitude all ride the same falling curve — but the current's sign flips, because charge now flows off the positive plate and around the loop the other way.

Two graph-reading crafts recur. The initial-slope trick: the tangent at $t = 0$ reaches the final value at exactly $t = \tau$ — a time constant read off a printed graph with a ruler, no logarithms. And the fraction check: one more $\tau$ always closes 63% of whatever gap remains. The curve has no memory of how long it has been running — a property the numericals gym exploits relentlessly.

The open-switch doctrine graduates into a full method the moment a capacitor sits inside a larger resistor network. In any settled DC circuit, no current flows through any capacitor branch. So: mentally delete every capacitor branch, solve the remaining pure-resistor circuit, then find the p.d. across the points where each capacitor connects — that p.d. is the capacitor's voltage, and $q = Cv$ finishes the job (the definition of capacitance belongs to Capacitance & the Parallel-Plate Capacitor). The same discipline handles the switching instant: replace each capacitor by a wire if uncharged (or by a small battery of its pre-existing voltage — voltage cannot jump), and solve for the initial currents. Start-circuit, end-circuit, exponential glide between: the complete anatomy of every single-capacitor problem.

Board level — one charging circuit, read three ways (A-Level / AP-2 pattern)

A 9.0 V battery of negligible internal resistance charges an initially uncharged 220 μF capacitor through a 150 kΩ resistor. Find (a) the time constant, (b) the initial current, (c) the capacitor voltage and the current at $t = 33$ s, and (d) the final charge.

Step 1. Part (a): $\tau = RC = (150 \times 10^3)(220 \times 10^{-6}) = \boxed{33\ \mathrm{s}}$. (Units: $\Omega \cdot \mathrm{F} = \mathrm{s}$ — state it.)

Step 2. Part (b): at $t = 0^+$ the uncharged capacitor is a wire, so Ohm's law rules the loop alone: $I_0 = \varepsilon/R = 9.0/(150 \times 10^3) = \boxed{60\ \mathrm{\mu A}}$.

Step 3. Part (c): $t = 33\ \mathrm{s}$ is exactly one time constant, so no calculator drama — invoke the landmarks. Voltage: $v_C = 9.0 \times (1 - e^{-1}) = 9.0 \times 0.632 = \boxed{5.7\ \mathrm{V}}$. Current: $i = 60 \times e^{-1} = 60 \times 0.368 = \boxed{22\ \mathrm{\mu A}}$. Cross-check with the loop rule: $iR + v_C = (22.1\ \mathrm{\mu A})(150\ \mathrm{k\Omega}) + 5.69 = 3.31 + 5.69 = 9.0\ \mathrm{V}$ ✓.

Step 4. Part (d): final charge $Q_0 = C\varepsilon = (220 \times 10^{-6})(9.0) = \boxed{1.98 \times 10^{-3}\ \mathrm{C} \approx 2.0\ \mathrm{mC}}$.

Competitive level — the two-instant doctrine in a network (JEE-Advanced pattern)

A 12 V battery with a series resistor $R_1 = 4\ \mathrm{k\Omega}$ feeds a junction from which two branches return to the battery: $R_2 = 8\ \mathrm{k\Omega}$, and a branch containing $R_3 = 2\ \mathrm{k\Omega}$ in series with an initially uncharged capacitor $C = 3\ \mathrm{\mu F}$. The switch closes at $t = 0$. Find (a) the battery current just after switching, (b) the battery current and the capacitor's charge after a long time, and (c) the time constant of the transient.

Step 1. Part (a), $t = 0^+$: the uncharged capacitor is a wire, so the capacitor branch is just $R_3$, sitting in parallel with $R_2$: $R_p = \dfrac{8 \times 2}{8 + 2} = 1.6\ \mathrm{k\Omega}$. Battery current: $I_0 = \dfrac{12}{4 + 1.6} = \boxed{2.14\ \mathrm{mA}}$. (If asked: the junction voltage is $12 \times 1.6/5.6 = 3.43\ \mathrm{V}$, so the capacitor branch momentarily carries $3.43/2 = 1.71\ \mathrm{mA}$.)

Step 2. Part (b), $t \to \infty$: the capacitor branch is an open switch — delete it. The circuit is $R_1$ in series with $R_2$: $I_\infty = \dfrac{12}{4 + 8} = \boxed{1.0\ \mathrm{mA}}$.

Step 3. Still part (b): the capacitor's voltage equals the p.d. across its branch's connection points. The branch spans $R_2$; with zero current through $R_3$ there is no drop across it, so the full branch p.d. lands on the capacitor: $v_C = I_\infty R_2 = 1.0 \times 8 = 8.0\ \mathrm{V}$. Charge: $q = Cv_C = 3 \times 8.0 = \boxed{24\ \mathrm{\mu C}}$.

Step 4. Part (c): the clock uses the resistance the capacitor actually sees — with the battery treated as a plain wire, the capacitor looks out through $R_3$ in series with ($R_1$ parallel $R_2$): $R_{\text{eff}} = 2 + \dfrac{4 \times 8}{12} = 2 + 2.67 = 4.67\ \mathrm{k\Omega}$, so $\tau = R_{\text{eff}}C = (4.67 \times 10^3)(3 \times 10^{-6}) = \boxed{14\ \mathrm{ms}}$. The battery current glides from 2.14 mA down to 1.0 mA on this clock. (This "resistance the capacitor sees" craft is drilled to reflex in ARCHETYPE: RC Transient Numericals.) Timing note: about 3 minutes as a full problem, under 60 seconds if only (a) and (b) are asked — which is the common MCQ form. The two-instant doctrine turns both instants into pure resistor circuits; the only step students fumble is Step 3's "no current through $R_3$ means no drop across it". Say that sentence to yourself every time.

Board scope first. No CBSE/JEE-Main/NEET tag: current NCERT-aligned syllabi stop at steady-state capacitor logic. JEE Advanced tests the full exponential machinery (and loves the two-instant doctrine); AP Physics 2 tests $\tau$, the curves, and semi-quantitative reasoning; CIE A-Level tests $\tau = RC$, the decay equations, and log-linear data handling. If you sit CBSE boards plus JEE Advanced, this chapter is Advanced-only material — budget revision accordingly.

τ is not the finish line. 63%, not 100% (see the warning box). And "fully charged" in a question is a licence to use the open-switch doctrine, not a claim about the mathematics.

Prefix crimes. $\mathrm{k\Omega} \times \mathrm{\mu F} = \mathrm{ms}$, $\mathrm{M\Omega} \times \mathrm{\mu F} = \mathrm{s}$. Two prefixes multiply; missing one shifts every answer a thousandfold. When rattled, convert everything to Ω and F.

The exponent eats a sign. Charging uses $(1 - e^{-t/\tau})$; discharging uses $e^{-t/\tau}$ alone. Writing the discharge law for a charging process gives a curve starting at 100% and "charging downward" — instantly wrong on a sketch, silently wrong in algebra.

Current direction on discharge. The discharge current runs opposite to the charging current through the circuit. Graph questions that plot signed current expect the flip below the axis; MCQ distractors show it upright.

Voltage cannot jump — current can. At a switching instant, carry the capacitor's voltage across unchanged, then let every current jump to whatever the new resistor circuit demands. Applying "cannot jump" to the current inverts the whole analysis (and will sabotage inductors later, where the protected quantity really is the current — a contrast previewed by "LR Circuits: Growth & Decay", coming in the induction chapter).

Zero current means zero drop. In steady state, a resistor in series with a capacitor drops nothing — the full branch p.d. sits on the capacitor. Forgetting this is the single most common error in steady-state network questions (Worked Example 2, Step 3).

Constants and conventions. $e^{-1} = 0.368$, $1 - e^{-1} = 0.632$; carry $\ln 2 = 0.693$ for the gym module. No $g$ appears here. A-Level data booklets supply $x = x_0 e^{-t/RC}$ — know which symbol is which before the exam, not during.

Texts. CIE 9702 A-Level, Topic 19 (capacitance: discharge equations and log-linear analysis) · AP Physics 2 CED, Unit 11 (circuits with capacitors) · H.C. Verma, Concepts of Physics Vol. 2, Ch. 32 (JEE-Advanced level transients) · Halliday, Resnick & Walker, Ch. 27.

Provenance. Downshifted from the UG treatment of RC transients (aq-em corpus), calculus replaced by the slope-proportional-to-gap argument and a numerical march; old-HS charging/discharging rows mined for hooks. A verifier should recompute: the Euler march ($\Delta t = 10^{-4}\tau$ gives $q(\tau) = 0.6321 Q_0$); Example 1 ($\tau = 33$ s, $I_0 = 60\ \mathrm{\mu A}$, $v_C(\tau) = 5.69$ V, $i(\tau) = 22.1\ \mathrm{\mu A}$, loop check $3.31 + 5.69 = 9.0$ V, $Q_0 = 1.98$ mC); Example 2 ($I_0 = 2.14$ mA, $I_\infty = 1.0$ mA, $v_C = 8.0$ V, $q = 24\ \mathrm{\mu C}$, $R_{\text{eff}} = 4.67\ \mathrm{k\Omega}$, $\tau = 14$ ms).

Prerequisites. Capacitance & the Parallel-Plate Capacitor · Resistors in Series & Parallel · Kirchhoff's Rules. Leads to. ARCHETYPE: RC Transient Numericals · LR Circuits: Growth & Decay (coming in the induction chapter) · RC Circuits (UG deep dive).