Charging and Discharging of RC Circuits
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Direct answer
Throw a switch on a series RC circuit and nothing jumps except the current: at t = 0 the uncharged capacitor behaves like a plain wire (current V/R at its largest), and as charge accumulates the current decays exponentially until the capacitor behaves like an open circuit (current zero, voltage V). During charging, q = CV(1 − e^−t/RC) and i = (V/R)e^−t/RC; during discharging through a resistor, q = q₀e^−t/RC. The time constant τ = RC sets every landmark: 63.2 per cent charged at t = τ, 36.8 per cent remaining, half-life t½ = τ ln 2 = 0.693τ, and 99.3 per cent done at 5τ. And the result examiners love most: charging a capacitor through any resistor from a battery stores ½CV² but dissipates an equal ½CV² in the resistance — half the energy is always lost, independent of R.
What you must remember
- Charging equations: q = CV(1 − e^−t/RC), i = (V/R)e^−t/RC, capacitor voltage V_C = V(1 − e^−t/RC); the current starts at V/R and dies, the capacitor voltage climbs from 0 to V.
- Discharging equations: q = q₀e^−t/RC, V_C = V₀e^−t/RC, i = (V₀/R)e^−t/RC with current reversed; everything decays with the same exponent.
- Time constant: τ = RC; units check: Ω × F = s; at t = τ, charge reaches 63.2% of final while current falls to 36.8% of initial.
- Landmark arithmetic: t½ = 0.693τ; 5τ ≈ 99.3% complete; after n time constants, remaining fraction = e^−n.
- Initial and final behaviour: uncharged capacitor = short circuit at t = 0 (i = V/R); fully charged capacitor = open circuit at t = ∞ (i = 0).
- Energy split: charging to q = CV stores ½CV² and dissipates ½CV² in R, totalling the battery's output CV² — the half-loss holds for every value of R.
- Continuity principle: capacitor voltage cannot jump instantly (that would demand infinite current), so V_C just after switching equals V_C just before — the anchor for two-stage problems.
The time constant in numbers
Take R = 2 MΩ and C = 5 μF: τ = RC = 10 s. Charging from a 20 V battery, the initial current is V/R = 10 μA and the final charge CV = 100 μC; after one τ, q = 63.2 μC, after two, 86.5 μC, with half-time 6.93 s. Halve R and τ halves — but the final charge and stored energy are untouched: R sets the pace, not the destination.
Now derive, because JEE Advanced asks for the equation itself. Kirchhoff's loop law during charging gives V = q/C + R(dq/dt), and integrating from 0 to q yields q = CV(1 − e^−t/RC); discharge (battery removed) is the same equation with V = 0, q = q₀e^−t/RC. Energy bookkeeping completes the picture: the battery does W = CV² of work, the capacitor banks ½CV², and the resistor burns the difference ½CV². Graphs: q versus t rises exponentially to the asymptote CV (the initial slope V/R reaches it at exactly t = RC), and i versus t decays with the same τ.
RC traps in exams
The steepest trap is the instant-of-switching: at t = 0 the uncharged capacitor has zero voltage across it, so the full battery voltage sits on R and i = V/R; giving i = 0 applies the final-state condition to the initial instant — the two limits swapped under pressure. Second, capacitor voltage is continuous, current is not: flipping a switch can jump the current discontinuously but never V_C, and questions specifically ask for values "just after" switching to test this. Third, the half-energy result: "a capacitor is charged through a resistance R; how does the energy dissipated change if R is halved?" — not at all, always ½CV², and any R-dependent answer is the planted distractor. Fourth, one megohm times one microfarad is exactly one second — a unit marriage worth memorising. Fifth, discharging through a different resistor: the τ in the decay formula is the discharge loop's RC, not the charging one. Main tests plug-ins and percentage recognition; Advanced builds switched two-loop circuits or extracts τ from an oscilloscope half-life.
Frequently asked questions
What is the time constant of an RC circuit and why is RC a time?
τ = RC; ohms (V/A) times farads (C/V) reduces to seconds — one megohm with one microfarad gives one second.
What fraction of full charge does a capacitor reach after one time constant?
63.2 per cent while charging, from q = CV(1 − e^−1); conversely 36.8 per cent of the initial charge remains after one time constant of discharge.
Why does an uncharged capacitor behave like a short circuit at the instant of switching?
Its voltage is zero and cannot change instantly, so at t = 0 the entire source voltage appears across the resistor, driving the maximum current V/R through the loop.
How much energy is lost in the resistor while charging a capacitor?
Exactly ½CV², equal to the energy stored: the battery supplies CV² in total, and the split is independent of R.
How long does a capacitor take to become practically fully charged?
About five time constants — 99.3 per cent of CV, the remaining gap being e^−5 ≈ 0.7 per cent.