Discharging an RC circuit

Charge the capacitor to q₀, then throw the switch so that the loop contains nothing but that capacitor and a resistor R. Walk round it: the capacitor lifts the potential by q/C from its negative plate to its positive one, and the resistor drops iR in the direction the current runs, so q/C − iR = 0. Read what that says — with no source, V_R = V_C at every instant, unlike charging where ε = V_R + V_C and the two terms trade off against each other. Now the current. It is carried by charge coming off the plates, so q is falling while i runs forward, and the honest bookkeeping is i = −dq/dt. Substituting gives R dq/dt + q/C = 0. Write i = +dq/dt out of reflex and the same algebra returns q = q₀ e^(+t/RC), a capacitor that charges itself out of an empty loop.

Write the equation as dq/q = −dt/(RC) and integrate each side between its own limits: ∫ from q₀ to q of dq/q equals −(1/RC) ∫ from 0 to t of dt, so ln(q/q₀) = −t/RC and q = q₀ e^(−t/RC). Dividing by C gives V_C = V₀ e^(−t/RC), and Ohm's law on the resistor gives i = (V₀/R) e^(−t/RC). The differential equation cannot supply the initial condition; physics does. A capacitor voltage cannot jump, because dV/dt = i/C and a jump would demand infinite current, so V_C just after the switch equals V_C just before it. The current is under no such rule and does jump: with R = 50 kΩ, C = 20 μF and V₀ = 9.0 V, the resistor current goes from zero to V₀/R = 0.18 mA in the instant of contact, and τ = RC = 1.0 s.

Check that τ = RC is a time: Ω × F = (V A⁻¹)(C V⁻¹) = C A⁻¹ = s. Nothing about the starting voltage appears in it, so a capacitor charged to 100 V and the same capacitor charged to 1 V empty on identical schedules — the curves differ only in height. The milestones follow from the exponent: 36.8% of the charge is left after τ, 13.5% after 2τ, 5.0% after 3τ and 0.67% after 5τ, which is why 5τ is the usual engineering shorthand for finished. A quicker reading needs no logarithms at all. Since dV/dt at t = 0 is −V₀/RC, the tangent to the curve at the switching instant would reach zero at exactly t = τ, so a ruler laid along the start of an oscilloscope trace measures the time constant. Mathematically the decay never ends; physically it does. Here q₀ = CV₀ = 1.8 × 10⁻⁴ C is about 1.1 × 10¹⁵ electrons, and e⁻⁴⁰ = 4.2 × 10⁻¹⁸ of them is less than one whole electron.

Form the ratio of the charge at two times separated by Δt: q(t + Δt)/q(t) = e^(−Δt/RC), with t gone from the answer. The decay has no memory of how far it has already fallen, so the time to lose any fixed fraction is the same wherever you start. Take the fraction to be a half and you get a half-life, t½ = RC ln 2 ≈ 0.693 RC: from 9.0 V the fall to 4.5 V and the later fall from 4.5 V to 2.25 V take the same time, even though the current in the second stage is half the size, because the charge left to move has halved with it. This is the same mathematics as radioactive decay, dN/dt = −λN, with the decay constant λ playing the part of 1/RC and the mean lifetime the part of τ. The analogy is in the equation only: nuclear decay is a statistical law over many nuclei, while a capacitor follows the curve deterministically.

The capacitor starts with U₀ = q₀²/2C = ½CV₀². There is no source in the loop and no other element, so every joule of that must end as heat in the resistor. The integral confirms it: ∫ from 0 to ∞ of i²R dt = (V₀²/R) ∫ e^(−2t/RC) dt = (V₀²/R)(RC/2) = ½CV₀². Two things in that line are worth keeping. The power carries e^(−2t/RC), so it decays with time constant RC/2, twice as fast as the voltage; one voltage half-life therefore removes three quarters of the energy, since the energy goes as V². And R has cancelled out of the total. Halve the resistance and the initial power V₀²/R doubles while τ halves, so the resistor runs hotter for half as long and receives exactly the same ½CV₀² either way.

A decay curve is data, so use it. Two points give τ directly: V₁/V₂ = e^{((t₂ − t₁)/RC)}, so τ = (t₂ − t₁)/ln(V₁/V₂), and no value of V₀ is needed. Better, plot ln(V_C/V) against t/s; the graph is a straight line of gradient −1/τ and intercept ln(V₀/V), every point contributes, and any departure from a single exponential shows up as curvature rather than hiding. Two practical warnings. A voltmeter of input resistance Rₘ sits in parallel with R, so the τ you measure belongs to the parallel pair — with R = 1.0 MΩ and Rₘ = 10 MΩ the effective resistance is 0.91 MΩ and the measured τ is 9% short. And the capacitance printed on an electrolytic can be 20% low or 80% high, so in practice measuring τ is how you find C, not how you check it.