RC Capacitor Discharge

A capacitor discharging through a resistor loses the same fraction of what remains in every equal interval, not the same number of volts. The reason is a short feedback loop: the voltage on the capacitor is what drives current through the resistor, that current is what removes charge, and removing charge is what lowers the voltage. As the voltage falls the current falls with it, so the discharge slows exactly in step with its own progress. Any quantity whose rate of decrease is proportional to itself decays as $e^{-t/\tau}$, and here $\tau = RC$. After one time constant 36.8% of the original voltage is left, after two 13.5%, after three 5%, after five 0.7%.

Real numbers make the point better than percentages. A camera flash or a switch-mode supply may hold 470 µF at 400 V, bled off through a 1 MΩ resistor. That gives $\tau = 470\ \text{s}$, close to eight minutes. Five minutes after the unit is unplugged the capacitor is still at $400\,e^{-300/470} = 211\ \text{V}$; after a full ten minutes it is at 111 V, which will still hurt you. Inverting the relation gives the useful form: the time to fall to a chosen voltage is $t = \tau \ln(V_0/V)$, so reaching a nominally safe 50 V from 400 V takes $470 \times \ln 8 = 977\ \text{s}$, sixteen minutes.

The mathematics is identical to radioactive decay, and the half-life language transfers directly: the voltage halves every $\tau \ln 2 = 0.693\tau$, regardless of where you start counting. That is often the easier mental model — six and a bit half-lives to reach 1%. The same exponential governs the RC charging curve, thermal cooling under Newton's law, and the settling of a pressure transient in a pipe; whenever a store discharges through a restriction, this is the shape you get.

The dangerous misunderstandings here are all about "empty". Mathematically the capacitor never reaches zero, so any statement that it is discharged is a statement about a threshold someone chose. Worse, a large capacitor that has been shorted out and released will climb back up on its own — often to tens of volts — as charge trapped in the dielectric relaxes out of it. That effect is called dielectric absorption, or soakage, and it is why service procedures call for a bleeder resistor left in place rather than a screwdriver across the terminals, and why you measure before you touch rather than assuming. Two smaller traps: the exponential assumes the only discharge path is $R$, so a real capacitor's own leakage and any parallel load shorten τ below the value you calculated; and when solving this page for $t$ or $\tau$ the logarithm demands $V$ below $V_0$, because a discharging capacitor only ever loses voltage.