RC Capacitor Charging

Close a switch and a capacitor does not jump to the supply voltage; it approaches it. The reason is a feedback loop running the opposite way from the discharge case: the current that charges the capacitor is set by the voltage across the resistor, which is the supply minus whatever the capacitor has already reached. The fuller the capacitor gets, the smaller that difference, the smaller the current, and the slower the filling. What is left to go shrinks by the same fraction in every equal interval, which is the signature of an exponential, and the curve is $V = V_s(1 - e^{-t/\tau})$ with $\tau = RC$.

The percentages are worth memorising because they turn up everywhere: 63.2% of the supply after one time constant, 86.5% after two, 95.0% after three, 98.2% after four and 99.3% after five. A 12 V supply charging through $\tau = 2\ \text{s}$ reaches $12(1 - e^{-1}) = 7.585\ \text{V}$ at $t = 2\ \text{s}$. Note what the companion discharge page gives for the same numbers — 4.415 V — and that the two sum to exactly 12. They are the same curve, measured from opposite ends.

Inverting for time is the form most often wanted in practice: $t = \tau \ln\!\big(V_s/(V_s - V)\big)$. Reaching half the supply takes $\tau \ln 2 = 0.693\tau$, which is why a 555 timer's astable period comes out with a $\ln 2$ in it. Reaching 99% takes $\tau \ln 100 = 4.6\tau$, which is why the trade rounds to five time constants and calls it charged.

The first thing to be honest about is that the capacitor never gets there. The exponential approaches $V_s$ asymptotically, so "fully charged" is always a threshold somebody chose, and solving this page for $t$ or $\tau$ with $V$ at or above $V_s$ correctly returns nothing — there is no finite time. Three practical limits follow. The curve assumes a constant supply and a single series resistance; a supply that sags under the inrush, or a load in parallel with the capacitor, both flatten it and the capacitor settles below $V_s$ rather than at it. It assumes the capacitor started at zero, which a part that was only partly discharged did not — the general form starts from $V_0$ and closes the remaining gap, and using this page on a partly charged capacitor understates the time. And the initial current is $V_s/R$ with nothing else limiting it, which for a large capacitor across a stiff supply is an inrush measured in tens or hundreds of amps; that is what destroys switch contacts and blows fuses on power-up, and why bulk capacitance gets a soft-start.