Last lesson a full capacitor emptied through a resistor. Now run the film backwards: an EMPTY capacitor is switched onto a DC supply through a resistor, and its voltage climbs. It climbs fast at first, because the whole supply voltage is pushing current through the resistor, and then slower and slower, because every volt already on the plates is a volt less doing the pushing.

$V = V_{s}\left(1 - e^{-t/\tau}\right)$, read aloud V equals V-s times one minus e to the minus t over tau. $V$ is the voltage on the capacitor after a time $t$, in volts. $V_{s}$ (say V-s, s for supply) is the voltage it is charging toward, also in volts. $t$ is the time since the switch closed and $\tau$ is the same time constant as before, $\tau = RC$, both in seconds, so the exponent comes out bare. $e$ is Euler's number, about 2.718, the base of the natural logarithm.

Set the two curves side by side. Discharging: $V_{0}\,e^{-t/\tau}$, starting full and falling to zero. Charging: $V_{s}(1 - e^{-t/\tau})$, starting at zero and rising to the supply. The exponential in the charging curve is the gap still to climb, and the gap decays exactly the way the discharge did. So the anchor facts carry over with the words swapped: after one τ the capacitor has reached 63% of the supply and 37% of the climb is left. After five τ it is charged, to within 1%. For the same supply and the same instant, the charging and discharging answers always add up to the supply. Picking the wrong curve is this lesson's named trap, and that sum is the quickest way to catch it.

Read backwards for the time, the relation becomes $t = \tau \ln\!\left(\dfrac{V_{s}}{V_{s} - V}\right)$: t equals tau times the natural log of V-s over V-s minus V, where $\ln$ is the natural logarithm, the key that undoes $e$. Note what sits underneath: the gap $V_{s} - V$, not the voltage itself. This is how every RC delay, from a relay timer to a 555, turns a threshold voltage into a number of seconds. Units guide, they do not confess: volts over volts inside the log is a bare number, as it must be, but the units will never tell you that you picked the falling curve for a rising capacitor.