RC steps: initial value, final value, time constant

Three facts determine a first-order step

For a linear first-order RC circuit, identify the capacitor voltage just before the change, its long-time final voltage, and the resistance seen by the capacitor with independent sources zeroed. The time constant is $\tau=R_{seen}C$. The response is $V_C(t)=V_f+(V_i-V_f)e^{-t/\tau}$. This form handles charging and discharging without choosing a separate memorized formula.

A 3.3 V ideal step drives a 10 kΩ resistor and 100 nF capacitor to ground. Initially the capacitor is uncharged. Final voltage is 3.3 V, and τ is 1 ms. At 1 ms, voltage is $3.3(1-e^{-1})=2.085998$ V. At 5 ms it is 3.277765 V, approximately 99.326% of the final value. The voltage has not mathematically reached its final value even after five time constants.

Loading affects both endpoint and speed

Add a 10 kΩ load in parallel with the capacitor. The final voltage is now 1.65 V. With the source shorted, the capacitor sees two 10 kΩ resistors in parallel, so τ becomes 0.5 ms. A lower endpoint and faster settling arise from the same extra branch. Keeping τ = 1 ms after adding the load would be an inconsistent model.

For discharge from 3.3 V through 10 kΩ, final voltage is zero and the same 1 ms time constant applies. At 1 ms the capacitor retains 1.214002 V. The source transition must be short compared with τ for the ideal-step approximation to be accurate.

Translate a settling requirement

The remaining error fraction is $e^{-t/\tau}$. Reaching within 1% requires $t\geq4.60517\tau$, about 4.61 ms in the unloaded example. This is an error relative to the step size, not a guarantee about instrument resolution or component tolerance. A measurement record should include both the final-voltage error and the time axis. If the actual transition is slow, compare with the driven waveform rather than fitting an instantaneous step to the wrong input.