RC charging and discharging

One second does not mean “finished”

With R = 10 kΩ and C = 100 µF, τ = 10000 × 0.0001 = 1 s. If the capacitor starts discharged and is connected to 5 V, it reaches about 3.161 V after 1 s: 63.2% of its final voltage.

After 5 s, the voltage is 4.966 V, about 99.3%. The curve approaches 5 V without reaching it exactly in a finite time. During discharge from 5 V, about 1.839 V remains after 1 s.

One time constant, two trajectories

Positive current is defined as entering the capacitor’s positive plate during charging. During discharge it is negative because it leaves that plate. The source is removed, and R completes a path between the plates.

Doubling R or C doubles τ and the time to reach any given percentage. The graph rescales its time axis to 5τ, so its shape stays the same while time labels change. Use the marker to compare precise values.

What the drawing leaves out

The resistor and capacitor are ideal. Leakage, ESR, tolerance, voltage rating and supply output resistance are not included. Switching between charge and discharge starts a separate experiment with the stated initial conditions; it does not continue the previous experiment’s state.

The animation covers five time constants in eight seconds of screen time. It is a visual aid, not real-time simulation. It starts paused and can be paused at any time.

Common mistakes

• Using µF as if they were F: 100 µF is 0.0001 F.
• Assuming the capacitor is fully charged after τ.
• Comparing curve shapes without checking the time scale.
• Mistaking negative discharge current for a calculation error.

Try it yourself