How this calculation works
Normalize ohms and farads, retain the initial capacitor voltage and evaluate exponential approach to the fixed source. A zero source with a nonzero initial voltage models discharge through the resistor.
τ = RC; Vc(t) = Vs + (Vi − Vs)e^(−t/τ); I(t) = (Vs − Vc)/RA worked example
10 kΩ and 100 µF give a one-second time constant. From 0 V toward a fixed 5 V source, capacitor voltage at one second is about 3.160603 V. At five time constants it has covered about 99.3262% of the initial voltage difference.
Good to know
One ideal resistor and capacitor with a fixed ideal DC source. No ESR, leakage, tolerances, component polarity/rating selection or safety assessment. Numerical exponential tails beyond 1,000τ are treated as zero.
Reference: Reference: model, definitions and assumptions (opens in a new tab) ↗
One time constant covers a fraction of the remaining voltage difference
The default 10 kΩ resistor and 100 µF capacitor give τ = 10,000 × 0.0001 = 1 second. Starting at 0 V with a fixed 5 V source, voltage at one second is 5(1 − e⁻¹), approximately 3.160603 V. About 36.788% of the original voltage difference remains.
At five time constants, the capacitor has covered about 99.3262% of the initial difference. Five time constants is not an exact finish time. The ideal exponential approaches its final value asymptotically, while a displayed difference may eventually round to zero.
The electrical power calculator evaluates DC voltage times current. The RC transient instead follows the evolving capacitor voltage and current after connection.
An explicit initial voltage handles charge and discharge on one page
Set source to 0 V and initial capacitor voltage to 5 V to describe discharge through the same resistor. After one time constant the capacitor has about 1.839397 V. Current is negative in the chosen into-capacitor reference direction because stored energy is being released through the resistor.
If initial and source voltage are equal, the ideal state does not change and current is zero. The 1%-error result then says it is already at the source voltage rather than inventing a wait. Signed references support the equation; they do not determine suitable polarity for a physical capacitor.
Normalize resistance and capacitance before multiplying
10 kΩ with 100 µF and 1 MΩ with 1 µF both give a one-second time constant. They do not give the same initial current for the same voltage difference, because current also depends on resistance. Equal RC products describe equal normalized voltage timing, not interchangeable component behavior. The capacitor network calculator addresses a static ideal equivalent; the RL transient calculator retains an inductor’s initial current in a different first-order model.
The Ohm’s law calculator can check the resistor voltage/current relationship at an instant. The electrical power calculator describes a separate DC voltage-times-current calculation. This page uses ideal components and does not select ratings, account for ESR or leakage, or assess an actual circuit’s electrical safety.
A few questions, answered
Does one time constant mean fully charged?
No. About 63.212% of the initial voltage difference is covered after one time constant under this ideal model.
How do I model discharge?
Use source voltage zero and a nonzero initial capacitor voltage, with the resistor providing the discharge path.
Why can current be negative?
The reference direction is into the capacitor’s positive reference terminal. Discharge from a positive initial voltage toward zero gives current in the opposite direction.
Is the final voltage reached exactly at a finite time?
The ideal nonzero exponential error approaches zero asymptotically. Rounding or the documented numerical-tail cutoff can display zero without exact finite-time settling.