RC Time Constant Calculator
Enter R, C, supply voltage Vs and time t to get τ=RC and the capacitor voltage Vc during charging and discharging. R=10 kΩ, C=100 µF → τ=1 s; at t=1 s a 12 V supply reaches ≈7.58 V (≈63.2%).
Input Data
Results
At a glance:The RC time constant (symbol τ, 'tau') characterises how fast a series resistor–capacitor circuit charges and discharges — one of the most fundamental parameters in analog electronics. RC circuits are everywhere: low/high/band-pass filters, 555-timer pulse width and oscillation, ADC sample-and-hold capacitors, power-supply decoupling caps, differentiating/integrating circuits, motor start caps, and even the cell-membrane time constant in biomedicine. The constant is simply τ=R·C (R in Ω, C in F, so τ in seconds because Ω·F = C/A = s). Physically, a capacitor initially uncharged, connected through R to an ideal source Vs, charges as Vc(t)=Vs·(1−e^(−t/τ)); a capacitor charged to V₀, discharged through R to 0 V, decays as Vc(t)=V₀·e^(−t/τ). The two curves are complementary: charge % + remaining % = 100%. Rules of thumb: at t=τ, ≈63.2% charged / 36.8% remains; 2τ → 86.5%/13.5%; 3τ → 95.0%/5.0%; 4τ → 98.2%/1.8%; 5τ → 99.3%/0.7%. The '5τ rule' says after about five time constants the capacitor is effectively full or empty (<1% error). Example: R=10 kΩ=10000 Ω, C=100 µF=1e-4 F → τ=1 s; at t=1 s with Vs=12 V: Vc=12(1−e⁻¹)≈7.585 V charged, remaining ≈4.415 V. Related ideas: (1) RC low-pass cutoff f_c=1/(2πRC)=1/(2πτ); (2) 555 one-shot pulse T≈1.1RC; (3) 555 astable: T1=0.693(RA+RB)C, T2=0.693RB·C; (4) decoupling caps short high-frequency supply noise to ground; (5) ADC sample-and-hold needs switch-on ≥7–10τ for <1 LSB error; (6) the same first-order equation models thermocouple response, drug half-life (0.693τ), and population decay.
Formula
Time constant: τ = R·C (s)
Charging: Vc(t) = Vs·(1 − e^(−t/τ))
Discharging: Vc(t) = V₀·e^(−t/τ)
RC low-pass cutoff: f_c = 1/(2πRC); 5τ rule
$$\tau = RC, \quad V_c(t) = V_s (1-e^{-t/\tau})$$How to Use
- Enter series resistance R (Ω) and capacitance C (F; mind µF conversion).
- Enter supply/initial voltage Vs (charging) or V₀ (discharging).
- Enter query time t (s); the tool outputs τ, charging Vc, and remaining Vc.
Case Studies
10 kΩ × 100 µF at t=τ=1 s
R=10000 Ω, C=100 µF=1e-4 F, Vs=12 V, t=1 s.
τ=R·C=10000×0.0001=1.000 s.
Charging Vc=12(1−e⁻¹)≈7.585 V; remaining=12·e⁻¹≈4.415 V.
Content review: Calculatorism Science Team. Results are for reference only; please refer to the relevant authorities for the official figures.