RC and RL Time Constant Calculator
RC and RL Time Constant Calculator
The time constant τ of a resistor-capacitor or resistor-inductor circuit, how far it has charged or discharged after each τ, the time to reach any percentage, and the voltage and current at any moment, with the charging curve drawn from 0 to 5τ.
Time constant and step response
RC, 10 kΩ and 100 nF charged from 5 V; the time to 90% and the voltage after 1 ms
First-order step response
- τ
- time constant, in seconds: ohms × farads, or henries ÷ ohms
- x
- the capacitor voltage (RC) or the inductor current (RL)
- Xfinal, X0
- the step voltage (RC) or V/R (RL)
- e
- 2.71828…; e−1 = 0.368, so one τ leaves 36.8% to go
Worked example
RC, 10 kΩ and 100 nF charged from 5 V; the time to 90% and the voltage after 1 ms
τ = 10,000 × 100 nF = 1 ms
Time to 90% = 1 ms × ln(1 ÷ 0.1) = 1 ms × 2.3026 = 2.303 ms
At 1 ms (one τ): 1 − e−1 = 63.21%, so the capacitor is at 5 × 0.6321 = 3.161 V
The current has fallen to 0.5 mA × e−1 = 183.9 µA
What the time constant tells you
Switch a voltage onto a resistor and a capacitor in series and the capacitor does not charge at once: the current, limited by the resistor, falls as the capacitor’s voltage rises to meet the supply. The voltage follows 1 − e−t/τ, with the time constant τ = R × C. Ohms times farads is seconds: 10 kΩ and 100 nF give 1 ms. An inductor in series with a resistor does the same with current instead of voltage, with τ = L ÷ R: its current rises towards V/R. 10 mH and 100 Ω give 100 µs, and after 100 µs on 12 V the current is 75.85 mA of the final 120 mA.
The percentages. After one time constant the change is 63.21% complete, after two 86.47%, three 95.02%, four 98.17% and five 99.33%. Discharging, what is left is the complement: 36.79% after one τ, 0.67% after five. Engineers usually treat five time constants as fully settled, but the approach is exponential and never quite finishes; for a precise figure use the time-to-percentage result, τ × ln(1 ÷ (1 − p/100)) when charging. Reaching 90% takes 2.303 τ; reaching 99% takes 4.605 τ.
In real circuits. The resistance is everything in the loop: the source’s output resistance, a capacitor’s ESR and an inductor’s winding resistance all add to R. Capacitors are rarely better than ±10% and electrolytics ±20%, so a timing circuit built on τ is only that accurate. Large resistors meet the capacitor’s leakage and a meter’s input resistance. For an inductor, never break the current abruptly: the inductor drives its voltage as high as it takes to keep the current flowing, which is why relay coils carry a freewheel diode.
How this was checked. The page’s formulas were compared with the circuit’s differential equation integrated in small steps, charging and discharging, for RC and RL. The same R and C also make a filter whose cut-off is 1 ÷ (2πτ): see the RC and RLC filter calculator. For resistor values, the Ohm’s law calculator.
Frequently asked questions
What is the RC time constant?
τ = R × C, in seconds when R is in ohms and C in farads. 10 kΩ with 100 nF gives 1 ms. After one τ a charging capacitor reaches 63.21% of the supply.
How long does a capacitor take to charge fully?
About five time constants, when it reaches 99.33%. Strictly it never finishes; to 90% takes 2.303τ and to 99% 4.605τ.
What is the time constant of an RL circuit?
τ = L ÷ R. 10 mH with 100 Ω gives 100 µs; the current reaches 63.21% of V/R after one τ.
How long does a capacitor take to discharge to half?
τ × ln 2, or 0.693τ. A 10 µF capacitor with a 1 MΩ bleed resistor (τ = 10 s) falls to half in 6.931 s.
Related calculators
References
- Horowitz P, Hill W. The Art of Electronics, 3rd ed. Cambridge University Press, 2015. Ch. 1 (RC circuits, the time constant, RC low-pass and high-pass filters).
- Alexander CK, Sadiku MNO. Fundamentals of Electric Circuits, 7th ed. McGraw-Hill, 2021. Ch. 7 (first-order RC and RL circuits), Ch. 14 (frequency response, series resonance, Q and bandwidth).