RL Time Constant Calculator

The RL time constant τ equals inductance in henries divided by resistance in ohms, and inductor current reaches about 63% of its final value in one τ.

Inductance unit
Ω
V
%
Time constant τ
100µs
Time to reach target current299.573 µs
Final (steady-state) current120 mA
Stored energy at final current72 µJ
Cutoff frequency (−3 dB)1.592 kHz

Show the math

τ = L ÷ R = 10 mH ÷ 100 Ω = 100 µs
Time to 95% = −τ × ln(1 − 0.95) = 299.573 µs
Final current = 12 V ÷ 100 Ω = 120 mA
Stored energy = ½ × 10 mH × (120 mA)² = 72 µJ
Cutoff f = R ÷ (2π L) = 1.592 kHz

Rounded the same way as the result above.

How it works

An inductor resists sudden changes in current. When you switch a voltage across an inductor and resistor in series, the current does not jump to its final value; it climbs exponentially. The time constant τ = L ÷ R sets how fast.

After one τ the current is 63.2% of its final value V ÷ R; after five τ it is above 99%. The energy the coil holds at the final current is ½ L I². That stored energy is why switching an inductive load off can make a voltage spike.

τ = L ÷ R I(t) = (V ÷ R) × (1 − e^(−t/τ)) t = −τ × ln(1 − fraction) Stored energy = ½ L I² Cutoff frequency f_c = R ÷ (2π L)

Worked example

A 10 mH coil in series with 100 Ω, switched onto 12 V, with a 95% target:

  1. τ = L ÷ R = 10 mH ÷ 100 Ω = 100 µs
  2. Time to 95% = −τ × ln(1 − 0.95) = 299.573 µs
  3. Final current = 12 V ÷ 100 Ω = 120 mA
  4. Stored energy = ½ × 10 mH × (120 mA)² = 72 µJ
  5. Cutoff f = R ÷ (2π L) = 1.592 kHz
InputValue
Inductance10
Inductance unitmH
Total series resistance100 Ω
Supply voltage12 V
Target current level95 %
ResultValue
Time constant τ100 µs
Time to reach target current299.573 µs
Final (steady-state) current120 mA
Stored energy at final current72 µJ
Cutoff frequency (−3 dB)1.592 kHz

Assumptions and limits

  • Ideal inductor with all resistance lumped into one series resistor, including coil winding resistance and the source.
  • Constant-voltage step input and an inductor that starts with zero current.
  • Iron-core coils saturate at high current, which makes the real inductance drop and the current rise faster than this ideal model.

Common questions

What is the time constant of an RL circuit?

τ = L ÷ R in seconds. A 10 mH coil with 100 Ω total resistance has τ = 100 µs.

How long until an inductor is fully energized?

About five time constants, when the current is within 1% of its final value.

Why does a relay coil make a spark when it turns off?

The coil's stored energy has nowhere to go, so the current change induces a high voltage across the switch. A flyback diode across the coil gives the current a path.

Sources

  • Standard first-order RL circuit theory: τ = L/R, i(t) = (V/R)(1 − e^(−t/τ)), energy = ½LI².

Updated 2026-09-30