RL Time Constant Calculator
Find the time constant τ of a resistor–inductor (RL) circuit from the inductance and resistance, and see how long the current takes to rise to 63.2%, 86.5%, 95% and (effectively fully) 99.3% of its final value.
Enter Values
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How to use this calculator
- Enter the inductance L in millihenries (mH).
- Enter the series resistance R in ohms (Ω).
- Read the time constant τ (in milliseconds and seconds) and the current-rise times at 1τ, 2τ, 3τ and 5τ.
How it works
The time constant of a series RL circuit is τ = L / R with L in henries and R in ohms. Here L is converted from mH (×1e-3). When a step voltage is applied the current builds up as i = I·(1 − e^(−t/τ)), where the final current I = V/R. The current reaches 63.2% of its final value after one time constant, ~86.5% after two, ~95% after three and ~99.3% after five — so the current is treated as fully established after about 5τ. The same τ controls how the current decays once the supply is removed.
Worked example
Worked example. For L = 100 mH and R = 10 Ω: τ = 0.1 H / 10 Ω = 0.01 s = 10 ms. The current reaches 63.2% of its final value after 10 ms and is effectively fully established after 5τ = 50 ms.
Common mistakes
- Entering L in henries instead of millihenries — this tool expects mH and converts internally.
- Forgetting that τ gets LARGER as resistance falls; a low-resistance coil takes longer to reach its final current, not shorter.
- Ignoring the voltage spike produced when an inductive circuit is switched off quickly — always allow for suppression (e.g. a flyback diode).
Frequently asked questions
Why does more resistance give a shorter time constant?
Because τ = L / R, increasing R divides the inductance by a bigger number, so the current reaches its (lower) final value in less time.
What is the final current in an RL circuit?
Once fully established the inductor behaves like a plain wire, so the steady current is simply I = V / R set by the supply voltage and the total series resistance.
Is the time constant the same for current decay?
Yes. When the source is removed the current decays as i = I·e^(−t/τ), falling to 36.8% after 1τ and about 0.7% after 5τ, using the same τ = L / R.
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