When an inductor and resistor share a circuit, the current can't change instantly — it rises or falls exponentially, and the RL time constant τ = L/R sets the pace. One time constant is the time to cover about 63.2% of the change; after roughly five, the current is within 1% of its final value and the circuit is treated as settled. Note the resistance is in the denominator here, so — unlike an RC circuit — more resistance makes an RL circuit respond faster.
Reviewed: June 20, 2026 · Author: Naveen P N, Founder — AI Calculator · Verified against: the RL transient relation τ = L/R and the exponential settling sequence.
The RL time constant equations
Convert inductance to henries first — millihenries are 10⁻³ H. Dividing L by the circuit resistance gives τ in seconds; multiply by 1000 for milliseconds. The current follows i(t) = I_final(1 − e^(−t/τ)) when rising, so each time constant closes about 63% of the remaining gap. Five time constants leave under 1%, which is the usual rule for "fully settled." To slow the circuit, raise L or lower R; to speed it, do the reverse.
Worked example — an inductor switching on
Scenario: A 10 mH inductor is in series with a 100 Ω resistor. How fast does the current settle?
The time constant is 0.1 ms, so the current reaches 63.2% of its final value in 0.1 ms and is effectively settled after 0.5 ms. Drop the resistance to 10 Ω and the circuit slows tenfold to τ = 1 ms (5τ = 5 ms); raise it to 1 kΩ and it speeds up to τ = 0.01 ms. This R-in-the-denominator behaviour is the opposite of an RC circuit, where adding resistance slows things down.
Frequently Asked Questions
τ = L/R. 10 mH with 100 Ω = 0.01/100 = 0.1 ms. Convert mH to H first.
Time to reach 63.2% of the final current (or fall to 36.8%). One characteristic step of the exponential.
After ~5τ the current is >99% of final. For τ = 0.1 ms, that's 0.5 ms.
RL: τ = L/R (more R = faster). RC: τ = R·C (more R = slower). Same 63.2%/5τ shape.
Yes — τ = L/R, so doubling L doubles τ. Bigger L or smaller R = slower response.