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RL time constant

RL time constant: the circuit drawn with the example's values100 Ω10mH12 V
Τ = 100 µs, I = 120 mA with 100 Ω and 10 mH

A coil is a flywheel for current: slow to start, and it fights if you stop it.

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The example, worked out

Given

Resistance R
100 Ω
Inductance L
10 mH
Step voltage
12 V
Level of interest
90 %
Time to cut the current
1 µs

Working

  1. henries over ohms are seconds
  2. once the coil looks like a wire
  3. to reach 90 % of it
  4. what the field holds, and must give back
  5. the coil keeps its current flowing, whatever voltage it takes

Results

Time constant τ100 µs
Final current V / R120 mA
To 90 % (108 mA)230 µs
Kick when cut in 1 µs1.2 kV
To 63 % (τ)100 µs
To 99 % (5 τ)500 µs
Cutoff of the RL low-pass1.59 kHz
Energy in the field72 µJ

The formulas

  1. τ = L / Rhenries over ohms are seconds
  2. i(t) = (V / R)(1 − e^(−t/τ))the current climbs like an RC voltage
  3. V = L di/dtcut the current fast and the coil answers with a spike

The usual traps

  • The current, not the voltage, is what rises slowly in a coil.
  • Opening a switch on a coil without a diode across it arcs: the kick is L × I / dt, hundreds of volts.
  • A bigger R makes τ shorter, the opposite of an RC.

Questions

How do I calculate the time constant of an RL circuit?

τ = L / R: henries over ohms are seconds. i(t) = (V / R)(1 − e^(−t/τ)): the current climbs like an RC voltage. V = L di/dt: cut the current fast and the coil answers with a spike.

What does it give for typical values?

Τ = 100 µs, I = 120 mA with 100 Ω and 10 mH. Time constant τ: 100 µs; Final current V / R: 120 mA.

What are the usual mistakes?

The current, not the voltage, is what rises slowly in a coil. Opening a switch on a coil without a diode across it arcs: the kick is L × I / dt, hundreds of volts. A bigger R makes τ shorter, the opposite of an RC.

In the app

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Results rounded to standard values (E12, E24, E96) and power ratings.

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Opens in the app on 1 December, with Circuit Notebook 1.0

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