- Inductance L (mH)
- 100mH
- Total series resistance R (Ω)
- 10Ω
10.00ms
Open with these values10.00ms
Result: 10.00 msMillihenries divided by ohms give milliseconds directly: a 100 mH coil in series with 10 Ω has a time constant of 10 ms. After one time constant the current has reached about 63 % of its final value, after five about 99 %. Use the total series resistance, coil winding included.
10.00ms
Open with these values0.47ms
Open with these values46.81ms
Open with these valuesτ = L ÷ R
| L (mH), R (Ω) | Steady state 5τ | τ in ms |
|---|---|---|
| 10, 2 | 25 ms | 5 |
| 47, 100 | 2.35 ms | 0.47 |
| 100, 10 | 50 ms | 10 |
| 220, 4.7 | 234.04 ms | 46.81 |
| 500, 1000 | 2.5 ms | 0.5 |
Divide the inductance by the resistance. In millihenries over ohms the answer comes out in milliseconds, so 100 mH and 10 Ω give 10 ms. That is how long the current takes to reach about 63 % of its final value after switch-on.
It sets how fast the current rises or falls. After 1τ the current is at about 63 % of its final value, after 2τ about 86 %, after 3τ about 95 % and after 5τ about 99 %. A larger inductance or a smaller resistance lengthens τ and slows the change.
After five time constants the current is within about 1 % of its final value, which engineers treat as settled. It is a working convention, not a physical boundary — the curve never quite arrives. For 100 mH and 10 Ω that means roughly 50 ms.
Yes, use the total series resistance: the inductor's winding resistance plus any external resistor. Leaving the coil out makes R too small and the computed time constant too long. Measure the coil's DC resistance and add it.
Millihenries. A 1 H coil is 1000 mH and a 100 µH coil is 0.1 mH. Entering raw henries would give an answer a thousand times too large.
Information, not professional advice.
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