Given i(t), find the inductance

Problem

In the closing transient of an RL circuit the current reaches $63\%$ of the steady-state value after $0{,}40 \ \mathrm{s}$. Knowing that $R = 20 \ \mathrm{\Omega}$, find the inductance $L$.

Solution

In the exponential growth $i(t) = i_\infty\left(1 - e^{-t/\tau}\right)$, the value $63\%$ corresponds to $1 - e^{-1} \approx 0{,}63$, i.e. to $t = \tau$. So the time constant is $\tau = 0{,}40 \ \mathrm{s}$. From $\tau = L/R$ we get: $L = \tau R = 0{,}40 \cdot 20 = \ev{8{,}0 \ \mathrm{H}}$

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Topics: Induzione elettromagnetica Concepts: Induttanza e autoinduzione · Circuito RC Skills: Impostazione simbolica