RL transient: closing the switch

Problem

A series circuit consists of a battery $\mathcal{E} = 12 \ \mathrm{V}$, a resistor $R = 4{,}0 \ \mathrm{\Omega}$ and an inductor $L = 0{,}20 \ \mathrm{H}$. At $t=0$ the switch $S$ closes. Calculate: (a) the time constant $\tau$; (b) the steady-state current $i_\infty$; (c) the current at the instant $t = 0{,}10 \ \mathrm{s}$.

Series RL circuit with battery, switch, resistor and inductor.

Solution

(a) Time constant. $\tau = \frac{L}{R} = \frac{0{,}20}{4{,}0} = \ev{0{,}050 \ \mathrm{s}}$

(b) Steady-state current. At steady state the inductor no longer opposes the flow ($\ddt{i}=0$), so only $R$ matters: $i_\infty = \frac{\mathcal{E}}{R} = \frac{12}{4{,}0} = \ev{3{,}0 \ \mathrm{A}}$

(c) Current at $t = 0{,}10 \ \mathrm{s}$. The current grows exponentially: $i(t) = i_\infty\left(1 - e^{-t/\tau}\right) = 3{,}0\left(1 - e^{-0{,}10/0{,}050}\right) = 3{,}0\left(1 - e^{-2}\right) = 3{,}0\,(1 - 0{,}135) \approx \ev{2{,}6 \ \mathrm{A}}$

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Topics: Induzione elettromagnetica Concepts: Induttanza e autoinduzione · Circuito RC · Legge di Ohm