RL transient: time constant

Problem

At $t = 0$ a switch closes a series circuit with $\mathcal{E} = 6{,}0 \ \mathrm{V}$, $R = 3{,}0 \ \mathrm{\Omega}$, $L = 0{,}6 \ \mathrm{H}$ (ideal inductor). (a) Write the equation governing the current and identify the time constant $\tau$. (b) Qualitatively sketch $i(t)$, indicating the tangent at $t=0$ and the asymptote. (c) After how long does the current reach $90\%$ of its final value?

Series RL circuit with battery, resistor and inductor.

Solution

(a) Equation and time constant. The loop equation gives $\mathcal{E} = R\,i + L\,\ddt{i}$, whose solution with $i(0)=0$ is: $i(t) = \frac{\mathcal{E}}{R}\left(1 - e^{-t/\tau}\right), \qquad \tau = \frac{L}{R} = \frac{0{,}6}{3{,}0} = \ev{0{,}2 \ \mathrm{s}}$

(b) Behaviour. Final value (asymptote): $i_{\max} = \frac{\mathcal{E}}{R} = \frac{6{,}0}{3{,}0} = \ev{2{,}0 \ \mathrm{A}}$ The initial tangent has slope $\ddt{i}\big|_0 = \dfrac{\mathcal{E}}{L} = 10 \ \mathrm{A/s}$ and meets the asymptote at $t=\tau$.

Exponential growth: initial tangent (blue) and asymptote $i_{\max}$.

(c) Time to 90%. $0{,}90 = 1 - e^{-t/\tau} \ \Rightarrow \ e^{-t/\tau} = 0{,}10 \ \Rightarrow \ t = \tau \ln 10 = 0{,}2 \cdot 2{,}303 \approx \ev{0{,}46 \ \mathrm{s}}$

Collegamenti

Argomenti: Electromagnetic induction Concetti: Inductance and self-induction · RC circuit · Ohm’s law