Reading the exponential RL graph

Problem

In an opening transient of an RL circuit, the current $i(t)$ is plotted on a semi-logarithmic scale: the graph of $\ln i$ versus $t$ is a straight line of slope $-2{,}0 \ \mathrm{s^{-1}}$. Knowing that $L = 0{,}50 \ \mathrm{H}$, find $R$.

Semi-logarithmic graph: $\ln i$ decreases linearly in time with slope $-1/\tau$.

Solution

When an RL circuit opens the current decays as $i(t) = i_0\,e^{-t/\tau}$. Taking the logarithm: $\ln i = \ln i_0 - \frac{t}{\tau}$ The line has slope $-1/\tau$. Equating to the value read off the graph: $-\frac{1}{\tau} = -2{,}0 \ \mathrm{s^{-1}} \ \Rightarrow \ \tau = \ev{0{,}50 \ \mathrm{s}}$ From $\tau = L/R$ we get the resistance: $R = \frac{L}{\tau} = \frac{0{,}50}{0{,}50} = \ev{1{,}0 \ \mathrm{\Omega}}$

Links

Topics: Induzione elettromagnetica Concepts: Induttanza e autoinduzione · Circuito RC Skills: Lettura dei grafici