\(i=\varepsilon\left\{1-e^{-t / 2}\right\}\)
\(i_{saturation}\) \(=\varepsilon\)
\(80 \%\) \(i_{saturastion}\) \(=0.8\, \mathrm{\varepsilon}\)
\(0.8 \varepsilon=\varepsilon\left\{1-e^{-t / 2}\right\}\)
\(0.8=1-\mathrm{e}^{-\mathrm{t/2} } \quad ; \quad \mathrm{e}^{-\mathrm{t/2} }=0.2\)
\(e^{t/L}=5\)
\(\mathrm{t}=\mathrm{L} \ln 5=10 \times 10^{-3} \times 1.6=16 \times 10^{-3}\)