From the relation \(\varepsilon=\mathrm{L} \frac{\mathrm{dI}}{\mathrm{dt}}\)
we have \(\mathrm{dt}=\mathrm{L} \frac{\mathrm{dI}}{\varepsilon}\)
Here \(\varepsilon\) is the \(\mathrm{emf}=380 \mathrm{~V}\);
L is the inductance \(=220 \mathrm{mH}\); change current \(\mathrm{dI}=2.8 \mathrm{~A}-1.1 \mathrm{~A}=1.7 \mathrm{~A}\)
Hence the minimum time required is
\(\mathrm{dt}=\left(220 \times 10^{-3} \mathrm{H}\right)\left(\frac{1.7 \mathrm{~A}}{380}\right)=0.9842 \times 10^{-3} \mathrm{~s}=0.9842 \mathrm{~ms}\)
Hence the tim e required is \(0.9842 \times 10^{-3} \mathrm{~s}\)
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