Question

The maximum allowable potential difference across a 220 mH inductor is 380 V . You need to raise the current through the inductor from 1.1 A to 2.8 A .

What is the minimum time you should allow for changing the current?
t = ______ seconds

Answer 1

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 \(0.9842 \times 10^{-3} \mathrm{~s}\)