When current changes from 4 \mathrm{~A} to 0 \mathrm{~A} in 0.1 \mathrm{~s} in an inductor the induced emf is found to be 100 \mathrm{~V}, then find the value of self-inductance is :
A 0.5 \mathrm{H}
B 2 \mathrm{H}
C 2.5 \mathrm{H}
D 1 \mathrm{H}
When current changes from 4 \mathrm{~A} to 0 \mathrm{~A} in 0.1 \mathrm{~s} in an inductor the induced emf is found to be 100 \mathrm{~V}, then find the value of self-inductance is :
A 0.5 \mathrm{H}
B 2 \mathrm{H}
C 2.5 \mathrm{H}
D 1 \mathrm{H}

Correct option is
C 2.5 \mathrm{H}

    \[\begin{array}{l} \Delta \mathrm{I}=(4-0)=4 \mathrm{~A} \\ \mathrm{t}=0.1 \mathrm{~s} \\ \mathrm{e}=100 \mathrm{~V} \\ \mathrm{~L}=\frac{-\mathrm{e}}{\frac{\mathrm{dI}}{\mathrm{dt}}}=\frac{-100}{\frac{4}{01}}=-\frac{10}{4}=2.5 \mathrm{H} \end{array}\]