The energy stored in a \(50 \mathrm{mH}\) inductor carrying a current of \(4 \mathrm{~A}\) is
The energy stored in a \(50 \mathrm{mH}\) inductor carrying a current of \(4 \mathrm{~A}\) is
\(0.4 \mathrm{~J}\)
\(4.0 \mathrm{~J}\)
\(0.8 \mathrm{~J}\)
\(0.04 \mathrm{~J}\)
Solution
Given,
\(\begin{aligned}
L=50 \mathrm{mH} & =5 \times 10^{-2} \mathrm{H} \\
I & =4 \mathrm{~A}
\end{aligned}\)
\(\therefore\) Energy stored in the inductor is given as
\(E=\frac{1}{2} L I^2=\frac{1}{2} \times 5 \times 10^{-2} \times 4^2=0.4 \mathrm{~J}\)