A capacitor of capacitance 500   μ F is charged completely using a de supply of 100   V . It…

Energy stored in capacitor

$E=\frac{1}{2}C{V}^2=\frac{1}{2}500\times {10}^{-6}\times {10}^4$

$\Rightarrow E=\frac{5}{2}\mathrm{J}$

Current will be maximum when whole energy of capacitor becomes energy of inductor.

$\Rightarrow \frac{1}{2}L{I}^2=\frac{5}{2}$

$\Rightarrow I=\sqrt{\frac{5}{L}}=\sqrt{\frac{5}{50\times {10}^{-3}}}=10 \mathrm{A}$