Assuming no losses, the energy in the capacitor will go into the inductor.
0.5 * C * V*V = 0.5 * L * I*I so I = V * sqrt(C/L)
Taking 300V, 200uF, E=9J, with L=1.5H, Ipeak=3.5A
The current will continue until the capacitor is charged in the reverse direction which is disastrous if it...