Charged Capacitor discharge into LC circuit (with Freewheeling diode)
When you close the switch you have energy stored in the Cap and none in the inductor, but you have started an ideal LC resonant circuit for 1/4 cycle until the Cap voltage oscillates negative and conducts the diode heavily with the same current now as the peak current in the inductor which dissipates almost all the energy in the next half-cycle.  This is going to be an unrealistic huge current even at 10V, forget about 100 to 1kV since you have ideal reactive parts. (unrealistic)
Analyze it with  Conservation of energy with an LC resonant half-wave followed by a negative exponential L/R =T decay towards zero.
\$Ec=½ CV^2=½ LI^2 ~thus~  I^2=\dfrac{CV^2}{L}\$  meaning the initial Cap voltage becomes a peak current wave.
\$f_o=\dfrac{1}{2\pi\sqrt{LC}}=3,577Hz=1/0.28ms\$ or 280us * 1/4 wave =70us (Vpeak to 0)
let's choose Vc=10V for initial conditions, I=0
\$I^2=180uF\cdot10^2/11uH=1,636  \$... I=40.5A
So current ramps up to 40A while voltage decays to 0V asa 1/4 sine wave then the R dissipates the current with an exponent T=L/R = 11u/85m= 129ms decay up towards zero while oscillating without damping on the positive side so that doubles the decay time to 258ms (asymptote.) Then x5 = 1 second for &lt;10% residual.
Now repeat that with practical values.