You can see according to this formula that the voltage is directly proportional to the derivative of the current. Since the derivative of a constant is equal to 0, if the current is a direct current (DC), the current across the inductor will be equal to 0. So if the current is a DC current, the current flowing through the capacitor will always be 0. This, again, is because the derivative of a constant is always equal to 0. A constant does not change. So if a user simply enters in a current such as 10A or 20A or 30A, the current will be 0, for all these values. This shows that no voltage can be across an inductor if it is connected to a DC power source. There is only voltage across an inducttor when it is connected to an AC source.
the frequency of resonance / Hz is 1/(2*pi*sqrt(L*C))
so if I apply 12v across your 11mH inductor ( taking the motor as a short for now) current will rise at 1091/amps second . In 1mS ( the pwm period) it will only get to 1.09 amps.Hence the inductor is far too large by a factor of at least 30 if we want to deal with 30 amp max.with 1kHz pwm.