Control an inductive load with VCO
The rule for inductors is V = L * di/dt.  So, at 100 KHz, when you turn on for the first cycle, you have V of 12 volts across the inductor.  With L = 0.1H, di/dt = 120 amps per second.  At 100 KHz and 90% duty cycle, your pulse width is 9 usec, so your current is a ramp starting at zero and increasing to 9 usec * 120 a/sec or about 1.08 milliamps peak.  At this low first-pulse current, the 2-ohm resistor has negligible effect.

Then the FET opens up, and the voltage on the drain will swing positive to 12.7 volts where the diode starts conducting.  Then the inductor will discharge through the 20-ohm resistance, and the current will drop until it either stops conducting (discontinuous mode) or the FET switch turns back on and the current ramps up from that point.

For an ideal inductor, you can calculate a point where the inductor current increase during the  "on" part of the cycle charged through the 2-ohm resistor will equal the decrease during the "off" cycle when the current in the inductor discharges through the diode and resistor.  The current waveform will have ripple at your switching frequency for this reason.