I wanted to know, if an inductor is large, why would the current through it be constant? If an inductor is large, it means that it would be an open at moderate frequencies, and at DC the inductor is a short anyway, so how does any of that explain the constant current? Does it have something to do with the fact that the current cannot change instantaneously in an inductor? If that's the case, what does the size of the inductor have to do with that particular observation (current cannot change instantaneously)? I've attached an image of the problem. Thank you.
The key is the instruction 'Assume L1 is large enough to act as an ac open circuit at the frequency of interest ...'
'The frequency of interest' means that for timescales related to changes in the signal voltage, L1 is so large that the current does not have time to change significantly. It will change, just not very much, not enough to change the bias conditions on the transistor.
When you are concerned with changes in current, the concept of 'DC' is not very useful, there are only lower and lower frequencies, slower and slower changes. Any voltage applied to an inductor will change the current by a significant amount given long enough.
When that circuit is switched on, the L1 current rises from zero in a time given by the applied voltage and its inductance. When M1 draws a varying current, if it does so fast enough, at frequencies of interest, then the L1 current changes so little during a cycle of modulation that essentially all of the current changes go through the load. If M1 draws a different current, and stays drawing that current, eventually the L1 current will change to match the new current. The larger L1 is, the longer this will take.
When the current slows or stops, the magnetic field around the inductor will collapse and induce the current to continue to flow. An inductor opposes changes in current with the energy built up in its magnetic field.
A capacitor charges like a very fast charging battery. When the voltage increases the cap will absorb it as a charge. When the voltage drops the cap will supplement with the energy it has stored.