Doubt regarding inductors and transformers
Well, a big feature of transformers is the huge primary inductance which opposes AC current 
Without it, the transformer would draw a lot of current even with no load. So it must have enough primary inductance to make no-load current very low.
Inductance is why 50Hz transformers are huge, but switching supplies' transformers which work at much higher frequency can be tiny.
But... how does it transform then?
Without load it works like an inductance. Primary current creates varying magnetic flux, which creates counter-emf in the primary (\$ e = - N_1 d\Phi/dt \$) which opposes mains voltage, and opposes current flow. This is how an inductance works. N is number of turns, \$ \Phi \$ is magnetic flux which is proportional to N times current.
The secondary is wound on the same core, so it gets the same flux and also produces a voltage \$ e = - N_2 d\Phi/dt \$. So you get a voltage transformer ratio \$ N_1/N_2 \$ according to number of turns.
And if current is drawn from the secondary, magnetic flux in the core is reduced in absolute value, which reduces counter-emf on the primary side, which draws in more current from mains into the primary, which brings flux back to where it was. When the transformer has a load on the secondary, the primary no longer behaves like an inductance at all. It's very simple, yet extremely useful.