RL circuit with switch
think that nature is lazy, so the current will allways take the shortest path, wich means the path with less resistance. 

In this case Elliot is right, you can forget about R2 before t=0 cause the switch is closed, and you can forget L to calcul the current (that a DC analysis).
Now if you are analysing this as a real circuit (AC and trasient analysis), once you will open the switch the energy stored in L will search to continue driving the same amount of current, that is the INERTIA of L, it goes againt changes in the circuit.

So in this case we have 10V/1000 ohm = 1mA, once you open the swich you will have 1mA going for a short amount of time, ( in a real life circuit is quite more difficult and unpredictable), and as crazy as it should sound 1mA over 100K will give you 100V over R2, for a short amount of time, it depends on the size of L.
you can use the law V = L . di/dt(variation of current)

this is the reason why we put a diode, or a transil over a load in inductive circuits, to protect against overvoltages when the circuits opens or close fast.
For exemple when using Relays 

for the time of realease, if my memories are good, for RL circtuits is around 5 x L/R, and because is a logarithmic equation( the complicated one) with 1 x L/R you allready discarche 65% so at 5 you can say is all gone