Inductors resist changes in current, so the current will take a while to build up from zero (with switch open) to its final value. So, just after the switch is closed there is negligible current in the circuit. First figure out what the voltage drop across the resistor is when the current is negligible :-) . Next, subtract that value from the battery voltage to get the voltage across the inductor. Good luck.
Do your own homework. Use some calculus skills.
For the circuit Vb Vr + Vl At t0+ the current in the resistor is zero (or nearly) so the voltage across the resistor is 0v. At time0 then VlVb At tinfinity the current has risen to Vb/R. This is the current in the inductor. The magnetic energy stored for the steady state is U 1/2 * L * i^2 (I get U 3.2E-4 J)
Just to supplement the first guys answer. at tinfinity, the circuit has reached steady state. In a steady state DC circuit, the inductors magnetic field is no longer changing so there is no induced emf. Therefore the inductor basically behaves as a short circuit so all the voltage Vb is across the resistor.
