Here's the task : calculate iL(t) after switching in the given circuit (key closes at t=0)

Data : e(t)=20(sqrt2)sin(t+45°)
R= 2Ω
L=1H
C=1/2 F

1. Initial conditions iL(0-) :

2. Steady state after switching iLU(0+) :

So first of all :

And now how to calculate the current iL ?Do we loose any current on the wire with key or not and iL will be the same as in the first step ?
The rest of the solution will follow when I'll get through this part.

 

Firstly your initial condition calculation will be incorrect as you have incorrectly written the effective pre-switch series impedance as 4+j/2.

Xc=1/(ωC)=1/(1*0.5)=2

So Z=Rtot+jXL-jXc=4+j-2j=4-j

For the transient analysis case at t>0, you can not use the steady state complex number analysis you have shown. You must obtain the transient solution either by using differential equations or Laplace transform. Also note there there are two initial conditions in the analysis for t>0 - the capacitor voltage and the inductor current at t=0.

 

I know that if the excitation is sinusoidal, we change approach of solving problem using Laplace method (so instead of initial and then transient as a whole we calculate initial, steady state after switching, natural response and final response). But I was pretty sure that with DC we always need initial, steady state after switching, then natural response after switching and elimination of all external sources and finally the final response. And from what I was able to find in my notes, differential equations are used to calculate natural response. And step two of my solution is just the steady state after switching.

 

Once the switch is closed it's a transient condition.

Your approach concerning the conditions at t=0+ is incorrect.

The conditions for iL and Vc in the remainder of the circuit isolated from the source by the switch closure, will be the same from t=0- to t=0+. You will have the values for Vc(0+) and iL(0+) from your analysis at t=0-. Remember that neither Vc nor iL can change instantaneously.

After t=0 the source only effects the current in the top left hand R.