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Without the resistor, when the transistor turns off, the current through the coil is I (t) = e^(-t*rcoil/L) (ideal diode)
As rcoil/L is generally small, the current fading time can be too long.
Yes, the voltage sources shorts . But that's a current source, which represents an open circuit.
Never say never:
1/Zth = 1/(30-j10) + 1/(40+j20) = (30+j10)/1000 + (40-j20)/2000 = 3/100 + j/100 + 2/100 - j/100 = 5/100 = 1/20
Of course, but has no effect on the current derivative at t=0+
Just think on the standard RLC: Vc + Vr + VL = 0
But VL = L I' = L Vr'/R
then Vc + Vr + (L/R)Vr' = 0
At t=0+ only the instantaneous voltage,current and L/R define Vr'
You start well:
At t=0+
Il = -Ix = 2A
Vc = 16V
Vx = -24V
Vr = 16V
(1) Vr - Vx - Vc + Vl = 0
That's OK.
After that, you derive by entering the second derivative (IL''), something that would lead you to solve the DEQ unnecessarily.
Furthermore, you have almost everything solved...
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