Please excuse my poorly drawn circuit. It has been in its current position for a long time, and at t=0 the switch is moved. I have to find the voltage of the capacitor and inductor at t=0, and the current of the inductor and capacitor at t=0, the voltage on the capacitor at t=infinity, and current on inductor at t=infinity. In other words, i need to find Vc(0), Ic(0), VL(0), IL(0), Vc(infinity), and IL(infinity). I'm pretty sure Vc(0)= 12V (equal to voltage across 4ohm resistor at t=0- since voltage cant change instantly in capacitors) and IL(0)= 3A (18V/(2+4), skip 5 ohm because L is treated as short circuit) If someone could tell me if those are right and how to get the rest that would be great, thanks!

 

Your assumptions so far are correct.

For VL(0):

At the instant the switch opens, the inductor current will initially be unchanged - i.e. 3A. The capacitor voltage is 12V at this instant. The next step in reasoning becomes a little more complicated. If the 5 ohm were not there, the assumption would be that the inductor voltage would be zero at t=0. Since, the drop across the 2 ohm would be 2x3 = 6V, the capacitor voltage is 12V and V2ohm(0) + Vc(0) =18V - so VL(0) would be zero as V2ohm(0) + Vc(0) just balances the 18V source.
But the 5 ohm is there, so does it have any effect at t=0?
If a current I were flowing in the 5 ohm away from the source (left-to-right) at t=0, then the voltage drop across the 2 ohm would be greater than 6V which would be inconsistent with the assumption of a current flowing in that direction - VC(0) still being 12V.
Conversely, if a current was flowing the other way (right-to-left) in the 5 ohm, then there would also be an inconsistency with the voltage drop across the 2 ohm being less than 6V. The only conclusion possible then is that the current in the 5 ohm at t=0 is zero and VL(0) is also zero.

For Ic(0):

Since the current in L is 3A and this same current can only flow into C at t=0 (with the switch now being open), then Ic(0)=3A.

For t=∞:

Current will flow into C until steady state conditions are again reached. Irrespective of the circuit damping (which is unknown but >0), the steady state voltage across C at t=∞ must be 18V. From that you should be able to deduce IL(∞)

 

Thank you so much for your help! I was taught that capacitors are to be treated as open circuits when they are fully charged, so at t=infinity, IL would be zero, since the circuit is incomplete, correct?

 

Thank you so much for your help! I was taught that capacitors are to be treated as open circuits when they are fully charged, so at t=infinity, IL would be zero, since the circuit is incomplete, correct?

Correct - IL(∞)=0

I would argue that the capacitor steady state voltage just balances the source voltage and no further charge transfer (or current flow) can take place.