This is similar to a problem I am working on for homework I

This is similar to a problem I am working on for homework. I do not need the solution to it, I am just hung up on a few things. Looking at (a), it occurs to me, that at steady state, when the switch is closed, the capacitor essentially functions like an open, and the inductor like a short. The Circuit would then act as (b). Because the entire right side is disconnected from the 24V, and is only connected to ground, it can then be disregarded for the steady state evaluation. This means the circuit would look like (c). I would also think that this would mean i₀(0^-) would be zero. Once this is found, the circuit must be evaluated with the switch open (without the 24V supply). This is confusing to me. If there was no current through i₀(t) before the switch was opened, and the voltage source was immediately disconnected, how will there be any current through i₀(t) now? If someone could better explain this/correct my errors in thinking, I would greatly appreciate it.

Sincerely,

A Frustrated Student

Solution

As you said first when switch close a) , capacitor and inductor is in steady state thus it is connected to 24 v supply this is time (0-) we have . in this time the current flowing through inductor which is short circuit taken as i(0-) it is not equal to zero as you mentioned in (c).... It must have current and acts as current source when totally disconnected from supply and connected to ground condition like (b).