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how long was the switch closed, before it was opened?
if it was a long time, then there was no current when the switch opened, and all of the stored energy was in the capacitor, which was at the battery voltage 16V.
therefore , your initial conditions below the schematic don't make sense.
once you open the switch, there is only 1 path, hence all components see the same current.
so what is happening before the switch opens, so know initial conditions?
Let's discuss what happens in the circuit before and after switching. Let's also assume that the switch was closed long enough so that the circuit was in steady state.
Before switching:
VL=0V meaning that iL was limited By R, so that iL = 16/R = 16/8 = 2A.
Vc was limited by the source voltage, thus Vc=16V meaning that Vx=0V and Ix=0A.
After switching:
The inductor wants to maintain the direction of current through it and the capacitor want to maintain its polarity. Realize that the inductor is a storage device and even when it had no voltage across it before switching, it had stored energy in form of a steady magnetic field around it. Now that 2A is required, the capacitor alone cannot provide it even for the shortest of time because its maximum current output is already limited by the sum of the two resistance which is 12+8 =20 thus it's maximum current would be 16/20 < 2A. This tells us that the inductor has to start discharging. So now we have both a decaying voltage source and a decaying current source in series with a 20-ohm resistor in the circuit. The current direction is now in the original direction of iL.
Reassess your work and solve again. Also observe the mutual dependence between the decay of Vc and inductor magnetic field.
Update your work so we see your progress and advise accordingly.
I would think that at the time of switching, the inductor has already attained its maximum flux for 2A. After switching, it would have to discharge to be able to sustain the 2A initially. So I believe that VL and Vc will have the same polarity after switching such that Vc +VL - VR = 0.
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.
The question required the solution at the instance immediately after switching. If it required expression for finding the parameter values at any instance down the line in time, then you would be seeing the exponential relationship. For the question, this solution suffices.
Why is the actual direction of the current through the inductor and the actual polarity of the voltage across the inductor does not agree with the passive sign convention?
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Why is the actual direction of the current through the inductor and the actual polarity of the voltage across the inductor does not agree with the passive sign convention?
Why is the actual direction of the current through the inductor and the actual polarity of the voltage across the inductor does not agree with the passive sign convention?
It does obey when it's storing energy. When it's releasing the energy, it behaves like a source.
Remember that an inductor is an energy storage device. It stored energy when the switch was closed. At the instance of consideration in this problem the inductor is actually a source too. It's converting its stored electromagnetic energy into electrical energy and supplying it to the rest of the circuit. Remember that any energy source releases current from its positive terminal, through load and back to its negative terminal.
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