Engineering Parallel RLC circuit confusion

AI Thread Summary
The discussion revolves around finding the voltage across an inductor (vL(0+)) in a parallel RLC circuit. The initial voltage across the capacitor is determined to be 2V, leading to the conclusion that the inductor voltage must be -2V due to opposite polarities. The challenge lies in understanding how the inductor maintains its current despite the capacitor's opposition, as it cannot allow instantaneous changes in current. It is clarified that the inductor acts like a current source, producing the necessary voltage to sustain the current momentarily, although it cannot do so indefinitely. The conversation highlights the importance of applying Kirchhoff's Current Law (KCL) to analyze the circuit dynamics at the switching instant.
ViolentCorpse
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Homework Statement


Find vL(0+)

Homework Equations


V=IR

The Attempt at a Solution



I have an exam tomorrow and so I'm in a bit of a hurry so please excuse the horrible drawing.

The problem I'm having with this question is a conceptual one. I've found vL in this this circuit by first finding Vc(0-), because Vc(0-)=Vc(0+)=vL(0+) using voltage-divider rule: V=12*(1)/(6)=2V.

Now, I think L and C have opposite polarities by looking at the diagram so if Vc=2 V, vL must be = -2 V

The trouble I'm having is that I'm unable to make a physical interpretation of what's happening here. The charged capacitor will try to force a current through the inductor in a direction opposite to that in which it is already flowing. How will the inductor offset this opposition from the capacitor, since it cannot allow any instantaneous change in its current?

Thanks a lot for your time! :)
 

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ViolentCorpse said:

Homework Statement


Find vL(0+)

Homework Equations


V=IR


The Attempt at a Solution



I have an exam tomorrow and so I'm in a bit of a hurry so please excuse the horrible drawing.

The problem I'm having with this question is a conceptual one. I've found vL in this this circuit by first finding Vc(0-), because Vc(0-)=Vc(0+)=vL(0+) using voltage-divider rule: V=12*(1)/(6)=2V.
Yes, that's good for the initial capacitor voltage. And the switch closure places the inductor directly across the capacitor so they must have the same potential difference. What about the resistor? What must be the initial current through the 1Ω resistor?
Now, I think L and C have opposite polarities by looking at the diagram so if Vc=2 V, vL must be = -2 V

The trouble I'm having is that I'm unable to make a physical interpretation of what's happening here. The charged capacitor will try to force a current through the inductor in a direction opposite to that in which it is already flowing. How will the inductor offset this opposition from the capacitor, since it cannot allow any instantaneous change in its current?
The capacitor is quite happy to sit there with its potential difference and supply no current if the nodes it's connected to are maintaining that potential difference thanks to the (initial) inductor current.

Apply KCL to the node where the inductor and capacitor meet (the top one). You know the initial potential of that node so you know the initial current through the resistor. You know the initial current arriving from the inductor. What then is the initial capacitor current?
 
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By KCL, the capacitor current should be zero, right?

But then, what is responsible for maintaining the inductor current at the switching instant? The capacitor is "supplying" no current, so the inductor must have self-induced some voltage across itself to maintain that current, right?

Thank you so much, gneill! You're always a great help! :)
 
ViolentCorpse said:
By KCL, the capacitor current should be zero, right?

But then, what is responsible for maintaining the inductor current at the switching instant? The capacitor is "supplying" no current, so the inductor must have self-induced some voltage across itself to maintain that current, right?
Correct. The inductor has stored energy in its magnetic field, and it is the source of, or 'motivation' for the EMF.
Thank you so much, gneill! You're always a great help! :)
Always happy to help!
 
Ah. I presume that the minus sign in the inductor voltage means that the inductor reverses polarities in order to maintain the same current for an instant? Would it be correct to say that it acts like a voltage-source in that one instant?
 
ViolentCorpse said:
Ah. I presume that the minus sign in the inductor voltage means that the inductor reverses polarities in order to maintain the same current for an instant? Would it be correct to say that it acts like a voltage-source in that one instant?
More like a current source... it will produce any amount of voltage required to try to maintain the current at its present value. Of course, since it has a limited amount of stored energy it can't maintain this current indefinitely, and in fact it begins to change after t = 0+.
 
gneill said:
More like a current source... it will produce any amount of voltage required to try to maintain the current at its present value. Of course, since it has a limited amount of stored energy it can't maintain this current indefinitely, and in fact it begins to change after t = 0+.

Oh right, of course.

Thanks again, gneill! :)
 

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