Parallel RLC circuit confusion

In summary, we are trying to find the value of vL at t = 0+, which can be found by first finding Vc(0-) using the voltage-divider rule. The initial capacitor voltage is 2V, which means that the initial inductor voltage is -2V due to their opposite polarities. The initial current through the 1Ω resistor can be found using KCL, and the initial capacitor current is 0A. At the switching instant, the inductor acts as a current source, producing any amount of voltage required to maintain the current at its present value. However, since it has a limited amount of stored energy, the current will eventually begin to change after t = 0+.
  • #1
ViolentCorpse
190
1

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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  • #2
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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  • #3
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! :)
 
  • #4
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!
 
  • #5
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?
 
  • #6
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+.
 
  • #7
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! :)
 

Related to Parallel RLC circuit confusion

What is a parallel RLC circuit?

A parallel RLC circuit is an electrical circuit that contains a resistor (R), an inductor (L), and a capacitor (C) connected in parallel. This means that the three components share the same two nodes, but each component has its own branch or path for the current to flow through.

How does a parallel RLC circuit work?

In a parallel RLC circuit, the resistor, inductor, and capacitor each have their own impedance, which is a measure of the opposition to the flow of current. The total impedance of the circuit is the sum of the individual impedances. When an AC current is applied to the circuit, the impedance of each component will vary based on the frequency of the current. The circuit will behave differently depending on the frequency, and this is known as resonance.

What is the difference between a series and a parallel RLC circuit?

In a series RLC circuit, the components are connected in a single loop and the current flows through each component in sequence. In a parallel RLC circuit, the components are connected in parallel, and the current splits into different branches. The total impedance of a series RLC circuit is the sum of the individual impedances, while the total impedance of a parallel RLC circuit is the reciprocal of the sum of the reciprocals of the individual impedances.

What is resonance in a parallel RLC circuit?

In a parallel RLC circuit, resonance occurs when the reactance of the inductor and capacitor cancel each other out, leaving only the resistance of the circuit. This causes the total impedance of the circuit to decrease, allowing more current to flow through the circuit. Resonance can only occur at a specific frequency, known as the resonant frequency, which is determined by the values of the inductor and capacitor.

What are the applications of parallel RLC circuits?

Parallel RLC circuits have a variety of applications, including power factor correction, filtering and tuning in radio and television receivers, and power supplies for electronic devices. They are also used in electronic filters to remove specific frequencies from a signal and in oscillators to generate a specific frequency. Parallel RLC circuits are also used in resonance imaging (MRI) machines in the medical field.

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