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Initial and final values for second order circuits

  1. Sep 28, 2015 #1
    1. The problem statement, all variables and given/known data

    I am attempting to understand this example shown below:

    example.PNG

    2. Relevant equations

    During stead state DC, the capacitor is an open circuit and the inductor is short circuited.

    3. The attempt at a solution

    The questions I have are really related to the concepts as I don't seem to understand it well enough even after reading through the chapter many times.
    1- How is Vc equal to -20V? Since Vc is an open circuit, how would there be a voltage across it?
    2- My understanding is that the inductor current (iL) is zero because there is no voltage source in the circuit during steady state DC. Please confirm if my understanding is correct.
    3- I have read through the book many times and I don't seem to understand how this is determined. If someone can provide a detailed explanation, I would really appreciate it!
     
  2. jcsd
  3. Sep 28, 2015 #2

    gneill

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    Staff: Mentor

    For (1) consider the circuit as drawn here with a voltmeter placed across the "open" capacitor position:

    Fig1.png

    The voltmeter leads are placed in such a way as to read the capacitor voltage as it is defined on the diagram. Essentially this makes the "-" lead of the capacitor the reference point for the reading. Now note the path marked out in blue: This comprises a single node since it's all connected by conductor. What is the potential of this node with respect to the voltmeter's reference point? So what will the voltmeter read?

    For (2) you are correct in this case. The capacitor is "open" so no current can flow from the 20 V source, and the current supply is zero so no current from there. With no sources of current there can be no current in the circuit.
     
  4. Sep 28, 2015 #3
    Dear gneill, that was a very clear explanation. From what I understand, the "-" lead is connected to the positive side of the voltage, which would make that point +20V. The "+" lead is connected to the single node, which is at a lower voltage. Therefore, the potential of this node with respect to the voltmeter's reference point is 20V. I was a little confused at where the negative sign comes from? I understand that I am able to get -20V if I perform KVL around the third loop (-10-Vc+40*iL=0, where iL equals 0 and therefore Vc=-10V). Is there another way of doing this?
     
    Last edited: Sep 28, 2015
  5. Sep 28, 2015 #4

    gneill

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    Staff: Mentor

    No, from the point of view of the meter, wherever the "-" lead is is the 0 V reference (for the meter).
    Do a "KVL walk" from the meter's "-" lead down through the voltage supply to the blue node. What's the potential of the blue node with respect to the lead?
     
  6. Sep 29, 2015 #5
    Thank you very much! I understand this now!
     
  7. Sep 30, 2015 #6
    I am attempting to solve another problem, which carries over the same concepts. The question I have is related to part (a) where the solution suggests "Since it is in series with the +10V source, together they represent a direct short at t=0+. This means that the entire 2A from the current source flows through the capacitor and not the resistor". Can you please explain this? I am confused about the direct short at t=0+.
     

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  8. Sep 30, 2015 #7

    gneill

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    Staff: Mentor

    At the instant t = 0+, the capacitor has a potential difference of -10 V. It's in series with the 10 V voltage source. Together they add up to 0 V. For that instant, the two components can be replaced with single zero volt voltage supply. A zero volt voltage supply is, for all intents and purposes, equivalent to a short circuit (a zero resistance path).
     
  9. Oct 1, 2015 #8
    Oh right! That makes sense! Thanks a lot, once again!
     
  10. Oct 8, 2015 #9
    Hi,
    I know this is a little late but I have been attempting to figure this out myself but had no success. In part (b) of that question, it is indicated that vR=vC+10. However, shouldn't vR=0 since all of the 2A goes to the short circuit (capacitor current)? This would make dvR/dt=0?
     
  11. Oct 8, 2015 #10

    gneill

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    Staff: Mentor

    Perhaps we're not looking at the same circuit? The one I see in the problem statement has a 3 A supply. When it turns on at time zero its current will be divided between the 2 Ω and 4 Ω resistors essentially instantaneously since, as you say, the capacitor looks like a short circuit to an instantaneous change. I'd expect that VR (or the point labelled "a" on the circuit shown as figure (b)) to pop up from zero to +4 V. From there I suppose there will be some "ringing" in the LC tank circuit which will also be seen on VR, I guess as a damped sinewave.

    Maybe this could be made more obvious if you replaced the current source and 2 Ω resistor with their Thevenin equivalent? You'd then have that node "a" in the middle of a voltage divider.
    Fig4.png
     
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