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KVL for a RLC circuit given current direction and polarity o

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  1. Feb 7, 2016 #1
    1. The problem statement, all variables and given/known data

    KVL for a RLC circuit given current direction and polarity of a resistor?

    Here is the diagram of the circuit:

    2.JPG

    2. Relevant equations
    KVL

    Sum of voltages equal zero

    3. The attempt at a solution

    I am actually looking for the voltage across the inductor (VL), so I decided the best approach in finding it is through Kirchoff's Voltage Law.
    Being given the polarity of the resistor, should I follow the passive sign convention for the other components in the circuit while doing KVL...if I decide to do a loop in the counterclockwise direction?

    I understand that current conventionally flows through the positive terminal to the negative.
    So would it be correct if I label the circuit like this, if I did a KVL mesh counterclockwise?:

    RLC.JPG


    Any help would be appreciated, thank you. :)
     
  2. jcsd
  3. Feb 7, 2016 #2

    cnh1995

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    What about voltage source? Where is it placed in the diagram? If RL branch and RC branch are in parallel to the voltage source, you can analyze them independently.
     
  4. Feb 7, 2016 #3
    Well, this is the original problem...I've done part a) already, but I'm at part b) trying to find the general equation for i_x(t)...and to do that I need to find the inductor voltage when the switch is open--since I'm solving for the initial conditions di(0+)/dt and that equals VL/L.

    Was just wondering if my polarities/KVL is correct in the diagram above? :frown:


    Capture.JPG
     
  5. Feb 7, 2016 #4

    cnh1995

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    If the switch was closed for a long time, the capacitor must be charged to 36V. Is the direction of Ix given in the problem or you assumed it?
    Edit:sorry I didn't see the diagram, it is given in the problem itself.
     
  6. Feb 7, 2016 #5
    Ix is given. I don't think the capacitor will be charged to 36V. t=0- , capacitor should be 16V.
     
  7. Feb 7, 2016 #6

    cnh1995

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    You should reverse the direction of Ix. It is not possible for given polarity of the resistor.
     
  8. Feb 7, 2016 #7

    cnh1995

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    Sorry, typo.
     
  9. Feb 7, 2016 #8
    Oh why is that...if you don't mind me asking? The resistor polarity was given as positive on top and negative on bottom. I figured if I did KVL and followed the polarity of that resistor, then I would be on the right track.
     
  10. Feb 7, 2016 #9

    cnh1995

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    After opening the switch, the capacitor will discharge. So, the current direction should be from +ve plate to -ve plate. Polarity of the resistor is correctly shown. You can change the direction of Ix or consider the magnitude of Ix as -Ix.
     
  11. Feb 7, 2016 #10

    gneill

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    Staff: Mentor

    Actually, since it's an LC circuit it's likely going to oscillate (you'll have to check the amount of damping provided by the resistances). So Ix will probably take on either direction over time. It should simply be seen as an instruction as to how to interpret the current direction.

    The initial current in the inductor is going to force the initial direction of current. Inductors are stubborn that way :smile:
     
  12. Feb 7, 2016 #11

    cnh1995

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    Right. But wouldn't the initial current direction be opposite to the given direction of Ix? I thought it would be easier to write KVL with the actual initial current direction. Hence I suggested to reverse the direction. After solving the equation, the response will be damped oscillations, but in the beginning, I thought it would be better to go with the actual direction of initial current.:smile:
     
  13. Feb 7, 2016 #12
    This was the solutions to the problem...but I just wanted to understsand the polarity of the circuit itself. I'm guessing as you mentioned, since capacitor is discharing, I_X goes the other direction ? Sorry I'm still a little unsure about this..

    sol.JPG
     
  14. Feb 7, 2016 #13

    gneill

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    Yes, the initial current must be clockwise as driven by the inductor's initial current. But that just makes Ix negative initially.

    It's usually not a good idea to tamper with predefined interpretation labels. It's too easy to forget what you did when it comes time to writing the result that's going to be submitted. But you can feel free to introduce your own currents, solve the problem, and then interpret the result in light of the defined currents/potentials.
     
  15. Feb 7, 2016 #14

    cnh1995

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    Right. Initial current goes in the other direction.
     
  16. Feb 7, 2016 #15
    So you're saying that we can completely ignore the defined polarity of the resistor that was given? And do KVL from there?
     
  17. Feb 7, 2016 #16
    So if this was the case...shouldn't the capacitor's polarity be (+) on the bottom and (-) on the top? Since current is flowing clockwise this time, and should flow from positive terminal of capacitor to negative. The solutions seem to have it the other way around.
     
  18. Feb 7, 2016 #17

    cnh1995

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    It is flowing from +ve terminal to -ve terminal.
     
  19. Feb 7, 2016 #18

    cnh1995

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    And the polarity of inductor voltage should be reversed.
     
  20. Feb 7, 2016 #19

    gneill

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    It's an interpretive label. At some point you may be asked to state what VR is at some particular time, or over time in general. You'd then use that label to interpret the result that you present. But you don't need that label or its implied polarity to solve the circuit for the current that flows.
     
  21. Feb 7, 2016 #20

    gneill

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    The potential and current labels on a diagram are merely there for interpretation purposes. They tell you how to (hypothetically) connect a meter to read a value for reporting a result. When you're solving the circuit you can introduce all your own currents and potentials and ignore the labels if you wish. When you report results that are to be interpreted in light of those labels, that's where they come into play.

    Just leave the interpretive labels alone and solve the circuit. Then interpret the result in terms of those definitions.
     
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