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Parallel RC with current source

  1. Apr 28, 2014 #1

    dwn

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    1. The problem statement, all variables and given/known data

    See images

    In the first image, I made a circuit with the switch closed and then two when the switch is opened.
    The bottom circuit is the Thev transform.

    2. Relevant equations



    3. The attempt at a solution

    Let me first explain my understanding of the circuit. Prior to t = 0, the dc source been suppling a current to capacitor for an lengthy amount of time in which case the cap would reach a 'steady state' of voltage. I decided to find this voltage by using a Thevenin transformation ( 90 V and 6Ω series).

    vc(0) = 90 V
    My concern with this is that when I created the new circuit in iCircuit app, the cap voltage only read 30 V..?

    vc(∞) = 0 V (intuitive understanding of the circuit)

    In order to find vc(t), I went back to the original circuit and decided to try:
    Code (Text):
    v[SUB]c[/SUB](t) = 1/C ∫i dt
    At this point, I just got lost and wasn't sure how to continue.
     

    Attached Files:

  2. jcsd
  3. Apr 28, 2014 #2

    gneill

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    Staff: Mentor

    Can you explain your Thevenin equivalent calculations? The result you get doesn't look right.
     
  4. Apr 28, 2014 #3

    dwn

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    Sure thing, I was doing this late in the evening/early morning depending on perspective, so that could be why.

    Starting with 18 Ohm and 5 A source and treating the rest as open circuit: V = IR = 18 * 5 = 90 V

    1/18 + 1/9 = 1/R R = 6
     
  5. Apr 28, 2014 #4

    gneill

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    Staff: Mentor

    Okay, what about the 9 Ω resistor? Won't it form a voltage divider with your new Thevenin model? Did you consider incorporating the 9 Ω into the Thevenin model to make things simpler?
     
  6. Apr 28, 2014 #5

    dwn

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    With a dc voltage source, isn't the capacitor shorted? Which would mean the resistors are in series?

    If I did use the voltage divider: V = 9(90/27) = 30V
     
  7. Apr 28, 2014 #6

    gneill

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    Staff: Mentor

    Nope. If you take only one of the resistors into your Thevenin equivalent the you have a voltage source in series with one resistor which together parallel the remaining resistor and capacitor.

    You know that you can change around the depicted order of components in parallel without altering the circuit (the same is true of components in series --- the sum of their resistances and potential drops are constants). If you rearrange and simplify your circuit before applying Thevenin then things will be easier:

    attachment.php?attachmentid=69174&stc=1&d=1398698412.gif
    Right. But it's better to simplify the resistor network first and reduce the complexity.
     

    Attached Files:

    • Fig2.gif
      Fig2.gif
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  8. Apr 28, 2014 #7

    dwn

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    Yes, I think I remember that now that you mention it.

    This stuff is killing all confidence I had in myself as a human being...may be time for a beer.
     
    Last edited: Apr 28, 2014
  9. Apr 28, 2014 #8

    dwn

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    Analyzing the circuit for t > 0

    RC equation : v(t) = V0e-t/RC = 30e-t/0.6

    Correct?
     
  10. Apr 28, 2014 #9

    gneill

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    Staff: Mentor

    Yes. You got it :approve:
     
  11. Apr 28, 2014 #10

    dwn

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    FINALLY! Are we allowed to make donations to you guys?!? If not, you should talk to someone about that!
     
  12. Apr 28, 2014 #11

    gneill

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    Staff: Mentor

    You can upgrade your Membership to become a paying member (QUICK LINKS ---> Upgrade Membership). Fees contribute to the upkeep of the site infrastructure (all staff are unpaid volunteers; we do it because promoting and encouraging science education is its own reward).
     
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