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Homework Help: Help with Series RLC circuit

  1. Nov 20, 2011 #1
    Edited for assignment purposes...
     

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  2. jcsd
  3. Nov 20, 2011 #2
    I think you need to know the frequency of the supply to get started.... do you have it?
     
  4. Nov 20, 2011 #3
    The source is just a DC source... so, there's no frequency.
     
  5. Nov 20, 2011 #4
    my apologies.... I thought it was an AC circuit
     
  6. Nov 20, 2011 #5

    gneill

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    Hi wiz0r,
    Presumably the initial conditions that you're looking for is the state of the circuit at the time t=0+ when the switch has just opened.

    Immediately before the switch opens, and assuming that the switch had been closed for "a long time", then your statement that the capacitor will look like an open circuit is correct -- no current will flow into or out of it. Similarly, the inductor will look like a short circuit.

    Can you determine what voltage should be across the capacitor at t=0-?
     
  7. Nov 20, 2011 #6
    Edited for assignment purposes...
     
    Last edited: Nov 20, 2011
  8. Nov 20, 2011 #7

    gneill

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    That looks fine. So when the switch opens the capacitor will have that voltage on it as an initial condition. And the current will still be zero because of the inductor.

    To find i(t) you'll have to either write and solve the differential equation for the new circuit configuration, or use a "canned" solution for the RLC circuit.
     
  9. Nov 20, 2011 #8
    Edited for assignment purposes...
     
    Last edited: Nov 20, 2011
  10. Nov 20, 2011 #9

    gneill

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    It's looking good!
     
  11. Nov 20, 2011 #10
    Edited for assignment purposes...
     
    Last edited: Nov 20, 2011
  12. Nov 20, 2011 #11

    gneill

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    The equation should yield current values, and Vc(0) is a voltage.

    At time t = 0 the current is zero, so K1 + K2 = 0. That is, K1 = -K2. You'll need to find another way to solve for the values. Perhaps if you could determine the initial rate of change of the current...
     
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  13. Nov 20, 2011 #12
    Edited for assignment purposes...
     
    Last edited: Nov 20, 2011
  14. Nov 20, 2011 #13

    gneill

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    Right :blushing:
    s1K1 + s2K2 won't be zero. It'll equal the initial rate of change of the current. Go back to the circuit diagram and see if you can't determine what that might be. You've an inductor that has suddenly had a voltage Vc impressed upon it... how will it react?
     
  15. Nov 20, 2011 #14
    Edited for assignment purposes...
     
    Last edited: Nov 20, 2011
  16. Nov 20, 2011 #15

    gneill

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    The equation of interest for the inductor is its defining equation: V = L dI/dt. In this case there's an impressed voltage Vc across inductor L=2.5H. What's dI/dt?
     
  17. Nov 20, 2011 #16
    Edited for assignment purposes...
     
    Last edited: Nov 20, 2011
  18. Nov 20, 2011 #17

    gneill

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    Right. At t=0 dI/dt = Vc/L = 10.286 A/s. So, given K1 = -K2,...
     
  19. Nov 20, 2011 #18
    Edited for assignment purposes...
     
    Last edited: Nov 20, 2011
  20. Nov 20, 2011 #19

    gneill

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    Looks good! Bravo!
     
  21. Nov 20, 2011 #20
    Woohoo... thank you so much!!

    One last thing... I'm told to make a plot of i(t).. to do that I used SPICE(attachment)... does the graph looks OK (attachment)?

    @all other people: If anyone reading is taking the class with Rosado please say hi!
     

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  22. Nov 20, 2011 #21

    gneill

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    Looks okay. I think if you made the maximum timestep a bit smaller you might resolve the peak a bit better (eliminate the "flat bottom").

    Here's the plot of the mathematical version:

    attachment.php?attachmentid=41094&stc=1&d=1321840842.jpg

    I've negated the current values ( -I(t) ) in order to match your Spice model current direction.
     

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  23. Nov 20, 2011 #22
    Perfect... thank you so much... I owe you big time... If I can repay you with something (except money.. I'm poor!) let me know!! :)
     
  24. Nov 20, 2011 #23

    gneill

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    You're quite welcome. Just go get an "A" in your course!
     
  25. Nov 20, 2011 #24
    I will definitely try my best. :)

    Thank you so much. Bye bye for now. :)

    Edwin
     
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