Urgent(capacitor with dependent source)

Discussion in 'Homework Help' started by Furki, Apr 12, 2009.

  1. Furki

    Thread Starter Member

    Mar 1, 2009
    10
    0
    Hello friends,

    The voltage source V2 is a unit step source which gives 100V after t = 0.
    I want to know the equation of the instantaneous voltage across the capacitor. voltage becomes 500V(approx) after 60-70 seconds.
    Ive tried to derive it but I dont think its correct.
    Heres the equation:
    vc = 100/((1/1 - e^t/RC) - 0.8)

    vc is the voltage across capacitor and it controls the current through the independent source. Heres the schematic.
    http://img12.imageshack.us/img12/8358/uploada.jpg

    Any help will be much appreciated.
    thanks.
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    What is the value of source V1 - 0V?
     
  3. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    If V1 is 0V - not sure why it's in the circuit if the case - then I have ...

    Vc = 500(1-exp(-71.429*t)) which partially agrees with your expected result.

    I would expect the value to be 500V after 60-70 milliseconds.
     
  4. Furki

    Thread Starter Member

    Mar 1, 2009
    10
    0
    Thank you for your reply,
    Yes, V1 is 0 volts. It is there because the old version of ICAP doesn't allow you to measure current without it. My equation only works when t is large. for small values of t like 5ms or 10ms, it fails.
     
  5. Furki

    Thread Starter Member

    Mar 1, 2009
    10
    0
    please tell me how you got that equation of yours.
     
  6. Furki

    Thread Starter Member

    Mar 1, 2009
    10
    0
    It would be really great if you tell me how to get to that equation from superposition or whatever method you used, especially the "-71.429" part. pleeeeeease!!!! :(
     
  7. Ratch

    New Member

    Mar 20, 2007
    1,068
    3
    Furki,

    Let me get this straight. According to the schematic, you have the capacitor physically grounded on one end, and virtually grounded at the other end by connecting it to the input terminals of a op amp? Does that make sense?

    Ratch
     
  8. Furki

    Thread Starter Member

    Mar 1, 2009
    10
    0
    Ratch,

    Yes, the control port of the dependent source is connected to the capacitor so that the capacitor voltage(Vc) controls the current of the dependent source(I).
    I = 0.02*Vc

    Capacitor is also connected to V2. (forget the V1 source, its zero volts)
    V2 = 100V
     
  9. SawabyPlus

    Member

    Feb 27, 2009
    14
    0
    I don't know if it would help, attached are the steps using Laplace transform.
    sorry, i see it's quite messy.
     
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  10. Furki

    Thread Starter Member

    Mar 1, 2009
    10
    0
    SawabyPlus,

    This helps a LOT!
    Thanks a lot mate. You're a life saver.:)
     
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