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Need Urgent help. Please someone (Circuits Capacitance, Inductance)

  1. Apr 18, 2009 #1
    The images tell it all. Please help someone.


    http://image.cramster.com/answer-board/image/200941884006337564080009162509346.jpg


    on this diagram, the expression IL(t) should read I(t).

    http://image.cramster.com/answer-board/image/2009418845296337564112974787505740.jpg
     
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. Apr 19, 2009 #2

    tiny-tim

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    Homework Helper

    Welcome to PF!

    Hi deyiengs! Welcome to PF! :wink:

    Show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:
     
  4. Apr 19, 2009 #3
    Thanks Tiny-tim for taking your time. I planned to post what i had but i forgot.
    On the first one, how do you get R- thevinin? If that's the way to go about it?.
    On the second one I need the initial and final conditions. I used -5mA and did a source transformation to get -2.1mA as the initial consition. I used 5mA for final and got 2.1mA as the final conditio for the inductor. Then used the formular:
    iL = (i(initial) - i(final))e^(-1/tc)+ i(final)
    My tc was 3.3333us
    Then used VL = L di/dt

    Is that close to the correct way or did I deviate
     
  5. Apr 19, 2009 #4
    On the following diagram, I have the following for finding Vo:

    node analysis on the top nope of the Vn
    http://image.cramster.com/answer-board/image/cramster-equation-2009419162557633757551572010000561.gif
    This would lead to
    http://image.cramster.com/answer-board/image/cramster-equation-20094191627356337575525582600002527.gif

    but i need to replace Vp by Vs.

    Hence for the node on the bottom part I have:
    http://image.cramster.com/answer-board/image/cramster-equation-200941916329633757555299510000920.gif
    I'm stuck here how can I factor out Vp on the above equation. is it possible if it it how.

    http://image.cramster.com/answer-board/image/2009419162211633757549313416250747.jpg

    If not is there another way of solving this equation.
     
    Last edited by a moderator: Apr 24, 2017
  6. Apr 19, 2009 #5
    In the first two equations of the opamp problem, shouldn't the Vp be Vn instead?
     
  7. Apr 19, 2009 #6
    yes but Vp = Vn
     
  8. Apr 20, 2009 #7
    Someday you may have to find a transfer function where the opamp isn't ideal. In that case, Vp#Vn, so it's good practice to use the symbol for the actual node that you're writing an equation for. Then at the end, if the opamp is ideal, substitute Vp=Vn. This way, you avoid mistakes in setting up your equations.
     
  9. Apr 20, 2009 #8
    The Electrician, could you help me solve it if you can please
     
  10. Apr 20, 2009 #9
    You can either solve your two simultaneous differential equations, or you can solve it in the Laplace domain. I would do the latter. Have you studied the use of the s-variable for solving AC circuits?

    Set up two equations:

    s*C*Vn + (1/R)*(Vn-Vs) = 0

    s*C*(Vp-Vs) + Vp/R = 0

    Now, you also know that Vp = R/(R + 1/(s*C)) * Vs by the voltage divider rule, and Vp=Vn.

    Substitute that that last equation, Vp = R/(R + 1/(s*C))*Vs, for both Vp and Vn in the first two equations. Then you will have two simultaneous equations you can solve with simple algebra.

    The solution is Vo = s*R*C*Vs, which in the time domain is Vo = R C d/dt(Vs). This ignores initial values for charge on C.
     
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