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First order Transient Circuit

  1. Jun 4, 2007 #1
    Please click the image to make it larger:

    To solve this circuit I'm going to use the differential equation approach. I'm concerned with the voltage across the capacitor at [tex]V_c(0^-)[/tex] and [tex]V_c(0^+)[/tex]

    At position 1 before the switch is thrown [tex]V_c(0^-)=0[/tex]

    At position 2 after switch is thrown I have the following from KCL:

    [tex]\frac{6-V(t)}{12k} = C\frac{dv}{dt} + \frac{V(t)}{6k}[/tex]

    This reduces to: [tex]\frac{dv}{dt} + 2.5V(t) = 5 [/tex]

    My problem is that I know that this is correct but I don't know how to put it all together.

    I know that the solution is of the form:

    [tex]K_1 +K_2e^{-t/t_c}[/tex]

    The answer is [tex]V(t)=1.33 -1.33e^{-2.5t}V[/tex]

    How do I extract this from my work?:rofl:
    Last edited: Jun 4, 2007
  2. jcsd
  3. Jun 5, 2007 #2
    Wow, I thought someone would understand this :cry:
  4. Jun 5, 2007 #3


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    Homework Helper

    Assuming that you have got the diff eqn correct (I can't see your image so can't check), then you need to either solve this by Integrating factor method or sub in the form
    [tex]K_1 +K_2e^{-t/t_c}[/tex]
    into your equation and then solve for K1, K2 and t_c. You will find that it a polynomial in [tex]e^{-t/t_c}[/tex], equating coefficient of [tex]e^{-t/t_c}[/tex] on both sides you get relation for K1 and t_c, and then use your initial condition to fix K2.

    i have got a feeling that
    RHS may be wrong
  5. Jun 5, 2007 #4
    I think you made a reduction error because you have the KCL equation correct.

    I get the same KCL, but mine reduced to:
    [tex]6 = 3v_c(t) + 1.2 \frac{dv_c(t)}{dt}[/tex]

    This is simply a differential equation that you need to solve for like any other diff eq would go. First find the homogeneous solution
    [tex]0 = 3v_c(t) + 1.2 \frac{dv_c(t)}{dt}[/tex]
    with separation of variables and then find the particular solution
    [tex]6 = 3v_c(t) + 1.2 \frac{dv_c(t)}{dt}[/tex]
    with a judicious guess. Add the two for your overall solution.
  6. Jun 5, 2007 #5
    Actually I see that my answer doesn't agree with the answer given to you, but I am pretty sure that mine is right because as a quick check you can look at the circuit given to you and for DC values replace capacitors with open circuits and inductors with short circuits.

    The DC solution (which happens to correspond with the particular solution because the DC source is driving the circuit) I get is the same through the Diff Eq. and the DC reduction method. Maybe you can verify this.
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