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Homework Help: AC Analysis

  1. Jan 25, 2008 #1
    1. The problem statement, all variables and given/known data
    I don't have access to a scanner so I will just say it, it's simple.

    An independent current source pointing up, in parallel with a dependent current source pointing up, in parallel with a capacitor. The independent current is providing the signal:

    [tex]i_s(t)= \sqrt{2} cos(4t)[/tex]

    The dependant current source is equal to:

    [tex]gV_x[/tex], where [tex]i_s(t)g = 0.5[/tex].

    Lastly, [tex]C = 0.125 F[/tex], and has voltage [tex]V_x[/tex] across it.

    I need to find the forced response of [tex]V_x[/tex], which is the voltage across the capacitor.

    The answer is : [tex]V_x = 2cos(4t-135\degree)[/tex], but I don't know how they got it.

    2. Relevant equations

    3. The attempt at a solution

    Since both current sources are pointing up, I added them to get a single current source in parallel with a capacitor:

    [tex]\sqrt{2} cos(4t) + gV_x[/tex]

    [tex]\sqrt{2} cos(4t) + \frac{1}{2}V_x[/tex]

    Now to find [tex]V_x[/tex], I just multiplied that by the impedance of the capacitor:

    [tex]V_x = (\sqrt{2} cos(4t) + \frac{1}{2}V_x)(\frac{1}{j\omega C})[/tex]

    [tex]V_x = (\sqrt{2} cos(4t) + \frac{1}{2}V_x)(\frac{1}{j0.5})[/tex]

    [tex]V_x = (\sqrt{2} cos(4t) + \frac{1}{2}V_x)(\frac{2}{j})[/tex]

    [tex]V_x = \frac{2\sqrt{2} cos(4t) + V_x}{j}[/tex]

    [tex]jV_x - V_x = 2\sqrt{2} cos(4t)[/tex]

    [tex]V_x(j - 1) = 2\sqrt{2} cos(4t)[/tex]

    [tex]V_x = \frac{2\sqrt{2} cos(4t)}{j-1}[/tex]

    [tex]V_x = \frac{2\sqrt{2} cos(4t)}{j-1} \frac{j+1}{j+1}[/tex]

    [tex]V_x = \frac{j2\sqrt{2} cos(4t) + 2\sqrt{2} cos(4t)}{1+1}[/tex]

    [tex]V_x = j\sqrt{2} cos(4t) + \sqrt{2} cos(4t)[/tex]

    I don't know where to go from here. Could someone please tell me where I need to go from here to get the answer?
  2. jcsd
  3. Jan 25, 2008 #2


    User Avatar

    You are mixing values in the time domain (the current from the source) with values in the frequency domain (the impedance of the capacitor). You must express your equations in only one domain.
  4. Jan 26, 2008 #3
    So what do I need to do in order to finish the problem?
  5. Jan 26, 2008 #4


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    Either you write all equations in the frequency domain, where [tex]cos\omega t = \omega[/tex] or you write everything in the time domain, where [tex]v_c(t)=v_c(0)+\frac{1}{C} \int_{0}^{t} i_c(\tau)d\tau [/tex]
  6. Jan 27, 2008 #5
    Oh now I understand. Thanks.
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