- #1
EugP
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Homework Statement
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.
Homework Equations
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?