Solve Electrical Circuit Problem: Find Forced Response Vx

[EugP]

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:

$$i_s(t)= \sqrt{2} cos(4t)$$

The dependant current source is equal to:

$$gV_x$$, where $$i_s(t)g = 0.5$$.

Lastly, $$C = 0.125 F$$, and has voltage $$V_x$$ across it.

I need to find the forced response of $$V_x$$, which is the voltage across the capacitor.

The answer is : $$V_x = 2cos(4t-135\degree)$$, 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:

$$\sqrt{2} cos(4t) + gV_x$$

$$\sqrt{2} cos(4t) + \frac{1}{2}V_x$$

Now to find $$V_x$$, I just multiplied that by the impedance of the capacitor:

$$V_x = (\sqrt{2} cos(4t) + \frac{1}{2}V_x)(\frac{1}{j\omega C})$$

$$V_x = (\sqrt{2} cos(4t) + \frac{1}{2}V_x)(\frac{1}{j0.5})$$

$$V_x = (\sqrt{2} cos(4t) + \frac{1}{2}V_x)(\frac{2}{j})$$

$$V_x = \frac{2\sqrt{2} cos(4t) + V_x}{j}$$

$$jV_x - V_x = 2\sqrt{2} cos(4t)$$

$$V_x(j - 1) = 2\sqrt{2} cos(4t)$$

$$V_x = \frac{2\sqrt{2} cos(4t)}{j-1}$$

$$V_x = \frac{2\sqrt{2} cos(4t)}{j-1} \frac{j+1}{j+1}$$

$$V_x = \frac{j2\sqrt{2} cos(4t) + 2\sqrt{2} cos(4t)}{1+1}$$

$$V_x = j\sqrt{2} cos(4t) + \sqrt{2} cos(4t)$$

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?

 

[CEL]

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:

$$i_s(t)= \sqrt{2} cos(4t)$$

The dependant current source is equal to:

$$gV_x$$, where $$i_s(t)g = 0.5$$.

Lastly, $$C = 0.125 F$$, and has voltage $$V_x$$ across it.

I need to find the forced response of $$V_x$$, which is the voltage across the capacitor.

The answer is : $$V_x = 2cos(4t-135\degree)$$, 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:

$$\sqrt{2} cos(4t) + gV_x$$

$$\sqrt{2} cos(4t) + \frac{1}{2}V_x$$

Now to find $$V_x$$, I just multiplied that by the impedance of the capacitor:

$$V_x = (\sqrt{2} cos(4t) + \frac{1}{2}V_x)(\frac{1}{j\omega C})$$

$$V_x = (\sqrt{2} cos(4t) + \frac{1}{2}V_x)(\frac{1}{j0.5})$$

$$V_x = (\sqrt{2} cos(4t) + \frac{1}{2}V_x)(\frac{2}{j})$$

$$V_x = \frac{2\sqrt{2} cos(4t) + V_x}{j}$$

$$jV_x - V_x = 2\sqrt{2} cos(4t)$$

$$V_x(j - 1) = 2\sqrt{2} cos(4t)$$

$$V_x = \frac{2\sqrt{2} cos(4t)}{j-1}$$

$$V_x = \frac{2\sqrt{2} cos(4t)}{j-1} \frac{j+1}{j+1}$$

$$V_x = \frac{j2\sqrt{2} cos(4t) + 2\sqrt{2} cos(4t)}{1+1}$$

$$V_x = j\sqrt{2} cos(4t) + \sqrt{2} cos(4t)$$

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?

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.

 

[EugP]

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.

So what do I need to do in order to finish the problem?

 

[CEL]

So what do I need to do in order to finish the problem?

Either you write all equations in the frequency domain, where $$cos\omega t = \omega$$ or you write everything in the time domain, where $$v_c(t)=v_c(0)+\frac{1}{C} \int_{0}^{t} i_c(\tau)d\tau$$

 

[EugP]

Either you write all equations in the frequency domain, where $$cos\omega t = \omega$$ or you write everything in the time domain, where $$v_c(t)=v_c(0)+\frac{1}{C} \int_{0}^{t} i_c(\tau)d\tau$$

Oh now I understand. Thanks.