Solving Impedance/Admittance for V_s w/ 2 Caps, 2 Inductors, 2 Res.

[VinnyCee]

Impedance and Admittance: Find Vs given Io in a circuit with 2 caps,2 inductors,2res

Homework Statement

Find $V_s$ if $I_0\,=\,2\angle0\deg$ A.

http://img237.imageshack.us/img237/6291/problem954zz0.jpg

Homework Equations

KCL, KVL

The Attempt at a Solution

I made a new diagram:

http://img201.imageshack.us/img201/9359/problem954part2gz3.jpg

But how do I combine the left hand equivalent impedance, so that the final circuit to work on would be this:

http://img201.imageshack.us/img201/1690/problem954part3fj1.jpg $$\frac{1}{Z_3}\,=\,\frac{1}{Z_1}\,+\,\frac{1}{j4\Omega}\,=\,\frac{1}{2\,+\,2j}\,+\,\frac{1}{4j}$$

Figuring $V_o$:

$$V_o\,=\,I_o\,Z_2\,=\,\left(2\angle0\right)\left(2\angle45\right)\,=\,4\angle45$$

$$I_L\,=\,\frac{V_0}{j2}\,=\,\frac{4\angle45}{\sqrt{2}\angle63.43}\,=\,\frac{4}{\sqrt{2}}\angle-18.43$$

$$I_1\,=\,-\left(I_L\,+\,I_0\right)\,=-\left[\,\left(\frac{4}{\sqrt{2}}\angle-18.43\right)\,+\,\left(2\angle0\right)\right]$$

$$V_1\,=\,I_1\,Z_1$$

I need this to find $V_s$.

$$V_s\,=\,V_1\,-\,V_0$$

How do I finish? Please help!

 

[VinnyCee]

I did the calculations and conversions of complex rectangular to polar and got this as a final answer:

$$V_S\,=\,9.581\,cos\left(t\,+\,29.70)$$

Does that look right?

 

[m2003]

I would suggest using a combination of KCL and KVL to solve this circuit. First, we can simplify the circuit by combining the two capacitors and two inductors into equivalent impedances. The equivalent impedance for two capacitors in parallel is given by:

Z_{eq}=\frac{1}{j\omega C_1}+\frac{1}{j\omega C_2}

Similarly, for two inductors in series, the equivalent impedance is given by:

Z_{eq}=j\omega L_1+j\omega L_2

Now, we can redraw the circuit with these equivalent impedances:

http://img201.imageshack.us/img201/9359/problem954part2gz3.jpg

Next, we can apply KCL and KVL to the circuit to find the voltage across the current source, V_s. Applying KCL at node A, we get:

\frac{V_s-V_1}{Z_1}+\frac{V_s}{Z_{eq}}=I_0

Applying KVL along the loop at the top, we get:

V_s+V_1+V_0=0

Substituting the value of V_1 from the first equation into the second equation, we get:

V_s+\left(I_0Z_1-Z_{eq}\right)+V_0=0

Solving for V_s, we get:

V_s=-I_0Z_1+Z_{eq}-V_0

Now, we can substitute the values for Z_1, Z_{eq}, and V_0 to get the final expression for V_s:

V_s=-2\angle0\left(2+j4\right)+j\left(\frac{1}{j\omega C_1}+\frac{1}{j\omega C_2}\right)-4\angle45

This can be simplified further by using the given values for the capacitors and inductors. I hope this helps to solve the problem. Keep in mind that there may be other approaches to solving this circuit, and it is always important to double check your calculations and assumptions.