# How to find the complex power for a current source?

The relevant circuit is shown below,

We should first find the complex power delivered by the independent source. I derived it but my answer was different than the answer given. So, I am wondering where I went wrong. My work is,

First thing we have to do is find $V_x$. We can see that the current through the inductor is $\dfrac{V_x}{j}$ so the current through the capacitor is then $5cis(30)-\frac{V_x}{j}$. Thus, the current in the 2 ohm resistor by KCL is $5cis(30)-\frac{V_x}{j} +2V_x$. Then doing KVL around the whole loop we get,

$-V_x-j(5cis(30)-\frac{V_x}{j})+2(5cis(30)-\frac{V_x}{j} +2V_x)=0$

Solving the equation for $V_x$, $$V_x= \frac{5cis(120)-10cis(30)}{3+3j}$$ $$S=V_xI^{*}$$ And to find the complex power we need only multiply by the conjugate of the current which is $5cis(-30)$ to get a final answer, $-4.2 +12.5j$ which is different than the given answer. Is my work wrong? I would appreciate it if someone can point out the mistake and/or offer a correction.

Here's how I might approach this problem:

Use superposition to write

$$V_x = 5\angle30\cdot j||(2 - j) + 2V_x \cdot \frac{2}{j - j + 2} \cdot j$$

Gather terms

$$V_x (1 - j2) = 5\angle30\cdot j||(2 - j)$$

Isolate the desired variable

$$V_x = 5\angle30\frac{j||(2 - j)}{(1 - 2j)}$$

Since the complex power delivered is

$$S = V_x \cdot 5\angle{-30}$$

See that

$$S = 25\frac{j||(2 - j)}{(1 - 2j)} = (-7.5 + j10)VA$$

• I found my mistake; it was a simple expansion error. Thanks! – user29568 Mar 22 '14 at 14:29