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The relevant circuit is shown below,

enter image description here

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.

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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

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

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

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