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.


1 Answer 1


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

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

  • \$\begingroup\$ I found my mistake; it was a simple expansion error. Thanks! \$\endgroup\$
    – user29568
    Mar 22, 2014 at 14:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.