1
\$\begingroup\$

A homework problem tells me to find the equivalent impedance, so I added the series and parallel impedances just like I would a resistor; however, I don't understand how does one go from \$100/(12-16j)\$ to \$3+4j \Omega\$. My book just jumps to that conclusion. Can anyone please explain?

\$\endgroup\$
1
  • \$\begingroup\$ multiply top and bottom by the complex conjugate of 12-16j and simplify. \$\endgroup\$
    – crgrace
    Commented May 7, 2015 at 3:10

1 Answer 1

0
\$\begingroup\$

To divide by a complex number, use the conjugate of the denominator. like so:

$$ \begin{align} {100\over (12 - 16j)} & = {100\over (12 - 16j)} {(12 + 16j)\over (12 + 16j)} \\ \\ & = {(1200 + 1600j)\over (144 + 256)} \\ \\ & = {(1200 + 1600j)\over 400} \\ \\ & = {(3 + 4j)\over 1} \\ \\ \end{align} $$

\$\endgroup\$

Your Answer

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

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