s=R/w^4L^2C^2+j(2/wc-1/w^3LC^2)

if the imaginary term equates to 0, show that the equation can be written as R=S/4

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- Jul 30th 2008, 07:46 AMjonlynn3936Desperately need help with random complex number problem
s=R/w^4L^2C^2+j(2/wc-1/w^3LC^2)

if the imaginary term equates to 0, show that the equation can be written as R=S/4 - Jul 30th 2008, 08:06 AMflyingsquirrel
Hello

We have $\displaystyle

S=\frac{R}{\omega^4L^2C^2}+\jmath\left( \frac{2}{\omega C}-\frac{1}{\omega^3 LC^2}\right)$ and we know that the imaginary part of $\displaystyle S$ equals 0. It gives an equation : $\displaystyle \frac{2}{\omega C}-\frac{1}{\omega^3 LC^2}=0$ from which you can get the value of $\displaystyle \omega$. (hint : factor by $\displaystyle \frac{1}{\omega C}$ to solve this equation) Once you've found $\displaystyle \omega$ you're done since $\displaystyle S=\frac{R}{\omega^4L^2C^2}=\ldots$ - Jul 30th 2008, 08:33 AMjonlynn3936
Thanks, but am still a bit confused. Do you treat the sum =0 as you would a normal fraction or do something else to it? Having a brain blip!!!

- Jul 30th 2008, 08:40 AMflyingsquirrel
You can treat it as a normal sum of fractions : bring the two fractions to the same denominator and then solve $\displaystyle \text{numerator}=0$. You can also follow the hint I've given : factoring by $\displaystyle \frac{1}{\omega C}$ should give you two equations (one of which has no solutions) which can be solved for $\displaystyle \omega$.

- Jul 31st 2008, 03:26 AMjonlynn3936thanks
Cheers for your help, was REALLY obvious and couldn't see the solution for looking at it. TOP MAN!!!