# [SOLVED] No idea where to start!

• Jul 24th 2008, 10:45 AM
ally79
[SOLVED] No idea where to start!
[Question]
A bridged T-network gives rise to the following equation,

S=R/omega^4L^2C^2 + J{2/omegaC - 1/omega^3LC^2} If the imaginary term
equates to zero, show that, the equation can be rewritten as R=S/4

I apologise if i've written it badly!

I have absolutely no idea of where to start with this problem
If i could get some direction in how to begin to tackle this problem that would be great

I'm sorry if i've posted in the wrong section

Regards

Ally
• Jul 24th 2008, 12:46 PM
Matt Westwood
If the imaginary part equals zero, that means that 2/(omega C) = 1/(omega^3) L C^2).

So you should be able to express omega^2 in terms of L and C. Plugging that into the real part of your equation it should drop into your lap.
• Jul 24th 2008, 12:57 PM
ally79
Many thanks, I'll give it a go and see how I get on.
• Jul 24th 2008, 01:14 PM
ally79
I cant seem to get omega^2 I can only get omega^-2. I end up with 2LC^2/C = omega^-2. I've went wrong somewhere, I just don't know where

Any help is SOOOOOO apprecciated!
• Jul 24th 2008, 01:27 PM
Matt Westwood
Flip it upside down then.
• Jul 24th 2008, 01:34 PM
ally79
so i end up with 2LC^2/C = 1/omega^2?
• Jul 24th 2008, 01:45 PM
Matt Westwood
Try omega^2 = 1/(2LC) (note you can also cancel out a C top and bottom of the other bit).

This is university level electronics? Long time since I did this...
• Jul 24th 2008, 01:49 PM
ally79
This is just some questions I have to do before i am allowed to do my university course, Many thanks fo helping me out with this. I've been stuck on it for two weeks just staring at it hoping divine inspiration would happen. It didn't
• Jul 24th 2008, 01:56 PM
Matt Westwood
But did this help? Did you get the answer in the end?
• Jul 24th 2008, 01:59 PM
ally79
I've not finished the whole thing yet, its 0100hrs here at the moment so i'm going hit the hay and try and finish it in the morning. I'm quite sure i'll be posting tomorrow morning, or should i say later today!
• Jul 25th 2008, 05:55 AM
ally79
I've finished it now, thanks to all who helped me
• Jul 25th 2008, 09:44 AM
Matt Westwood
Please feel free to press the "thanks" button attached to any of the helpers' messages that have helped you. Also, you'll find in "thread tools" at the top of the thread an opportunity to mark the thread as "Solved". This alerts others to the fact that the problem is in fact solved and they won't need to inspect it.