Hey guys... I've decided to finally tackle this monster of a module. So far I've managed to get half way through one question lul... if I didn't laugh I'd cry... no seriously

Anywho... I'm tacking question 2 of the tutorial, I put on the other thread... the math is attached here with basic schematic. I've managed to logically understand most of it.
However see this part over on the right hand side? Which begins:-

W^2 = (R^2 C - L) / (R^2 C^2 L)

Can someone please explain the next stage of the equation.
I'd have thought for example that the R^2 cancel.
The denominator should have C^1
Who does the 1 belong to?
The Ls seem to completely disappear then reappear a stage later.

WTH???

So confused... think I'm taking a allergic reaction.

 

(R^2 C - L) / (R^2 C^2 L)

The R^2 don't cancel because of the '-L' part.
If you divide top and bottom by R^2 you get: (C - L/R^2) / (C^2 L)
For the same reason you can't cancel the C either.

 

Doh sorry brain dead here atm...

Ok so using that method here's what I get

 

Sorry please ignore this a moment. Need to rethink

 

K sorry. Here' what I get

 

(R^2 C - L) / (R^2 C^2 L)

Try splitting it like this:
(R^2 C) / (R^2 C^2 L) - L / (R^2 C^2 L)
then you can do some cancelling.

 

Hmm the resistor ends up cancelling itself

 

Dang... that's not right either. Such a muppet. Please ignore those pics above.

 

This is as far as I can get everything else after it is a path of madness.

I probably need to sleep tbh.

 

Hi,

Why start with w^2?

 

Because I'm trying to follow something the lecturer did. Check out the original post

 

This is as far as I can get everything else after it is a path of madness.

I probably need to sleep tbh.

View attachment 146393

If you multiply both of these by CL, you also need to multiply the ω^2 by CL as well, but I don't think you will find that helpful.

 

Because I'm trying to follow something the lecturer did. Check out the original post

Hi,

Oh ok, well the "other thread" could be any thread so we dont know

All you have to do though is combine impedances.

 

I made sure all the required info is on this thread MrAI. I just mentioned the other thread because I have all the questions for the module there. It was q2.
I spoke to the lecturer today, he seemed to indicate that the 2 parts are seperate things with no relation. And he has pulled the value = zab from an earlier part of the equation. He asked me did I understand now? and I said yes... But I don' really tbh. I just didn't want to sound even more stupid.

Sigh...4 hours of a waste... trying to arrive at an impossible calculation... lul.

It does look remarkably similar though... Perhaps it' a combination.

 

I made sure all the required info is on this thread MrAI. I just mentioned the other thread because I have all the questions for the module there. It was q2.
I spoke to the lecturer today, he seemed to indicate that the 2 parts are seperate things with no relation. And he has pulled the value = zab from an earlier part of the equation. He asked me did I understand now? and I said yes... But I don' really tbh. I just didn't want to sound even more stupid.

Sigh...4 hours of a waste... trying to arrive at an impossible calculation... lul.

It does look remarkably similar though... Perhaps it' a combination.

Hi,

Ok well the images are a little hard to read. If you could write it out in text it would be nice.
It's up to you though if you want to take the time.

From your first post all it looks like you have to do it divided top and bottom by C and you'll get C^1 in the denominator.
It's hard to see what you are really after here though, maybe it is just algebra.

 

Ha don't chat to me about his writing, believe me you're preaching to the converted. You're probably in a much better position than me, to try and interpret exactly what it means.
But anyways - Never worry about it MrAI. I think I've resigned myself to the fact that my marks in this module will be low.
Now I freaking hate low marks! believe me... although I seem bad at this, I study Very hard.
This module is going to collapse my overall grade for the year. :Angry smiley:

Shush... I can already feel my blood pressure spiking! Breatheeeeee.... yes be as one and all that jazz.

Maybe you could test me instead give me some practice? The hardest section of the module is deriving first order differential equations.

Only if you want... if you haven't the time it's no problem.

 

Ha don't chat to me about his writing, believe me you're preaching to the converted. You're probably in a much better position than me, to try and interpret exactly what it means.
But anyways - Never worry about it MrAI. I think I've resigned myself to the fact that my marks in this module will be low.
Now I freaking hate low marks! believe me... although I seem bad at this, I study Very hard.
This module is going to collapse my overall grade for the year. :Angry smiley:

Shush... I can already feel my blood pressure spiking! Breatheeeeee.... yes be as one and all that jazz.

Maybe you could test me instead give me some practice? The hardest section of the module is deriving first order differential equations.

Only if you want... if you haven't the time it's no problem.

Hi again,

You mean first order diff equations for electrical circuits?
Sure i could provide a few for you for exercises if you like.

 

Yes that' the ones exactly MrAI. This is an electronics module I'm studying. Now I must warn you I'm completely new to derivations and will probably ask what seem to be the most stupid questions ever, to start with. I have a weird ability to comprehend the complicated and be lost on the simple. But if you walk me through one or two, I'll probably be fine.

 

@KevinEamon what you need to do is to replace that \(\displaystyle{\omega^{2}}\) in the \(Z_{AB}\) equation.

 

K I'll have a look at that when I get home. Is that value I have in post 9 correct?