Engineering Finding the parallel resonance formula

[arhzz]

Homework Statement

Find the formula for the parallel resonance

Relevant Equations

The Admittance

Hello! I'm having trouble with getting the right result in this litle example. Consider this admittance

$$ C + Cs - w^2_{pr} CCs $$ Now to get the resonance we need to set the imaginary part of the admittance 0.I did that like this

$$0 = C + Cs - w^2_{pr} CCs $$ Now I need to get $w^2$ I've added the $-w^2 CCs$ to both sides of the equation. Now I have this

$$ w^2 LCCs = C+Cs $$ Now I put the LCCs on the other side of the equation
$$ w^2 = \frac{C+Cs}{LCCs} $$ Now get rid of the square on the left and we get this
$$ w = \sqrt{\frac{C+Cs}{LCCS}} $$ And now I'd assume some simplifications are made because the result looks like this

$$ w = \frac{1}{\sqrt{LC}}\sqrt{ 1 + \frac{C}{Cs}} $$

I've tried splitting the fraciton inside the root to simplify but I am not getting there.Some help would be great,thank you!

 

[Jody]

Would it help if you could do something like this?
$$\frac{1}{\sqrt{8}} = \frac{1}{\sqrt{2 \cdot 4}} = \frac{1}{\sqrt{2}} \frac{1}{\sqrt{4}}$$

 

[arhzz]

Would it help if you could do something like this?
$$\frac{1}{\sqrt{8}} = \frac{1}{\sqrt{2 \cdot 4}} = \frac{1}{\sqrt{2}} \frac{1}{\sqrt{4}}$$

Well it certainly would but I am not seeing such a scenario here? I don't have 1 over something and I don't know if I can split anyting the way you split the root of 8.

 

[Jody]

Well, in your final answer or what you're trying to massage it to you have a $1/\sqrt{LC}$. Is it possible from the step you have before that... to pull out a $1/\sqrt{LC}$?

Just a reminder of what you currently have in your original post, right?
$$\omega = \sqrt{ \frac{C + C_{S}}{LC \cdot C_{S}} } = \sqrt{ \frac{1}{LC} \frac{C + C_{S}}{C_{S}}}$$

 

[arhzz]

Well, in your final answer or what you're trying to massage it to you have a $1/\sqrt{LC}$. Is it possible from the step you have before that... to pull out a $1/\sqrt{LC}$?

Just a reminder of what you currently have in your original post, right?
$$\omega = \sqrt{ \frac{C + C_{S}}{LC \cdot C_{S}} } = \sqrt{ \frac{1}{LC} \frac{C + C_{S}}{C_{S}}}$$

$$ \omega = \sqrt{\frac{C+C_{S}}{LC \cdot C_{S}}} = \frac{1}{\sqrt{LC}} \cdot \sqrt{ 1 + \frac{C}{C_{S}}} $$

This is what I need to "show" or achieve if you will.Your post is close,we could split the root but we would have $\frac{C+Cs}{Cs}$ and it should be $1 + \frac{C}{Cs}$

 

[Jody]

Is there another way you can write?
$$\frac{C + C_S}{C_S}$$
You're almost there.

 

[LvW]

I must admit that I do not understand your equations at all.
Is "C" an admittance? And what about "w²CCs" ? Also an admittance? Both appear within one formula...

 

[DaveE]

I know a lot about resonant circuits, yet your presentation of this problem has greatly exceeded my interest in deciphering your work. You'll get better answers if you ask better questions.

There appears to be some problems with dimensions, or with my interpretation of what L, C, ω, or s are. If you use those the way most EEs do then your equations don't make sense. For example, I would bet that no one reading this knows what $ω^2_{pr}$ is; although you imply it's also equal to $ω^2L$, whatever that is.

 

[arhzz]

Is there another way you can write?
$$\frac{C + C_S}{C_S}$$
You're almost there.

Ah I got it $$ \frac{C}{Cs} + \frac{Cs}{Cs} $$ the second fraction will cancel out leaving me with

$$\frac{C}{Cs} + 1 $$

Thanks !

 

[arhzz]

I know a lot about resonant circuits, yet your presentation of this problem has greatly exceeded my interest in deciphering your work. You'll get better answers if you ask better questions.

There appears to be some problems with dimensions, or with my interpretation of what L, C, ω, or s are. If you use those the way most EEs do then your equations don't make sense. For example, I would bet that no one reading this knows what $ω^2_{pr}$ is; although you imply it's also equal to $ω^2L$, whatever that is.

Maybe I rushed it a little,this is just the last part of a pretty big example (with a circuit ).I was not sure what to put in the title of my question because the only problem I was facing is just the math to get the simplification.

 

[Jody]

I think you were fine. I'm glad you solved the problem :) great job