Can't simplify equation into form (RLC)

[CoolDude420]

Homework Statement

The answer in (algebraic form) is

Homework Equations

The Attempt at a Solution

I'm not sure how to simplfy into the form required from here.

 

[Charles Link]

Suggestion: To start, $1/(j \omega C)=-j/(\omega C )$. What yo have is correct so far. The rest is just algebra.

 

[CoolDude420]

Suggestion: To start, $1/(j \omega C)=-j/(\omega C )$. What yo have is correct so far. The rest is just algebra.

 

[Charles Link]

You are almost there. When you have a $1/j$ multiply by $j/j$ and you get $-j$. That gives you $-j \omega L/(\omega^2 LC-1)$. Then just multiply through upstairs and downstairs by $1-\omega^2 LC$.

 

[CoolDude420]

You are almost there. When you have a $1/j$ multiply by $j/j$ and you get $-j$. That gives you $-j \omega L/(\omega^2 LC-1)$

Yeah, that's what I got too now. Not sure how I'm going to get all those terms on the top like in the final answer. The R is still in the denomintor so I can't even flip and multiply

 

[Charles Link]

See my edited post 4. It's simple.

 

[CoolDude420]

See my edited post 4. It's simple.

Sorry about this. I still can't seem to get it through my head. Do you mean multiply the entire fraction(including the Vs) above and below by that or do I just multiply the fraction in the denomiator by that. I tried both and I'm still getting stuck :(

OR

 

[Charles Link]

The second way, and it immediately gives you the correct answer ! $- j \omega L (1-\omega^2LC)/(\omega^2 LC-1)=j \omega L$.

 

[CoolDude420]

The second way, and it immediately gives you the correct answer ! $- j \omega L (1-\omega^2LC)/(\omega^2 LC-1)=j \omega L$.

Ah! I see. Thank you! It's always the algebra problems..