Fraction Simplification: Solving Equations with Constants C, L, and R

AI Thread Summary
The discussion focuses on simplifying two equations involving constants C, L, and R to achieve a cleaner form. The user struggles with substituting V1(s) = V2(s)(1+sCR) into the first equation, resulting in a complex expression. They seek assistance in simplifying their work, particularly in the equation for i(s). The goal is to manipulate the equations to reach a more straightforward representation. Help is requested for clearer steps in the simplification process.
jkface
Messages
16
Reaction score
0
I have the following two equations

5bCv09x.png

3Eq3tTM.png


C, L, and R are all constants.

I need to somehow get the above two equations and get it to look like this:

ieBherg.png


But I can't get mine to look that nice. I have V1(s) = V2(s)(1+sCR) and I substituded it into equation 1 and it ends up looking messy. Any help would be appreciated.
 
Physics news on Phys.org
Show us your work and you may get some helpful suggestions.
 
i(s) = V2(s)(1+sCR) / sL + (V2(s)(1+sCR) - V2(s)) / R
i(s) = V2(s)( (1+sCR) / sL + sCR / R)
V2(s) = i(s) (sL / (1+sCR) + 1 / sC) = i(s) ((s^2 * LC + 1 + sCR) / sC(1+sCR))
V2(s) = i(s) ((s^2 * LC + 1 + sCR) / (sC+s^2 * C^2 * R))

this is obviously not what I'm supposed to have.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
Back
Top