May 9th, 2018, 08:55 PM  #11 
Global Moderator Joined: Dec 2006 Posts: 20,089 Thanks: 1902 
Where did "t" come from, given that there's no "t" in the image? Also, "that point in the pic, the middle bit" is too vague to be clear. How come you give an equation with Z in the denominator even though Z appears only as the lefthand side of equations in the image? 
May 10th, 2018, 06:02 AM  #12  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,782 Thanks: 970  Quote:
So we have: 1 / (ab) = c + 1/R and you want the (b) on the RHS, so leaving LHS = 1/a :YOKAY?! Here we go: 1 / (ab) = c + 1/R change RHS to a fraction: 1 / (ab) = (cR+1) / R crisscross multiplication: R = (ab)(cR+1) Continue... R = acR + a  bcR  b R = a(cR+1)  b(cR+1) R + b(cR+1) = a(cR+1) [R + b(cR+1)] / a = cR+1 1 / a = (cR+1) / [R + b(cR+1)] That's it...you can substitute back in if you wish... Q: what d'you call an Irishman that bounces off the walls? A: Rick O'Shea  
May 10th, 2018, 11:49 AM  #13 
Global Moderator Joined: Dec 2006 Posts: 20,089 Thanks: 1902 
If the initial equation is incorrect, that doesn't help.

May 10th, 2018, 12:05 PM  #14 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,782 Thanks: 970 
True nuff. Well, at least Kevin can "see" a few ways/tips to "rearrange": I noticed that the ones he showed are not ending up correctly. 
May 10th, 2018, 06:22 PM  #15 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,621 Thanks: 117 
The impedance of an RLC parallel circuit is: $\displaystyle \frac{1}{Z}=\frac{1}{R}+\frac{1}{j\omega C}+j\omega L$ Solve for Z in the form Z = A + jB. EE's use j instead of i (current). Is that what this is all about? This is a trivial problem in complex algebra. You need to know how to add, subtract, and divide complex numbers. EDIT: $\displaystyle a+\frac{1}{jb}+jc=a +\frac{1+jbjc}{jb}=a+\frac{1bc}{jb}\cdot \frac{j}{j}=a+\frac{bc1}{b}j$ Last edited by skipjack; May 13th, 2018 at 09:10 PM. 
May 13th, 2018, 06:58 PM  #16 
Newbie Joined: May 2018 From: Mt Eliza Victoria Australia Posts: 1 Thanks: 0 
Kevin, the answer to your question is: Z = wR^2 LC/(R^2 + w C)^2 + j w^2 R L C^2/(R^2 + wC)^2 Firstly, Z = w^2 RLC/(R + JwC). Now multiply numerator and denominator by (R  jwC), which is the complex conjugate of (R + jwC). The denominator now becomes (R^2 + wC)^2 which is real since j^2 = 1. You now have Z in the form A + j B I hope that helps. Good wishes, Tim06 Last edited by skipjack; May 13th, 2018 at 09:14 PM. 

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