May 9th, 2018, 07:55 PM  #11 
Global Moderator Joined: Dec 2006 Posts: 19,513 Thanks: 1744 
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, 05:02 AM  #12  
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,112 Thanks: 912  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, 10:49 AM  #13 
Global Moderator Joined: Dec 2006 Posts: 19,513 Thanks: 1744 
If the initial equation is incorrect, that doesn't help.

May 10th, 2018, 11:05 AM  #14 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,112 Thanks: 912 
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, 05:22 PM  #15 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,433 Thanks: 105 
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 08:10 PM. 
May 13th, 2018, 05:58 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 08:14 PM. 

Tags 
problem, transposition, tricky 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Tricky Problem.. Help  Philip56901  Math  2  October 4th, 2015 06:29 PM 
Transposition problem  adamfoley  Algebra  8  February 12th, 2015 04:40 AM 
Tricky problem I need help with  artursm4  Elementary Math  8  January 11th, 2014 08:35 PM 
Some tricky Problem Sum  Need urgent help please !  hellopanda  Elementary Math  1  February 5th, 2012 08:09 PM 
Tricky Algebra Problem  pksinghal  Algebra  2  October 31st, 2010 02:01 PM 