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 Linear Algebra Linear Algebra Math Forum

 May 9th, 2018, 07:55 PM #11 Global Moderator   Joined: Dec 2006 Posts: 20,373 Thanks: 2010 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 left-hand side of equations in the image? May 10th, 2018, 05:02 AM   #12
Math Team

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From: Ottawa Ontario, Canada

Posts: 14,124
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Quote:
 Originally Posted by Kevineamon 1/(Zt-jwl) = jwc + (1/R) All I want to do is get that (-jwl) over to the RHS
Let a = Zt, b = jwl, c = jwc (this makes process EASIER/SHORTER)

So we have:
1 / (a-b) = c + 1/R
and you want the (-b) on the RHS, so leaving LHS = 1/a :YOKAY?!

Here we go:
1 / (a-b) = c + 1/R
change RHS to a fraction:
1 / (a-b) = (cR+1) / R
criss-cross multiplication:
R = (a-b)(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: 20,373 Thanks: 2010 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: 14,124 Thanks: 1003 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 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 124 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: $\displaystyle a+\frac{1}{jb}+jc=a +\frac{1+jbjc}{jb}=a+\frac{1-bc}{jb}\cdot \frac{j}{j}=a+\frac{bc-1}{b}j$ Thanks from topsquark 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 Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Philip56901 Math 2 October 4th, 2015 06:29 PM adamfoley Algebra 8 February 12th, 2015 04:40 AM artursm4 Elementary Math 8 January 11th, 2014 08:35 PM hellopanda Elementary Math 1 February 5th, 2012 08:09 PM pksinghal Algebra 2 October 31st, 2010 02:01 PM

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