My Math Forum Tricky transposition problem

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 May 9th, 2018, 07:55 PM #11 Global Moderator   Joined: Dec 2006 Posts: 19,162 Thanks: 1638 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
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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: 19,162 Thanks: 1638 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: 12,751 Thanks: 860 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: 1,390 Thanks: 100 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.

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