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 Complex Analysis Complex Analysis Math Forum

 June 18th, 2017, 08:58 AM #1 Newbie   Joined: Jun 2017 From: N/A Posts: 4 Thanks: 0 obtaining a solution using phasors I feel good about the basics transformation of complex numbers to polar and ec and vice versa. I have hit a wall in trying to solve the equation by hand. I feel like it should just be adding the fractions and if the complex values are to be multiplied then change to polar but must be wrong or my calculations are wrong 1/(10-j40)+1/120-1/12(1-j4) the ans is 91.2 -j38.4 Thank you for any help June 18th, 2017, 10:02 AM #2 Senior Member   Joined: Sep 2015 From: USA Posts: 2,549 Thanks: 1399 I don't get anything like your answer so please repost the expression using parentheses to make it's interpretation unique. Thanks from topsquark June 19th, 2017, 11:50 AM   #3
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How we did in class was different from the text but maybe this will help, and the equation was
It=Vt[(1/(10-j40))+(1/120)-(1/12(1-j4))]
the Zth=Vt/It = 91.2 -j38.4

Sorry I messed that question up. Attached Images forum pic 1.jpg (20.1 KB, 2 views) forum pic 2.jpg (15.3 KB, 3 views) June 19th, 2017, 06:32 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,966 Thanks: 2216 1/(10 - 40j) + 1/120 - 1/(12(1 - 4j)) = (10 + 40j)/1700 + 1/120 - (1 + 4j)/204 = 1/170 + 1/120 - 1/204 + (4/170 - 1/51)j = (12 + 17 - 10)/2040 + ((12 - 10)/510)j = 19/2040 + 1/255j = 1/2040(19 + 8j) Hence 1/(1/(10 - 40j) + 1/120 - 1/(12(1 - 4j))) = 2040/(19 + 8j) = 2040(19 - 8j)/425 = 48/10(19 - 8j) = 91.2 - 38.4j Thanks from bdewbac June 19th, 2017, 06:46 PM #5 Senior Member   Joined: Sep 2015 From: USA Posts: 2,549 Thanks: 1399 ok so you want the reciprocal of the sum of those 3 terms. $Z^{-1}=\dfrac {1}{10}\dfrac{1}{1-j4} + \dfrac{1}{120} -\dfrac{1}{12(1-j4)}$ $Z^{-1} = \dfrac{1}{10}\dfrac{1+j4}{17}+\dfrac{1}{120}-\dfrac{1+j4}{12\cdot 17}$ $Z^{-1} = \left(\dfrac{1}{170}+\dfrac{1}{120}-\dfrac{1}{204}\right) + j \left(\dfrac{4}{170} - \dfrac{1}{51}\right)$ $Z^{-1} = \dfrac{19}{2040} + j\dfrac{1}{255}$ $Z^{-1} = \dfrac{1}{24 \sqrt{17}}e^{\arctan\left(\frac{8}{19}\right)}$ $Z = 24\sqrt{17} e^{-\arctan\left(\frac{8}{19}\right)}$ $Z = \dfrac{456}{5}-j\dfrac{192}{5}$ $Z \approx 91.2\, -j38.4$ Thanks from bdewbac Tags obtaining, phasors, solution 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 KaiL Calculus 3 February 16th, 2017 05:39 AM kgiboin Calculus 2 March 24th, 2013 05:15 PM Chikis Advanced Statistics 56 July 30th, 2012 03:43 AM senogles100 Calculus 5 May 1st, 2012 01:25 AM senogles100 Algebra 3 December 31st, 1969 04:00 PM

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