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June 18th, 2017, 09:58 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/(10j40)+1/1201/12(1j4) the ans is 91.2 j38.4 Thank you for any help 
June 18th, 2017, 11:02 AM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,262 Thanks: 1198 
I don't get anything like your answer so please repost the expression using parentheses to make it's interpretation unique.

June 19th, 2017, 12:50 PM  #3 
Newbie Joined: Jun 2017 From: N/A Posts: 4 Thanks: 0  how about this
How we did in class was different from the text but maybe this will help, and the equation was It=Vt[(1/(10j40))+(1/120)(1/12(1j4))] the Zth=Vt/It = 91.2 j38.4 Sorry I messed that question up. 
June 19th, 2017, 07:32 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,089 Thanks: 1902 
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 
June 19th, 2017, 07:46 PM  #5 
Senior Member Joined: Sep 2015 From: USA Posts: 2,262 Thanks: 1198 
ok so you want the reciprocal of the sum of those 3 terms. $Z^{1}=\dfrac {1}{10}\dfrac{1}{1j4} + \dfrac{1}{120} \dfrac{1}{12(1j4)}$ $Z^{1} = \dfrac{1}{10}\dfrac{1+j4}{17}+\dfrac{1}{120}\dfrac{1+j4}{12\cdot 17}$ $Z^{1} = \left(\dfrac{1}{170}+\dfrac{1}{120}\dfrac{1}{204}\right) + j \left(\dfrac{4}{170}  \dfrac{1}{51}\right)$ $Z^{1} = \dfrac{19}{2040} + j\dfrac{1}{255}$ $Z^{1} = \dfrac{1}{24 \sqrt{17}}e^{\arctan\left(\frac{8}{19}\right)}$ $Z = 24\sqrt{17} e^{\arctan\left(\frac{8}{19}\right)}$ $Z = \dfrac{456}{5}j\dfrac{192}{5}$ $Z \approx 91.2\, j38.4 $ 

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obtaining, phasors, solution 
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