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June 18th, 2017, 08:58 AM   #1
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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
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June 18th, 2017, 10:02 AM   #2
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I don't get anything like your answer so please repost the expression using parentheses to make it's interpretation unique.
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June 19th, 2017, 11:50 AM   #3
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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/(10-j40))+(1/120)-(1/12(1-j4))]
the Zth=Vt/It = 91.2 -j38.4

Sorry I messed that question up.
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June 19th, 2017, 06:32 PM   #4
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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
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June 19th, 2017, 06:46 PM   #5
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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 $
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