How is Zp solved for in this problem? (Answer given)

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Adam Bourque
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Given:
Rs = 0.20 Ω
R'r = 0.17 Ω
Xs = 0.50 Ω
X ' r = 0.60 Ω
Rc = 200 Ω
Xm = 15 Ω
Solution:
at Nr = 1165, slip = (Ns- Nr)/Ns = 0.02916 ~2.92%slip
at Nr = 1195, slip = (Ns- Nr)/Ns = 0.00416 ~0.42%slip
Zp = j Xm || ( (R'r / s) + j X'r )
therefore for i) slip s = 0.02916
Zp = 5.2787 <26.36' or 4.7295 + 2.344j
Ztotal = Zs + Zp = ( 0.2 + 0.5j) + (4.7295 + 2.344j) = 5.69<29.98'
What do the double vertical lines mean in this case, and how is the Zp equation solved? Please show work, thank you. Also how was that equation created? (Trying to study, thanks guys)
 
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Adam Bourque said:
Given:
Rs = 0.20 Ω
R'r = 0.17 Ω
Xs = 0.50 Ω
X ' r = 0.60 Ω
Rc = 200 Ω
Xm = 15 Ω
Solution:
at Nr = 1165, slip = (Ns- Nr)/Ns = 0.02916 ~2.92%slip
at Nr = 1195, slip = (Ns- Nr)/Ns = 0.00416 ~0.42%slip
Zp = j Xm || ( (R'r / s) + j X'r )
therefore for i) slip s = 0.02916
Zp = 5.2787 <26.36' or 4.7295 + 2.344j
Ztotal = Zs + Zp = ( 0.2 + 0.5j) + (4.7295 + 2.344j) = 5.69<29.98'
What do the double vertical lines mean in this case, and how is the Zp equation solved? Please show work, thank you. Also how was that equation created? (Trying to study, thanks guys)
Welcome to the PF.

Double vertical or // lines usually means "in parallel with".

Can you post the circuit, and say how you think the problem was solved? Thanks.
 
Adam - do you really expect that somebody can understand how the various symbols are formed into a circuit?
 
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Adam Bourque said:
Zp = 5.2787 <26.36' or 4.7295 + 2.344j
That is not a "less than" sign, it's the angle of the "vector" whose length is 5.2787
An equivalent way to write a vector is as its x-component and y-component, and in this case it is shown as a complex number, 4.7295 + ##j## 2.344
 
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