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AAGR2013

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When I use MVA method instead of PU method, the result is different. Do I miss something? 

[SIZE=16.0pt]ILoad = 120[/SIZE]∠0° / (10+j8) = [SIZE=16.0pt]120[/SIZE]∠0° / 12.8∠38.7° = 9.4∠-38.7°

Q = I2Z = 9.4∠-38.7° * 9.4∠-38.7° * (1+j3+10+j8) = 88.4∠-77.4° * 11√2∠45° = 1375.2∠-32.4° = 1161.1-j736.9. So the active power is 1161.1 and reactive power is 736.9. Which is different from PU method 969+j969.

 
you forget conjugate of current

S=VI*=(IZ)I*=(9.4∠-38.7°)(1+j3+10+j8) (9.4∠+38.7°)=(9.4∠-38.7°)(11*1.414∠45°) (9.4∠+38.7°)=1374.56∠45°=971+j971

 
nice! When calculate apparent power, S=V∠a◦ x I∠b◦, the final result is UI∠ (a-b)◦

 
Last edited by a moderator:
I have a question over this why did they not add the 2 impedances as series so it would be Ztotal = 11+j11 prior to solving out? When I do this I get 927 W and 927 VAR.

 
if you add 2 impedance firstly, how to get the current? 120V is the 10+j8 voltage, not for both.

 

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