Single Phase Generator (Power Factor)

Discussion in 'Homework Help' started by jegues, Sep 24, 2011.

  1. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
    735
    43
    See figure attached for my attempt at part b).

    What am I doing wrong? My voltage for the generator is slighty off.
     
  2. mik3

    Senior Member

    Feb 4, 2008
    4,846
    63
    I has an angle equal to -arccos(PF).
     
  3. Zazoo

    Member

    Jul 27, 2011
    114
    43
    You chose to use the load current as your reference (0°), but it looks like in calculating -Vg + VTL + VL = 0 you switched to load voltage as the reference (VL = 860<0°).
     
  4. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
    735
    43
    That was it!

    Thanks Zazoo!
     
  5. nyasha

    Active Member

    Mar 23, 2009
    90
    1
    jegues, how did you get the impedance of the transmission to be 0.2+0.4j
     
  6. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
    735
    43
    (2 + j4)mΩ/km

    There's 100km,

    (2 + j4)mΩ/km * 100km = (0.2 + j0.4)Ω
     
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