3-phase circuits

Discussion in 'Homework Help' started by MareBear, Jan 25, 2009.

  1. MareBear

    Thread Starter Member

    Oct 29, 2008
    22
    0
    im doing the homework my professor assigned on friday for 3-phase circuits and im going over my notes but i dont know how he got the phase current. he divided phase voltage (254) by 10+j15 and got 14.09 A. how did he get 14.09, when i put it in my calculator i get it in polar form. am i missing something?
     
  2. Wendy

    Moderator

    Mar 24, 2008
    20,766
    2,536
    Try drawing the problem as a vector, and solve it using trig.
     
  3. MareBear

    Thread Starter Member

    Oct 29, 2008
    22
    0
    i got it, thanks
     
  4. mhean_ee

    Member

    Aug 25, 2008
    10
    0
    You can also compute for 14.09A by changing first the impedance of 10+15j in a polar form. Which will be resulted to 18.02776 angle 56.31 degrees. but since the voltage will be cosidered at 0' angle, your impedance lags, which makes up a -56.31 degrees.
    Angle -56.31 degrees only denoted the magnitude or direction (which can explain by vector). The 254V now can be divided by 18.02776. Which resulted to 14.09A.
     
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