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find phasor current

Discussion in 'Electronics Homework Help' started by Ledwardz, Dec 24, 2010.

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  1. Ledwardz

    Ledwardz

    43
    0
    Dec 21, 2010
    Hi, we have been given exam papers to revise from with no solutions....... yes i know its stupid and means i don't know if what i am doing is right.


    ive attached the question and here is my answer


    4.46A with a phase angle of 30.38.

    but if im doing it wrong i need to know........

    my working is as follows:

    Vt = 25v to angle of 53 + 33v with no angle
    = 58v to angle 53

    The magnitude of impedance = 13 (square root of 5 +12 squared)
    the angle of impedance = 22.62 degrees. (arctan 5/12)

    then i = V/Z

    so i get 4.46 < 30.38 A

    if any one could confirm that id appreciate it. Thanks, Lee.
     

    Attached Files:

  2. barathbushan

    barathbushan

    223
    0
    Sep 26, 2009
    The whole method of solving the problem is right and the calculation of IMPEDENCE is also right , but the voltage computation is absolutely
    wrong !!

    i wont re-solve the entire problem , but i will tell you that , you just cant add the voltage magnitudes and angles like you have done , as it is clearly shown here

    http://answers.yahoo.com/question/index?qid=20090421045522AAnABsZ

    so learn your complex algebra well , and you will be able to tackle such problems
     
  3. Ledwardz

    Ledwardz

    43
    0
    Dec 21, 2010
    soo

    i should get Voltage = (58cos53 + j53sin53)
    and impedence = (13cos22.62 + j13sin22.62)

    then i got rid of sin n cos

    (34.9 + j46.3) / (12 + j5)

    then multiply by complex conjugate (12 - j5)

    which = 3.85 + j2.25

    magnitude = 4.46
    angle = 30.3

    so answer = 4.46 to angle of 30.3 A ???
     
  4. Ledwardz

    Ledwardz

    43
    0
    Dec 21, 2010
    ahh crap i got same answer...... nm thats obviously wrong ill have another look. AHHHH.... i get it ha ha ha ha yeah okay u mean adding the voltages at the start is wrong h aha ha ha well i look like an idiot now. o well.
     
    Last edited: Dec 25, 2010
  5. Ledwardz

    Ledwardz

    43
    0
    Dec 21, 2010
    okay ill try again

    voltage magnitude = 52V
    and angle = 22.6

    I think that should be right?
     
  6. barathbushan

    barathbushan

    223
    0
    Sep 26, 2009
    no its wrong , let me give you the answer its
    52.63<24.15 volts

    one more method of solving the addition is converting both the voltages
    25<53 and 33<0
    to rectangular form , then add the real and imaginary parts
    after that re-convert them to the polar form

    use the above method is used if you have difficulty with polar form , but it's
    not a good way to solve , please go through basic complex algebra
     
  7. Ledwardz

    Ledwardz

    43
    0
    Dec 21, 2010
    i really dont know where i am going wrong as far as i am aware adding in polar goes something like this. i tried using example but exchanged the sin and cos for their actual numbers so when i squared it was easier.

    =(25cos53 + j25sin53) + (33cos0)
    = 15.04.. + j 19.966.. + 33
    = 48.04.. + j19.966..
    then magnitude is square root of em squared
    n arctan for angle.

    have i got it all wrong? It seems near the answer, maybe i have just done something stupid?
     
  8. barathbushan

    barathbushan

    223
    0
    Sep 26, 2009
    yes that's the answer just convert 48.04 + 19.996j into polar form , then you will get the right answer of 52.63<24.15 volts

    magnitude = sqrt (48.04^2 +19.996^2 )=52.02
    angle = tan-1(19.996/48.04) =22.56

    ie 52.02<22.56 (approx to my answer of 52.63<24.15 )
     
  9. Ledwardz

    Ledwardz

    43
    0
    Dec 21, 2010
    Thanks! i appreciate the help.
     
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