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Find V2 in Single Phase Transformer

  1. Dec 7, 2016 #1
    media%2F91c%2F91c7c515-c3ed-4661-8dfb-fcb78ae40a8a%2FphpA4NkXn.png 1. The problem statement, all variables and given/known data
    Find V2. Everything else is available in the picture

    2. Relevant equations
    r1=r1+a^2*r2
    x1=x1+a^2*r2
    z2=a^2*r2
    I1=vp/z2+r1+x1

    3. The attempt at a solution
    So I have used the first three formulas but it's the last one I am getting stuck on.
    Since my x1 is an imaginary number how do I do this?
    I'm currently stuck at I1=1200/192+4+j16
    Can someone help me?

    EDIT: VP=1200 NOT 1000 LIKE THE PICTURE STATES. EVERYTHING ELSE IS THE SAME
     
  2. jcsd
  3. Dec 7, 2016 #2

    cnh1995

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    The impedance 192+4+j16 can be written as 196+j16.
    Convert this into polar form. For multiplication and division of two phasors, you need their polar forms.
     
  4. Dec 8, 2016 #3
    You can also solve as:
    I1 = 1200/(196 + j6). Multiply numerator and denominator by 196 - j6.
    This will remove 'j' from denominator and then you can proceed.
    j * j = -1.
     
  5. Dec 8, 2016 #4
    Okay I got.
    But then wouldn't I still have "j" in my numerator?
     
  6. Dec 8, 2016 #5

    cnh1995

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    Yes. If you want to keep the answer in complex form, you should keep it in the form
    Ireal+jIreactive. Or you can simply convert it into polar form.
     
  7. Dec 8, 2016 #6
    Okay. It's my first time working with Complex numbers. Just to confirm.
    In order to do 1200/196+j16 I have to convert the numerator into polar form first. Reason for this is because 1200 is basically 1200∠0 which is polar form. You can not divide polar by rectangular.

    Can you divide rectangular by polar?
     
  8. Dec 8, 2016 #7

    cnh1995

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    Right.
    Yes and no.
     
  9. Dec 8, 2016 #8
    T
    Thank you, been a big help?
     
  10. Dec 8, 2016 #9
    It's easier to convert into polar form when dividing or multiplying, and to convert to rectangular form when adding or subtracting. But if the denominator has simple terms like 1 + i or 1 - i then multiplying num and den by 1- i or 1 + i respectively is quicker than using calculator.
     
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