Finding phase angle of circuit using phasors.

Discussion in 'Homework Help' started by danielb33, Feb 12, 2015.

  1. danielb33

    Thread Starter Member

    Aug 20, 2012
    105
    0
    See circuit in file attached (and problem statement).

    I first found the voltage across the inductor - I * Z = 5.6*<-28.
    Next I simply used V1 - delta V = V2. I found that 100 - 5.6*cos(-28)+ i*sin(-28)*5.6 = ... and simplified the expression and brought back to phasor notation. I did not get the right answer...at all!

    Note that my answer in the image attached was my first attempt. The way I solved above gave a different answer.

    Any help as to where I am going wrong? Thanks!
     
  2. WBahn

    Moderator

    Mar 31, 2012
    17,715
    4,788
    What's the angle of the impedance of the inductor?
     
  3. shteii01

    AAC Fanatic!

    Feb 19, 2010
    3,378
    494
    i would change all the values to the complex notation

    sum all the voltages is equal to zero: (real v1+ imaginary v1) + (real v inductor+ imaginary v inductor) + (real v2+ imaginary v2)=0

    solve for real v2 and imaginary v2

    use trig to find magnitude and angle of v2
     
  4. danielb33

    Thread Starter Member

    Aug 20, 2012
    105
    0
    shteii01 - 100 + 0.5<90 + V2<theta = 0 --> 100 + 0.5*cos(90) + i*sin(90)0.5+V2*cos(theta)+i*sin(theta)*V2 = 0

    How doe this help me?

    Wbahn - impedence of inductor is 0.5<90
     
  5. WBahn

    Moderator

    Mar 31, 2012
    17,715
    4,788
    So now find the voltage across the inductor a bit more carefully.
     
  6. shteii01

    AAC Fanatic!

    Feb 19, 2010
    3,378
    494
    Those are not complex numbers.

    You can do your own thing... invent your own math... if you like. I aint gonna stop u.
     
Loading...