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Phasor voltage

  1. Apr 8, 2005 #1
    Hello,
    now we are in phasors and I am missing something in this problem:
    given a circuit of indep current source (5cos(wt)), R, L (with vL(t)) all in series, no values for them,
    although voltage across inductor vL(t) = 3cos(wt + theta), we need to find theta.

    So, by using impedance for L: Z = jwL, vL(t) = Z*I,
    so I got vL(t) = jwL5 /_ 0 which is also = 3/_theta
    And now what? The answer is 90 degrees, but how do I get it?

    Thanks a lot.
     
    Last edited: Apr 8, 2005
  2. jcsd
  3. Apr 9, 2005 #2
    The reason its 90 degrees is simple... the voltage is indeed ZI, and I has a phase of zero--- but you would do well to convince yourself that the impedance Z is purely imaginary, so you have a phase for your impedance equivalent to the arctan of Im[Z]/Re[Z] where Re[Z] = 0, and thus you have an arctan of infinity, which is undoubtedly equal to 90 degrees.

    Therefore you have V = ZI which yields a phase of 0 + 90, the angular frequency and the amplitudes of Z and I are immaterial for this problem.
     
  4. Apr 9, 2005 #3
    By the by, do you intro EE students still happen to make use of Nilsson and Riedel by any chance?
     
  5. Apr 9, 2005 #4
    Huh???
    I guess, not.

    Thanks for reply!
     
  6. Apr 9, 2005 #5
    Oh sorry, it's just a textbook that I found really great for my intro EE courses.
     
  7. Apr 11, 2005 #6
    Thanks, I might check it out as complementary source. We are using DeCarlo and Lin, since DeCarlo is actually a professor here, I doubt they will change our texbook any time soon :rolleyes:
     
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