Capacitor Correcting Power Factor

Discussion in 'Homework Help' started by jegues, Sep 24, 2011.

  1. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
    735
    43
    I feel fairly comfortable with this question, however I have some concerns about the signs for certain values throughout the problem.

    First concern, the negative sign on the angle θ from the PF, does it matter that I put it to -36.87°?

    The voltage across the load sits at 0°, and since the PF is lagging, the current should be behind the voltage, hence my reason for the negative sign.

    Note that I did this for the angle for the new PF in part 2), which brings my next question.

    Since I made θ and α negative angles, my reactive portions of the power come out as negative, which is okay but how do I get rid of the j in the impeadance for the capacitor in the expression,

    Q_{cap} = V^{2}(j\omega C)

    Are we only considering the magnitude?

    If that's the case then the j disapeers, but how do I get rid of the negative on the LHS?

    Is this the right way of thinking about it?

    EDIT: I realized I've made mistakes in the first portion of the question.
     
    • QZ1.JPG
      QZ1.JPG
      File size:
      16.4 KB
      Views:
      10
    • QZ1'.JPG
      QZ1'.JPG
      File size:
      50.4 KB
      Views:
      10
    Last edited: Sep 24, 2011
  2. mik3

    Senior Member

    Feb 4, 2008
    4,846
    63
    Note that the angle of I is negative but not the angle used in cos() and sin() since S=VI*.

    To get rid of j you can set it as a 90 degrees angle.
     
Loading...