RLC circuit conversions and angles

Discussion in 'Homework Help' started by notoriusjt2, Jan 27, 2011.

  1. notoriusjt2

    Thread Starter Member

    Feb 4, 2010
    209
    0
    http://www.allaboutcircuits.com/worksheets/ac_sp.html

    I am trying to solve the questions from the link above. I need a clarification on how to find the impedence angles

    Question #6.) This one I can solve OK
    XL=942.48Ω
    Z=1771.5Ω
    R-parallel=2092Ω
    X-parallel=3329.74Ω
    therefore
    L=1.326H

    Zt=1771.5Ω
    θ=tan^-1(R/XL)=32.14 degrees

    Question #7.)
    Xc=1178.93Ω
    Z=1066.55Ω
    R-series=455Ω
    X-series=964.88Ω
    therefore
    C=3.29uF

    Zt=1066.55Ω
    θ=tan^-1(Xc/R)=64.75

    the only reason I was able to find θ on question #7 is because I looked at the answer, then tried to figure out what method they used to get it. But why is...
    Question #6.)θ=tan^-1(R/XL)
    and
    Question #7.)θ=tan^-1(Xc/R)

    what is the correct method in finding the angles?
     
  2. Papabravo

    Expert

    Feb 24, 2006
    10,135
    1,786
    In the z-plane we plot resistance on the horizontal axis and we plot reactance on the vertical (jω) axis. In all cases the angle, measured from the positive horizontal axis, is computed as the arctangent of the reactance over the resistance. The answer key is wrong or there is something else going on.
     
  3. amilton542

    Active Member

    Nov 13, 2010
    494
    64
    Last edited: Jan 27, 2011
  4. notoriusjt2

    Thread Starter Member

    Feb 4, 2010
    209
    0
    so both of these angles should be calculated using the following?

    θ=tan^-1(X/R)
     
  5. amilton542

    Active Member

    Nov 13, 2010
    494
    64
    the easiest method to calculate a vector angle for complex impedances is using polar and rectangular notation which also consists of trigonometry.have you not gone through the links i gave you its all in there
     
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