Please help me understand this.

Discussion in 'General Electronics Chat' started by tjterblanche, Jun 23, 2009.

  1. tjterblanche

    Thread Starter New Member

    Jun 23, 2009
    3
    0
    Hi All,

    Referring to Question 9 of the worksheet titled "Series and parallel AC circuits" (http://www.allaboutcircuits.com/worksheets/ac_s_p.html),
    I get the phase angle at -42.08 by doing the following steps:

    1.) Calculated Xc = 0 - j1354.51Ω
    2.) Calculated the phase difference at -42.08° using θ = arctan Xc/R

    The worksheet answer is -47.9°, which seems like it's a complimentary
    angle to -42.08°

    Could anybody please help me understand why the phase difference is
    not -42.08° as calculated in step 2 above?

    Sincerely,
    tjterblanche
     
  2. GetDeviceInfo

    Senior Member

    Jun 7, 2009
    1,571
    230
    because you are basically measuring across the capacitor, you will be referring to the reactive component voltage, which for the angle you would then use arcsine(Xc/Z).

    oops, that's wrong, you would use arcsine(R/Z)
     
    Last edited: Jun 23, 2009
  3. tjterblanche

    Thread Starter New Member

    Jun 23, 2009
    3
    0
    It seems like you're calculating the complementary angle to 42.08° when you use θ=arcsine R/Z?

    IMHO there are two ways to draw the phasor triangle for any circuit with a single reactive component.

    1.) You start by drawing the reactive component's phasor first (from the origin); You then draw the resistive phasor with the end of the reactive phasor as its origin;

    2.) You draw the resistive phasor first; You then draw the reactive phasor with the end of the resistive phasor as its origin.

    Both these phasor diagrams will have the exact same hypotenuse (magnitude and angle), but their right angles are at opposite sides of the hypotenuse.

    Which phasor diagram do you use and why, because it determines the angle that θ points to?
     
  4. tjterblanche

    Thread Starter New Member

    Jun 23, 2009
    3
    0
    I wanted to include this as an example:

    The VERY FIRST phasor diagram on the Series Resistor-Capacitor Circuits page (http://www.allaboutcircuits.com/vol_2/chpt_4/3.html) shows the phase angle between E_{R} and E_{T} to be -79.3°.

    The way I understand it, θ=arcsine R/Z would calculate the phase angle between E_{C} and E_{T}
     
  5. GetDeviceInfo

    Senior Member

    Jun 7, 2009
    1,571
    230
    I think your reading too much into it. There is only one reactive component, and by convention, the right angle is to the left.
     
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