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Help with a Series RL network and Power factor

  1. Oct 2, 2012 #1
    1. The problem statement, all variables and given/known data
    In a 60 Hz series RL network, the source is supplying 400 W of real power. If R = 15.08 Ω and L = 40 mH, find
    a. The power factor. (Indicate if it is leading or lagging)
    b. The apparent power supplied.
    c. The reactive power supplied.


    2. Relevant equations
    None were given


    3. The attempt at a solution
    Ive been trying to google about power factor and reading up on it for quite some time, but I am still stuck. I know I can get PF by taking cosθ, but I am unsure of how to find θ for this particular problem. I believe it can be done by taking cos^(-1) (Real Power/Apparent Power), but again I was not sure of how to find apparent power.

    I guess the biggest problem for me so far is been trying to relate the 60Hz to the rest of the problem. If someone could give me a hint on this I would greatly appreciate it. I had a funeral to attend to so I missed the class discussing most of these things, and its been difficult trying to read up on it online. Any help is appreciated!
     
  2. jcsd
  3. Oct 2, 2012 #2

    gneill

    User Avatar

    Staff: Mentor

    The power factor is the cosine of an angle. This angle is the offset between the current and voltage waveforms. As such, it also happens to be the angle of the impedance vector. If you find the impedance of the series circuit in polar form, the resulting angle will be the one you want. Find its cosine.

    If you are as yet unfamiliar with complex valued impedance, then the same can be accomplished using resistance and reactance. Form the impedance triangle with resistance on the base leg and reactance on the vertical leg. The legs make a right angle. The angle of the hypotenuse (the impedance Z) with the resistance leg is the angle you're after.

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  4. Oct 2, 2012 #3
    I was working on this a bit more after doing some research and this is what Ive found.

    P=(I^2)R, using my values gives me √(400/15.08)= I, which is I = 5.15A.

    Using the equation Z= √(R^2 + (ωL)^2) = Z = √(15.08^2 + (120∏*.040)^2) = Z = 21.33Ω.

    I then use Apparent Power = (I^2)Z which gave me App. Power = 565.72 VA.

    I used PF = (Real Power)/(App. Power) which gave me .707 and its lagging.

    I then used the PF to find θ≈45°.

    I used Reactive Power = 400 tan(45) to see Rea. Power = 400VAR.

    I believe I was doing this correctly, do you all see anything that might have gotten me to be doing this incorrectly? Again, the equations and methods I am using are based off researching from google. The assignment is due tonight so I dont have a lot of time to make sure I am doing this correctly.

    Thanks again!
     
  5. Oct 2, 2012 #4

    gneill

    User Avatar

    Staff: Mentor

    You've found a path to the correct values, so that's fine.

    My own approach would have been to use the given resistor and inductor values to calculate the impedance first, and then take the power factor angle directly from that. With the angle in hand, the power triangle is similar to the impedance triangle (in the geometric sense) and so the apparent and reactive powers are simple to find.
     
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