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Value of parallel capacitance necessary to achieve a Power F. of unity

  1. May 2, 2013 #1
    1. The problem statement, all variables and given/known data

    The electrical model of an operating AC induction motor yields an impedance, Z, of 18.65+j15.106Ω . What value of parallel capacitance is necessary to achieve a power factor of unity if the operating frequency is 60Hz?

    2. Relevant equations

    V=IR
    w=2pif
    P=VI

    3. The attempt at a solution

    I know the reactance of a capacitor is 1/jwC. And impedances add like 1 over the sum of reciprocals.
     
  2. jcsd
  3. May 2, 2013 #2

    rude man

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    Homework Helper
    Gold Member

    Put the complex impedance of the motor in parallel with a capacitor. Then force the parallel combination Z (or Y) to be real by appropriate selection of the capacitance.

    PS if you haven't had phasors you'll have to write the differential equation for the total current given a sinusoiudal voltage excitation, and solving for the value of C that will give zero phase shift between excitation voltage and total current.
     
  4. May 2, 2013 #3
    so far I have the impedance of the parallel combination as:

    [itex]\frac{1}{\frac{1}{Z}-\frac{ωC}{j}}[/itex]

    where Z is the given complex impedance of the motor. What can I do from here?
     
  5. May 2, 2013 #4
    Just to clarify, I think when it says "power factor of unity" it means a ratio of 1:1
     
  6. May 2, 2013 #5

    rude man

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    Gold Member

    "power factor of unity" means zero imaginary component of the total impedance (current in phase with the voltage).

    Your expression for total Z is correct. Now substitute the given expression for motor Z, compute C such that the imaginary part of the total Z = 0. Remember what w is.
     
  7. May 3, 2013 #6
    [itex]\frac{1}{0.032377-j0.026225+377jC}[/itex]

    So.....if I do.....

    -0.026225+377C=0

    and solve for C...

    C=69.56μF

    Math correct?
     
  8. May 3, 2013 #7

    gneill

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    Staff: Mentor

    The value that you found is correct.

    Another approach might be to first turn the motor's impedance into its equivalent admittance; Y = (1/Z). Parallel admittances simply add (like conductances do), rather than all that inverse of the sum of the inverses stuff you need for impedances. That means you can pick out the magnitude of the required admittance of the capacitor directly from the admittance of the motor -- it's the imaginary term. Capacitor admittance is just ##j2\pi f C = j \omega C##.
     
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