Capacitor c to have power factor equal to .707

Discussion in 'Homework Help' started by led23, Apr 28, 2009.

  1. led23

    Thread Starter New Member

    Mar 10, 2009
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    for the attachement i am trying to determine
    1. the value of c to have a power factor equal to .707
    2. for this value of capacitance, determine the real power transferred to the 10 K ohm load
    3. what is the apparent power demanded from the source
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    782
    Not much interest in this problem it seems.

    I have ...

    1. C = 39.79 nF

    2. Power = 10 mW

    3. Apparent Power = 14.14 mVar
     
  3. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    782
    Oops - Maybe I should have

    3. Apparent Power = 14.14 mVA

    :rolleyes:
     
  4. led23

    Thread Starter New Member

    Mar 10, 2009
    6
    0
    what is the formula you used to come to c= 39.79 nF? if I may ask
     
  5. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    782
    To achieve a power factor of 0.707 (cos 45°) would require the capacitive reactance to equal the load resistance. If you are not sure why this is the case, draw a load (or power) triangle representing the resistive and reactive terms.

    The required capacitive reactance magnitude is then known from the problem statement as

    |Xc| = Rload = 10k

    Given |Xc| you should be able to then calculate C at 400Hz.
     
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