Capacitor Correcting Power Factor

Discussion in 'Homework Help' started by jegues, Sep 13, 2011.

  1. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
    735
    43
    Everything in this problem is working out fine in my brain except for the last part where he determines the capcitance.

    Q_{cap} = 2.551 kVAR

    but how does

    Q_{cap} = V^{2} \omega_{c} \quad \text{?}

    I'm thinking,

    V_{c} = \frac{Q}{C}

    What is he doing here?

    I know that \omega_{c} = \omega since we assume steady state but where is that coming from?

    Thanks again!
     
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  2. Peytonator

    Active Member

    Jun 30, 2008
    105
    3
    At steady state:

    Qc = (Vc^2)/Zc = 2.551 kVAR.

    Now,

    Zc = 1/(2*pi*f*C)

    Remember that

    w = 2*pi*f

    Thus,

    2.551*10^3 = (2*pi*f*C) * V^2 = V^2 * w * C

    Solve for C and you get the answer.
     
    Last edited: Sep 13, 2011
  3. mrmount

    Active Member

    Dec 5, 2007
    59
    7
    The Qcap he has referred to in the formula is reactive power (not charge).
     
  4. nyasha

    Active Member

    Mar 23, 2009
    90
    1
    I am doing a similar question from the textbook. However, l am having a hard time coming up with a relationshiop between the reactive power of the capacitor and the reactive power of the inductor. When the power factor is unity it simply means the reactive powers cancel each other. But when the power factor to lagging improves by 0.95 l don't know how to calculate that.
     
  5. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    5,448
    782
    If you post the problem - perhaps as a new thread then you will most likely get some good advice.
     
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