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Designing a Load

  1. Oct 3, 2012 #1
    1. The problem statement, all variables and given/known data
    Design a load which draws an average power of 25 W at a leading PF of 0.88, with Vrms = 120 V @ 60 Hz.


    2. Relevant equations

    P = VIcos([itex]\theta[/itex]-[itex]\phi[/itex])
    Q = VIsin([itex]\theta[/itex]-[itex]\phi[/itex])
    S = P + jQ
    |S| = VI

    3. The attempt at a solution
    Since it has a leading PF, the load should be capacitive.

    [itex]\theta[/itex] = 0°
    [itex]\phi[/itex] = cos-1(0.88) = 28.36°
    I = [itex]\frac{P}{Vcos(\theta-\phi)}[/itex] = [itex]\frac{25}{120(0.88)}[/itex] = 237 mA
    Q = 120(0.237)(sin([itex]\theta[/itex]-[itex]\phi[/itex] = -13.51 VAR
    S = 25 - j13.51 VA or 28.4∠-28.3°

    I probably do not need Q or S, but where do I go from here, since I don't know Z, X, or R?
    Could I do something like this to solve for Z?:

    P = [itex]\frac{V^{2}}{R}[/itex]cos([itex]\theta[/itex]-[itex]\phi[/itex])

    If I do that, Z = 576 Ω, R = 506.88 Ω, and X = 273.585 Ω. I could then use X to find capacitance C. Would this be the correct route to take?

    Also, am I right to say that the resistor and capacitor would be in parallel?

    Thanks!
     
  2. jcsd
  3. Oct 3, 2012 #2
    No, you have:
    [tex]S=VI^*=V\left(\frac{V}{Z}\right)^*=\frac{VV^*}{Z^*}=\frac{|V|^2}{|Z|}e^{j\varphi}\\
    P=Re\,S=|S|\cos\varphi=\frac{|V|^2}{|Z|}\cos \varphi
    [/tex]
     
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