Designing Load Homework Statement: 25 W @ 0.88 PF, 120 V

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Homework Statement


Design a load which draws an average power of 25 W at a leading PF of 0.88, with Vrms = 120 V @ 60 Hz.


Homework Equations



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

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!
 
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hogrampage said:
Could I do something like this to solve for Z?:

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

No, you have:
[tex]S=VI^*=V\left(\frac{V}{Z}\right)^*=\frac{VV^*}{Z^*}=\frac{|V|^2}{|Z|}e^{j\varphi}\\<br /> P=Re\,S=|S|\cos\varphi=\frac{|V|^2}{|Z|}\cos \varphi[/tex]