1. Oct 3, 2012

### hogrampage

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($\theta$-$\phi$)
Q = VIsin($\theta$-$\phi$)
S = P + jQ
|S| = VI

3. The attempt at a solution

$\theta$ = 0°
$\phi$ = cos-1(0.88) = 28.36°
I = $\frac{P}{Vcos(\theta-\phi)}$ = $\frac{25}{120(0.88)}$ = 237 mA
Q = 120(0.237)(sin($\theta$-$\phi$ = -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 = $\frac{V^{2}}{R}$cos($\theta$-$\phi$)

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. Oct 3, 2012

### szynkasz

No, you have:
$$S=VI^*=V\left(\frac{V}{Z}\right)^*=\frac{VV^*}{Z^*}=\frac{|V|^2}{|Z|}e^{j\varphi}\\ P=Re\,S=|S|\cos\varphi=\frac{|V|^2}{|Z|}\cos \varphi$$