# Power Factor Correction

In an assignment I was asked to improve the power factor of a certain circuit to 1. The supply is 240V peak-to-peak @ 60Hz. Rs = 58 ohm, Rl = 2 ohm and L = 8 mH. Below are my calculations but I can't seem to get the right answer. Indicating where I am going wrong would be a huge help :)

• $X_L = 2 \pi f L = 3.0159j\Omega$
• $Z = 60 + 3.0159j = 60.0758 \angle 2.8776$
• $I = \Large{\frac{240}{60.0758}} = \normalsize 3.995 A$

• True Power = $I^2 * R = 15.9596 * 60 = 957.576 W$

• Reactive Power = $I^2 * X = 15.9596 * 3.0159 = 48.1326 VAR$
• Apparent Power = $I^2 * Z = 15.9596 * 60.0758 = 958.7857 VA$

Then for the capacitor:

• $X_C = \Large\frac{V^2}{Q} = \frac{57600 }{ 48.1326 }= \normalsize1196.6941 \Omega$
• $C = \Large\frac{1}{ 2 \pi f X_C} = \frac{1 }{ 2 \pi * 60 * 1196.6941} = \normalsize2.3557 nF$

But this value for the capacitor seems way too small :/ Thanks for any help!

• A schematic would help :) Commented Feb 26, 2012 at 11:07
• @clabacchio: there is no need for a schematic, since there is only one inductor... Commented Feb 26, 2012 at 12:25
• @CountZero it's not necessary, but don't you agree that helps reading the question? Commented Feb 26, 2012 at 12:29
• @OlinLathrop: I agree that the resistances are unnecessary information (as I said in my answer). I believe it is simply $X_L = X_C$ (that's my understanding of the problem). The only element that contributes to the reactive power in this circuit is L, no matter how it is connected. If you follow the calculations of the OP, you can see there are several flaws in it anyway... Commented Feb 26, 2012 at 13:20

• Impedance for C is $\dfrac{1}{j \omega C}$, so $C = \dfrac{1}{(2 \pi f)^2 \cdot L}$ = 880$\mu$ F Commented Feb 26, 2012 at 14:54
• That's the result I got, too. ;) Commented Feb 26, 2012 at 17:40
The last equation contains a blatant mistake: the result is $2.35\times 10^{-6}$, which is clearly not nF... Plus, I don't really like the fact that you are mixing RMS values and peak values in your calculations... Oh, and one more thing! The voltage given is peak-to-peak! This means you need to halve it to get the amplitude and then work with that value!
By the way, there is a simple way to calculate the necessary capacitance, there is a lot of superfluous data there. All you need is set the reactance of the inductor equal with that of the capacitor. But I'm too lazy to do that for you. :P (Basically you have a resonant circuit at PF = 1.)