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!

  • \$\begingroup\$ A schematic would help :) \$\endgroup\$
    – clabacchio
    Commented Feb 26, 2012 at 11:07
  • \$\begingroup\$ @clabacchio: there is no need for a schematic, since there is only one inductor... \$\endgroup\$
    – Count Zero
    Commented Feb 26, 2012 at 12:25
  • \$\begingroup\$ @CountZero it's not necessary, but don't you agree that helps reading the question? \$\endgroup\$
    – clabacchio
    Commented Feb 26, 2012 at 12:29
  • \$\begingroup\$ @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... \$\endgroup\$
    – Count Zero
    Commented Feb 26, 2012 at 13:20

2 Answers 2


Apparently you have a inductive load to a AC power supply, and want to cancel the reactive part of the load by putting a capacitor accross it. You mention values Rs and Rl, but since you provided no definition for these, we can't tell how they are hooked up and therefore their relevance.

However, if I understand the problem correctly, this is mostly about finding the capacitance that cancels the inductance. From basic circuits you know that the impedance magnitude of a capacitance is 1/ωC, and for a inductance is ωL. Setting these two equal and solving for C yields C = 1/ω²L = 1/(2πf)²L = 880 µF.

  • \$\begingroup\$ Impedance for C is \$\dfrac{1}{j \omega C}\$, so \$ C = \dfrac{1}{(2 \pi f)^2 \cdot L} \$ = 880\$\mu\$ F \$\endgroup\$
    – stevenvh
    Commented Feb 26, 2012 at 14:54
  • \$\begingroup\$ @stevenvh: Doh! Fixing right now. \$\endgroup\$ Commented Feb 26, 2012 at 16:00
  • \$\begingroup\$ That's the result I got, too. ;) \$\endgroup\$
    – Count Zero
    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.)


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