Power Factor Correction

  • Thread starter timfoster
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In an assignment I was asked to improve the power factor of a certain circuit to 1. The supply is 240V peek-to-peek @ 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 :)

http://ajubi.com/pf.png

Xl = 2 * pi * f * l = 3.0159j Ω
Z = 60 + 3.0159j = 60.0758 phase-angle 2.8776

I = 240 / 60.0758 = 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:

Xc = V^2 / Q = 57600 / 48.1326 = 1196.6941 Ω;
C = 1 / ( 2 * pi * f * Xc) = 1 / ( 2 * pi * 60 * 1196.6941) = 2.3557 nF

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

Answers and Replies

  • #2
NascentOxygen
Staff Emeritus
Science Advisor
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You need to make the load appear resistive. The impedance of the load does not include the 60 Ohms.
 

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