Improve Power Factor of Circuit to 1: Calculations and Troubleshooting | AJUBI

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To improve the power factor of the circuit to 1, the calculations indicate that the load impedance needs to be adjusted. The circuit has a supply voltage of 240V peak-to-peak at 60Hz, with a resistance of 58 ohms and an inductance of 8 mH. The calculated inductive reactance is approximately 3.0159j Ω, leading to a total impedance of around 60.0758 Ω. The user derived a capacitor value of 2.3557 nF, which seems too small for the application. The discussion emphasizes that the load must appear resistive, and the 60 ohms should not be included in the load impedance calculations.
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!
 
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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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