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!
Best Answer
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.