Electronic – Help with complex power and power factor correction problem

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.

Power factor correction of a linear inductive load is exactly the same as tuning a parallel circuit of a capacitor and inductor. You pick a capacitor that works with the value of the transformer's magnetizing inductance to satisfy this: -

60Hz = \$\dfrac{1}{2\Pi\sqrt{L_M.C}}\$

Where \$L_M\$ is the transformer's magnetizing inductance and C is the capacitor chosen to "neutralize" the current taken by the coil.

For a pure lossless inductance and capacitance, the resultant current taken from the supply is zero.

If your transformer primary indicates an inductive reactance at 60Hz of 300 ohms, the magnetizing inductance value is this divided by \$2\Pi\times 60\$ = 0.8 henries.

The capacitive reactance required is 300 ohms and this is 8.8 uF. As a sanity check: -

F = \$\dfrac{1}{2\Pi\sqrt{0.8\times 8.8\times 10^{-6}}}\$ = 59.98Hz