accapacitivepower-factor-correction
When we want to correction power factor of a motor(inductive load) its simply can done by adding an capacitor in parallel with the motor but its not possible to correction power factor of an capacitive load such as LED lamp by adding an coil in series and have to use a special circuit to do that, why?
simulate this circuit – Schematic created using CircuitLab
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
The power factor of a circuit is the ratio of real or useful power P (something we get out of circuit - heat, light, mechanical) to apparent power S. It is a kind of efficiency. The closer the pf is to 1 the better.
$$pf = \frac PS = cos\ \theta $$
Most loads are inductive in nature. They have a lagging power factor. This means that the source has to provide more current than is needed to drive the load (real power P).
If you look at the blue triangle. Source must provide S1. Q1 (\$ Q_L \$) is lagging inductive reactive power in VARs. I lags \$ V_S \$ by \$ \phi_1 \$.
Power factor correction adds a capacitor in parallel with the load. This adds leading reative power to circuit. The Qc. P and Q1 still exist.
The capacitor acts like a source decreasing the reactive power the source must provide (Q2 = Q1 - Qc). This means S2 will be less than S1, so the current required will be less. I lags \$ V_S \$ by \$ \phi_2 \$.
The power factor has improved. Best case scenario is when \$ Q_L = Q_C \$. The triangle disappears and pf = 1.
Best Answer
An LED lamp is a non-linear load not a capacitive load. Non-linear loads cause current waveform distortion. The distortion has the effect of adding harmonic currents to the load current. Harmonic currents do not transmit power, so they have the same effect as reducing the power factor. However harmonic currents can not be compensated using conventional power factor correction techniques. Harmonic filters are required.