Power factor correction

Thread Starter

banzai

Joined Jan 3, 2014
71
Hi

could some one kindly explain how a capacitors are able to correct power factor,

How i look at it is that when the voltage wave starts to fall from the peak the capacitor will release its current in the same direction as the inductor was feed from the generator which will cause the inductor to maintain its stored energy which is in a from of magnetism, and this will cause the inductor to still have impedance on the current flow from the generator and lag the voltage, i cant see how this can improve the power factor ?

I see it as in a LC Circuit, where the current is oscillating.

Would be very thankful for an explanation :)
 

Papabravo

Joined Feb 24, 2006
21,225
I'm assuming that you know what power factor is; the cosine of the phase angle between the voltage and current. IT DEPENDS ENTIRELY ON THE IMPEDANCE OF THE LOAD. So from this we can posit certain rules of thumb:

1. For a purely resistive load the power factor is always 1, and no correction is possible or desired. Adding a capacitor is guaranteed to make the power factor worse!
2. For any CONSTANT reactive load, a reactance of the opposite type may be used bring the power factor back towards unity. Component tolerances will probably prevent you from achieving it.
3. Most reactive loads on the AC mains are motors which is why capacitors may be used to compensate for the inductive load.
4. For ANY reactive load that is not constant, a fixed compensation scheme (adding a capacitor) will be next to useless at least some of the time.

In residential and non-commercial applications it is seldom a worthwhile pursuit.
 

DerStrom8

Joined Feb 20, 2011
2,390
Hopefully you know that impedance works on a complex plane, where the x-axis is real and the y-axis is imaginary. The real part is the resistance of the load and the imaginary part is the reactance. Many loads tend to be primarily inductive, which means their reactance is a positive j value. This reactance draws what is referred to as reactive power, which doesn't really supply anything to the circuit--it's just wasted power. By adding a capacitor, which has a -j value reactance, it cancels out the +j reactance of the inductor, thus causing the circuit to draw only real power.
 
Last edited:

crutschow

Joined Mar 14, 2008
34,444
Basically the current leads the voltage in a capacitor and the current lags the voltage in an inductor (which is a phase-shift between current and voltage). This is reactive current when dissipates no power. Typically the mains power total load has an excessive of inductive load (such as from large motors and transformers) as compared to any capacitive loads, which thus gives a lagging phase shift (lagging power factor) at the generator. The goal is a power factor of one (no reactive current, thus looking like a pure resistance to the generator), so you want to add lead current from a capacitor that's large enough to cancel most (ideally all) of the lag current from the inductor. That's power factor correction.

Make sense?
 
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