power factor is basically just measuring how efficiently the lighting is applying the power it is using
Power factor is how much power the ballast is actually using vs the load on the supply
For example :
You plug in an 1200W convection space heater, which is essentially a Steel wire with total resistance of 12 Ohm, assembled in a device that lets it heat up and dissipate this heat into the air
120V / 12 Ohm = 10A
120V * 10A = 1200W
The space heater uses 1200W of power, and draws 10A from the line
Now take a 220uF capacitor
Xc = 1 / ( 2pi f c ) = 1 / ( 2pi * 60Hz * 220uF ) = 12 Ohm
120V / 12 Ohm = 10A
120V * 10A = 1200W
Plug in the 220uF capacitor into 120V. Just the capacitor, with nothing else. It will pass 10A, but won't do anything with power (i.e. it won't heat up, converting electrical power to heat power)
The capacitor is drawing 10A, same as if it would be 1200W heater, but it does not actually use the power. It does not take the power and convert it to heat, light, mechanical work or anything like that. It does not get wasted to losses either. All it does is charge and discharge back into the line on every half cycle of the AC, 120 times/sec
So, is the capacitor actually drawing 1200W or not ?
If we measure the current, we see that there are 10A flowing. Apparently it looks like if 120V * 10A = 1200W load is connected and drawing power. This is Apparent power
- The wiring, circuit breakers, and all that have to be sized to handle the 10A, because 10A really flow in the circuit
If we measure the power used (and not returned into the line), it is 0. This is the real power
- The meter does not spin for the capacitor
- The power plant's turbine load (so by extension, the fuel use. pollution etc) are not increased by the capacitor's presence
The relation between the power really used (real power) and the "would be" power that is calculated from the current x voltage (apparent power) is the power factor
For a resistor like the heater, the real power = apparent power, PF = 1
For the capacitor, the real power is 0, PF = 0
For something else, it is something inbetween
How efficiently the ballast uses the power (ie. the part delivered to the lamp vs. the part in the losses) is efficiency, this have nothing to do with power factor. There can be a combination of any efficiency with any power factor
also assuming zero ballast losses
In case of Electronic ballast driving a lamp rated for Magnetic ballast, this is allmost the case. There are ballast losses, but at high frequency less power is needed to get the same output from the lamp
So for example, a 40W lamp powered by say 36W at HF will be as bright as 40W lamp powered by 40W from magnetic ballast. This gives 4W of "hidden headroom" for ballast losses, bc/ for example if the HF ballast have 4W losses, the complete lamp+ballast system will use 40W and light like proper 40W. so it behaves as if the ballast indeed have no losses
Most Electronic ballasts are not quite that efficient but the difference amounts to single watts