# Application of power factor correction

If the reactive power that is induced by an inductor that is connected to a resistor (a lamp) is corrected by a capacitor, does that mean that I get more light out of a lamp?

The way I see it is that, in this case the real power stays unchanged and the power factor goes to zero then the apparent power gets equal to the real power.

You don't have any leading or lagging power with a purely resistive load. In fact, adding capacitance will only move the power factor away from unity (which is ideal).

• at first I have an inductor and a resistor and then in order to get rid of the reactive power I am adding a capacitor to the circuit to move close to unity.
– Jack
Feb 8, 2016 at 16:31
• If you mean the R and L are in series across the supply, and C is then also connected across the supply (and in parallel with R~L), then you can improve the power factor but you still have the same supply voltage across R~L, so the power in R doesn't change. This is the connection used in industry since it's impossible to split the R and L components of an inductive load, e.g. a motor.
– Chu
Feb 8, 2016 at 16:42
• @Chu, so in this case there is no use of correcting the power factor if all I care about is increasing the power of the resistor.
– Jack
Feb 8, 2016 at 16:44
• Not unless the power supply company charges extra for poor power factor. You could connect the capacitor in series and tune it to get resonance with the L, that would place all of the supply voltage across the R.
– Chu
Feb 8, 2016 at 16:48

If the inductor is fully 'corrected' by connecting the capacitor to the inductor, then their total reactance will go to zero, or infinity, depending on whether it's a series or parallel connection. Their reactive power in this case falls to zero.

The output from a lamp connected with them will be the same (assuming ideal, non-lossy L and C) as if it were connected directly to the supply. With zero reactive power, the power factor goes to 1.

That is more light from the lamp than if it was connected to the supply through either the inductor by itself, or the capacitor alone.

With real, lossy, components, there will be some residual R left, mostly from inductor losses, that will dim the lamp slightly with respect to a direct connection to the supply.

However, my assumptions about circuit diagrams may be incorrect, there is another possibility.

If you have a resistor and an inductor in series, then add the capacitor across the supply, to improve the power factor that the supply sees, the brightness of the lamp will remain unchanged, as the voltage on the capacitor, and the voltage across the R+L, is controlled by the low impedance of the supply. What the capacitor will do is to draw current from the supply to cancel out the reactive current drawn by the inductor. When fully corrected for reactive power, the real power drawn by the circuit will be equal to the power used in the lamp. This will be less than had the lamp been connected directly to the supply.

• So if I buy a 100 Watts amp, its real power is not always 100 Watts and it depends on the circuitry, right?
– Jack
Feb 8, 2016 at 16:35
• A 100w lamp will only deliver 100w if you connect it to a suitable supply that will run it at 100w. If you connect odd components in series, it may run at less than 100w on the same supply. Feb 8, 2016 at 21:43

If the inductor is in parallel with the resistor across an ideal voltage source then adding a parallel capacitor will not change the power consumed by the resistor. However, it will reduce the overall current taken from the supply if it tunes with the inductor to resonance.

If the inductor is in series with the resistor then putting a capacitor (also in series) can cancel the impedance of the inductor and allow more power to be consumed by the resistor. Again, it's down to resonant tuning which is of course another name for power factor correction.