16-08-2014, 07:06 PM
So let's just remind ourselves how reactive volt-amps (VAr), volt-amps (VA) and watts (W) are related.
1. Watts are the result of the product of voltage across a load and the current through it - and that current is in-phase with that voltage. Hence, the load is resistive.
2. Volt-amps reactive is the product of voltage across a pure reactance and the current through it - and that current is in quadrature with that voltage.
3. Volt-amps is the product of the voltage across a load (resistive and reactive) and the apparent current through that load. That current will make an angle with the voltage - i.e., it is out-of-phase.
Mathematically, we have:
(Watts)² = (Volt-amps)² - (VAr)² . . . by Pythagoras Theorem.
So, what does all that amount to?
For the size of a given resistive load, if the reactive component is reduced, the amount of volt-amps decreases. (Draw a right-angled triangle, based on the above formula, if you in any doubts about that). And if I am measuring the current through a resistive + reactive load, that current will decrease. But the amount of watts that I am consuming stays the same. And it is watts that I am paying for. If I could ultimately reduce the number of reactive volt-amps to zero, the number of watts would be the same as the number of volt-amps. I then have all the current in my load now in-phase with the voltage across that load - i.e., I have achieved a power factor of unity, and the load is now purely resistive - or looks that way to the supplying source. It is by connecting power-factor correction capacitors across an inductive load (or inductors across a capacitive load, e.g. PSUs in PCs) that I can make my load appear to be pure resistive to that source. And that benefits the supplier, not the consumer.
Al.
1. Watts are the result of the product of voltage across a load and the current through it - and that current is in-phase with that voltage. Hence, the load is resistive.
2. Volt-amps reactive is the product of voltage across a pure reactance and the current through it - and that current is in quadrature with that voltage.
3. Volt-amps is the product of the voltage across a load (resistive and reactive) and the apparent current through that load. That current will make an angle with the voltage - i.e., it is out-of-phase.
Mathematically, we have:
(Watts)² = (Volt-amps)² - (VAr)² . . . by Pythagoras Theorem.
So, what does all that amount to?
For the size of a given resistive load, if the reactive component is reduced, the amount of volt-amps decreases. (Draw a right-angled triangle, based on the above formula, if you in any doubts about that). And if I am measuring the current through a resistive + reactive load, that current will decrease. But the amount of watts that I am consuming stays the same. And it is watts that I am paying for. If I could ultimately reduce the number of reactive volt-amps to zero, the number of watts would be the same as the number of volt-amps. I then have all the current in my load now in-phase with the voltage across that load - i.e., I have achieved a power factor of unity, and the load is now purely resistive - or looks that way to the supplying source. It is by connecting power-factor correction capacitors across an inductive load (or inductors across a capacitive load, e.g. PSUs in PCs) that I can make my load appear to be pure resistive to that source. And that benefits the supplier, not the consumer.
Al.
