Hello all - just wanted to add a little more to the thread (the already posted replies are very good examples, and this message is just some "FYI" data);

In simple terms, what makes the difference between Volt-Amps (VA) and Wattage (Watts) is the type of load.

Basic example:

An Incandescent Lamp - such as a Quartz-Halogen Lamp, draws all Power as Wattage; whereas an HID Lamp + Ballast - such as a Metal Halide HID Luminaire, would draw Power as Volt-Amps (VA).

Wattage = "True Power",

Volt-Amps = "Apparent Power".

Apparent Power contains a certain level of True Power (wattage) flowing within it, and the remainder is comprised of the "Cosine" component - called Reactive Power, known as "Volt-Amps Reactive" (VARs).

Some additional basic examples:

*** 1: Quartz-Halogen Lamp. ***

This "_Pure Resistance_" device has a power Factor of 100% (1.0 PF). All Power is True Power (Watts), and as a result, E×I=P with "P" being True Power / Watts.

If the Lamp was a 400 Watt Lamp, and it is connected to a Circuit with a Voltage of 120 VAC, the load current draw would be 3.334 Amperes.

Multiply the 3.334 Amps by the system's Voltage of 120 VAC, and you get 400.08 Watts of True Power.

(the .08 watt is from rounding off the Amperes to 3.334, instead of using the full string of 3.33333333333...)

*** 2: Metal Halide HID Luminaire. ***

This setup incorporates a "Ballast" device to regulate the Current flow through the HID Lamp.

The Ballast is a Reactor Coil, which works against the load current via Inductive principles. The Ballast results in an Inductive Reactance being introduced into the circuit.

The Lamp is an Arc Tube, which works via Ionizing principles - to form a Plasma - which produces the output Light.

The Lamp has a Capacitive feature in the Arc Tube, due to the way the current flows in it, but also has an Inductive component.

The Lamp introduces Capacitive Reactance to the circuit.

(FYI: the Lamp exibits what is known as "Negative Resistance" when it's connected to a Power Source. This means it will draw more and more load current as the light output gets higher and higher - which happens very rapidly! So to keep the Lamp from rapid self destruction, the amount of current it can draw is "Ballasted" - or restricted; hence the use of the ballast).

With the HID assembly connected to an Alternating Current source, it will draw Power in an envelope of Apparent Power - or Volt-Amps.

The VA "Package" contains the Wattage consumed by the Lamp for light output + by any of the Heat produced in the Lamp and the Ballast; along with the resulting VARs per the Power Factor of the complete assembly.

In this case, if the Lamp is 400 Watt, and the power Factor is 83.4% (0.834 PF), the resulting Apparent Power will be 480 VA.

The VA package will carry the 400 Watts of true power for the lamp (figuring all the wattage is for light output - no losses), and the "extra 80" is part of the total VARs (Reactive Power) for the Reactive component.

The VARs in this situation will be 265.3 VARs

Together they result in 480 VA of Apparent Power.

The total load current for this device, if connected to a system with an AC voltage of 120 VAC, would be 4.0 Amperes - equaling an apparent power of 480 VA.

This is the reasoning behind rating Reactive Devices per VA, not Wattage.

BTW, you could rate a Wattage as VA if the PF is 100%. Just FYI

Figuring the PF and VARs is something for another topic posting, but for a quick example, it's done the same way an Impedance is figured!

(Triangle formula

...)

Scott35

FYI: Edited this message to correct some blunders!

Must have been ½ asleep, or had a Cranial insertion between both Gluteus Maximus points for a prolonged amount of time!

Me Bee Dumm-EE!, or just call me "Sofa King"

Sofa King Stu Pid(say it really fast a few times, and you will understand the "Saof King" reference!)

Scott

[This message has been edited by Scott35 (edited 07-04-2004).]