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#128714 11/06/03 01:13 AM
Joined: Aug 2001
Posts: 545
aldav53 Offline OP
Is VA (volt/amps) the same as wattage on a transformer or other? If so, why do they use VA?

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#128715 11/06/03 02:54 AM
Joined: Jun 2001
Posts: 642
Watts are used to indicate the power availble for work.
VA is used to indicate apparent power. Not all of this may be availble for work.
W=VA x pf (power factor) in single phase.

for 3 phase loads
W=1.73 x E x I x pf
VA=1.73 x E x I

(These formulas were copied from 13th edition of American Electricians hand book)

Typically generators and transformers are rated in VA.
Hope this helps.

#128716 11/06/03 07:49 AM
Joined: Aug 2001
Posts: 7,520
I'll add that the difference between watts and VA is caused by the reactive portion of a circuit, i.e. the capacitive and/or inductive portions.

Power is only consumed by the resistive part of the load; current flowing in the reactive part does not result in power being dissipated.

However the transformer needs to be sized to allow for the current flowing in the reactive part of the circuit, thus the VA rating instead of watts.

This is why it is considered a good idea to keep the power factor as close to unity as possible, so as to minimize the "wattless current" which would otherwise represent a substantial percentage of the total.

#128717 11/07/03 03:02 AM
Joined: Oct 2000
Posts: 2,723
Likes: 1
Broom Pusher and
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 [Linked Image]...)


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!
[Linked Image]

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!)


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

Scott " 35 " Thompson
Just Say NO To Green Eggs And Ham!
#128718 11/07/03 10:25 AM
Joined: Sep 2003
Posts: 650
I am going to add one more point: whenever you talk about voltage, current, power factor, etc. for an AC circuit, the numbers that you are using are various types of _averages_ taken over time.

For example, when we say '120V', this is really the 'root mean square' average taken over at least one AC cycle. The actual _instantaneous_ voltage is continuously changing, going from roughly -170V to +170V. Similarly, we report a single number for current, which is really a form of average.

Okay, for _DC_ circuits, and for _instantaneous_ measurements, power is _always_ simply volts times amps. No power factor, no RMS, no nuttin'. But when the voltage and the current are constantly changing, the only way to figure out power is to multiply _instantaneous_ voltage and _instantaneous_ current to get _instantaneous_ power...then you have to average that power result over time, and you are back to the averages used to describe AC circuits.

If you look at an AC circuit with a pure resistive load, you will find that the current is precisely in phase with the voltage. The instantaneous power is simply the product of the two at each moment in time...but since the voltage and current are proportional, you will find that the instantaneous power is proportional to the _square_ of each. So for a pure resistive load you can simply multiply the RMS voltage and the RMS current.

But when the circuit has reactive components, you will find that the voltage and current are no longer in phase. The peak voltage point in the cycle does not correspond to the peak current point, and for portions of the cycle, the voltage is positive when the current is negative, or vice-versa. What this means is that for portions of the cycle, the power is _negative_.

Now don't get scared by 'negative' power; this is really just accounting. Negative power is still real juice flowing; it simply means that the net power is flowing in the opposite direction, from the device that is nominally the load back to the device that is nominally the source. This happens all the time, for example, with an induction motor attached to an overhauling load, where the motor becomes a generator and supplies power back to the mains.

When you have a load with reactive components, what happens is that for portions of the AC cycle, power flows to the load. Some of it gets used up in the load, and some of it gets stored in the reactive portion of the load (building up a magnetic field or charging a capacitor plate). During other portions of the AC cycle power actually flows back from the load to the source, being supplied by this stored energy. The net result is that more total current is flowing in the wires than needed to deliver power to the load itself.

We use the term power factor to correct the results that we get from multiplying the _averaged_ voltage and current (using RMS average) in order to find the true power being delivered to the load.

If you remember that the values for voltage/current/power factor/crest factor/etc. are all tools to let you use _average_ values to describe continuously varying quantities, then the whole thing of power factor becomes much less mysterious.

For example, above I described reactive power factor. But the term power factor is actually used whenever the product of RMS voltage and RMS current does not yield power delivered to the load.

With many power supplies, the input rectifier only draws current during the peak of the AC voltage cycle. During this 'on' period, the current is _much_ higher than average. The heating in the wires is accurately determined by the RMS current flow, but the power delivered to the rectifier is not correctly given by RMS current times RMS voltage. Again power factor could be used to describe this. Note, however, that this is neither a capacitive nor an inductive power factor, and normal power factor correction techniques wouldn't help. Thus a _different_ correction factor is sometimes used: 'crest factor' which describes the _shape_ of the current waveform.


#128719 01/05/04 04:58 AM
Joined: Jul 2002
Posts: 8,443
Likes: 3
Hope that this hasn't already been mentioned, but the reactive component of current in a Reactive circuit, is caused by the Magnetising/Demagnetising Current, also known as Hysteresis Current or B/H currents.
The reason why transformers are rated in VA and not Watts, is because the Secondary side Power Factor (cos phi), is often unknown,this really makes a difference at HV levels, where as a small transformer in a radio or TV may have little loss, a 500kVa 33kV transformer may be totally different, with respect to power factor. [Linked Image]

#128720 02/09/04 11:48 AM
Joined: Dec 2002
Posts: 110
If I may add to the above discussion.
In general, the reactive power associated with any circuit VI Sin (Theta). The symbol for reactive power is Q and its unit of measure is ' Volt-Amps-reactive' . The Q is derived from the quadrature (90 degrees) relationship between the various powers.
Therefore if the average power is zero and the energy supplied is returned within one cycle, why is reactive power of any significance ?
At every instant of time along the power curve(ie a sine-wave)that the curve is above the positive axis, energy must be supplied to the inductor, even tho it will be returned during the negative portion of the cycle.
This power requirement during the positive portion of the cycle requires that the generating plant provide this energy during that inteval.
The effect of reactive elements such as the inductor can be to raise the power requirement of the generating plant even though the reactive power is not dissapated but simply 'borrowed'. The increased power demand during these intervals is a cost factor that must be passed on to the industrial consumer. In fact most larger users of electrical energy pay for the apparent power demand rather than the watts dissipated since the VA used are sensitive to the reactive power requirement. The closer the power factor of the plant is to one, the more efficient is the plants operation since it is limiting its use of 'borrowed' power.
The net flow of power to the pure(ideal) Inductor is zero over a full cycle, and no energy is lost in the transaction. Therefore any energy lost in this transaction is not due to the inductor rather than to the accompanying resistance associated with the inductance.
As far as transformers go, The magnetization current in the transformer is not sinusoidal. The higher frequency components in the Mag current are due to magnetic saturation in the transformer core.
Once peak flux reaches the saturation point in the core a small increase in peak flux requires a very large increase in the peak magnetization current.
The core-loss current is nonlinear because of the nonlinear effects of hysteresis. the core loses are modeled as a resistance rather than an inductor, because of Hysteresis and eddy current losses.
So the total EXCITATION current is Iex = I(hysteresis + eddy current) + Im (the magnetization current). These are added as phasors and Im can be found by using the(as mentioned) square root sum of the squares and
the angle of the Im is the Arctan(X/R) or the (reactive component/real component)
Hope this helps

#128721 02/09/04 06:44 PM
Joined: Apr 2002
Posts: 2,527
Adding a comment to wocult's... this makes AC circuits and application of Ohm's Law a lot more interesting than DC circuits.

#128722 07/04/04 11:49 AM
Joined: Oct 2000
Posts: 2,723
Likes: 1
Broom Pusher and

... this makes AC circuits and application of Ohm's Law a lot more interesting than DC circuits.

LOL!!! [Linked Image] [Linked Image]

Where else would a Circuit with a 4 Ohm Resistor and a 3 Ohm Inductor (Coil) in Series, add up to 5 Ohms!

DC value for this circuit = 7 Ohms.
AC value for this circuit = 5 Ohms (Impedance).


Scott " 35 " Thompson
Just Say NO To Green Eggs And Ham!
#128723 07/12/04 11:00 AM
Joined: Jul 2004
Posts: 2
Junior Member
All are reply in deep but I would like to add some more.
Treansformer have two type losses one is called Core losses and other is called cooper losses,The core losses is depend on voltage of transformer and cooper losses depend current of transformer there is no involevment of phase angle that is why the rating of transformer in VA.

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