# Why are some ratings in VA and others in W?

Discussion in 'General Electronics Chat' started by Tonyr1084, Aug 20, 2016.

1. ### Tonyr1084 Thread Starter Active Member

Sep 24, 2015
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I'm looking at a remote dimmer control for my ceiling fan / light fixture. It's rated to handle 300 Watts Tungsten or 300 VA Ballast.

I know watts is a function of multiplying volts and amps, so aren't the two the same thing? So why do they designate a tungsten load in watts while they rate a ballast in volt/amps?

Jul 18, 2013
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3. ### Tonyr1084 Thread Starter Active Member

Sep 24, 2015
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Hey, thanks Max. I kind of suspected that was the case. The link you provided added greatly to my understanding. That's all I wanted to know - why the different terminology.

4. ### crutschow Expert

Mar 14, 2008
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3,361
Power is indeed volts times amps, but only at the point in time where the two waveforms are in phase.
Inductance or capacitance in the load causes a phase shift between voltage and current which must be allowed for.
In other words real power is the sum of the instantaneous voltage times the instantaneous current at each instant in time averaged over one cycle (the integral).

For undistorted voltage and current sinewaves you can multiply the RMS voltage times the RMS current times the cosine of the phase angle between them to get the real power.

5. ### MrAl Distinguished Member

Jun 17, 2014
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514

Hi there,

The main reason why some things are rated in VA instead of in Watts is so that the manufacturer can specify a quantity that reflects the true capability of the equipment. If they were to state it in Watts only sometimes the user can burn up the device and come back and say, "Gee, i was only using it at 900 watts and it's rated at 1000 watts, so what went wrong?".

What went wrong is the manufacturer did not rate the device in VA they rated it in Watts. Let's take a quick look and see why this happened...

If we calculate the Watts from the VA like this we have:
P1=V*A

and that may not be the true wattage.

But if we use the correct formula (assuming sine waves that is) we have:
P2=V*A*cos(a)

where 'a' is the angle of lead or lag.

Now let's throw some numbers in there, working at what we will call 1000 watts...

At 1000 watts, both P1 and P2 are 1000, so we have:
P1=1000=V*A
P2=1000=V*A*cos(a)

and let's assume 100 volts for simplicity, then calculate the current...

P1=1000=100*A
P2=1000=100*A*cos(a)

Now let's solve for the current A in both of these...

P1=1000=100*A, so A=10
P2=1000=100*A*cos(a), so A=10/cos(a)

and let's assume an angle a=45 degrees, then we have:
For P1: A=10 amps as before
For P2: A=10/0.7071=14.1 amps

So by rating it in Watts the user thinks he can get P2=1000 watts and so he pumps 14.1 amps through it at 100 volts, which is 41 percent higher current. Any wire used in the design could overheat and burn the unit out because of this. And that's not really the end of the story. Imagine we had an angle of 89 degrees (just for an extreme case). Then we get:
P2=1000=100*A*cos(a)
1000/100/cos(a)=A
10/cos(a)=A
and with angle 'a' equal to 89 degrees we get:
A=10/0.0174525=573 amps !!!

How long do you think the device would last with 573 amps when it's made for only 10 amps? Answer: Not long

So now you see why VA is an important specification and remember in all the examples above the WATTAGE was still 1000 watts, that never changed, but in one case it runs fine (at 10 amps) and in the other cases it burns out.

A place you will see VA a lot is in transformer sales. If they do it right they will rate it in VA but sometimes you will see it rated in watts and then you've got to understand that the current is assumed to be in phase or only slightly out of phase. If the current is substantially out of phase then the user has to understand that they cant actually get the rated wattage from that device and has to derate it based on the phase angle.

Last edited: Aug 20, 2016
6. ### Tonyr1084 Thread Starter Active Member

Sep 24, 2015
628
109
Mr Al; thanks for the attempt. This has thoroughly confused me. Give me time to digest it and I'm sure I'll get a grasp. I just need to play with the numbers.

The conclusion I draw is that if I take a coil (or cap) (90˚ out of phase) and power it using AC (lets say 1 volt) it can draw infinite amperage. Makes my head spin. If so, couldn't I use the current to generate a voltage? Could that voltage be higher than 1 volt?

For the moment, I'm lost (again). Fear not, I'll get this sorted out. Right now I am embroiled in a "Honey-do" project. Gotta love the wife. No, really, you GOTTA love the wife. But I do. She's pretty great. Still, sometimes she comes up with projects that are easy in her mind but to me - it means cutting holes in ceilings and boxing in ceiling joists, re-plastering, priming and painting - THEN the "Honey-do" can be done.

I'm sure I'll get there. Just not today. Or this weekend. Maybe next weekend. Maybe.

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7. ### WBahn Moderator

Mar 31, 2012
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If you tell me that a circuit component has 10 Vrms across it and a current of 10 Arms through it, I can't tell you very much about the real power being dumped into (or draw out of) that component except that it could be anywhere between -100 W and +100 W. It could well be zero. That's because you haven't told me the phase relationship between the voltage and the current. As a result, the product of the RMS voltage and current just "appears" to be power (in fact, we call it "apparent power") and so I don't want to use "watts" to describe. Instead I use "volt-amps", or "VA" just to emphasize that while the units are consistent with power, it is just apparent power.

8. ### Tonyr1084 Thread Starter Active Member

Sep 24, 2015
628
109
Thanks WB, I understand a little about phase angle. I've been reding up on that and understand how phase can change a lot of things. What Mr. Al said somewhat threw me for a loop. I haven't given much time to solving (or understanding) how an 89˚ shift can result in such high currents. Especially when nearly no work is being done.

What started out as a simple question - to understand why some things have ratings in VA while others are rated in W, which Max answered quite nicely has become somewhat more perplexing than I had hoped.

Yesterday after my post I had a large get-together with several friends and family for a day out at the train museum. By the time we got home I was not thinking about VA's & W's. So I plan on giving some devotion to the subject perhaps later today.

I know I'll get a handle on it. But for now it seems like a very greasy handle and my grip is somewhat compromised. Still, I'm confident I'll get there. What seems so difficult for me to understand is child's play for most of you. But I'll catch up soon enough. For now I'm not designing any AC circuits. My current project (on hold) is to drive a Laser Diode using audio frequency and amplitude. A simple OpAmp will serve that project well enough. There are more parameters to it, but it has nothing to do with my current question. Actually it's a ceiling fan remote module. I noticed the 300 W Tungsten and the 300 VA Ballast designation and started wondering what the difference was.

But thanks. I really have to read all the posts again and again quite carefully and sit down with pencil and paper and do the math. The hard way. I learn best when I can see the relationships.

9. ### wayneh Expert

Sep 9, 2010
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Every time I study the watts versus VA issues, I read until it clicks. I get it. Then times go by, and poof, I have to go back and read it again.

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10. ### KeepItSimpleStupid Well-Known Member

Mar 4, 2014
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When you talk about generators VA comes to play as well. motors are a VA type of load and because of the inductance I generally is larger for starting. The generator MUST be able to supply that lathe initial current to get the motor running.

You have power dissipated (Watts), reactive power (VARS) and VA. The utility company wants the power factor to be close to unity because it really doesn't want to have to oversize the conductors. Generally, you get a 100, 150 or 200 A service and not a 100(240V) 24,0000 Watt service. You need to stay within the 240 and 100 A (V*A) service.

Jul 18, 2013
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Study the phase relationship between the inductive current and voltage and see the result of the two at any given time, IOW the peaks of each do not co-inside as they do with a restive load.
Max.

12. ### DGElder Member

Apr 3, 2016
347
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The KVA rating of a transformer is determined by the maximum temperature rise the unit can withstand without damage. This is determined by voltage and the current through the coils. The KVA rating will be at a specified rated voltage. From that you can derive a maximum current (KVA / Vrated). If you operate at a lower voltage the KVA rating must be derated in proportion to the lower voltage, i.e. at half the voltage you must draw half the KVA. In practice this means, when operated at the rated voltage, the maximum power, watts, the transformer can deliver to a load is the KVA. This will only be the case when the load is purely resistive. The power it can deliver at the specified KVA will decrease as the capacitive or inductive component of the load increases with respect to the resistive component of the load.

In other words the transformer doesn't care how much power the load uses, it cares about the maximum voltage and the maximum current through its coils. Since the transformer has no idea what the power factor of the load will be it can not specify delivered watts.

Last edited: Aug 21, 2016
13. ### MrAl Distinguished Member

Jun 17, 2014
2,550
514

Hi,

Oh yes good idea, play with the numbers, that can give you deep insight. Do a few examples, calculate the Watts and the VA, then imagine you manufacture a device that can work with phase angles out of phase

To make a long story short, the two are just different ratings like other things we see although they are a little more closely related to each other.

Things with no chance of lead or lag phase angle are usually rated in watts because there's no chance anyone could do anything wrong with that. Like a high power LED, which is predominantly resistive ... so if you pump in 1 amp you get 3 volts (or whatever) every time you do that. A resistor is the best example though, or a piece of wire. There is no chance the phase angle can be other than zero through those two so we can rate resistors in watts. The time when it really matters is when there is a complex load that is made up of a resistor part and a reactive part like an inductor. The current is out of phase with the voltage, but in the resistor it's not, so the resistor heats up. In fact, that's a good example to play around with i think because an inductor in series with a resistor is one of the simplest loads that has an out of phase current.

If you like we can do some examples right here with some calculations.