VA or not VA

[mphillipps]

I'm trying to better understand VA.
When is VA = (V*A=W) or (IE=p) basic Ohm's law?
And is volt-ampere ever equal to something more or different?

I've read Wikipedia (quoted below). It didn't help. Can someone simplify the explanation and apply VA (other than watts) to a real world electrical situation.

"A volt-ampere, in electrical terms, is the amount of apparent power in a circuit equal to the product of voltage and current. While equal to the electrical power measured in watts for Direct Current (DC) circuits the apparent power may differ from the real power for Alternating Current (AC) circuits, where voltage and current may be out of phase. The real power is also defined as apparent power multiplied by the power factor."

 

[zog]

In this case the Wikipedia definition is pretty good.

Simply put, VA=W in a DC circuit or a AC circuit where the Power Factor =1.0

 

[charlie b]

Here is a long-winded explanation that I typed up some time ago. Its subject is "power factor," which is not quite exactly what you had asked about. But I think it covers what you want to know.

-------------------------------------------------------------

There are three different kinds of electrical power. There is ?real power,? and there is ?reactive power,? and the two add up (in a trigonometric sort of way) to give you ?apparent power.? What causes ?reactive power? to come into the picture are things that involve magnetic fields (such as the windings of motors, ballasts of fluorescent lights, and welding machines) and things that involve electric fields (such as capacitors and synchronous generators).

I don?t really like the ?beer and foam? analogy for power factor. But here it is anyway. Real power is like the liquid that fills most of a glass of beer. Reactive power is like the foam at the top of the glass. What you really want is the liquid, but the foam is there and you can?t avoid it being there and it doesn?t change the amount of liquid at the bottom. Suppose you start with a glass with some amount of liquid, and a little foam at the top. In order to contain both, the glass has to be a bit larger than necessary for the liquid alone. Suppose you can generate some extra foam, and suppose you pour it into the glass. You haven?t changed the amount of liquid, but you need a bigger glass to contain it all.

Now back to electrical stuff, and a bit of trigonometry. Take a sheet of paper, and draw a horizontal line about one inch long. Label the left hand end of the line as point ?1? and the right hand end as point ?2.? From point 2, draw a vertical line upwards one inch. Label the top of that line as point ?3.? Finally connect points 1 and 3 to form a triangle. Line 1-2 represents the real power, and it is a measure of the ?real work? being done by the load equipment. Line 2-3 represents reactive power, and is a measure of the energy being exchanged (you might even say ?wasted?) between the magnetic field of the utility generator and the magnetic field of the customer?s motors and other inductive loads. Line 1-3 represents apparent power, and is a measure of the power that the utility has to provide, in order for the customer to run the motors.

Power factor is a measure of the angle located at point 1. In terms of trigonometry, PF is defined as the cosine of that angle. If that angle is zero degrees (which happens if line 2-3 has no length at all, so that line 1-2 and line 1-3 lie one on top of the other), then PF is unity. If that angle is large (which happens if line 2-3 is long), the PF is considered ?poor.?

What would happen to that triangle, if you drew the vertical line as 2 inches long, instead of 1 inch? The real work, line 1-2, (like the liquid portion of the glass of beer) does not change. But the apparent power provided by the utility (line 1-3) is now much longer. Thus, in order to allow the customer?s motors to do the same amount of work, the utility has to provide more power (i.e., you need a larger glass to hold the same amount of liquid with a larger amount of foam).

What would happen to that triangle, if you drew the vertical line as 1/2 inch long, instead of 1 inch? Here again, the real work, line 1-2, does not change. But the apparent power provided by the utility (line 1-3) is now much shorter. Thus, in order to allow the customer?s motors to do the same amount of work, the utility has to provide less power (i.e., you can get by with a smaller glass, since there is less foam to contain). You can see why the utility does not like the customer to have a lot of reactive power in their loads. You can see why the utility will charge some customers (large industrial plants, for example) a penalty for having a poor power factor.

What a power factor correction device does is to reduce the length of line 2-3. It does that by adding a reactive power load of a type that is opposite to that in the customer?s loads. Usually, a customer?s loads are motors and other inductive loads. To that, you add a capacitive load. What that does to the triangle is like starting at point 3, and drawing a line downwards towards (but not as far as) line 2. Call the end of this new line point ?4.? Here?s what you get: Line 2-3 is the customer?s inductive loads. Line 3-4 is the power factor correction capacitors. Line 2-4 is the new value of total reactive power. Line 1-4 (I didn?t tell you to draw that yet, did I?) is the new value of total apparent power. You will note that line 1-4 is shorter than line 1-3. So the amount of power supplied by the utility has gone down.

What is physically happening in the power factor correction device is that, where you used to have an exchange of energy between the between the magnetic field of the utility generator and the magnetic field of the customer?s motors and other inductive loads, you now have an exchange of energy between the magnetic field of the customer?s motors and the electric field of the power factor correction capacitors. The utility company meter does not record that energy exchange, because it takes place entirely within the customer?s facility.

One final note. The units of measure for the three lines described above are fundamentally the same. They all relate to the rate of use of energy, and can all be expressed in terms of ?joules of energy per second of time.? But to keep track of them separately, they are given separate names. The ?real power? unit (line 1-2) is expressed in ?watts (W)? or ?kilowatts (KW).? The ?reactive power? unit (line 2-3) is expressed in ?volt-amperes reactive (VAR)? or kilo-volt-amperes reactive (KVAR).? The ?apparent power? unit (line 1-3) is expressed in ?volt-amperes (VA),? or kilo-volt-amperes (KVA).?

 

[charlie b]

One more tidbit. "Ohm's Law" is about the relationship between current, voltage, and resistance. It does not address the relationship between current, voltage, and power.

 

[zog]

One more tidbit. "Ohm's Law" is about the relationship between current, voltage, and resistance. It does not address the relationship between current, voltage, and power.

And it is not really a "law", more of a rule of tumb.

 

[mphillipps]

Yes Zog I'm a certified electrician. Fairly new in the trade and use this forum to grow in my understanding of the trade.

I'm working on a 12v lighting transformer. I was a bit surprized to find 200VA rather than 200W on the information plate. Does 200VA imply DV output?

 

[winnie]

The key thing that you need to keep in mind when talking about voltage and current for AC circuits is that a single number is used to represent a constantly changing value. A true representation of the voltage on a wire would require a graph of 'instantaneous voltage versus time', but for most electrical work this is not done because it would be both true and essentially meaningless. Generally the single number is far more meaningful and useful.

The single 'voltage' number that you use as an electrician is the 'root mean square' of the continuous value. This particular single number is meaningful because of the way it corresponds to the DC voltage. AC of a particular RMS voltage will deliver the same average power to a given resistive load as DC of the same continuous voltage.

You can also measure 'RMS current'.

Now go back to measuring power. If you take the graph of instantaneous voltage versus time and multiply it by the graph of instantaneous current versus time, you will get the graph of instantaneous power versus time. This applies equally to AC and DC circuits, but only if you look at _instantanous_ measurements.

Unfortunately, the 'root mean square' data reduction throws away too much information, and RMS current * RMS voltage does not equal average power. This is because the RMS technique throws away all information about the _timing_ of the voltage and current waveforms. You could imagine the voltage waveform has such relative timing to the current waveform that even with high voltage and current values, the net power delivered to the load is _zero_ or even negative.

It turns out that for sinusoidal voltage and current flow, there are various techniques which can be used to calculate power from RMS values. For sinusoidal current flow P(ave) = I(RMS) * E(RMS) * PF , where PF is a factor related to the difference in timing of the two waveforms.

So P=I*E for instantaneous values, but as soon as you are looking at 'average' values you need to consider timing and P=I*E no longer applies unless you compensate for timing.

-Jon

 

[al hildenbrand]

When is VA = (V*A=W) or (IE=p) basic Ohm's law?

For the electrician working with single phase services where all of the connected load is resistance heat and incandescent lamps (no motors or capacitors or electronic ballasts or electronic power supplies) then the simple Volts times Amps equals Watts works.

The assumption about the Voltage and the Current coming in from the Power Company through the service is, that, the Voltage and Current are "ideal", that is, they are together, in phase, with each other. The Voltage and Current are sine waves that cross zero at the same instant. The total load, in this case, is called Resistive Load.

None of the resistance heat or incandescent lamps cause the sine waves of the current or voltage to shift "apart" from each other to different zero crossing times. The zero crossing remains together, regardless of how much or how little load is running.

The power used in a totally resistive load is only Watts, or, Real Power.

And is volt-ampere ever equal to something more or different?

Yes.

When current and voltage sine waves don't cross zero at the same instant then part of the product of voltage times current is not "real" power. Part of the product of voltage times current is "reactive" power.

I always liked the idea that was used when I was in school for this. The phrase was "imaginary power". Imaginary power can't do "real" work, but the power company has to use the fuel to create the "Total Power" that includes both the imaginary and real power.

Non resistive loads, loads like motors, capacitors, switching power supplies for electronic appliances or for fluorescent lights use both "real" power and "imaginary" power.

The "total power", that is, the sum (it is a vector sum) of real and imaginary power, is given in VA.

 

[LarryFine]

How about an even simpler explanation, as I understand it, anyway:

When the voltage waveform peaks, the current through a resistive load peaks at the same time. That means that a watt-meter, which measures both voltage and current simultaneoulsy, will indicate the power in watts.

When the load is capacitive, the current actually peaks ahead of the voltage, because the capacitance charging current is at its highest rate while the voltage waveform is changing from plus to minus or minus to plus.

The voltage changing rate is the highest when the voltage itself is lowest, i.e., when near the zero (crossover) point. The capacitance current is high before the voltage is, so we say the current leads the voltage.

An inductance has the opposite effect. It's current is the highest after the voltage peaks, when on the way back towards the zero point, instead of before it does, so we say the inductive current lags the voltage.

In both cases, the current peak does not coincide with the voltage peak, so a meter that measures both the voltage and current in order to read out the power, will not see the two separate peaks at the same time.

A watt-meter will showt the power to be less than volts x amps. Watts are the voltage x the current at any given instant. Volt-amps are the peak voltage x the peak current, regardless of their relative timing.

Even though the power supply system must be sized to carry the highest current as well as the highest voltage, not all of the total voltage and the total current will be indicated by the watt-meter.

What's worse, most loads can only utilize the same energy that the watt-meter reads, so the portion of the total current that is not in synch with the voltage does not go towards productive load output.

The power-company's supply system, has to be sized to carry both the usable power and the unusable power, just as the customer's system must, yet the customer only benefits from the usable portion.

When a customer has a mix of resistive loads as well as capacitive and/or inductive (collectively known as reactive) loads, the proportion is what we call the power factor. When everything is in synch, the PF is 1.

When there are reactive components in the load, the PF will be less than 1, unless the capacitive and inductive effects happen to be exactly equal. PF-correction devices do exactly that: counteract the low PF.

When PF-correction means are used, the don't eliminate the low PF, they merely reduce how mcuh of the supply system must carry the resulting unusable current. They sort-of 'shunt' the wasted current.

The closer to the offending load the corrective device is, the less of the system has to carry the extra current. Otherwise, the unusable current thravel through the entire system, back to the POCO's generators.

The POCO, when saddled with extra power the meter doesn't resond to (this not allow them to bill for), yet they must carry, they tack on a low-PF penalty. So, it benefits the customer to improve the PF.

However, since homes aren't billed for low PF, there's no monetary advantage for a homeowner to buy and use the "power-saving" devices being sold these days. Besides, the correction has to be sized for the PF.

Added: Keep in mind that PF can never exceed 1, and that Watts can never exceed VA. I deall, they're equal, but when they're not, PF and VA will be lower. Remember that and you won't mess up on test questions.

One more tidbit. "Ohm's Law" is about the relationship between current, voltage, and resistance. It does not address the relationship between current, voltage, and power.

That's what I lovingly call "Watts' Law."

 

[LarryFine]

Added: Keep in mind that PF can never exceed 1, and that Watts can never exceed VA. I deall, they're equal, but when they're not, PF and VA will be lower. Remember that and you won't mess up on test questions.

Correction: Ideally, they're equal, but when they're not, PF and watts will be lower.

 

[cadpoint]

The OP should study at http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html there's plenty of various appliable aspects of and for understanding presented there!

Click on the Electricity and Magnetism note just the Web of inter-linked terms!

Note that it is a *.EDU site

 

[mphillipps]

Cadpoint... great site (from what I've seen so far). They've laid and linked thing so very nicely and all in one spot.

I'll take some time this weekend reading through all this info.

Thanks everyone... This is why I love this forum.

 

[zog]

Yes Zog I'm a certified electrician. Fairly new in the trade and use this forum to grow in my understanding of the trade.

I'm working on a 12v lighting transformer. I was a bit surprized to find 200VA rather than 200W on the information plate. Does 200VA imply DV output?

OK, there is a question I can understand someone asking.

Things that convert electrical energy into or from a different type of energy (Mechanical, heat, etc) are rated in Watts (Or HP). Watts are units of work. Motors, generators, lights, heaters, etc.

Transformers are rated in VA because they convert electrical energy into electrical energy with different quanities. No "work" is performed. A transformer needs to account for not only the watts (Real power) but also the reactive power when considering its ratings.

That help?

 

[gar]

091210-2151 EST

mphillipps:

The amount of power that can be transfer thru a transformer is a function of the maximum temperature internal to the transformer due to losses. Broadly there are two types of losses in a transformer --- core losses and copper winding wire resistance loss.

A transformer might be designed for equal losses in the core and copper at full rated load. The core losses won't change much with load. The copper losses are approximately the DC winding resistance times the square of RMS load current. The phase angle of the load current relative the supply voltage has no effect on the winding resistance losses.

If you had a resistive load load on the transformer of I amps, then the transformer resistive losses would be Rint*I^2, and the output power would be V*I.

Change to a pure capacitive load with the same load current. The transformer losses will be unchanged. However, the output power is essentially zero.

.

 

[mivey]

A transformer needs to account for not only the watts (Real power) but also the reactive power when considering its ratings.

I think that is the simplest way to put it.

 

[weressl]

One more tidbit. "Ohm's Law" is about the relationship between current, voltage, and resistance. It does not address the relationship between current, voltage, and power.

If so then what is I(SQRD)*R?

http://www.flickr.com/photos/11903290@N08/4188475116/

 

[charlie b]

If so then what is I(SQRD)*R?

Power. But it is not Ohm's Law that gets you there. That diagram is very useful, but it has more information than just Ohm's Law shown on it. So the person who first created it and gave it the title "Ohm's Law" took a bit of liberty with the facts.

 

[weressl]

Power. But it is not Ohm's Law that gets you there. That diagram is very useful, but it has more information than just Ohm's Law shown on it. So the person who first created it and gave it the title "Ohm's Law" took a bit of liberty with the facts.

I agree, but don't you agree that it is actually Ohm's law that 'gets' you there? Ohm just did not define or realise the other relationships?

 

[charlie b]

Agreed. I don't know how much he knew about the other relationships, but his name was tagged only with the E-IR part.