Volt*Amps and Watts on my home power meter

I am in the UK. A home power meter I bought says this in its Chinese-English instructions. Can anyone make sense of it?
Press 'WATT' display Watt meter, then press key to display VA meter, and active power for 1 second each automatically.
'WATT/VA' Key is a toggle function Key, the LCD will show Watt is the active power, and VA is the apparent Power. (VA = Vrms Arms)
There seems to be some difference between V multiplied by A and Watts?
It mentions "apparent power" but what is it?
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This is the meter:
http://www.prodigit.com/img/product/2000mu-01.jpg
These are the instructions (see #5): http://www.prodigit.com/pdf/2000mu-01.pdf
It looks quite like the US's "Kill-A-Watt" Electric Usage Monitor http://store.servomagazine.com/image.php?type=T&id502
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Allie-
This isn't necessarily a problem translating Chinese. Its one of those things where if you already knew the answer, it would be quite obvious!
Some equipment such as electric motors, have inductance (coils of wire wound around a metal core). Inductance has reactance that delays current, so the peak current does not coincide with peak voltage.
If you simply measure voltage and current, the VA product would be apparent power. For equipment that has reactance, actual power (Watts) is less than VA, and the ratio of Watts divided by VA, is called the power factor (PF).
If there is no reactance, such as the case of a heating element in a stove burner, VA = Watts and PF = 1.0.
Fred
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How hard is it to type "apparent power" into google?
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I asked several connected questions ... (1) How to read the display of the meter. (2) "There seems to be some difference between V multiplied by A and Watts?" (3) "It mentions "apparent power" but what is it?"
When I had keyed "apparent power" into Google I got gobbledegoogle back. For example, the Wikipedia had said
"Apparent Power (|S|) ... is, the absolute value of complex power S - unit: volt-ampere (VA)" and "Apparent power is conventionally expressed in volt-amperes (VA) since it is the product of rms voltage and rms current." and "While real power and reactive power are well defined in any system, the definition of apparent power for unbalanced polyphase systems is considered to be one of the most controversial topics in power engineering"
Heh? I'm not a specialist, you know! I want to know in ordinary English. :-)
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On AC circuits, V * A has units of watts but is conventionally called "volt-amps" to distinguish it, apparent power, from "real power" which is (also) in watts but gets to keep the name as well. For "real power" (and for all DC circuits) the current and voltage is in phase. Pure imaginary power has current and voltage 90-degrees out of phase; no work gets done there.
Apparent power includes both real power, on the x-axis, and imaginary power, on the y-axis. It's imaginary since it doesn't do any real work, it just sloshes back and forth between the reactive elements (inductive, capacitive) in the power system. Apparent power is the hypotenuse between them (and the product of V and A). The "power factor" is the cosine of the angle that the apparent power makes with the x-axis. Note that's the same as the ratio of the real power (adjacent) to the apparent power (hypotenuse).
Utilities prefer that things stay near a unity power factor, since even though the "imaginary" power doesn't do real work in your house, it does show up as resistive losses on the power lines, so you're kind of paying for less power (at your end) than the utility has to pump into the lines at their end.
Don't worry about "unbalanced polyphase systems," since you'll probably mostly (entirely?) be working with single phase household power.
--
Rich Webb Norfolk, VA

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Let me try to give a simple explanation :
when you measure voltage and current in a device, then you multiply Volts by Amps, you get apparent power.
because the Current and voltage may not be exactly in phase.
for a heater, a resistor, they are in phase, no problem
for a transformer, the current is always "late" regarding to voltage, this is due to the coil. The best example is a microwave (big transformer inside) or a motor.
As the current is not at the maximum at the same time than voltage, if you mesure maybe every millisecond the current and voltage and then multiply, you measure real power.
Your device is able to mde both measurements, apparent power and real power.
The issue is that if your apparent power is too big, you will get higher current in your wires, even if the real power is not so high !
at home, this is a minor problem, but in plants, when you play with hundreds of kilowatts, it makes a sense to check both.
JJ
Allie a crit :

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jj wrote:

This is almost correct. However, if you measure the product of voltage and current at microsecond intervals, AND take the average over a cycle, then you get the "real" power. Consider that use of rms values of voltage and current implies averaging over a cycle.
I may be nit picking but .....
As an aside for the original questioner- the WIKI article does assume some knowledge of complex number theory as applied to AC circuits- there are simpler references. look up power factor. one such is
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/powfac.html
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Allie submitted this idea :

Apparent Power (VA) is the voltage measured * the current measured. True Power (Watts) is the actual amount of power being used. If the load is purely resistive, then the Apparent power and true power will be the same. However if you introduce a capacitive or inductive load you get reluctance. So know you have 3 factors: VA, Watts, and Q (reactive power). Thier relationship is called the power factor. Power factor is true power divided by apparent power Apparent power = Sq.Rt. of (VA sq. + Watts sq.)
Think of it as a right triangle, the vertical leg is reactive power. The Horizontal leg is true power and the hypotenuse is VA or apparent power.
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Arlowe used his keyboard to write :

I need to make a correction... Apparent power= sq. rt of ("Q" sq + watt sq.)
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