"John" <[log in to unmask]> wrote in message
news:94khnr024n0@enews1.newsguy.com...
> The simplest (and cheapest) way is to look at the label on the
> appliance.  It will normally tell you the power consumed in watts
> (sometimes it's labelled VA but the difference isn't very significant for
> our purposes).

For AC circuits with purely resistive loads (like incandescent light bulbs,
ovens, etc.),  power is equal to the product of Volts*Amps (VA).  However,
for appliances with induction motors or other  loads with a reactive
component, the formula becomes P=V*A*COS(~), where ~ is the phase angle
between the voltage and current.   A true power meter performs this
calculation before displaying the result.   The phase angle between voltage
and current is 0 degrees for a purely resistive load, +90 deg. for a purely
inductive load, and -90 degrees for a purely capacitive load.  By
inspection, if the load is purely reactive (inductive or capacitive), then
the power is zero, although the V*A product might be large.  Most real-world
loads have both resistive and reactive components (Impedance), with a phase
angle somewhere between 0 and 90 deg.  The COSINE term is usually called the
"Power Factor".   Actual power (P) is expressed in Watts.  Since most
Induction motors used in large appliances (Air Conditioners,  Pumps, Fans,
etc.) have a significant Inductive component, using V*A without considering
the power factor can produce large errors.  Apparant Power (V*A) is always
going to be equal or greater than the actual power consumed.  Also, some
electronic equipment using switching power supplies may have a fairly low
power factor and lead to significant measurement errors.  Flourescent lights
generally have an inductor (ballast) in series to limit current after
"ignition", and this leads to a low power factor as well.   So if you want
to measure power with any degree of accuracy, you need to consider the load
before relying on a simple V*A product.   Also, the amount of heat and/or
useful work  done at the load is equivalent to the power supplied.  One
horsepower is equivalent to ~746 Watts, for example.  The difference between
the apparant power and the actual power is returned ("reflected") to the
source -- minus losses in the wiring (circulating Amps^2 * Resistance).
This last "feature" is why power companies don't like low power-factor loads
(wire size has to be increased to handle the total circulating current).
Industrial customers with giant AC induction motors often have to install
huge capacitors or synchronous motors in parallel with their inductive motor
loads to bring the phase angle closer to zero and increase the power factor
to required limits.  Otherwise, the transmission line losses would be
excessive and breakers/fuses could blow.  The amount of power that has to be
supplied by the generating plant is equal to the actual power consumed at
the load minus the transmission network losses.   Of course, the typical
generating plant and transmission system is probably running at about 33%
overall efficiency,  so it might take about 300 Watts of heat at the power
plant to furnish enough electrical energy to power a 100 Watt light bulb at
your house ..........

Winston K.