I'm not aware of a formula. You need to know stuff about the motor as
well as the load that is on it. Unless you mean volts x amps = watts?
You're better off getting a clamp on amp meter and noting the usage
while the appliance is running.
There is no formula for figuring out appliances like a refrigerator, a
Kill-A-Watt meter works well and is accurate, [although i just bought
a defect a HD] my old KAW works well. New apliances like friges can be
75% more efficent than units made maybe 20 years ago, with defrost
cycles and old compressors pulling more at end of life you need
something to measure consumption over a day at least to get an
accurate overall picture. For a motor use a clamp on amp meter.
For single-phase motors,
HP = V I Eff Pf/746
where V = input voltage (V),
I = current drawn (A),
Eff = efficiency (fraction), and
Pf = power factor (fraction)
For theoretical purposes you can use nameplate values and nominal
voltage and get estimate of current draw assuming the nameplate has
necessary data for Eff/Pf; (if they aren't given it's often because
they're low... :) )
If you're after normal operation consumption you'll have to measure the
inputs as nameplate ratings are at max and you'll undoubtedly not know
the other factors precisely...
My formula is- (plug in Kill-o-watt meter), read display.
In the case of refrigerators, freezers, washing machines, or anything
that doesn't use a constant amount of electricity, I leave it on for a
few days and note high usage & overall usage.
Is this just an intellectual exercise, or do you have a specific
OK, wise guy: since you brought it up (or ran with it in this thread),
try this on for size:
Try to sing the following well-known song that's been phase-changed just
a little bit (a music teacher friend used to use this to teach his
grade-school kids simple music theory):
Me out to the ball game, take
Me out to the game, buy
Me some peanuts and crackerjacks, I
Don't care if I never get back, 'cause it's
Root, root, root for the home team if
They don't win it's a shame 'cause
It's one, two, three strikes you're out of
The old ball game ...
Not easy, eh?
(Some of the words may not be exactly right, but the meter is basically
You were wrong, and I'm man enough to admit it.
Mike; the amount of electricity used by anything electrical is
measured in watts.
For example a 100 watt electric light bulb switched on for one hour
consumes 100 watt hours which is equal to one tenth of a kilowatt
hour. If there are ten 100 watt bulbs on for the hour they will use 10
x 100 = 1000 watt hours = One kilowatt hour.
When it comes to electric motors, fridges etc. the rating (under
normal load conditions) will be either in Horsepower or the volts and
amps will be on the nameplate. A table saw for example may have a half-
horsepower motor? One horsepower (HP) is very close to 750 watts, so
again if the half-horse motor runs for one hour it will use 0.375
kilowatt hours of electricity.
A coffee maker may be rated at at say 3 amps; amps times the voltage =
watts. So if the coffeemaker bubbles away plugged into a regular 115
volt North American wall outlet, continuously for one hour (strong
coffee eh!) it will use 115 x 3 = 345 watt hours or 0.345 kilowatt
hours of electrcity.
We pay about ten cents per kilowatt hour for electrcity here; your
rate/mileage may vary!
So the 100 watt bulb would cost me one cent per hour.
Ten 100 watt bulbs would cost me ten cents per hour.
The half-horse motor will cost about 3.75 cents per hour; of course
if I'm cutting oak instead of pine the harder wood will probably use
The coffee maker about 3.45 cents per hour.
Not everything runs by the hour of course, so if I only run my table
saw for six minutes (one tenth of an hour) ............. Or if it
takes twenty minutes (one third of an hour) to make
coffee .................... Well work it out.
Kwh is exactly what it says. Kw (watts/1000) X hours run.Watts is
only Amps X Volts in the case of resistive loads (heating or tunsten/
If you have a inductive load, Kw = volts X amps X the cosine of the
phase angle between them. The cosine of the angle between them is
known as the power factor, it may be marked on the motor, it's usually
With resistive loads, the power factor is 1 (ie Volts and Amps are in
phase) and so is ignored.
So, if you ran a 500w (0.5Kw) motor for 2 hours it would use 1 Kwh.
If you ran a 750w (0.75Kw) motor for 1 hour it would use 0.75Kwh.
Assuming they were fully loaded up that is, the Kw given is the
maximum power they can draw.
Electricity is purchased by measuring kilowatt-hours. For example, a
1000 watt appliance (on high) for one hour will have used 1
kilowatt-hour, or a 500 watt appliance on for 2 hours will also use 1
kilowatt-hour, etc. If you have several items to check, get a
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