How can I measure power draw?


Is there a simple way to measure how much power my components are drawing from the wall? My interest is prompted by experimenting with the PS Audio Power Plant and various conflicting info about how much it draws. (The answer I'd love to get is that there is a Radio Shack box I can stick in between the DUT and the wall, but you tell me...)
Ag insider logo xs@2xdrubin
Drubin, Go to doubleed.com and check out their "wattsup" device. This unit plugs into the wall and then your amp plugs into this unit. Turn on the "Wattsup" and then turn on your amp. This unit measures the power consumption of your amp. It can also calcuate the cost of electricity. Pretty need stuff. I saw one at Fry's electronics for $80.00. Have fun.
The Wattsup sounds like just the ticket. I'll get one and report back. Thanks. -Dan
An accurate power draw cannot be obtained without considering the power factor. In essence this means measuring the phase angle of voltage vs current (as well as measuring voltage vs current). The true RMS power figure is obtained by: P = VI cos phi. The cosine of the phase angle is called the power factor. This formula holds for all sinewave sources of power (such as AC from the wall).
As a more simple method to the above suggestions, you can measure the power consumption by watching the meter at the power distribution box for your house. I'm currently in Japan, and I'm from Ireland, but both countries use the same type of meter; so the following probably also applies in America. My meter, has a rotating disc, one full revolution is 1/100 on a kWHr. You can calculate the average power consumption, over a given period of time, by turning on your equipment, and counting the number of revolutions over that time. The power in Watts is given by the following equation P = 600 X (#Revolutions/Time in Minutes) This will directly give the power for which you will pay your Electricity Supply company. Make sure to turn off all other appliances in your house before making the measurement. During the day is the best time to do this experiment!! Brian Kearns
Sorry, I checked my meter this morning, and I have to make a correction to the formula above. One full revolution is 1/300 kWHr, so the correct value is three times smaller, ie: P = 200 X (#Revolutions/Time in Minutes). Brian Kearns.